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The 2507 statements of Algebra, by chapter and §. Pick a kind, filter by a word, or put the most cited first.

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I. Algebraic structures

§ 1. Laws of composition; associativity; commutativity

Definition 1 Let E be a set. A mapping f of E × E into E is called a law of composition on E. The value f(x, y) of f for an ordered pair (x, y) ∈ E × E is called the composition of x and y unde… ↩ 1 055B
Example (1) The mappings (X, Y) ↦ X ∪ Y and (X, Y) ↦ X ∩ Y are laws of composition on the set of subsets of a set E. 055C
Definition 2 Let E be a magma and τ denote its law of composition. The law of composition (x, y) ↦ y τ x on E is called the opposite of the above. The set E with this law is called the opposite… 055D
Definition 3 A mapping f of E into E' such that relation (1) holds for every ordered pair (x, y) ∈ E × E is called a homomorphism, or morphism, of E into E'; if E = E', f is called an endomorph… ↩ 1 055E
Definition 4 Let (x_α)_α ∈ A be an ordered sequence of elements in a magma E whose indexing set A is non-empty. The composition (under the law T) of the ordered sequence (x_α)_α ∈ A, denoted by… ↩ 1 055F
Remark Note that there is a certain arbitrariness about the definition of the composition of an ordered sequence; the induction we introduced proceeds "from right to left". If we proceede… 055G
Definition 5 A law of composition (x, y) ↦ x top y on a set E is called associative if, for all elements x, y, z in E, (x top y) top z = x top (y top z). A magma whose law is associative is cal… 055H
Example (1) Addition and multiplication of natural numbers are associative laws of composition on N (Set Theory, III, § 3, no. 3, Corollary to Proposition 5) (2) The laws cited in Examples… 055I
Theorem 1 (Associativity theorem) Let E be an associative magma whose law is denoted by top. Let A be a totally ordered non-empty finite set, which is the union of an ordered sequence of non-empty subsets (Bᵢ)_i ∈… ↩ 5 055J
Definition 6 A subset A of a set E is called stable under the law of composition T on E if the composition of two elements of A belongs to A. The mapping (x, y) ↦ x T y of A × A into A is then… 055K
Proposition 1 Let E and F be two magmas and f a homomorphism of E into F. (i) The image under f of a stable subset of E is a stable subset of F. (ii) The inverse image under f of a stable subset… ↩ 2 055L
Proposition 2 Let E be an associative magma and X a subset of E. Let X' be the set of x₁ top x₂ top ⋯ top xₙ, where n ≥ 1 and where xᵢ ∈ X for 1 ≤ i ≤ n. The stable subset generated by X is equa… 055M
Example (1) In the set N of natural numbers the stable subset under addition generated by {1} is the set of integers ≥ 1; under multiplication the set {1} is stable. 055N
Definition 7 Let E be a magma whose law is denoted by top. Two elements x and y of E are said to commute (or to be permutable) if y top x = x top y. 055O
Definition 8 A law of composition on a set E is called commutative if any two elements of E commute under this law. A magma whose law of composition is commutative is called a commutative magma… 055P
Example (1) Addition and multiplication of natural numbers are commutative laws on N (Set Theory, III, § 3, no. 3, Corollary to Proposition 5). 055Q
Definition 9 Let E be a magma and X a subset of E. The set of elements of E which commute with each of the elements of X is called the centralizer of X. ↩ 1 055R
Proposition 3 Let E be an associative magma whose law is denoted by top. If an element x of E commutes with each of the elements y and z of E, it commutes with y top z. 055S
Corollary Let E be an associative magma. The centralizer of any subset of E is a stable subset of E. ↩ 1 055T
Definition 10 The centralizer of a magma E is called the centre of E. An element of the centre of E is called a central element of E. 055U
Proposition 4 Let E be an associative magma, X and Y two subsets of E. If every element of X commutes with every element of Y every element of the stable subset generated by X commutes with ever… 055V
Corollary 1 If x and y are permutable under the associative law top so are topᵐ x and topⁿ y for all integers m > 0 and n > 0; in particular topᵐ x and topⁿ x are permutable for all x and all… 055W
Corollary 2 If all pairs of elements of a subset X are permutable under an associative law T, the law induced by T on the stable subset generated by X is associative and commutative. ↩ 2 055X
Theorem 2 (Commutativity theorem) Let T be an associative law of composition on E; let (x_α)_α ∈ A be a non-empty finite family of elements of E which are pairwise permutable; let B and C be two totally ordered set… ↩ 2 055Y
Theorem 3 Let T be an associative law on E and (x_α)_α ∈ A a non-empty finite family of elements of E which are pairwise permutable. If A is a union of non-empty subsets (Bᵢ)_i ∈ I which are… ↩ 1 055Z
Proposition 5 Let E and F be magmas whose laws are denoted by T and let f and g be homomorphisms of E into F. Let f T g be the mapping x ↦ f(x) T g(x) of E into F. If F is associative and commut… ↩ 2 0560
Definition 11 Let E be a set. A law of composition T and an equivalence relation R on E are said to be compatible if the relations x ≡ x' (mod R) and y ≡ y' (mod R) (for x, x', y, y' ∈ E) imply… 0561
Proposition 6 Let E and F be two magmas and f a homomorphism of E into F. Let R[x, y] denote the relation f(x) = f(y) between elements x, y of E. Then R is an equivalence relation on E compatibl… 0562
Proposition 7 Let E be a magma and R an equivalence relation on E compatible with the law on E. For an equivalence relation S on E/R to be compatible with the quotient law it is necessary and su… 0563
Proposition 8 Let E be a magma, A a stable subset of E and R an equivalence relation on E compatible with the law on E. The saturation B of A with respect to R (Set 0564
Proposition 9 Preserving the above notation, let f be a homomorphism of M into a magma such that f(u_α) = f(v_α) for all α ∈ I. Then f is compatible with R. ↩ 2 0565

§ 2. Identity element; cancellable elements; invertible elements

Definition 1 Under a law of composition top on a set E an element e of E is called an identity element if, for all x ∈ E, e top x = x top e = x. ↩ 1 056M
Definition 2 A magma with an identity element is called a unital magma. If E, E' are unital magmas, a homomorphism of the magma E into the magma E' which maps the identity element of E to the i… 056N
Example (1) In the set N of natural numbers 0 is an identity element under addition and 1 is an identity element under multiplication. Each of these two laws gives N a commutative monoid s… 056O
Definition 3 Let E be a unital magma. The identity element of E is called the composition of the empty family of elements of E. 056P
Definition 4 Given a law of composition top on a set E, the mapping x ↦ a top x (resp. x ↦ x top a) of E into itself is called left translation (resp. right translation) by an element a ∈ E. 056Q
Proposition 1 If the law T is associative, then for all x ∈ E and y ∈ E ↩ 2 056R
Definition 5 An element a of a magma E is called left (resp. right) cancellable (or regular) if left (resp. right) translation by a is injective. A left and right cancellable element is called… ↩ 1 056S
Example (1) Every natural number is cancellable under addition; every natural number ≠ 0 is cancellable under multiplication. 056T
Proposition 2 The set of cancellable (resp. left cancellable, resp. right cancellable) elements of an associative magma is a submagma. ↩ 2 056U
Definition 6 Let E be a unital magma, T its law of composition, e its identity element and x and x' two elements of E. x' is called a left inverse (resp. right inverse, resp. inverse) of x if x… ↩ 1 056V
Example (1) An identity element is its own inverse. 056W
Proposition 3 Let E be a monoid and x an element of E. ↩ 2 056X
Remark Let E be a monoid and x an element of E. If x is left invertible it is left cancellable; for, if x' is a left inverse of x, then 056Y
Proposition 4 Let E be a monoid and x and y invertible elements of E with inverses x' and y' respectively. Then y' top x' is the inverse of x top y. ↩ 7 056Z
Corollary 1 Let E be a monoid; if each of the elements x_α of an ordered sequence (x_α)_α ∈ A of elements of E has an inverse x'_α, the composition ∏_α ∈ A x_α has inverse ∏_α ∈ A' x'_α, where… ↩ 1 0570
Corollary 2 In a monoid the set of invertible elements is stable. 0571
Proposition 5 If in a monoid x and x' are inverses and x commutes with y, so does x'. ↩ 3 0572
Corollary 1 Let E be a monoid, X a subset of E and X' the centralizer of X. The inverse of every invertible element of X' belongs to X'. ↩ 1 0573
Corollary 2 In a monoid the inverse of a central invertible element is a central element. 0574
Lemma 1 In E × S' the relation RRx, yR defined by: ↩ 1 0575
Definition 7 Let E be a commutative monoid, S a subset of E and S' the submonoid of E generated by S. The quotient monoid (E × S')/R, where the equivalence relation R is as described in Lemma 1… ↩ 1 0576
Proposition 6 The notation is that of Definition 7 and ε denotes the canonical homomorphism of E into E_S. (i) Let a and b be in E; in order that ε(a) = ε(b), it is necessary and sufficient that… ↩ 4 0577
Theorem 1 Let E be a commutative monoid, S a subset of E, E_S the monoid of fractions associated with S and ε : E → E_S the canonical homomorphism. Further let f be a homomorphism of E into… ↩ 6 0578
Corollary Let E and F be two commutative monoids, S and T subsets of E and F respectively, f a homomorphism of E into F such that f(S) ⊂ T and ε : E → E_S, η : F → F_T the canonical homomorp… 0579
Remark (1) Theorem 1 can also be expressed by saying that (E_S, ε) is the solution of the universal mapping problem for E, relative to monoids, monoid homomorphisms and homomorphisms of E… 057A
Lemma 2 Let E be a monoid and x an element of E. (i) There exists a unique homomorphism f of N into E with f(1) = x and f(n) = n/1 x for all n ∈ N. (ii) If x is invertible, there exists a… ↩ 1 057B

§ 3. Actions

Definition 1 Let Ω and E be two sets. A mapping of Ω into the set Eᴱ of mappings of E into itself is called an action of Ω on E. 057T
Example (1) Let E be an associative magma written multiplicatively. The mapping which associates with a strictly positive integer n the mapping x ↦ xⁿ of E into itself is an action of N^*… 057U
Definition 2 A subset A of a set E is called stable under an action α ↦ f_α of Ω on E if f_α(A) ⊂ A for all α ∈ Ω. An element x of E is called invariant under an element α of Ω if f_α(x) = x. 057V
Remark Let E be a magma with law denoted by τ. It should be noted that a subset A of E which is stable under the left action on E on itself is not necessarily stable under the right actio… 057W
Example Take the magma E to be the set N with multiplication. The set 1 is stable under the internal law of N, but the stable subset under the action of N on itself generated by 1 is the w… 057X
Definition 3 Let α ↦ f_α be an action of Ω on E and A a stable subset of E. The mapping which associates with an element α ∈ Ω the restriction of f_α to A (considered as a mapping of A into its… 057Y
Definition 4 Let α ↦ f_α be an action of a set Ω on a set E. An equivalence relation R on E is said to be compatible with the given action if, for all elements x and y of E such that x ≡ y pmod… 057Z
Definition 5 Let E₁, …, Eₙ and F be sets and u a mapping of E₁ × ⋯ × Eₙ into F. Let i ∈ 1, n. Suppose Eᵢ and F are given the structures of magmas. u is said to be distributive relative to the i… ↩ 1 0580
Example Let E be a monoid (resp. group) written multiplicatively. The mapping (n, x) ↦ xⁿ of N × E (resp. Z × E) into E is distributive with respect to the first variable by the equation x… 0581
Proposition 1 Let E₁, E₂, …, Eₙ and F be commutative monoids written additively and let u be a mapping of E₁ × ⋯ × Eₙ into F, which is distributive with respect to all the variables. For i = 1… ↩ 2 0582
Remark If u(a₁, …, aᵢ₋₁, 0, aᵢ₊₁, …, aₙ) = 0 for i = 1, 2, …, n and aⱼ ∈ Eⱼ (j ≠ i), then formula (2) remains true for families (xᵢ, λ)_λ ∈ Lᵢ of finite support. 0583
Definition 6 An action α ↦ f_α of a set Ω on a magma E is said to be distributive if, for all α ∈ Ω, the mapping f_α is an endomorphism of the magma E. 0584
Definition 7 Let T and ⊥ be two internal laws on a set E. The law ⊥ is said to be distributive with respect to the law T if 0585
Example (1) In the set P(E) of subsets of a set E, each of the internal laws ∩ and ∪ is distributive with respect to itself and the other. This follows from formulae of the form 0586

§ 4. Groups and groups with operators

Definition 1 A set with an associative law of composition, possessing an identity element and under which every element is invertible, is called a group. 058H
Example In any monoid E the set of invertible elements with the structure induced by that on E is a group. In particular, the set of bijective mappings of a set F onto itself (or set of pe… 058I
Definition 2 Let Ω be a set. A group G together with an action of Ω on G which is distributive with respect to the group law, is called a group with operators in Ω. 058J
Example In a commutative group G, written multiplicatively, (xy)ⁿ=xⁿyⁿ for all n∈Z (§ 2, no. 8, equation (1)); the action n↦(x↦ xⁿ) of Z on G therefore defines, together with the group law… 058K
Definition 3 Let G and G' be groups with operators in Ω. A homomorphism of groups with operators of G into G' is a homomorphism of the group G into the group G' such that 058L
Definition 4 Let G be a group with operators. A stable subgroup of G is a subset H of G with the following properties: 058M
Proposition 1 Let G be a group with operators and H a subset of G which is stable under the homotheties of G. The following conditions are equivalent: ↩ 4 058N
Remark (1) Similarly it can be shown that condition (b) is equivalent to the condition (c') H ≠ ∅ and the relations x ∈ H and y ∈ H imply y⁻¹x ∈ H. (2) For every subgroup H of G there are… 058O
Proposition 2 Let X be a non-empty subset of a group with operators G and hatX the stable subset under the action of Ω on G generated by X. The stable subgroup generated by X is the stable subse… ↩ 1 058P
Corollary 1 Let G be a group with operators and X a subset of G which is stable under the action of Ω. The subgroup generated by X and the stable subgroup generated by X coincide. 058Q
Corollary 2 Let G be a group and X a subset of G consisting of pairwise permutable elements. The subgroup of G generated by X is commutative. ↩ 2 058R
Corollary 3 Let f : G → G' be a homomorphism of groups with operators and X a subset of G. The image under f of the stable subgroup of G generated by X is the stable subgroup of G' generated b… ↩ 2 058S
Proposition 3 Let G be a group with operators, P a subset of G and H a stable subgroup G not meeting P. The set of stable subgroups of G containing H and not meeting P has a maximal element. ↩ 1 058T
Theorem 1 Let R be an equivalence relation on a group with operators G; if R is left (resp. right) compatible (§ 3, no. 3) with the group law on G and compatible with the action of Ω, the eq… ↩ 1 058U
Proposition 4 Let H and K be two subgroups of a group G such that H ⊂ K. Then (G:H) = (G:K)(K:H). ↩ 2 058V
Corollary If G is a finite group of order g and H is a subgroup of G of order h, then h · (G:H) = g (in particular, the order and index of H are divisors of the order of G). ↩ 2 058W
Definition 5 Let G be a group with operators. A stable subgroup H of G is called a normal (or invariant) stable subgroup of G if xHx⁻¹ = H for all x ∈ G. ↩ 1 058X
Theorem 2 Let G be a group with operators. For an equivalence relation R on G to be compatible with the group law and the action of Ω, it is necessary and sufficient that it be of the form x… ↩ 1 058Y
Definition 6 The quotient of a group with operators G by the equivalence relation defined by a normal subgroup H of G, with the quotient structure, is called the quotient group with operators o… 058Z
Proposition 5 Let f : G → G' be a homomorphism of groups with operators and H and H' normal stable subgroups of G and G' respectively such that f(H) ⊂ H'. The mapping f is compatible with the eq… ↩ 3 0590
Remark (1) If A is any subset of a group G and H is a normal subgroup of G, then AH = HA; this set is obtained by saturating A with respect to the relation x ≡ y pmodH. 0591
Definition 7 A group with operators G is called simple if G ≠ e and there exists no normal stable subgroup of G other than G and e. ↩ 4 0592
Proposition 6 Let G be a group with operators and G' a magma with an action by Ω, written exponentially. Let f : G → G' be a homomorphism of the magna G into the magma G' such that, for all α ∈… ↩ 1 0593
Definition 8 Let f : G → G' be a homomorphism of groups with operators. The inverse image of the identity element of G' is called the kernel of f. 0594
Theorem 3 Let f : G → G' be a homomorphism of groups with operators. (a) Ker(f) is a normal stable subgroup of G; (b) Im(f) is a stable subgroup of G'; (c) the mapping f is compatible with t… ↩ 2 0595
Proposition 7 Let G and H be two groups with operators, f a homomorphism of G into H and N the kernel of f. (a) Let H' be a stable subgroup of H. The inverse image G' = f⁻¹(H') is a stable subgr… ↩ 3 0596
Corollary 1 Suppose that f is surjective. Let G (resp. G') be the set of stable (resp. normal stable) subgroups of G containing N and ℌ (resp. ℌ') the set of stable (resp. normal stable) subgr… ↩ 2 0597
Corollary 2 Let f : G → H be a homomorphism of groups with operators, N the kernel of f, G' a stable subgroup of G and L a normal stable subgroup of G'. Then LN, L.(G' ∩ N) and f(L) are normal… ↩ 1 0598
Corollary 3 Let f:G→ H be a homomorphism of groups with operators, N its kernel, X a subset of G such that f(X) generates H and Y a subset of N which generates N. Then X∪ Y generates N. 0599
Remark In the notation of Proposition 7, the fact that the inverse image of a subgroup of H is a subgroup of G follows from the following more general fact. 059A
Proposition 8 Let G be a group with operators and A and B two stable subgroups of G. Suppose that the relations a∈ A and b∈ B imply aba⁻¹∈ B^* (in other words, A normalizes B)^*. Then AB=BA is a… ↩ 1 059B
Theorem 4 Let G be a group with operators and N a normal stable subgroup of G. ↩ 15 059C
Definition 9 A composition series of a group with operators G is a finite sequence (Gᵢ)_0 ≤ i ≤ n of stable subgroups of G, with G₀ = G and Gₙ = e and such that Gᵢ₊₁ is a normal subgroup of Gᵢ… ↩ 1 059D
Theorem 5 (Schreier) Given two composition series Σ₁, Σ₂ of a group with operators G, there exist two equivalent composition series Σ'₁, Σ'₂, finer respectively than Σ₁ and Σ₂. ↩ 2 059E
Lemma 1 (Zassenhaus) Let H and K be two stable subgroups of a group with operators G and H' and K' normal stable subgroups of H and K respectively; then H'.(H \cap K') is a normal stable subgroup of H'… 059F
Definition 10 A Jordan-Hölder series of a group with operators G is a strictly decreasing decomposition series \Sigma such that there exists no strictly decreasing decomposition series distinct… 059G
Proposition 9 For a decomposition series of G to be a Jordan-Hölder series it is necessary and sufficient that all the quotients of the series be simple. ↩ 2 059H
Theorem 6 (Jordan-Hölder) Two Jordan-Hölder series of a group with operators are equivalent. ↩ 6 059I
Corollary Let G be a group with operators in which there exists a Jordan-Hölder series. If Σ is any strictly decreasing composition series of G, there exists a Jordan-Hölder series finer tha… ↩ 2 059J
Remark A group with operators does not always possess a Jordan-Hölder series; an example is given by the additive group Z of rational integers: the sequence (2ⁿ · Z)_n ≥ 0 is a strictly d… 059K
Definition 11 Let G be a group with operators; the length of G is the upper bound of the integers n such that there exists a strictly decreasing composition series of G (Gᵢ)_0≤ i≤ n. ↩ 1 059L
Proposition 10 Let G be a group with operators and H a normal stable subgroup of G. The length of G is the sum of the lengths of H and G/H. ↩ 1 059M
Corollary Let G be a group with operators and (Gᵢ)_0≤ i≤ n a composition series of G. The length of G is the sum of the lengths of the Gᵢ/Gᵢ₊₁, 0≤ i≤ n-1. 059N
Definition 12 The group with operators G = ∏_i ∈ I Gᵢ defined above is the product group with operators of the Gᵢ. The mappings prᵢ : G → Gᵢ are called the projection homomorphisms. 059O
Proposition 11 Let (φᵢ : Hᵢ → Gᵢ)_i ∈ I be a family of homomorphisms of groups with operators and let Φ = ∏_i ∈ I φᵢ. Then: 059P
Definition 13 Let G be a group with operators and (Hᵢ)_i ∈ I a family of normal stable subgroups of G. Let pᵢ : G → G / Hᵢ be the canonical homomorphism. G is called the internal product (or pro… 059Q
Proposition 12 Let (φᵢ: Gᵢ → K)_i ∈ I be a family of homomorphisms of groups with operators such that, for all i ∈ I and j ∈ I with i ≠ j, x ∈ Gᵢ, y ∈ Gⱼ, the elements φᵢ(x) and φⱼ(y) of K commut… ↩ 3 059R
Definition 14 Let G be a group with operators and (Hᵢ)_i ∈ I a family of stable subgroups of G. G is called the internal restricted sum (or restricted sum) of the family of subgroups (Hᵢ) if eve… 059S
Proposition 13 Let G be a group with operators and (Hᵢ)_i ∈ I a family of stable subgroups of G such that every element of Hᵢ is permutable with every element of Hⱼ for j ≠ i. For G to be the res… 059T
Proposition 14 Let G be a group with operators and (Hᵢ)_i ∈ I a finite family of stable subgroups of G. For G to be the restricted sum of the family of subgroups (Hᵢ), it is necessary and suffici… 059U
Proposition 15 Let G be a group with operators and (Hᵢ)_1 ≤ i ≤ n a sequence of normal stable subgroups of G such that (H₁ H₂ … Hᵢ) ∩ Hᵢ₊₁ = e for 1 ≤ i ≤ n-1, the set H₁ H₂ … Hₙ is a normal stab… ↩ 5 059V
Proposition 16 Let a be an integer > 0. The integers r such that 0 leqslant r < a form a system of representatives of the equivalence relation x ≡ y pmoda on Z. ↩ 1 059W
Corollary Let a be an integer > 0. The group Z/aZ of rational integers modulo a is a group of order a. ↩ 1 059X
Proposition 17 Let H be a subgroup of Z. There exists one and only one integer a geqslant 0 such that H = aZ. ↩ 2 059Y
Definition 15 A group is called monogenous if it admits a system of generators consisting of a single element. A finite monogenous group is called cyclic. ↩ 1 059Z
Proposition 18 A finite monogenous group of order a is isomorphic to Z/aZ. An infinite monogenous group is isomorphic to Z. ↩ 2 05A0
Proposition 19 Let a be an integer 0. Let H be a subgroup of Z/aZ, b the order of H and c its index in Z/aZ. Then a = bc, H = cZ/aZ and H is isomorphic to Z/bZ. ↩ 3 05A1
Corollary Every subgroup of a monogenous group is monogenous. 05A2
Definition 16 An integer p > 0 is called prime if p ≠ 1 and it admits no divisor 1 other than p. ↩ 2 05A3
Proposition 20 An integer p > 0 is prime if and only if the group Z/pZ is a simple group. ↩ 1 05A4
Corollary Every commutative simple group is cyclic of prime order. ↩ 1 05A5
Remark A finite group G of prime order is necessarily cyclic. G admits no subgroup other than G and e and hence it is generated by every element ≠ e. 05A6
Lemma 2 Let a be an integer > 0. By associating with every composition series (Hᵢ)_0 ≤ i ≤ n of the group Z/aZ the sequence (sᵢ)_1 ≤ i ≤ n, where sᵢ is the order of Hᵢ₋₁/Hᵢ, a one-to-one c… ↩ 1 05A7
Theorem 7 (decomposition into prime factors) Let a be a strictly positive integer. There exists one and only one family (vₚ(a))_p ∈ 𝔓 of integers > 0 such that the set of p ∈ 𝔓 with vₚ(a) ≠ 0 is finite and ↩ 2 05A8
Corollary Let a and b be two integers > 0. Then vₚ(ab) = vₚ(a) + vₚ(b). For a to divide b, it is necessary and sufficient that vₚ(a) ≤ vₚ(b) for every prime number p. 05A9
Proposition 21 In a finite group G of order n, xⁿ = e for all x ∈ G. 05AA

§ 5. Groups operating on a set

Definition 1 Let M be a monoid, with law written multiplicatively and identity element denoted by e, and E a set. An action α ↦ f_α of M on E is called a left (resp. right) operation of M on E… 05B1
Example (1) Let E be a set; the canonical action of Eᴱ on E (§ 3, no. 1, Example 3) is a left operation. 05B2
Definition 2 Let M be a monoid and E and E' M-sets. A mapping f of E into E' such that, for all x ∈ E and all α ∈ M, f(α.x) = α.f(x) is called an M-set homomorphism (or M-morphism, or mapping c… 05B3
Definition 3 Let M be a monoid operating on a set E and A and B subsets of E. The set of α ∈ M such that α A ⊂ B (resp. α A = B) is called the transporter (resp. strict transporter) of A to B… 05B4
Proposition 1 Let M be a monoid operating on a set E and A a subset of E. (a) The stabilizer, strict stabilizer and fixer of A are submonoids of M. (b) Let α be an invertible element of M; if α… 05B5
Corollary Let G be a group operating on a set E and A be a subset of E. The strict stabilizer S and the fixer F of A are subgroups of G and F is a normal subgroup of S. 05B6
Proposition 2 Let G be a group operating on a set E and, for all x ∈ E, let Sₓ be the stabilizer of x. For all α ∈ G, S_α x = α Sₓ α⁻¹. ↩ 5 05B7
Proposition 3 Let G be a group. For every element x of G, the mapping Int(x): y ↦ xyx⁻¹ of G into itself is an automorphism of G. The mapping Int: x ↦ Int(x) of G into Aut(G) is a group homomorp… 05B8
Definition 4 Let G be a group and x ∈ G. The automorphism y ↦ xyx⁻¹ is called the inner automorphism of G defined by x and is denoted by Int x. 05B9
Proposition 4 Let G be a group, H a characteristic (resp. normal) subgroup of G and K a characteristic subgroup of H. Then K is a characteristic (resp. normal) subgroup of G. 05BA
Remark (1) The normalizer (resp. centralizer) of A is the strict stabilizer (resp. fixer) of A when G operates on itself by inner automorphisms. In particular the centralizer is a normal… 05BB
Definition 5 Let G be a group, E a G-set and x ∈ G. An element y ∈ E is conjugate to x under the operation of G if there exists an element α ∈ G such that y = α x. The set of conjugate elements… 05BC
Example (1) Let G be a group and consider the operation of G on itself by inner automorphisms. Two elements of G which are conjugate under this operation are called conjugate under inner a… 05BD
Definition 6 Let G be a group. An operation of G on a set E is called transitive if there exists an element x ∈ E whose orbit is E. A G-set E is called homogeneous if the operation of G on E is… 05BE
Example If E is a G-set, each orbit of E, with the induced operation, is a homogeneous set under G. 05BF
Proposition 5 With the above notation, G/H is a homogeneous G-set. The mapping φ induces an isomorphism of (N/H)⁰ onto the group of automorphisms of the G-set G/H. ↩ 1 05BG
Remark (1) Let G be a group, H a subgroup of G and φ : G → S_G/H the homomorphism corresponding to the operation of G on G/H. The kernel of φ is the intersection of the conjugates of H (n… 05BH
Proposition 6 Let G be a group, E a homogeneous G-set, a ∈ E, H the stabilizer of a and K a subgroup of G contained in H. There exists one and only one G-morphism f of G/K into E such that f(e.K… ↩ 1 05BI
Theorem 1 Let G be a group. (a) Every homogeneous G-set is isomorphic to a homogeneous G-set of the form G/H where H is a subgroup of G. (b) Let H and H' be two subgroups of G. The G-sets G/… ↩ 2 05BJ
Example (1) Let E be a non-empty set. The group S_E operates transitively on E. If x and y are two elements of E, the mapping τ : E → E such that τ(x) = y, τ(y) = x and τ(z) = z for z ≠ x… ↩ 3 05BK
Definition 7 Let G be a group. An operation of G on a set E is called simply transitive if there exists an element x of E such that the orbital mapping defined by x is a bijection. A set E, tog… ↩ 1 05BL
Example (1) Let G operate on itself by left (resp. right) translation. Thus a left (resp. right) homogeneous principal G-set structure is defined on G, which is sometimes denoted by Gₛ (re… 05BM
Definition 8 Let E be a finite set, ζ ∈ 𝔖_E a permutation of E and barζ the subgroup of 𝔖_E generated by ζ. ζ is called a cycle if, under the operation of barζ on E, there exists one and only o… 05BN
Lemma 1 Let (ζᵢ)_i ∈ I be a family of cycles whose supports are pairwise disjoint. Then the ζᵢ are pairwise permutable. Let σ = ∏_i ∈ I ζᵢ and barσ be the subgroup generated by σ. Then σ(x… ↩ 1 05BO
Proposition 7 Let E be a finite set and σ a permutation of E. There exists one and only one finite set C of cycles satisfying the two following conditions: ↩ 1 05BP
Definition 9 A cycle of order 2 is called a transposition. 05BQ
Proposition 8 Let E be a finite set. The group 𝔖_E is generated by the transpositions. ↩ 3 05BR
Proposition 9 Let n be an integer ≥ 0. The group 𝔖ₙ is generated by the transpositions (τ_i, i+1)_1 ≤ i ≤ n-1. ↩ 4 05BS
Lemma 2 p ≠ 0 and σ p = (-1)^ν(σ) for σ ∈ 𝔖ₙ. ↩ 1 05BT
Theorem 2 Let E be a finite set. There exists one and only one homomorphism ε from 𝔖ₙ to the multiplicative group -1, +1 such that ε(τ) = -1 for every transposition τ. ↩ 1 05BU
Definition 10 In the notation of Theorem 2, the number ε(σ) (also denoted ε_σ) is called the signature of the permutation σ. The kernel of the homomorphism ε is called the alternating group of E… 05BV
Example If σ is a cycle of order d, then ε(σ) = (-1)ᵈ⁻¹. The number of inversions of the permutation (1, 2, 3, …, d) ↦ (d, 1, 2, …, d-1) is equal to d-1. 05BW

§ 6. Extensions, solvable groups, nilpotent groups

Definition 1 Let F and G be two groups. An extension of G by F is a triple E = (E, i, p), where E is a group, i an injective homomorphism of F into E and p a surjective homomorphism of E onto G… 05CQ
Proposition 1 Let E: F xrightarrowi E xrightarrowp G and E': F xrightarrowi' E' xrightarrowp' G be extensions of G by F. If u: E → E' is a morphism of E into E', u is an isomorphism of E onto E'… ↩ 1 05CR
Proposition 2 Let E: F xrightarrowi E xrightarrowp G be an extension of G by F. The following conditions are equivalent: 05CS
Definition 2 Let F and G be two groups and τ a homomorphism of G into the automorphism group of F. Write τ(g)(f) = gf for g ∈ G and f ∈ F. The set F × G with the law of composition ↩ 1 05CT
Proposition 3 The external semi-direct product F ×_τ G is a group. The mappings i: F → F ×_τ G defined by i(f) = (f, e), p: F ×_τ G → G defined by p(f, g) = g, and s: G → F ×_τ G defined by s(g)… ↩ 1 05CU
Proposition 4 With the above notation, there exists one and only one isomorphism u of E_τ onto E' such that u ∘ s = s'. ↩ 2 05CV
Remark The definition of the operation τ by formula (2) depends on the extension E' and the section s'. When F is commutative, the operation τ does not depend on s'. For Int(s'(g)) | i'(F… 05CW
Corollary Let G be a group and H and K two subgroups of G such that H is normal, H ∩ K = e and H.K = G. Let τ be the operation of K on H by inner automorphisms of G. The mapping (h, k) ↦ hk… ↩ 1 05CX
Example (1) Let G be a group and E a homogeneous principal G-set; let Γ denote the automorphism group of G. Let A be the set of permutations f of E with the following property: There exist… 05CY
Definition 3 Let G be a group and x and y two elements of G. The element x⁻¹y⁻¹xy of G is called the commutator of x and y. 05CZ
Proposition 5 Let A, B, C be three subgroups of G. ↩ 1 05D0
Definition 4 Let G be a group. The subgroup generated by the commutators of elements of G is called the derived group of G. 05D1
Proposition 6 Let f : G → G' be a group homomorphism. Then f(D(G)) ⊂ D(G'). If f is surjective, the homomorphism of D(G) into D(G') the restriction of f is surjective. ↩ 2 05D2
Corollary 1 The derived group of a group G is a characteristic subgroup of G. In particular it is a normal subgroup of G. 05D3
Corollary 2 Let G be a group. The quotient group G/D(G) is commutative. Let π : G → G/D(G) be the canonical homomorphism. Every homomorphism f of G into a commutative group G' can be expressed… ↩ 2 05D4
Corollary 3 Let H be a subgroup of G. The following conditions are equivalent: ↩ 3 05D5
Corollary 4 Let G be a group and X a subset of G which generates G. The group D(G) is the normal subgroup of G generated by the commutators of elements of X. ↩ 1 05D6
Remark (1) Corollary 2 can also be expressed by saying that G/D(G), together with π, is a solution of the universal mapping problem for G, relative to commutative groups and homomorphisms… 05D7
Example (1) If G is a non-commutative simple group, then D(G) = G. Therefore every homomorphism of G into a commutative group is trivial. 05D8
Definition 5 Let G be a group. The lower central series of G is the sequence (Cⁿ(G))_n ≥ 1 of subgroups of G defined inductively by: C¹(G) = G, Cⁿ⁺¹(G) = (G, Cⁿ(G)). ↩ 1 05D9
Definition 6 A group G is called nilpotent if there exists an integer n such that Cⁿ⁺¹(G) = e. The least integer n such that Cⁿ⁺¹(G) = e is called the nilpotency class of a nilpotent group G. ↩ 1 05DA
Example (1) A group is nilpotent of class 0 (resp. ≤ 1) if and only if it consists of the identity element (resp. is commutative). 05DB
Proposition 7 Let G be a group and n an integer. The following conditions are equivalent: ↩ 4 05DC
Corollary A central extension of a nilpotent group (by a necessarily commutative group) is nilpotent. 05DD
Proposition 8 Let G be a nilpotent group of class ≤ n and let H be a subgroup of G. There exists a sequence of subgroups ↩ 1 05DE
Corollary 1 Let G be a nilpotent group and H a subgroup of G. If H is distinct from G, the normalizer N_G(H) of H in G is distinct from H. ↩ 1 05DF
Corollary 2 Let G be a nilpotent group and H a subgroup of G. If H is distinct from G, there exists a normal subgroup N of G, containing H, distinct from G and such that G/N is commutative. ↩ 1 05DG
Corollary 3 Let G be a nilpotent group and H a subgroup of G. If G = H.(G, G), then G = H. ↩ 2 05DH
Corollary 4 Let f: G' → G be a group homomorphism. Suppose that (a) G is nilpotent. (b) The homomorphism f₁: G'/(G', G') → G/(G, G), derived from f by passing to the quotients, is surjective… 05DI
Proposition 9 Let G be a nilpotent group of class ≤ n and let N be a normal subgroup of G. There exists a series of subgroups ↩ 1 05DJ
Corollary 1 Let G be a nilpotent group, Z the centre of G and N a normal subgroup of G. If N ≠ {e}, then N ∩ Z ≠ {e}. ↩ 1 05DK
Corollary 2 Let f be a homomorphism from a nilpotent group G to a group G'. If the restriction of f to the centre of G is injective, f is injective. 05DL
Definition 7 Let G be a group. The derived series of G is the series (D^n(G)){n \in \mathbf{N}} defined inductively by: 05DM
Definition 8 A group G is called solvable if there exists an integer n such that Dⁿ(G) = e. If G is a solvable group, the least integer n such that Dⁿ(G) = e is called the solvability class of… 05DN
Example (1) A group is solvable of class 0 (resp. ≤ 1) if and only if it is reduced to e (resp. is commutative). 05DO
Proposition 10 Let G be a group and n an integer. The following conditions are equivalent: ↩ 1 05DP
Corollary Let G be a finite group and G = G⁰ ⊃ G¹ ⊃ … ⊃ Gⁿ = e a Jordan-Hölder series of G For G to be solvable, it is necessary and sufficient that the quotients Gᵏ/Gᵏ⁺¹ be cyclic of prime… 05DQ
Definition 9 A finite group whose order is a power of p is called a p-group. ↩ 1 05DR
Proposition 11 Let E be a finite set and G a p-group operating on E. Let Eᴳ denote the set of x ∈ E such that gx = x for all g ∈ G (the fixed points). Then Card(Eᴳ) ≡ Card(E) pmodp. ↩ 5 05DS
Corollary Let G be a p-group. If G is not reduced to e, its centre is not reduced to e. ↩ 3 05DT
Theorem 1 Let G be a p-group and pʳ its order. There exists a sequence of subgroups of G ↩ 4 05DU
Corollary Every p-group is nilpotent. ↩ 1 05DV
Proposition 12 Let G be a p-group and H a subgroup of G distinct from G. Then: ↩ 3 05DW
Corollary Let G be a p-group. Every subgroup of G of index p is normal. 05DX
Definition 10 Let G be a finite group. A Sylow p-subgroup of G is any subgroup P of G satisfying the two following conditions: 05DY
Example (1) In the group 𝔖ₚ let ζ be a cycle of order p. The subgroup generated by ζ is a Sylow p-subgroup of 𝔖ₚ for p does not divide (p - 1)!. (2) Let k be a finite field of characterist… 05DZ
Theorem 2 Every finite group contains a Sylow p-subgroup. ↩ 3 05E0
Lemma Let n = pʳ m, where m is an integer which is not a multiple of p. Then binomnpʳ not≡ 0 pmodp. 05E1
Corollary If the order of G is divisible by p, the group G contains an element of order p. 05E2
Remark For every prime number q dividing Card(G), let Pq be a Sylow q-subgroup of G. Then the subgroup H of G generated by the Pq is of order a multiple of Card(Pq) for each q and of orde… 05E3
Theorem 3 Let G be a finite group. (a) The Sylow p-subgroups of G are conjugate to one another. Their number is congruent to 1 mod. p. (b) Every subgroup of G which is a p-group is contained… ↩ 2 05E4
Corollary 1 Let P be a Sylow p-subgroup of G, let N be its normalizer in G and let M be a subgroup of G containing N. The normalizer of M in G is equal to M. ↩ 1 05E5
Corollary 2 Let f : G₁ → G₂ be a homomorphism of finite groups. For every Sylow p-subgroup P₁ of G₁ there exists a Sylow p-subgroup P₂ of G₂ such that f(P₁) ⊂ P₂. 05E6
Corollary 3 (a) Let H be a subgroup of G. For every Sylow p-subgroup P of H there exists a Sylow p-subgroup Q of G such that P = Q ∩ H. 05E7
Corollary 4 Let N be a normal subgroup of G. The image in G/N of a Sylow p-subgroup of G is a Sylow p-subgroup of G/N and every Sylow p-subgroup of G/N is obtained in this way. 05E8
Theorem 4 Let G be a finite group. The following conditions are equivalent: ↩ 6 05E9
Remark (1) Let G be a finite group and p a prime number. By no. 6, Theorem 3 (a) and no. 6, Theorem 2, the following conditions are equivalent: (i) there exists a normal Sylow p-subgroup… 05EA
Example The group 𝔖₃ is of order 6. It contains a normal Sylow 3-subgroup of order 3: the group A₃. It contains three Sylow 2-subgroups of order 2: the groups e, τ, where τ is a transposit… 05EB

§ 7. Free monoids, free groups

Proposition 1 Let M be a magma. Every mapping f of X into M may be extended in a unique way to a morphism of M(X) into M. ↩ 2 05FH
Proposition 2 Let u : X → Y be a mapping. If u is injective (resp. surjective, bijective), so is M(u). 05FI
Proposition 3 Let M be a monoid. Every mapping f of X into M extends uniquely to a homomorphism of Mo(X) into M. ↩ 1 05FJ
Proposition 4 (a) For all i ∈ I, the mapping φᵢ is a homomorphism of Mᵢ into M and φᵢ ∘ hᵢ = h for all i ∈ I. Further, M is generated by ⋃_i ∈ I φᵢ(Mᵢ). ↩ 3 05FK
Proposition 5 Under hypothesis (A) every element x of M admits a unique reduced decomposition σ. Every decomposition σ' ≠ σ of x satisfies l(σ') > l(σ). ↩ 2 05FL
Lemma 1 Let i₁, …, iₙ be in I and p_α in P_i_α for 1 ≤ α ≤ n. For all a ∈ A there exists a' ∈ A and a sequence (p'_α)_1 ≤ α ≤ n with p'_α ∈ P_i_α such that φ_i₁(p₁) … φ_iₙ(pₙ)h(a) = h(a')φ… 05FM
Corollary Under hypothesis (A) the homomorphisms φᵢ and h are injective. For i ≠ j in I, φᵢ(Mᵢ) ∩ φⱼ(Mⱼ) = h(A). ↩ 1 05FN
Lemma 2 Let M be the monoidal sum of the family (Mₓ)_x ∈ X defined by Mₓ = N for all x ∈ X and let φₓ denote the canonical homomorphism of Mₓ into M. The mapping x ↦ φₓ(1) of X into M exte… ↩ 1 05FO
Proposition 6 Let w be an element of Mo(X). (a) There exist an integer n ≥ 0, elements x_α of X and integers m(α) > 0 (for ↩ 2 05FP
Proposition 7 Let g be an element of the free group F(X). There exist an integer n ≥ 0 and a sequence (x_α, m(α))_1 ≤ α ≤ n determined uniquely by the relations x_α ∈ X, x_α ≠ x_α+1 for 1 ≤ α <… ↩ 2 05FQ
Proposition 8 Let G be a group and f a mapping of X into G. There exists one and only one homomorphism barf of F(X) into G which extends f. ↩ 1 05FR
Example (1) The group defined by the presentation ⟨ τ; τ^q = e ⟩ is cyclic of order q. 05FS
Proposition 9 Let G be a group, t = (tᵢ)_i ∈ I a generating family of G and r = (rⱼ)_j ∈ J a family of relators of t. The following conditions are equivalent: 05FT
Proposition 10 Let M be a commutative monoid (resp. group) and f a mapping of X into M. There exists one and only one homomorphism of N^(X) (resp. Z^(X)) into M such that g(δₓ) = f(x) for all x ∈… ↩ 2 05FU
Remark Let M be the multiplicative monoid of strictly positive integers and let 𝔓 be the set of prime numbers (§ 4, no. 10, Definition 15). By Proposition 10 there exists a homomorphism u… 05FV
Proposition 11 The canonical homomorphism ν of Mo(X) into N^(X) is surjective. Let w = x₁ … xₙ and w' = x'₁ … x'ₘ be two elements of Mo(X); in order that ν(w) = ν(w'), it is necessary and suffici… ↩ 1 05FW
Remark Let S be a subset of X. Recall that we have identified M(S) with a submagma of M(X), Mo(S) with a submonoid of Mo(X) and N^(S) with a submonoid of N^(X). Then M(S) = λ⁻¹(Mo(S)). Cl… 05FX

§ 8. Rings

Definition 1 A ring is a set A with two laws of composition called respectively addition and multiplication, satisfying the following axioms: 05H2
Proposition 1 Let A be a commutative ring and (x_λ)_λ ∈ L a finite family of elements of A. For every family of positive integers β = (β_λ)_λ ∈ L, let |β| = ∑_λ ∈ L β_λ. Then ( ∑_λ ∈ L x_λ )ⁿ =… 05H3
Corollary 1 (binomial formula) Let x and y be two elements of a commutative ring A. Then: ↩ 2 05H4
Corollary 2 Let A be a commutative ring, X a set, u = (uₓ)_x ∈ X and v = (vₓ)_x ∈ X two families of elements of A. Let u + v denote the family (uₓ + vₓ)_x ∈ X. For all λ ∈ N^(X) we write λ! =… ↩ 2 05H5
Proposition 2 Let A be a ring, x₁, …, xₙ elements of A and I = 1, 2, …, n. For H ⊂ I, we write x_H = ∑_i ∈ H xᵢ. Then ↩ 4 05H6
Definition 2 Let A and B be two rings. A morphism, or homomorphism, of A into B is any mapping f of A into B satisfying the relations: 05H7
Example (1) Let A be a ring. It is immediately seen that the mapping n ↦ n · 1 of Z into A is the unique homomorphism of Z into A. In particular, the identity mapping of Z is the unique en… 05H8
Definition 3 Let A be a ring. A subring of A is any subset B of A which is a subgroup of A under addition, which is stable under multiplication and which contains the unit of A. 05H9
Example (1) Every subgroup of the additive group Z which contains 1 is equal to Z. Thus Z is the only subring of Z. 05HA
Definition 4 Let A be a ring. A subset a of A is called a left (resp. right) ideal if it is a subgroup of the additive group of A and the relations a ∈ A, x ∈ a imply ax ∈ a (resp. xa ∈ a). a i… 05HB
Example (1) Let A be a ring. The set A is a two-sided ideal of A; so is the set consisting of 0, which is called the zero ideal and sometimes denoted by 0 or (0) instead of 0. 05HC
Definition 5 Let A be a ring. By an abuse of language, a left ideal a is said to be maximal if it is a maximal element of the set of left ideals distinct from A. 05HD
Theorem 1 (Krull) Let A be a ring and a a left ideal of A distinct from A. There exists a maximal ideal m of A containing a. ↩ 2 05HE
Proposition 3 Let A be a ring, (x_λ)_λ ∈ L a family of elements of A and a (resp. b) the set of sums ∑_λ ∈ L a_λ x_λ where (a_λ)_λ ∈ L is a family with finite support of elements of A (resp. ∑_λ… 05HF
Proposition 4 Let A be a ring and (a_λ)_λ ∈ L a family of left ideals of A. The left ideal generated by ⋃_λ ∈ L a_λ consists of the sums ∑_λ ∈ L y_λ where (y_λ)_λ ∈ L is a family with finite sup… 05HG
Definition 6 Let A be a ring and a a two-sided ideal of A. The quotient ring of A by a, denoted by A/a, is the quotient set of A by the equivalence relation x ≡ y(a), with addition and multipli… 05HH
Theorem 2 Let A be a ring and a a two-sided ideal of A. (a) The canonical mapping π of A onto A/a is a ring homomorphism. (b) Let B be a ring and f a homomorphism of A into B. If f(a) = 0, t… ↩ 8 05HI
Theorem 3 Let A and B be rings and f a homomorphism of A into B. (a) The kernel a of f is a two-sided ideal of B. (b) The image B' = f(B) of f is a subring of B. (c) Let π : A → A/a and i… ↩ 1 05HJ
Proposition 5 Let A and A' be two rings, f a homomorphism of A into A' and a the kernel of f. (a) Let B' be a subring of A'. Then B = f⁻¹(B') is a subring of A containing a. If f is surjective… ↩ 1 05HK
Remark In the above notation, θ and η are ordered set isomorphisms (Φ and Φ' being ordered by inclusion). 05HL
Corollary Let A be a ring and a a two-sided ideal of A. (a) Every left (resp. right, two-sided) ideal of A/a can be written uniquely in the form b/a, where b is a left (resp. right, two-side… 05HM
Proposition 6 Let a, b₁, …, bₙ be two-sided ideals of A. If A = a + bᵢ for all i, then A = a + b₁ b₂ … bₙ = a + (b₁ ∩ b₂ ∩ ⋯ ∩ bₙ). ↩ 4 05HN
Proposition 7 Let b₁, …, bₙ be two-sided ideals of A such that bᵢ + bⱼ = A for i ≠ j. Then b₁ ∩ b₂ ∩ ⋯ ∩ bₙ = ∑_σ ∈ Sₙ b_σ(1) b_σ(2) ⋯ b_σ(n). In particular, if A is commutative, b₁ ∩ b₂ ∩ ⋯ ∩ b… ↩ 2 05HO
Proposition 8 Suppose I is finite. If b is a left or right ideal of A, b is the direct sum of the b ∩ aᵢ. ↩ 3 05HP
Proposition 9 Let A be a ring and (b₁, …, bₙ) two-sided ideals of A such that bᵢ + bⱼ = A for i ≠ j. The canonical homomorphism of A into ∏ᵢ₌₁ⁿ (A/bᵢ) is surjective of kernel ⋂ᵢ₌₁ⁿ bᵢ = ∑_σ ∈ Sₙ… ↩ 5 05HQ
Definition 7 Let A be a ring. A finite family (bᵢ)_i ∈ I of two-sided ideals of A such that the canonical homomorphism of A into ∏_i ∈ I (A/bᵢ) is an isomorphism is called a direct decompositio… ↩ 1 05HR
Proposition 10 Let A be a ring, A' its centre and (bᵢ)_i ∈ I a finite family of two-sided ideals of A. The following conditions are equivalent: ↩ 3 05HS
Remark Let A be a ring. Let (aᵢ)_i ∈ I be a finite family of subgroups of the additive group A⁺ of A such that A⁺ is the direct sum of the aᵢ. Suppose aᵢ aᵢ ⊂ ai for i ∈ I and aᵢ aⱼ = 0 f… 05HT
Example Ideals and quotient rings of Z 05HU
Proposition 11 Let n₁, …, n_d be integers geqslant 1 which are relatively prime in pairs and n = n₁ … nᵣ. The canonical homomorphism of Z into the product ring ∏ᵢ₌₁ʳ Z/nᵢZ is surjective of kernel… ↩ 1 05HV
Theorem 4 Let A be a commutative ring and S a subset of A. Let A_S be the monoid of fractions of A (provided only with multiplication) with denominators in S (§ 2, no. 4). Let ε : A → A_S be… ↩ 1 05HW
Definition 8 The ring defined in Theorem 4 is called the ring of fractions associated with S, or with denominators in S, and is denoted by A[S⁻¹]. 05HX
Theorem 5 Let A be a commutative ring, S a subset of A, B a ring and f a homomorphism of A into B such that every element of f(S) is invertible. There exists one and only one barf of A[S⁻¹]… ↩ 1 05HY

§ 9. Fields

Definition 1 A ring K is called a field if it does not consist only of 0 and every non-zero element of K is invertible. 05IF
Example (1) We shall define in no. 4 the field of rational numbers; In General Topology the field of real numbers will be defined (General Topology, IV, § 1, no. 3), also the field of comp… 05IG
Proposition 1 Let K be a field. For every subset X of K, the centralizer (§ 8, no. 5, Example 3) X' of X is a subfield of K. 05IH
Corollary The centre of a field K is a (commutative) subfield of K. 05II
Theorem 1 Let A be a ring. The following conditions are equivalent: ↩ 7 05IJ
Remark In Theorem 1, left ideals may be replaced by right ideals. In Chapter VIII, § 5, no. 2, we shall study non-zero rings A which have no two-sided ideal distinct from 0 and A; such ri… 05IK
Corollary 1 Let A be a ring and a a two-sided ideal of A. For the ring A/a to be a field, it is necessary and sufficient that a be a maximal left ideal of A. ↩ 3 05IL
Corollary 2 Let A be a commutative ring which is not 0. There exists a homomorphism of A onto a commutative field. 05IM
Corollary 3 Let a be an integer geqslant 0. For the ring Z/aZ to be a field, it is necessary and sufficient that a be prime. 05IN
Theorem 2 Let K be a field and A a non-zero ring. If f is a homomorphism of K into A, then the subring f(K) of A is a field and f defines an isomorphism of K onto f(K). ↩ 2 05IO
Definition 2 A ring A is called an integral domain (or a domain of integrity) if it is commutative, non-zero, and the product of two non-zero elements of A is non-zero. 05IP
Proposition 2 If A is an integral domain, the total ring of fractions K of A is a commutative ring. 05IQ
Proposition 3 Let B be a non-zero ring and A a commutative subring of B such that every non-zero element of A is invertible in B. (a) A is an integral domain. (b) Let A' be the field of fraction… 05IR
Proposition 4 Let A be a commutative ring and 𝔭 an ideal of A. The following conditions are equivalent: (a) the ring A/𝔭 is an integral domain; (b) A ≠ 𝔭 and, if x ∈ A - 𝔭 and y ∈ A - 𝔭, then xy… ↩ 1 05IS
Definition 3 In a commutative ring A, an ideal p satisfying the conditions of Proposition 4 is called a prime ideal. ↩ 2 05IT
Example (1) Let A be a commutative ring. If m is a maximal ideal of A, m is prime; the ring A/m is a field (no. 1, Corollary 1 to Theorem 1). 05IU
Definition 4 The field of fractions of the ring Z of rational integers is called the field of rational numbers and is denoted by Q. The elements of Q are called rational numbers. 05IV

§ 10. Inverse and direct limits

Definition 1 An inverse system of magmas relative to the indexing set I is an inverse system of sets (E_α, f_αβ) relative to I, each E_α having a magma structure and each f_αβ being a magma hom… 05JH
Definition 2 A direct system of magmas relative to the indexing set I is a direct system of sets (E_α, f_βα) relative to I, each E_α having a magma structure and each f_βα being a magma homomor… 05JI
Proposition 1 There exists on E one and only one magma structure for which the mappings f_α : E_α → E are homomorphisms. If the magmas E_α are associative (resp. commutative), so is E. If the ma… ↩ 1 05JJ
Proposition 2 Let (E_α, f_β α) be a direct system of magmas and let E be its direct limits f_α : E_α → E the canonical homomorphisms. Suppose a magma F and homomorphisms u_α : E_α → F are given… ↩ 1 05JK
Proposition 3 (a) If the A_α are non-zero, A is non-zero. (b) If the A_α are integral domains, A is an integral domain. (c) If the A_α are fields, A is a field. 05JL

II. Linear algebra

§ 1. Modules

Definition 1 Given a ring A, a left module over A (or left A-module) is a set E with an algebraic structure defined by giving: ↩ 1 05JO
Example (1) Let φ be a homomorphism of a ring A into a ring B; the mapping (a, x) ↦ φ(a)x (resp. (a, x) ↦ xφ(a)) of A × B into B defines on B a left (resp. right) A-module structure. In pa… 05JP
Definition 2 A left (resp. right) vector space over a field K is a left (resp. right) K-module. 05JQ
Example (4) A field is both a left and a right vector space with respect to any of its subfields. 05JR
Definition 3 An element x of an A-module E is said to be a linear combination with coefficients in A of a family (aᵢ)_i ∈ I of elements of E if there exists a family (λᵢ)_i ∈ I of elements of A… 05JS
Definition 4 Let E and F be two (left) modules with respect to the same ring A. A linear mapping (or A-linear mapping, or homomorphism, or A-homomorphism) of E into F is any mapping u : E → F s… ↩ 1 05JT
Remark When E and F are two commutative groups considered as modules over the ring Z (no. 1), every homomorphism u of the group E (without operators) into the group F (without operators)… 05JU
Example (1) Let E be an A-module and a an element of E; the mapping λ ↦ λ a of the A-module Aₛ into E is a linear mapping θₐ such that θₐ(1) = a. (2) Let I be an open interval of the real… 05JV
Proposition 1 Let E, F, G be three A-modules, u a linear mapping of E into F and v a linear mapping of F into G. Then the composite mapping v ∘ u is linear. ↩ 2 05JW
Example (1) In any module E the set consisting of 0 is a submodule (the zero submodule, often denoted by 0, by an abuse of notation). (2) Let A be a ring. The submodules of Aₛ (resp. A_d)… 05JX
Remark Let M be a submodule of an A-module E and φ : E → E/M the canonical homomorphism. For an A-linear mapping u : E → F to be of the form v ∘ φ, where v is a linear mapping of E/M into… 05JY
Definition 5 Let F, G, H be three A-modules; let f be a homomorphism of F into G and g a homomorphism of G into H. The ordered pair (f, g) is called an exact sequence if 05JZ
Remark (1) If the ordered pair (f, g) is an exact sequence, then g ∘ f = 0; but of course this property does not characterize exact sequences, for it only means that the image of f is con… 05K0
Proposition 3 (a) For 0 longrightarrow E xrightarrowf F to be an exact sequence, it is necessary and sufficient that f be injective. ↩ 1 05K1
Remark (2) To say that there is an exact sequence ↩ 1 05K2
Remark (5) The definition of an exact sequence extends immediately to groups which are not necessarily commutative; in this case of course multiplicative notation is used with 0 replaced… 05K3
Proposition 4 Let E = ∏_i ∈ I Eᵢ be the product of a family of A-modules (Eᵢ)_i ∈ I. For every A-module F and every family of linear mappings fᵢ : F → Eᵢ there exists one and only one mapping f… ↩ 2 05K4
Proposition 5 (i) Let (Eᵢ)_i ∈ I, (Fᵢ)_i ∈ I be two families of A-modules with the same indexing set I; for every family of linear mappings fᵢ : Eᵢ → Fᵢ (i ∈ I), the mapping f : (xᵢ) → (fᵢ(xᵢ))… ↩ 3 05K5
Corollary Under the conditions of Proposition 5, (i), ↩ 1 05K6
Proposition 6 Let (Eᵢ)_i ∈ I be a family of A-modules, M an A-module and, for all i ∈ I, let fᵢ : Eᵢ → M be a linear mapping. Then there exists one and only one linear mapping g : ⨁_i ∈ I Eᵢ → M… ↩ 11 05K7
Corollary 1 Let (Eᵢ)_i ∈ I, (F_λ)_λ ∈ L be two families of A-modules. The mapping Hom_A ( ⨁_i ∈ I Eᵢ, ∏_λ ∈ L F_λ ) → ∏_(i, λ) ∈ I × L Hom_A(Eᵢ, F_λ) which associates with each g ∈ Hom_A ( ⨁_i… ↩ 8 05K8
Corollary 2 Let (Eᵢ)_i ∈ I be a family of A-modules, F an A-module and, for each i ∈ I, let fᵢ : Eᵢ → F be a linear mapping. For f = ∑_i ∈ I fᵢ to be an isomorphism of E = ⨁_i ∈ I Eᵢ onto F, i… ↩ 3 05K9
Proposition 7 (i) Let (Eᵢ)_i ∈ I, (Fᵢ)_i ∈ I be two families of A-modules with the same indexing set I; for every family of linear mappings fᵢ : Eᵢ → Fᵢ (i ∈ I), the restriction to ⨁_i ∈ I Eᵢ of… ↩ 4 05KA
Corollary 1 Under the conditions of Proposition 7, (i), Ker(f) = ⨁_i ∈ I Ker(fᵢ), Im(f) = ⨁_i ∈ I Im(fᵢ) and there are canonical isomorphisms Coim(f) → ⨁_i ∈ I Coim(fᵢ), Coker(f) → ⨁_i ∈ I Cok… ↩ 3 05KB
Corollary 2 Let (Eᵢ)_i ∈ I, (E'ᵢ)_i ∈ I, (F_λ)_λ ∈ L, (F'_λ)_λ ∈ L be four families of A-modules and, for each i ∈ I (resp. each λ ∈ L), uᵢ : E'ᵢ → Eᵢ (resp. v_λ : F_λ → F'_λ) a linear mapping… 05KC
Remark Recall that the definition of direct sum extends immediately to a family (Eᵢ)_i ∈ I of groups which are not necessarily commutative, multiplicative notation of course replacing the… 05KD
Proposition 8 If a family (Mᵢ)_i ∈ I of submodules of E has intersection reduced to 0 E is canonically isomorphic to a submodule of ∏_i ∈ I (E / Mᵢ). 05KE
Proposition 9 The submodule generated by a family (aᵢ)_i ∈ I of elements of an A-module E is the set of linear combinations of the family (aᵢ). ↩ 4 05KF
Corollary 1 Let u : E → F be a linear mapping, S a subset of E and M the submodule of E generated by S. Then u(M) is the submodule of F generated by u(S). ↩ 1 05KG
Remark If u(x) = 0 for all x ∈ S, then also u(x) = 0 for all x ∈ M. We shall sometimes refer to this result as the "principle of extension of linear identities" or "principle of extension… 05KH
Corollary 2 The submodule generated by the union of a family (Mᵢ)_i ∈ I of submodules of a module E is identical with the set of sums ∑_i ∈ I xᵢ, where (xᵢ)_i ∈ I runs through the set of famil… ↩ 1 05KI
Corollary 3 If (M_λ)_λ ∈ L is a right directed family of submodules of an A-module E, the sum ∑_λ ∈ L M_λ is identical with the union ⋃_λ ∈ L M_λ. 05KJ
Corollary 4 Let 0 → E xrightarrowf F xrightarrowg G → 0 be an exact sequence of A-modules, S a generating system of E, T a generating system of G. If T' is a subset of F such that g(T') = T, T… 05KK
Corollary 5 In an exact sequence 0 → E → F → G → 0 of A-modules, if E and G are finitely generated, so is F. ↩ 1 05KL
Proposition 10 Let M, N be two submodules of an A-module E. Then there are two exact sequences ↩ 3 05KM
Definition 6 An A-module E is said to be the direct sum of a family (M_ι)_ι ∈ I of submodules of E if the canonical mapping ⨁_ι ∈ I M_ι → E (no. 6) is an isomorphism. ↩ 1 05KN
Remark (1) Let (M_ι)_ι ∈ I, (N_ι)_ι ∈ I be two families of submodules of a module E, with the same indexing set; suppose that E is both the direct sum of the family (M_ι) and of the famil… ↩ 2 05KO
Proposition 11 Let (M_ι)_ι ∈ I be a family of submodules of an A-module E. The following properties are equivalent: ↩ 1 05KP
Definition 7 An endomorphism e of an A-module E is called a projector if e ∘ e = e (in other words, if e is an idempotent in the ring End(E)). In End(E) a family (e_λ)_λ ∈ L of projectors is ca… ↩ 1 05KQ
Proposition 12 Let E be an A-module. ↩ 4 05KR
Corollary For every projector e of E, E is the direct sum of the image M = e(E) and the kernel N = e⁻¹(0) of e; for all x = x₁ + x₂ ∈ E with x₁ ∈ M and x₂ ∈ N, x₁ = e(x); 1 − e is a projecto… ↩ 2 05KS
Remark (2) Let E, F be two A-modules such that E is the direct sum of a finite family (Mᵢ)_1 ≤ i ≤ m of submodules and F the direct sum of a finite family (Nⱼ)_1 ≤ j ≤ n of submodules. Th… 05KT
Definition 8 In an A-module E, two submodules M₁, M₂ are said to be supplementary if E is the direct sum of M₁ and M₂. 05KU
Proposition 13 Let M₁, M₂ be two supplementary submodules in an A-module E. The restriction to M₁ of the canonical mapping E → E/M₂ is an isomorphism of M₁ onto E/M₂. ↩ 3 05KV
Corollary If M₂ and M₂' are two supplements of the same submodule M₁ of E, the set of ordered pairs (x, x') ∈ M₂ × M₂' such that x - x' ∈ M₁ is the graph of an isomorphism of M₂ onto M₂'. ↩ 2 05KW
Definition 9 A submodule M of an A-module E is called a direct factor of E if it has a supplementary submodule in E. 05KX
Proposition 14 For a submodule M of a module E to be a direct factor, it is necessary and sufficient that there exist a projector of E whose image is M or a projector of E whose kernel is M. ↩ 3 05KY
Proposition 15 Given an exact sequence of A-modules (33) 0 longrightarrow E xrightarrowf F xrightarrowg G longrightarrow 0 the following propositions are equivalent: (a) The submodule f(E) of F i… ↩ 9 05KZ
Corollary 1 Let u : E → F be a linear mapping. For there to exist a linear mapping v : F → E such that u ∘ v = 1_F (the case where u is said to be right invertible and v is said to be the righ… ↩ 1 05L0
Corollary 2 Let u : E → F be a linear mapping. For there to exist a linear mapping v : F → E such that v ∘ u = 1_E (the case where u is said to be left invertible and v is said to be the left… ↩ 3 05L1
Remark (1) Let M, N be two supplementary submodules in an A-module E, p, q the projectors of E onto M and N respectively, corresponding to the decomposition of E as a direct sum of M and… 05L2
Proposition 16 Let M be an A-module and N a submodule of M; for M to be of finite length, it is necessary and sufficient that N and M/N be so, and then long(N) + long(M/N) = long(M). The proof ha… ↩ 6 05L3
Corollary 1 Let M be an A-module of finite length; for a submodule N of M to be equal to M, it is necessary and sufficient that long(N) = long(M). ↩ 1 05L4
Corollary 2 Let u : M → N be an A-module homomorphism. If M or N is of finite length, so is Im(u). If M is of finite length, so is Ker(u) and long(Im(u)) + long(Ker(u)) = long(M). If N is of f… ↩ 2 05L5
Corollary 3 Let (Mᵢ)_0 ≤ i ≤ n be a finite family of A-modules of finite length. If there exists an exact sequence of linear mappings 0 longrightarrow M₀ xrightarrowu₀ M₁ xrightarrowu₁ M₂ long… ↩ 5 05L6
Corollary 4 Let M and N be two submodules of finite length of an A-module E; then M + N is of finite length and long(M + N) + long(M ∩ N) = long(M) + long(N). It suffices to apply Corollary 3… 05L7
Corollary 5 Let M be an A-module the sum of a family (Nᵢ) of submodules of finite length. Then M is of finite length and long(M) ≤ ∑ᵢ long(Nᵢ). ↩ 2 05L8
Corollary 6 Let M and N be two submodules of an A-module E such that E/M and E/N are modules of finite length; then E/(M \cap N) is of finite length and 05L9
Corollary 7 Let (Mᵢ) be a finite family of submodules of an A-module E such that the E/Mᵢ are modules of finite length. Then E/(⋂ᵢ Mᵢ) is of finite length and 05LA
Remark With the exception of no. 7, Proposition 9, all the results of nos. 2 to 10 are valid for arbitrary commutative groups with operators, submodules (resp. quotient modules) being rep… 05LB
Proposition 17 For every A-module E and every mapping f : T → E, there exists one and only one A-linear mapping g : Aₛ⁽ᵀ⁾ → E such that f = g ∘ φ. ↩ 5 05LC
Corollary 1 Let T, T' be two sets, g : T → T' a mapping. Then there exists one and only one A-linear mapping f : A⁽ᵀ⁾ → A^(T') which renders commutative the diagram beginarrayccc T & oversetgl… 05LD
Corollary 2 For a family (aₜ)_t ∈ T of elements of an A-module E to be a generating system of E, it is necessary and sufficient that the linear mapping Aₛ⁽ᵀ⁾ → E determined by this family be s… ↩ 4 05LE
Definition 10 A family (aₜ)_t ∈ T of elements of an A-module E is called a free family (resp. a basis of E) if the linear mapping Aₛ⁽ᵀ⁾ → E determined by this family is injective (resp. bijectiv… ↩ 2 05LF
Corollary 3 Let E be a free A-module, (aₜ)_t ∈ T a basis of E, F an A-module and (bₜ)_t ∈ T a family of elements of F. There exists one and only one linear mapping f : E → F such that f(aₜ) =… ↩ 11 05LG
Proposition 18 For a family (aₜ)_t ∈ T of a module E to be free, it is necessary and sufficient that every finite subfamily of (aₜ)_t ∈ T be free. 05LH
Proposition 19 Let E be an A-module, the direct sum of a family (M_λ)_λ ∈ L of submodules. If, for each λ ∈ L, S_λ is a free subset (resp. generating set, basis) of M_λ, then S = ⋃_λ ∈ L S_λ is a… 05LI
Remark (1) By Definition 10, if A ≠ 0 and (aₗ)_l ∈ I is a free family, no element a_κ can be equal to a linear combination of the aₗ of index l ≠ κ. But conversely, a family (aₗ) satisfyi… ↩ 1 05LJ
Remark (2) A free module can have elements ≠ 0 which are not free: for example, the A-module Aₛ is free but the right divisors of zero in A are not free elements of Aₛ. 05LK
Proposition 20 Every A-module E is isomorphic to a quotient module of a free A-module. ↩ 4 05LL
Proposition 21 Every exact sequence of A-modules ↩ 10 05LM
Remark (5) Let (aᵢ)_1 ≤ i ≤ n be a basis of an A-module E and let (bᵢ)_1 ≤ i ≤ n be a family of elements of E given by the relations 05LN
Definition 11 The annihilator of a subset S of an A-module E is the set of elements α ∈ A such that α x = 0 for all x ∈ S. ↩ 1 05LO
Definition 12 A module is called monogenous if it is generated by a single element. 05LP
Example (1) Every monogenous group, being commutative (I, § 4, no. 10, Proposition 18), is a monogenous Z-module. (2) If A is a commutative ring, the monogenous submodules of the A-module… 05LQ
Proposition 22 Let A be a ring. Every quotient module of Aₛ is monogenous. Conversely, let E be a monogenous A-module, c a generator of E and a its annihilator; the linear mapping ξ ↦ ξ c defines… ↩ 2 05LR
Corollary Every submodule of a monogenous A-module E is isomorphic to a quotient module b/a where a and b are two left-ideals of A such that a ⊂ b. Every quotient module of a monogenous A-mo… 05LS
Proposition 23 Let E be an A-module, the direct sum of an infinite family (Mi){i \in I} of non-zero submodules. For every generating system S of E, Card(S) \geq Card(I). ↩ 2 05LT
Corollary 1 Under the hypotheses of Proposition 23, suppose that each Mi is monogenous and that E is the direct sum of a second family (N\lambda){\lambda \in L} of non-zero monogenous submodul… 05LU
Corollary 2 If a module E admits an infinite basis B, every generating system of E has cardinal \geq Card(B) and every basis of E is equipotent to B. ↩ 1 05LV
Proposition 24 Let A, B be two rings and ρ : B → A a ring homomorphism. 05LW
Proposition 25 Let A, B be two rings, ρ : B → A a ring homomorphism and E an A-module. Let (α_λ)_λ ∈ L be a generating system (resp. free family of elements, basis) of A considered as a left B-mo… ↩ 11 05LX
Corollary If A is a finitely generated left B-module and E a finitely generated left A-module, ρ_*(E) is a finitely generated left A-module. 05LY
Remark (1) Let K be a field and A a subring of K with the following property: for every finite family (ξᵢ)_1 ≤ i ≤ n of elements of K, there exists a γ ∈ A which is non-zero and such that… 05LZ
Example It has been seen (no. 1) that a homothety h_α : x ↦ α x on an A-module E is not necessarily a linear mapping. But if α is invertible, h_α is a semi-linear mapping (which is moreove… 05M0
Definition 13 Let (A_λ)_λ ∈ L,\ (B_μ)_μ ∈ M be two families of rings; an ((A_λ),\ (B_μ))-multimodule (or multimodule over the families of rings (A_λ)_λ ∈ L,\ (B_μ)_μ ∈ M) is a set E with, for ea… 05M1
Example (1) On a ring A the module structures of Aₛ and A_d are compatible and A can therefore be considered canonically as an (A, A)-bimodule. ↩ 2 05M2
Remark (1) Let F be an A-module and C the centre of the ring A; as central homotheties commute with all homotheties, F has a bimodule structure whose rings of left operators are A and C… ↩ 1 05M3

§ 2. Modules of linear mappings. Duality

Theorem 1 Let A be a ring, E', E, E'' three A-modules and u : E' \to E, v : E \to E'' two homomorphisms. For the sequence ↩ 8 05MV
Corollary For an A-linear mapping u : E → F to be surjective (resp. bijective, resp. zero), it is necessary and sufficient that, for every A-module G, the mapping Hom(u, 1_G) : Hom(F, G) → H… 05MW
Proposition 1 If the exact sequence of linear mappings ↩ 1 05MX
Theorem 2 Let A be a ring, F', F, F'' three A-modules and u: F' → F, v: F → F'' two homomorphisms. For the sequence ↩ 3 05MY
Remark (1) Theorem 2 allows us, for every submodule F' of F, to identify Hom(E, F') with a sub-Z-module of Hom(E, F). When this identification is made, then, for every family (M_λ) of sub… 05MZ
Corollary For an A-linear mapping u : E → F to be injective, it is necessary and sufficient that, for every A-module G, the mapping Hom(1_G, u) : Hom(G, E) → Hom(G, F) be injective. 05N0
Proposition 2 If the exact sequence ↩ 1 05N1
Definition 1 An A-module P is called projective if, for every exact sequence F' → F → F'' of A-linear mapping, the sequence ↩ 2 05N2
Proposition 3 For an A-module P, the direct sum of a family of submodules (Mi), to be projective, it is necessary and sufficient that each of the Mi be projective. ↩ 2 05N3
Corollary Every free A-module is projective. 05N4
Proposition 4 Let P be an A-module. The following properties are equivalent: ↩ 18 05N5
Corollary 1 For an A-module to be projective and finitely generated, it is necessary and sufficient that it be a direct factor of a free A-module with a finite basis. ↩ 4 05N6
Corollary 2 Let C be a commutative ring and E, F two finitely generated projective C-modules; then Hom_C(E, F) is a finitely generated projective C-module. 05N7
Definition 2 For every right A-module E, the right A-module Hom_A(E, Aₛ) is called the dual module of E (or simply the dual† of E) and its elements are called the linear forms on E. ↩ 1 05N8
Example On the vector space (with respect to the field R) of continuous real-valued functions on an interval [a, b] of R, the mapping x ↦ ∫ₐᵇ x(t) dt is a linear form. 05N9
Proposition 5 For every ring A, the mapping which associates with every ξ ∈ A the linear form η ↦ η ξ on Aₛ is an isomorphism of A_d onto the dual of Aₛ. ↩ 1 05NA
Definition 3 Let E be an A-module and E its dual; an element x ∈ E and an element x ∈ E are called orthogonal if ⟨x, x⟩ = 0. 05NB
Definition 4 Given a subset M of E (resp. a subset M' of E), the set of x ∈ E (resp. the set of x ∈ E) which are orthogonal to M (resp. M') is called the submodule totally orthogonal to M (resp… ↩ 1 05NC
Proposition 6 Let M, N be two subsets of E such that M ⊂ N; if M' and N' are the submodules of E orthogonal to M and N respectively, then N' ⊂ M'. ↩ 1 05ND
Proposition 7 Let (Mi) be a family of subsets of E; the submodule orthogonal to the union of the Mi is the intersection of the submodules M'i which are respectively orthogonal to the Mi; this su… 05NE
Definition 5 For every linear mapping u of an A-module E into an A-module F, the linear mapping y ↦ y ∘ u of the dual F of F into the dual E of E is called the transpose of u and is denoted by… ↩ 1 05NF
Remark From (17) and (18) it follows that if u is left (resp. right) invertible, ᵗ u is right (resp. left) invertible. 05NG
Proposition 8 Let u : E → F be an A-linear mapping, M a submodule of E and M' the orthogonal of M in E^*; the orthogonal of u(M) in F^* is the inverse image ᵗ u⁻¹(M'). 05NH
Corollary The orthogonal of the image u(E) in F^* is the kernel ᵗ u⁻¹(0) of ᵗ u. ↩ 2 05NI
Definition 6 Given an isomorphism u of an A-module E onto an A-module F, the transpose of the inverse isomorphism of u (equal to the inverse isomorphism of the transpose of u) is called the con… ↩ 1 05NJ
Proposition 9 Let E', E, E'' be A-modules and 05NK
Corollary Let M be a submodule of an A-module E and φ : E → E/M the canonical homomorphism. Then ᵗ φ is an isomorphism of the dual of E/M onto the submodule M' of E^* orthogonal to M. ↩ 3 05NL
Proposition 10 Let (Eᵢ)_i ∈ I be a family of A-modules and for all i ∈ I let jᵢ : Eᵢ → E = ⨁_i ∈ I Eᵢ be the canonical injection. Then the product mapping x^* ↦ (ᵗ jᵢ(x^*)) ↩ 4 05NM
Corollary Let M, N be two supplementary submodules in an A-module E and p : E → M, q : E → N the corresponding projectors; then ᵗp + ᵗq : M^* ⊕ N^* → E^* is an isomorphism and ᵗp (resp. ᵗq)… ↩ 2 05NN
Proposition 11 The dual of a free module with a basis of n elements is a free module with a basis of n elements. ↩ 1 05NO
Definition 7 If E is a free module with a finite basis (eₜ), the basis (eₜ^*) of the dual E of E defined by relations (23) is called the dual basis of (eₜ). 05NP
Corollary The dual of a finitely generated projective module is a finitely generated projective module. 05NQ
Proposition 12 Let E be an A-module and (aₜ)_t ∈ T a generating system of E. The following conditions are equivalent: (a) E is a projective A-module. (b) There exists a family (aₜ^*)_t ∈ T of lin… ↩ 5 05NR
Proposition 13 If E is a free module (resp. a free module with a finite basis), the canonical mapping c_E : E → E^** is injective (resp. bijective). 05NS
Corollary 1 Let E be a free A-module with a finite basis; for every basis (eₜ) of E, (c_E(eₜ)) is the dual basis of the basis (eₜ^*) of E^* dual to (eₜ). 05NT
Corollary 2 If E is a free A-module with a finite basis, every finite basis of E is the dual basis of a basis of E. 05NU
Corollary 3 Let E, F be two A-modules each with a finite basis, E (resp. F) being canonically identified with its bidual E (resp. F). Then, for every linear mapping u : E → F, tᵗ u = u. 05NV
Corollary 4 If P is a projective module (resp. a finitely generated projective module) the canonical mapping c_P : P → P** is injective (resp. bijective). ↩ 5 05NW
Lemma 1 Let M, N be two supplementary submodules in an A-module E and i : M → E, j : N → E the canonical injections. Then the diagram 05NX
Proposition 14 If x₀ is a solution of the linear equation u(x) = y₀, the set of solutions of this equation is equal to the set of elements x₀ + z, where z runs through the set of solutions of the… ↩ 1 05NY
Proposition 15 Let u be a linear mapping of a module E into a module F. If the equation u(x) = y₀ has at least one solution, y₀ is orthogonal to the kernel of ᵗ u. ↩ 1 05NZ
Remark (1) Let E be an A-module, (Fᵢ)_i ∈ I a family of A-modules and for all i ∈ I let uᵢ : E → Fᵢ be a linear mapping. Every system of linear equations uᵢ(x) = yᵢ (i ∈ I) where the yᵢ ∈… 05O0

§ 3. Tensor products

Definition 1 The tensor product of the right A-module E and the left A-module F, denoted by E ⊗_A F or E ⊗_A F (or simply E ⊗ F if no confusion is to be feared) is the quotient Z-module C/D (th… ↩ 3 05OR
Proposition 1 (a) Let g be a Z-linear mapping of E ⊗_A F into a Z-module G. The mapping (x, y) ↦ f(x, y) = g(x ⊗ y) of E × F into G is Z-bilinear and satisfies conditions (1). ↩ 8 05OS
Corollary 1 Let H be a Z-module and h : E × F → H a Z-bilinear mapping satisfying conditions (1) and such that H is generated by h(E × F). Suppose that for every Z-module G and every Z-bilinea… ↩ 1 05OT
Corollary 2 Let E⁰ (resp. F⁰) denote the module E (resp. F) considered as a left (resp. right) module over the opposite ring A⁰; then there exists one and only one Z-module isomorphism σ : E ⊗… ↩ 3 05OU
Remark The tensor product of non-zero modules may be zero: for example, taking the two Z-modules E = Z/2Z and F = Z/3Z, 2x = 0 and 3y = 0 for all x ∈ E and y ∈ F; therefore, in E ⊗_Z F, x… 05OV
Proposition 2 Let A, B be two rings, ρ : B → A a ring homomorphism and E (resp. F) a right (resp. left) A-module. Then there exists one and only one Z-linear mapping (6) φ : ρ_*(E) ⊗_B ρ_*(F) →… ↩ 1 05OW
Corollary Let 𝔍 be a two-sided ideal of A such that 𝔍 is contained in the annihilator of E and in the annihilator of F, so that E (resp. F) has a canonical left (resp. right) (A/𝔍)-module st… ↩ 1 05OX
Proposition 3 Let G be a ((B'i), (B''h); (C'j), (C''k))-multimodule. ↩ 2 05OY
Scholium An element of E ⊗_A F may in general be written in several ways in the form ∑ᵢ (xᵢ ⊗ yᵢ), where xᵢ ∈ E and yᵢ ∈ F; but to define a linear mapping g of the multimodule E ⊗_A F into… 05OZ
Proposition 4 If E is a right A-module, the mapping h : x ↦ x ⊗ 1 of E into E ⊗_A (sA_d) is a right A-module isomorphism, whose inverse isomorphism g is such that g(x ⊗ λ) = xλ for x ∈ E, λ ∈ A. ↩ 16 05P0
Remark (1) Let A, B be two commutative rings, ρ : B → A a ring homomorphism and E and F two A-modules; then the canonical mapping (6) of no. 3 is a B-linear mapping 05P1
Proposition 5 Let E, E', E'' be right A-modules, F a left A-module and ↩ 8 05P2
Corollary Let F, F', F'' be left A-modules, E a right A-module and beginarrayccccc F' & xrightarrows & F & xrightarrowt & F'' longrightarrow 0 endarray (16) an exact sequence of linear mappi… ↩ 2 05P3
Remark Note that in general, if E' is a submodule of a right A-module E and j : E' \to E the canonical injection, the canonical mapping j ⊗ 1_F : E' ⊗ F → E ⊗ F is not necessarily injecti… 05P4
Proposition 6 Given two exact sequences (14) and (16), the homomorphism v \otimes t : E \otimesA F \to E'' \otimesA F'' is surjective and its kernel is equal to Im(u ⊗ 1_F) + Im(l_E ⊗ s) ↩ 4 05P5
Corollary 1 Let E' be a submodule of a right A-module E, F' a submodule of a left A-module F and Im(E' ⊗_A F) and Im(E ⊗_A F') the sub-Z-modules of E ⊗_A F, the respective images of the canoni… ↩ 2 05P6
Corollary 2 Let a be a right ideal of A, F a left A-module and aF the sub-Z-module of F generated by the elements of the form λ x, where λ ∈ a and x ∈ F. Then there is a canonical Z-module iso… ↩ 1 05P7
Corollary 3 Let A be a commutative ring, a an ideal of A and E and F two A-modules such that a is contained in the annihilator of F. Then the (A/a)-modules E ⊗_A F and (E/aE) ⊗_A/a F are canon… ↩ 1 05P8
Corollary 4 Let a, b be two ideals in a commutative ring C; the C-module (C/a) ⊗_C (C/b) is then canonically isomorphic to C(a + b). ↩ 1 05P9
Proposition 7 The canonical mapping (23) is bijective. ↩ 15 05PA
Corollary 1 If the left A-module F admits a basis (b_μ)_μ ∈ M, every element of E ⊗_A F can be written uniquely in the form ∑_μ (x_μ ⊗ b_μ), where x_μ ∈ E and the family (x_μ) has finite suppo… ↩ 9 05PB
Corollary 2 If E and F are two free modules over a commutative ring C and (aλ) (resp. (bμ)) is a basis of the C-module E (resp. F), then (aλ ⊗ bμ) is a basis of the C-module E ⊗C F. ↩ 9 05PC
Corollary 3 Let (Eλ){λ ∈ L} be a family of right A-modules and F a free (resp. finitely generated free) left A-module. Then the canonical mapping (22) ↩ 2 05PD
Corollary 4 Let A be a ring with no divisor of zero, E a free right A-module and F a free left A-module. Then the relation x ⊗ y = 0 in E ⊗_A F implies x = 0 or y = 0. ↩ 1 05PE
Corollary 5 Let E be a right A-module, F a left A-module, M a submodule of E and N a submodule of F. If M is a direct factor of E and N a direct factor of F, the canonical homomorphism M ⊗_A N… ↩ 4 05PF
Corollary 6 Let P be a projective left A-module and E, F two right A-modules. For every injective homomorphism u : E → F, the homomorphism u ⊗ 1_P : E ⊗_A P → F ⊗_A P is injective. ↩ 4 05PG
Corollary 7 Let C be a commutative ring. If E and F are two projective C-modules, E ⊗_C F is a projective C-module. 05PH
Remark 2 Under the hypotheses of Proposition 7, let E'_λ be a submodule of E_λ, F'_μ a submodule of F_μ and let E' = ⨁_λ ∈ L E'_λ, F' = ⨁_μ ∈ M F'_μ. Let Im(E' ⊗_A F') (resp. Im(E'λ ⊗A F'μ)… 05PI
Proposition 8 Let A, B be two rings, E a right A-module, F an (A, B)-bimodule and G a left B-module. Then E ⊗A F is a right B-module, F ⊗B G a left A-module and there exists one and only one Z-l… ↩ 6 05PJ
Remark (1) The above remarks concerning the case of modules over a commutative ring generalize as in no. 5, Remark 2 when there is a tensor product E₁ ⊗_A₁ E₂ ⊗ ⋯ ⊗_Aₙ₋₁ Eₙ where the ring… 05PK

§ 4. Relations between tensor products and homomorphism modules

Proposition 1 (a) Let E be a (B, A)-bimodule, F a left A-module and G a left B-module. For every mapping g ∈ Hom_B(E ⊗_A F, G), let g' be the mapping of F into Hom_B(E, G) defined by (g'(y))(x)… ↩ 4 05PP
Corollary If C is a commutative ring and E, F, G three C-modules, then the C-modules Hom_C(E ⊗_C F, G), Hom_C(E, Hom_C(F, G)), Hom_C(F, Hom_C(E, G)), L₂(E, F; G) are canonically isomorphic. ↩ 1 05PQ
Proposition 2 (i) When F is a projective (resp. finitely generated projective) B-module, the canonical homomorphism (7) is injective (resp. bijective). ↩ 7 05PR
Lemma 1 For every homomorphism u : F → F', the diagram ↩ 1 05PS
Lemma 2 Let M, N be two supplementary submodules in F and i : M → F, j : N → F the canonical injections. The diagram ↩ 1 05PT
Lemma 3 Under the hypotheses of Lemma 2, for ν_F to be injective (resp. surjective), it is necessary and sufficient that ν_M and ν_N be so. 05PU
Lemma 4 For every homomorphism v : E → E', the diagram 05PV
Lemma 5 Let M and N be two supplementary submodules in E and p : E → M, q : E → N the canonical projections. The diagram 05PW
Corollary (i) When F is a projective (resp. finitely generated projective) module, the canonical homomorphism (11) is injective (resp. bijective). (ii) When E is a finitely generated project… ↩ 3 05PX
Proposition 3 Let C be a commutative ring, E, F two finitely generated projective C-modules and u : E → F and v : F → E two linear mappings; then ↩ 3 05PY
Corollary If u₁, …, uₚ are endomorphisms of E, then 05PZ
Proposition 4 When one of the ordered pairs (E₁, E₂), (E₁, F₁), (E₂, F₂) consists of finitely generated projective C-modules, the canonical homomorphism (21) is bijective. ↩ 3 05Q0
Corollary 1 If one of the modules E, F is projective and finitely generated, the canonical homomorphism (23) is bijective. ↩ 2 05Q1
Corollary 2 Let E₁, E₂ be two finitely generated projective C-modules, u₁ an endomorphism of E₁ and u₂ an endomorphism of E₂; then 05Q2
Remark If E, F, G are any three C-modules, it is immediately verified that the diagram 05Q3

§ 5. Extension of the ring of scalars

Proposition 1 For every left A-module E, the mapping φ : x ↦ 1 ⊗ x of E into the A-module ρ_*(ρ^*(E)) is A-linear and the set φ(E) generates the B-module ρ^*(E). Further, for every left B-module… ↩ 5 05QD
Remark (1) Proposition 1 shows that the ordered pair consisting of E_(B) and φ_E is a solution of the universal mapping problem (Set Theory, IV, § 3, no. 1), where Σ is the species of lef… ↩ 1 05QE
Corollary Let E, E' be two left A-modules; for every A-linear mapping u : E → E', v = l_B ⊗ u is the unique B-linear mapping which renders commutative the diagram 05QF
Proposition 2 Let ρ : A → B,\ σ : B → C be ring homomorphisms. For every left A-module E, there exists one and only one C-homomorphism ↩ 9 05QG
Proposition 3 Let A, B be two commutative rings, ρ : A → B a ring homomorphism and E, E' two A-modules. There exists one and only one B-homomorphism ↩ 3 05QH
Proposition 4 Let E be an A-module admitting a basis (a_λ)_λ ∈ L; if φ : x ↦ 1 ⊗ x is the canonical mapping of E into ρ^*(E), then (φ(a_λ))_λ ∈ L is a basis of ρ^*(E). If ρ is injective, so is φ… ↩ 9 05QI
Corollary For every projective A-module E, the B-module ρ^*(E) is projective. If further ρ is injective, the canonical mapping of E into ρ^*(E) is injective. ↩ 5 05QJ
Proposition 5 Let E be a left A-module and F a left B-module; the composite mappings ↩ 1 05QK
Corollary The mappings ρ^*(φ_E) and φ_ρ_*(F)^* are injective and respectively identify ρ^*(E) with a direct factor of ρ^*(ρ_*(ρ^*(E))) and ρ_*(F) with a direct factor of ρ_*(ρ^*(ρ_*(F))). 05QL
Proposition 6 Let E be a left A-module and F a right B-module. There exists one and only one Z-homomorphism ↩ 3 05QM
Proposition 7 Let A be a commutative ring, B a ring, ρ : A → B a central homomorphism and E, F two A-modules. ↩ 1 05QN
Proposition 8 If one of the A-modules E, ρ_* (Bₛ) is projective and finitely generated, the homomorphism v_E is bijective. ↩ 4 05QO
Proposition 9 Let B be a ring, A a subring of B and P a projective left A-module. Then, if P_(B) is a finitely generated B-module, P is itself a finitely generated A-module. ↩ 1 05QP

§ 6. Inverse and direct limits of modules

Proposition 1 Let (E_α, f_αβ),\ (E'_α, f'_αβ),\ (E''_α, f''_αβ) be three inverse systems of A_α-modules and (u_α),\ (v_α) two inverse systems of A_α-linear mappings such that the sequences ↩ 1 05QX
Remark (1) Proposition 1 and its proof are valid for arbitrary groups, except for change of notation. ↩ 1 05QY
Proposition 2 For every inverse system (E_α, f_α β) of A-modules and every A-module F, the canonical mapping u ↦ (f_α ∘ u) is a Z-module isomorphism ↩ 1 05QZ
Corollary For every A-module homomorphism v: F → F', the 05R0
Proposition 3 Let (E_α, f_β α), (E'_α, f'_β α), (E''_α, f''_β α) be three direct systems of A_α-modules and (u_α), (v_α) two direct systems of A_α-linear mappings such that the sequences 05R1
Proposition 4 Let (E_α, f_β α) be a direct system of A_α-modules, E = lim → E_α its direct limit and φ_α : A_α → A and f_α : E_α → E the canonical mappings for all α ∈ I. If, for all α ∈ I, S_α… ↩ 2 05R2
Proposition 5 With the hypotheses and notation of Proposition 4, suppose that for all α ∈ I, E_α is the direct sum of a family (M^λ_α)_α ∈ L of submodules (the indexing set L being independent o… ↩ 3 05R3
Corollary Let (P_α) be a direct system of subsets of E_α and let P = lim overrightarrowP_α. If, for all α ∈ I, P_α is a free subset (resp. basis) of E_α, then P is a free subset (resp. basis… ↩ 1 05R4
Proposition 6 For every direct system (E_α, f_β α) of A-modules and every A-module F, the canonical mapping u ↦ (u ∘ f_α) is a Z-module isomorphism d_F : Hom_A(lim overrightarrowE_α, F) → lim ←… 05R5
Corollary 1 For every A-module homomorphism v : F → F', the barv_α = Hom(1_E_α, v) : Hom(E_α, F) → Hom(E_α, F') 05R6
Corollary 2 If (E_α, f_β α) is a direct system of left A-modules and E = lim longrightarrow E_α, (E_α^*, t f_β α) is an inverse system of right A-modules and lim ← E_α^* is canonically isomorp… 05R7
Remark Let E be an A-module and (M_α)_α ∈ I an increasing family of submodules of E such that E is the union of the M_α; if j_β α: M_α → M_β (for α ≤ β) and j_α: M_α → E are the canonical… 05R8
Proposition 7 The Z-linear mapping (7) is bijective. ↩ 8 05R9
Corollary 1 Let (E'_α, f'_β α) (resp. (F'_α, g'_α β)) be another direct system of right (resp. left) A_α-modules; for all α ∈ I, let u_α : E_α → E'_α (resp. v_α : F_α → F'_α) be an 05RA
Corollary 2 Let (E_α, f_βα) be a direct system of right A_α-modules and let E'_α = E_α ⊗_A_α A be the A-module obtained by extending the ring of scalars to A = lim A_α by means of the canonica… 05RB
Corollary 3 Let A be a ring, (E_α, f_βα) a direct system of right A-modules and F a left A-module. Then the Z-modules lim (E_α ⊗_A F) and (lim E_α) ⊗_A F are canonically isomorphic. 05RC
Corollary 4 Let M be a right A-module, N a left A-module, (xᵢ)_1 ≤ i ≤ n a family of elements of M, (yᵢ)_1 ≤ i ≤ n a family of elements of N, such that ∑ᵢ (xᵢ ⊗ yᵢ) = 0 in M ⊗_A N. Then there… ↩ 1 05RD

§ 7. Vector spaces

Theorem 1 Every vector space over a field K is a free K-module. ↩ 5 05RI
Theorem 2 Given a generating system S of a vector space E over a field K and a free subset L of E contained in S, there exists a basis B of E such that L ⊂ B ⊂ S. ↩ 11 05RJ
Lemma 1 Let (ai){i \in I} be a free family of elements of E; if b \in E does not belong to the subspace F generated by (ai), the subset of E consisting of the ai and b is free. ↩ 1 05RK
Corollary For a subset B of a vector space E, the following properties are equivalent: (a) B is a basis of E. (b) B is a maximal free subset of E. (c) B is a minimal generating system of E… ↩ 1 05RL
Example Given a ring A and a subfield K of A, A is a (right or left) vector space over K and therefore admits a basis; in particular, every extension field of a field K has a basis as a le… 05RM
Remark For a family (aᵢ)_i ∈ I of elements of a vector space E over a field K to be free, it is necessary and sufficient that, for all κ ∈ I, a_κ belong to no subspace of E generated by t… 05RN
Theorem 3 Two bases of the same vector space E over a field K are equipotent. ↩ 2 05RO
Definition 1 The dimension of a vector space E over a field K, denoted by dim_K E or [E:K] (or simply dim E) is the cardinal of any of the bases of E. If M is a subset of E, the rank of M (over… ↩ 1 05RP
Corollary For every subset M of E, the rank of M is at most equal to dim E. 05RQ
Proposition 1 (i) For a left vector space over K to be of finite dimension n, it is necessary and sufficient that it be isomorphic to Ks^n. ↩ 2 05RR
Proposition 2 Let (Et){t \in I} be a family of vector spaces over K. Then ↩ 3 05RS
Remark (1) Examples can be given of modules admitting two finite bases not having the same number of elements (§ 1, Exercise 16(c)). However: ↩ 1 05RT
Proposition 3 Let A be a ring such that there exists a homomorphism ρ of A into a field D; then for every free A-module E, any two bases of E are equipotent. ↩ 3 05RU
Corollary If A is a commutative ring ≠ 0 and E a free A-module, any two bases of E are equipotent. ↩ 2 05RV
Remark (2) When a free A-module E is such that any two bases of E are equipotent, the cardinal of an arbitrary basis of E over A is also called the dimension or rank of E and denoted by d… 05RW
Proposition 4 Every subspace F of a vector space E is a direct factor of E and ↩ 7 05RX
Corollary 1 If E, F, G are vector spaces over a field K, every exact sequence of linear mappings 0 → E → F → G → 0 splits. 05RY
Corollary 2 Let (Eᵢ)_0 ≤ i ≤ n be a finite family of vector spaces over a field K. If there exists an exact sequence of linear mappings ↩ 3 05RZ
Corollary 3 If M and N are two subspaces of a vector space E, then ↩ 1 05S0
Corollary 4 For every subspace F of a vector space E, dim F ≤ dim E; if E is finite dimensional, the relation dim F = dim E is equivalent to F = E. ↩ 4 05S1
Corollary 5 If a vector space E is the sum of a family (Fᵢ) of vector subspaces, then 05S2
Definition 2 Given a vector space E, the codimension (with respect to E) of a subspace F of E, denoted by codim_E F, or simply codim F, is the dimension of E/F (equal to that of any supplement… 05S3
Proposition 5 Let F, F' be two subspaces of a vector space E, such that F ⊂ F'. Then codim_E F' ≤ codim_E F ≤ dim E. If codim_E F is finite, the relation codim_E F' = codim_E F implies F = F'. 05S4
Proposition 6 If M and N are two subspaces of a vector space E, then codim(M + N) + codim(M ∩ N) = codim M + codim N. (10) It suffices to apply Corollary 2 of Proposition 4 to the exact sequence… 05S5
Proposition 7 If (Fᵢ) is a finite family of subspaces of a vector space E, then codim(⋂ᵢ Fᵢ) ≤ ∑ᵢ codim Fᵢ. If F = ⋂ᵢ Fᵢ, E/F is isomorphic to a subspace of the direct sum of the E/Fᵢ (§ 1, no… 05S6
Proposition 8 In a vector space E over a field K, every vector subspace F is the intersection of the hyperplanes which contain it. ↩ 1 05S7
Remark Most of the properties proved in this no. for subspaces of a vector space do not hold for submodules of a free module whose dimension (§ 2, no. 7, Remark 2) is defined. For example… 05S8
Definition 3 Let E, F be two vector spaces over a field K. For every linear mapping u of E into F, the dimension of the subspace u(E) of F is called the rank of u and denoted by rg(u). ↩ 1 05S9
Proposition 9 Let E, F be two vector spaces over a field K and u : E → F a linear mapping. (i) rg(u) ≤ inf(dim E, dim F). (ii) Suppose that E is finite-dimensional; in order that rg(u) = dim E… ↩ 2 05SA
Corollary Let E be a vector space of finite dimension n and u an endomorphism of E. The following properties are equivalent: (a) u is bijective; (b) u is injective; (c) u is surjective; (d)… ↩ 2 05SB
Theorem 4 The dimension of the dual E^* of a vector space E is at least equal to the dimension of E. For E^* to be finite-dimensional, it is necessary and sufficient that E be so, and then d… ↩ 2 05SC
Corollary For a vector space E, the relations E = 0 and E^* = 0 are equivalent. ↩ 1 05SD
Theorem 5 Given two exact sequences of vector spaces (over the same field K) and linear mappings beginalign* 0 &→ E' → E → E'' → 0 \\ 0 &→ F' → F → F'' → 0 endalign* and two vector spaces G… ↩ 1 05SE
Corollary For every exact sequence 0 longrightarrow E' xrightarrowu E xrightarrowv E'' longrightarrow 0 of vector spaces over the same field K and linear mappings the sequence 0 longrightarr… ↩ 4 05SF
Theorem 6 For every vector space E over a field K, the canonical mapping c_E : E → E^** (§ 2, no. 7) is injective; for it to be bijective, it is necessary and sufficient that E be finite-dim… ↩ 6 05SG
Proposition 10 Let E, F be two vector spaces over the same field K and u : E \to F a linear mapping. ↩ 3 05SH
Theorem 7 Let E be a vector space over a field K, F a subspace of E and F' the orthogonal of F in E. ↩ 15 05SI
Remark An infinite-dimensional subspace G' of E is not necessarily the orthogonal of a subspace of E, in other words, if F is the orthogonal of G' in E, the orthogonal F' of F in E may be… 05SJ
Corollary 1 Let (xᵢ^*)_1 ≤ i ≤ p be a finite sequence of linear forms on E and let F be the subspace of E consisting of the x such that ↩ 4 05SK
Corollary 2 (i) Let (xᵢ^*)_1 ≤ i ≤ p be a finite sequence of linear forms on E. For (xᵢ^*) to be a free system, it is necessary and sufficient that there exist a sequence (xᵢ)_1 ≤ i ≤ p of ele… ↩ 4 05SL
Corollary 3 Let S be a set and V a vector subspace of the right vector K-space K_d^S of mappings of S into K. In order that dim V ≥ p (where p is an integer), it is necessary and sufficient th… 05SM
Corollary 4 Let E be a vector space and M, N two subspaces of E of finite codimension; if M', N' are the orthogonals of M and N in E, the orthogonal of M ∩ N in E is M' + N'. 05SN
Corollary 5 Let E be a vector space of finite dimension n. For every subspace F of E of dimension p, the orthogonal F' of F in E is of dimension n − p. For every subspace G' of E of dimension… 05SO
Proposition 11 For every hyperplane H in a vector space E, there exists a linear form x₀^* on E such that H = x₀^*-1(0). Given such a form x₀^*, for a linear form x^* on E to satisfy H = x^*-1(0)… ↩ 1 05SP
Proposition 12 Let E, F be two vector spaces over a field K and u : E → F a linear mapping. For the linear equation ↩ 1 05SQ
Definition 4 The dimension of the subspace of E^* generated by the family (xᵢ^*) is called the rank of the system (21). 05SR
Proposition 13 For the system (21) to be of finite rank r, it is necessary and sufficient that the linear mapping u : x ↦ (⟨ x, xᵢ^* ⟩) of E into Kₛᴵ be of rank r. 05SS
Theorem 8 Let ↩ 2 05ST
Remark When the rank of the system (21) is infinite, the condition of Theorem 8 is no longer sufficient. For example, suppose that the xᵢ^* are the coordinate forms on the space E = Kₛ⁽ᴵ⁾… 05SU
Corollary 1 A system of scalar linear equations in a vector space, consisting of a finite number of equations whose left hand sides are linearly independent forms, always admits solutions. 05SV
Corollary 2 For a homogeneous linear system (22) of equations in n unknowns with coefficients in a field K to admit non-trivial solutions consisting of elements of K, it is necessary and suffi… 05SW
Corollary 3 For a linear system (22) with coefficients and right hand sides in a field K, consisting of n equations in n unknowns, to have one and only one solution consisting of elements of K… 05SX
Proposition 14 Given an exact sequence ↩ 9 05SY
Corollary Let K be a field, E a right vector space over K, F a left vector space over K, (M_α)_α ∈ A a family of vector subspaces of E and (N_β)_β ∈ B a family of vector subspaces of F. Then… ↩ 3 05SZ
Proposition 15 If (E_λ)_λ ∈ L is a family of right vector spaces over a field K and (F_μ)_μ ∈ M a family of left vector spaces over K, the canonical mapping ( ∏_λ ∈ L E_λ ) ⊗_K ( ∏_μ ∈ M F_μ ) →… ↩ 3 05T0
Corollary Let F be a left vector space over K; for every set X, the left vector space K_d^X ⊗_K F is identified with the subspace of the space F^X of all mappings of X into F, consisting of… 05T1
Proposition 16 (i) Let K, L be two fields, E a left vector space over K, F a left vector space over L and G a (K, L)-bimodule. Then the canonical Z-homomorphism ↩ 5 05T2
Corollary 1 Let E and F be two vector spaces over a field K; the canonical mapping 05T3
Corollary 2 Let E be a right vector space and F a left vector space over the same field K; the canonical mapping 05T4
Remark (1) Let K be a commutative field, E, F two vector spaces over K, (a\lambda) a basis of E and (b\mu) a basis of F; then (a\lambda \otimes b\mu) is a basis of the vector K-space E \o… 05T5
Proposition 17 Let u be an element of E ⊗_K F and M ⊂ E and N ⊂ F its associated subspaces. For every expression u = ∑ᵢ₌₁ˢ xᵢ ⊗ yᵢ of u, where xᵢ ∈ E and yᵢ ∈ F for 1 ≤ i ≤ s, the subspace M (res… ↩ 2 05T6
Corollary 1 The rank of u is the smallest integer s such that there exists an expression u = ∑ᵢ₌₁ˢ xᵢ ⊗ yᵢ, where xᵢ ∈ E and yᵢ ∈ F for 1 ≤ i ≤ s. 05T7
Corollary 2 Let K be a commutative field, E, F two vector spaces over K and L a commutative extension field of K. Let u be an element of E ⊗_K F, M and N its associated subspaces and u' the ca… 05T8
Proposition 18 Let φ be a homomorphism of a field K into a ring A. For every exact sequence of vector K-spaces and K-linear mappings 05T9
Corollary For every K-linear mapping f : E' → E, Im(f_(A)) = (Im(f))_(A), Ker(f_(A)) = (Ker(f))_(A), Coker(f_(A)) = (Coker(f))_(A), to within canonical isomorphisms. ↩ 3 05TA
Proposition 19 Let φ be an injective homomorphism of a field K into a ring A. For every left vector space E over K, the canonical mapping φ : E → φ^*(E) = A ⊗_K E is injective. Moreover, for ever… ↩ 8 05TB
Corollary Let φ be an injective homomorphism of a field K into a ring A. For a K-linear mapping f : E → F (where E and F are two vector spaces over K) to be injective (resp. surjective, zero… 05TC
Proposition 20 Let ρ be a homomorphism of a field K into a ring A. For every left vector space E over K, the canonical right A-module homomorphism 05TD
Proposition 21 Let K be a field and L an extension field of K. 05TE
Proposition 22 Let K be a commutative field, ρ : K → A an injective central homomorphism and E, F two vector spaces over K. Then the canonical homomorphism 05TF
Proposition 23 In a module E over an integral domain A, the set T of non-free elements is a submodule of E. ↩ 1 05TG
Remark Let E be a module over any commutative ring A. If x is a non-free element of E, every element of the submodule Ax is non-free. On the other hand, if A contains divisors of 0, the s… 05TH
Definition 5 In a module E over an integral domain A, the torsion submodule of E is the submodule of E consisting of the non-free elements (also called the torsion elements of E). 05TI
Proposition 24 Let A be an integral domain. For every A-module E, let T(E) denote the torsion submodule of E. Let f : E → E' be an A-linear mapping, E and E' being A-modules. 05TJ
Corollary 1 For every A-module E, E/T(E) is torsion-free. 05TK
Corollary 2 For every exact sequence of A-modules and A-linear mappings 05TL
Proposition 25 Let A be an integral domain and (Eᵢ) a family of A-modules; then ↩ 1 05TM
Proposition 26 Let A be an integral domain, K its field of fractions, E an A-module and E_(K) = K ⊗_A E the vector space over K obtained by extending the ring of operators; let φ denote the canon… ↩ 4 05TN
Remark Let α^{-1}φ(x), β^{-1}φ(y) be two elements of E{(K)}, with α ∈ A, β ∈ A, x ∈ E, y ∈ E, αβ ≠ 0. In order that α^{-1}φ(x) = β^{-1}φ(y), it is necessary and sufficient that βx − αy be… 05TO
Corollary 1 If E is a torsion-free A-module, the canonical mapping φ : E → E{(K)} is injective. ↩ 1 05TP
Corollary 2 Let f be an A-linear mapping of E into a vector space F over K; if Ker(f) ⊂ T(E), the K-linear mapping vecf associated with f is injective. ↩ 1 05TQ
Corollary 3 Let E be an A-module and g an A-linear mapping of E into a vector space F over K such that g(E) generates F and Ker(g) ⊂ T(E). Then the K-linear mapping barg associated with g is a… ↩ 1 05TR
Corollary 4 Every finitely generated A-module is of finite rank. 05TS
Proposition 27 For every exact sequence ↩ 2 05TT
Corollary 1 If E' is a submodule of E, E'_(K) is canonically identified with a vector subspace of E_(K) and (E/E')_(K) with E_(K)/E'_(K). ↩ 1 05TU
Corollary 2 For every A-linear mapping f : E → F, Ker(f_(K)) = (Ker(f))_(K), Im(f_(K)) = (Im(f))_(K), Coker(f_(K)) = (Coker(f))_(K) to within canonical isomorphisms. In particular, for f_(K) t… ↩ 1 05TV
Corollary 3 Let E be an A-module and (x_λ)_λ ∈ L a family of elements of E. For (x_λ) to be a free family, it is necessary and sufficient that in the vector K-space E_(K) the family (1 ⊗ x_λ)… 05TW

§ 8. Restriction of the field ofscalars in vector spaces

Proposition 1 Let V be a right vector space over K and V' a subset of V which is a vector subspace over K'. The following conditions are equivalent: ↩ 4 05V0
Definition 1 Let V be a right vector space over a field K and K' a subfield of K. Every vector sub-K'-space V' of V satisfying the equivalent conditions of Proposition 1 is called a K'-structur… ↩ 3 05V1
Example Let B be a basis of V over K. For every subfield K' of K, the vector sub-K'-space of V generated by B admits B as a basis over K' and hence is a K'-structure on V. For example, if… 05V2
Definition 2 Let V be a right vector space over K, with a K'-structure V'. A vector of V is said to be rational over K' if it belongs to V'. A vector sub-K-space W of V is said to be rational o… ↩ 1 05V3
Proposition 2 Let V be a right vector space over K and V' a K'-structure on V. The mapping W' \mapsto W'.K is a bijection of the set of vector sub-K'-spaces of V' onto the set of vector sub-K-sp… ↩ 4 05V4
Corollary 1 Every sum and every intersection of vector sub-K-spaces of V which are rational over K' is a rational subspace over K'. ↩ 1 05V5
Corollary 2 Every basis of V over K which is rational over K' is a basis of V' over K'. 05V6
Definition 3 Let V₁, V₂ be two right vector spaces over K with K'-structures V'₁, V'₂ respectively. A K-linear mapping f : V₁ → V₂ is said to be rational over K' if f(V'₁) ⊂ V'₂. 05V7
Proposition 3 Let V₁, V₂ be two right vector spaces over K and V'₁, V'₂ K'-structures on V₁, V₂ respectively. V₁ (resp. V₂) is canonically identified with V'₁ ⊗_K' K (resp. V'₂ ⊗_K' K) (no. 1, P… ↩ 5 05V8
Corollary 1 In the notation of Proposition 3, Im(f) = (Im(f'))_(K), Ker(f') = (Ker(f'))_(K), Coker(f) = (Coker(f'))_(K). In particular, for f to be injective (resp. surjective, zero), it is ne… ↩ 2 05V9
Corollary 2 Let f : V₁ → V₂ be a K-linear mapping rational over K'. For every vector sub-K-space W₁ of V₁ (resp. W₂ of V₂) rational over K', f(W₁) (resp. f⁻¹(W₂)) is a vector sub-K-space of V₂… ↩ 3 05VA
Proposition 4 For a K-linear mapping f : V₁ → V₂ to be rational over K', it is necessary and sufficient that its graph Γ be rational over K' for the product K'-structure on V₁ × V₂. ↩ 2 05VB
Proposition 5 Let V be a right vector space over K, V' a K'-structure on V and W a vector sub-K-space of V. For W to be rational over K', it is necessary and sufficient that there exist a set H… ↩ 1 05VC
Proposition 6 (i) Given a system of homogeneous linear equations ↩ 2 05VD
Proposition 7 (i) Let V be a right vector K-space with a K'-structure V'. For every vector x ∈ V (resp. every vector sub-K-space W of V), the set of subfields L of K containing K' and such that… 05VE
Lemma 1 Let V be a right vector K-space with a K'-structure V' and W a vector sub-K-space of V. There exist two vector sub-K-spaces W₁, W₂ of V, rational over K', such that V is the direct… ↩ 1 05VF
Theorem 1 Let M be a subset of End_K'(K) and L = χ(M) the subfield of K the centralizer of M. 05VG

§ 9. Affine spaces and projective spaces

Definition 1 Given a left (resp. right) vector space T over a field K, an affine space attached to T is any homogeneous space E of the additive group T (I, § 5, no. 5) such that 0 is the only o… 05VM
Remark The definitions of this no. and some of the results which follow extend immediately to the case where, instead of a vector space T, we consider an arbitrary commutative group with… 05VN
Proposition 1 Let (xᵢ)_i ∈ I be a family of points in an affine space E and (λᵢ)_i ∈ L a family of elements of K of finite support such that ∑_i ∈ I λᵢ = 1 (resp. ∑_i ∈ I λᵢ = 0). If a is any po… 05VO
Definition 2 Given an affine space E, a subset V of E is called an affine linear variety (or simply a linear variety or an affine subset of E) if, for every family (xᵢ)_i ∈ I of points of V and… 05VP
Proposition 2 For a non-empty subset V of an affine space E to be a linear variety, it is necessary and sufficient that V be a vector subspace for the vector space structure on E obtained by tak… 05VQ
Proposition 3 Given a family (aᵢ)_i ∈ I of points of an affine space E, the set V of barycentres ∑_i ∈ I λᵢ aᵢ ((\lambdai) of finite support, ∑_i ∈ I λᵢ = 1) is a linear variety of E. 05VR
Proposition 4 For a non-empty family (aᵢ)_i ∈ I of points in an affine space E to be affinely related, it is necessary and sufficient that there exist a family (λᵢ)_i ∈ I of elements not all zer… 05VS
Proposition 5 For a non-empty family (aᵢ)_i ∈ I of points of an affine space E to be affinely free, it is necessary and sufficient that, for every index κ ∈ I, a_κ do not belong to the linear va… 05VT
Definition 3 Given two affine spaces E, E' attached to two vector spaces T, T' over the same field K, a mapping u of E into E' is called an affine linear mapping (or an affine mapping) if, for… 05VU
Proposition 6 Let u be an affine mapping of E into E'. There exists one and only one linear mapping v of T into T' such that 05VV
Definition 4 Given a left (resp. right) vector space V over a field K, the left (resp. right) projective space derived from V, denoted by P(V), is the quotient of the complement V - 0 of 0 in V… ↩ 1 05VW
Remark In a projective space P(V) over a field K, let H = P(W) be a hyperplane. There exists a bijective linear mapping f of V onto Kₛ × W such that f(W) = W; let g be the projective mapp… 05VX

§ 10. Matrices

Definition 1 Let I, K, H be three sets; a matrix of type (I, K) with elements in H (or a matrix of type (I, K) over H) is any family M = (m_iκ)_(i, κ) ∈ I × K of elements of H whose indexing se… 05WH
Definition 2 The transpose of a matrix M = (m_ι κ)_(ι, κ) ∈ I × K, denoted by ᵗM, is the matrix (m'_κ ι)_(κ, ι) ∈ K × I over H given by m'_κ ι = m_ι κ for all (ι, κ) ∈ K × L. 05WI
Remark The above formulae extend to a more general situation. To be precise: 05WJ
Proposition 1 Let A, B be two rings and M = (mᵢₖ)_(i,k) ∈ I × K and M' = (m'ᵢₖ)_(i,k) ∈ I × K two matrices with finite indexing sets over an (A, B)-bimodule G. Suppose that for every matrix unit… ↩ 2 05WK
Definition 3 In each of the above two cases, the matrix of u with respect to the bases (eᵢ), (fₖ) is the matrix M(u) = (uₖᵢ)_(k, i) ∈ K × I such that 05WL
Proposition 2 Suppose I and K are finite. For every element x ∈ E, the matrix M(u(x)) with respect to the basis (fₖ) is given by the formula ↩ 2 05WM
Corollary Let E, F, G be three right (resp. left) modules over a ring A, (eᵢ)_i ∈ I, (fₖ)_k ∈ K, (gₗ)_l ∈ L respective finite bases of E, F, G, u : E → F, v : F → G two linear mappings, M(u)… ↩ 1 05WN
Proposition 3 Let E, F be two right (resp. left) A-modules and (eᵢ)_i ∈ I, (fₖ)_k ∈ K finite bases of E and F respectively. For every linear mapping u of E into F, let M(u) be the matrix of u wi… ↩ 6 05WO
Remark (2) Let E and F be two left A-modules with bases (eᵢ)_i ∈ I and (fₖ)_k ∈ K respectively. For every A-linear mapping u : E → F, by (14 G), u(eᵢ) = ∑ₖ uₖᵢ fₖ; these relations can als… 05WP
Remark Let A be a ring and σ an anti-endomorphism of A (no. 3); consider the two following situations: 05WQ
Definition 4 A matrix whose rows and columns have the same indexing set is called a square matrix. 05WR
Remark It should be noted that a matrix for which the indexing sets of the rows and columns have the same cardinal but are not identical, must not be considered as a square matrix; in par… ↩ 1 05WS
Proposition 4 Let E be a right A-module with finite basis (eᵢ)_1 ≤ i ≤ n of n elements. For a family of n elements e'ᵢ = ∑ⱼ₌₁ⁿ eⱼ aⱼᵢ (1 ≤ i ≤ n) to be a basis of E, it is necessary and sufficie… ↩ 1 05WT
Proposition 5 Let (eᵢ), (e'ᵢ) be two bases of n elements of E and P the matrix of passage from (eᵢ) to (e'ᵢ). If (eᵢ^*) and (e'ᵢ^*) are the respective dual bases of (eᵢ) and (e'ᵢ), the matrix of… ↩ 2 05WU
Proposition 6 Let E and F be two right A-modules, (eᵢ) and (e'ᵢ) two bases of E with n elements, (fⱼ) and (f'ⱼ) two bases of F with m elements, P the matrix of passage from (eᵢ) to (e'ᵢ) and Q t… ↩ 2 05WV
Corollary 1 If u is an endomorphism of E and M(u) and M'(u) its matrices with respect to the bases (eᵢ) and (e'ᵢ) respectively, then ↩ 1 05WW
Corollary 2 If M(x) and M'(x) are the matrices with one column of the same element x ∈ E with respect to the bases (eᵢ) and (e'ᵢ) respectively, then 05WX
Remark (1) We now start with a left A-module E with two bases (eᵢ), (e'ᵢ), each with n elements; if we write e'ᵢ = ∑ⱼ₌₁ⁿ aⱼᵢ eᵢ, P = (aⱼᵢ) is also called the matrix of passing from (eᵢ) t… 05WY
Definition 5 Two matrices X, X' with m rows and n columns over a ring are called equivalent if there exists an invertible square matrix P of order m and an invertible square matrix Q of order n… 05WZ
Definition 6 Two square matrices X, X' of order n over a ring A are called similar if there exists an invertible square matrix P of order n such that ↩ 1 05X0
Remark (1) Two square matrices which differ only in the order of their rows (or the order of their columns) are equivalent, but in general not similar. A matrix similar to a square matrix… 05X1
Proposition 7 Let C be a commutative ring and for every matrix P ∈ Mₙ(C) let f_P be the linear form X ↦ Tr(PX) on Mₙ(C); the mapping P ↦ f_P is a C-linear bijection of Mₙ(C) onto its dual. ↩ 1 05X2
Proposition 8 If g is a linear form on the C-module Mₙ(C) such that g(XY) = g(YX) for all matrices X, Y in Mₙ(C), there exists one and only one scalar c ∈ C such that g(X) = c · Tr(X) for every… 05X3
Definition 7 Given a matrix X with m rows and n columns over a field K, the dimension of the subspace of K_dᵐ generated by the n columns of X is called the rank of X with respect to K and denot… ↩ 2 05X4
Proposition 9 If the elements of a matrix X with m rows and n columns belong to a subfield K₀ of a field K, the rank of X with respect to K₀ is equal to the rank of X with respect to K. 05X5
Proposition 10 The rank of a matrix X over a field K is equal to the rank of its transpose ᵗ X over the opposite field K⁰. 05X6
Proposition 11 Let X be a square matrix of order n over a field K. The following properties are equivalent: ↩ 2 05X7
Proposition 12 For a system of m linear equations in n unknowns ↩ 1 05X8
Proposition 13 Let E, F be two finite-dimensional vector spaces over a field K. If u : E → F is a linear mapping of rank r, there exist bases of E and F such that, with respect to these bases, ↩ 3 05X9
Corollary For two matrices over a field, of type (m, n), to be equivalent, it is necessary and sufficient that they have the same rank. 05XA
Lemma 1 Let X = (ξᵢⱼ) be a matrix of type (m, n) over a ring A. Suppose that m geqslant 2 and that there exists an element ξᵢ₁ in the first column of X which is invertible in A. Then there… ↩ 2 05XB
Proposition 14 Let X be a matrix of type (m, n) over a field K. If X is of rank r, there exist two invertible square matrices P ∈ Mₘ(K), Q ∈ Mₙ(K) such that P (resp. Q) is a product of matrices o… ↩ 2 05XC
Corollary 1 If X is an invertible square matrix of order n over a field K, there exist three invertible matrices P, Q, D of order n such that X = PDQ, P and Q being products of matrices of the… ↩ 2 05XD
Corollary 2 For every field K, the group of invertible matrices GL(n, K) is generated by the permutation matrices (no. 7, Example 2), the diagonal matrices diag(a, 1, …, 1) (a ≠ 0 in K) and th… ↩ 1 05XE
Remark (1) In Chapter III, we shall see that, if m = n = r and K is commutative, then, for all choices of P and Q satisfying the conditions of Proposition 14, the element δᵣ is always the… 05XF

§ 11. Graded modules and rings

Definition 1 Given a commutative group G written additively and a set Δ, a graduation of type Δ on G is a family (G_λ)_λ ∈ Δ of subgroups of G, of which G is the direct sum. The set G, with the… ↩ 1 05XX
Example (1) Given any commutative monoid Δ (with identity element 0) and a commutative group G, a graduation (G_λ)_λ ∈ Δ is defined on G by taking G₀ = G and G_λ = 0 for λ ≠ 0; this gradua… 05XY
Definition 2 Given a ring A and a graduation (A_λ) of type Δ on the additive group A, this graduation is said to be compatible with the ring structure on A if 05XZ
Proposition 1 If every element of Δ is cancellable and (A_λ) is a graduation of type Δ compatible with the structure of a ring A, A₀ is a subring of A (and in particular 1 ∈ A₀). ↩ 2 05Y0
Definition 3 Let A be a graded ring of type Δ, (A_λ) its graduation and M a left (resp. right) A-module; a graduation (M_λ) of type λ on the additive group M is compatible with the A-module str… 05Y1
Example (1) On any ring A the trivial graduation of type Δ is compatible with the ring structure. If A is graded by the trivial graduation, for a graduation (M_λ) of type Δ on an A-module… 05Y2
Remark (1) The graduations most often used are of type Z or of type Zⁿ; when we speak of graded (resp. bigraded, trigraded, etc.) modules and rings without mentioning the type, it is unde… 05Y3
Definition 4 Let A, A' be two graded rings of the same type Δ and (A_λ), (A'_λ) their respective graduations. A ring homomorphism h : A → A' is called graded if h(A_λ) ⊂ A'_λ for all λ ∈ Δ. ↩ 1 05Y4
Example (6) If M is a graded A-module and M(λ₀) is a graded A-module obtained by shifting (no. 2, Example 3), the Z-linear mapping of M(λ₀) into M which coincides with the canonical inject… ↩ 3 05Y5
Proposition 2 Let A be a graded ring of type Δ, M a graded A-module of type Δ, (M_λ) its graduation and N a sub-A-module of M. The following properties are equivalent: ↩ 7 05Y6
Corollary 1 If N is a graded submodule of M and (xᵢ) is a generating system of N, the homogeneous components of the xᵢ form a generating system of N. ↩ 1 05Y7
Corollary 2 If N is a finitely generated submodule of M, N admits a finite generating system consisting of homogeneous elements. 05Y8
Proposition 3 Let A be a graded ring of type \Delta, M, N two graded A-modules of type \Delta and u: M \to N a graded A-homomorphism of degree \delta. Then: (i) Im(u) is a graded submodule of N… ↩ 2 05Y9
Corollary Let A, B be two graded rings of type \Delta and u: A \to B a graded homomorphism of graded rings. Then Im(u) is a graded submodule of B, Ker(u) a graded two-sided ideal of A and th… 05YA
Proposition 4 Let A be a graded ring of type \Delta and M a graded A-module of type \Delta. (i) Every sum and every intersection of graded submodules of M is a graded submodule. (ii) If x is a h… ↩ 1 05YB
Remark 2 Let M be a graded A-module and E a submodule of M; it follows from Proposition 4(i) that there exists a largest graded submodule N' of M contained in E and a smallest graded submod… 05YC
Proposition 5 Let A be a graded ring of type Δ. If every element of Δ is cancellable, then, for every homogeneous element a ∈ A, the centralizer of a in A (I, § 1, no. 5) is a graded subring of… ↩ 1 05YD
Corollary If every element of Δ is cancellable, the centralizer of the graded subring B of A (and in particular the centre of A) is a graded subring of A. ↩ 1 05YE
Proposition 6 Let Δ be an ordered commutative group, A a graded ring of type Δ, (A_λ) its graduation, M a graded A-module of type Δ and (M_λ) its graduation. Suppose that the relation A_λ ≠ 0 im… ↩ 4 05YF
Corollary 1 With the hypotheses on Δ and A of Proposition 6, if M is a finitely generated graded A-module such that ℑ₀ M = M, then M = 0. ↩ 2 05YG
Corollary 2 With the hypotheses on Δ and A of Proposition 6, let M be a finitely generated graded A-module and N a graded submodule of M such that N + 𝔍₀ M = M; then N = M. 05YH
Corollary 3 With the hypotheses on Δ and A of Proposition 6, let u : M → N be a graded homomorphism of graded right A-modules, where N is assumed to be finitely generated. If the homomorphism ↩ 1 05YI
Remark It follows from the proof of Corollary 1 that Corollaries 1 and 2 (resp. Corollary 3) are still valid when, instead of assuming that M (resp. N) is finitely generated, the followin… 05YJ
Proposition 7 Suppose that Δ = Z. With the hypotheses on A and M of Proposition 6, consider the graded A₀-module N = M/𝔍₀ M and suppose the following conditions hold: (i) each of the N_λ conside… ↩ 4 05YK
Lemma 1 For a commutative group Δ to be such that there exists on Δ a total ordering compatible with the group structure of Δ, it is necessary and sufficient that Δ be torsion-free. 05YL
Proposition 8 Let Δ be a torsion-free commutative group and A a graded ring of type Δ. If the product in A of two homogeneous elements ≠ 0 is ≠ 0, the ring A has no divisor of 0. ↩ 2 05YM
Remark When A has the trivial graduation (no. 1, Example 1), (M ⊗_A N)_λ is then simply the direct sum of the sub-Y-modules M_μ ⊗_A N_ν of M ⊗_A N such that μ + ν = λ. 05YN
Remark If M and N are graded left A-modules, Homgr_A(M, N) is in general distinct from Hom_A(M, N). However these two sets are equal when M is a finitely generated A-module. For M is then… 05YO

III. Tensor algebras, exterior algebras, symmetric algebras

§ 1. Algebras

Definition 1 Let A be a commutative ring. An algebra over A (or an A-algebra, or simply an algebra, when no confusion is to be feared) is a set E with a structure defined by giving the followin… 05YR
Example (1) Every commutative ring A can be considered as an (associative and commutative) A-algebra. (2) Let E be a pseudo-ring (I, § 8, no. 1). Multiplication on E and the unique Z-modul… 05YS
Definition 2 Given two algebras E, E' over a commutative ring A, a homomorphism of E into E' is a mapping f : E → E' such that (1) f is an A-module homomorphism; (2) f(xy) = f(x)f(y) for all x… ↩ 1 05YT
Proposition 1 For every A-algebra E, the canonical mapping φ_E : x ↦ x ⊗ 1 of E into E_(B) is an A-homomorphism of algebras. Moreover, for every B-algebra F and every A-homomorphism f : E → F, t… ↩ 1 05YU
Corollary Let E, E' be two A-algebras; for every A-homomorphism of algebras u : E → E', u ⊗ 1_B is the unique B-homomorphism of algebras v : E ⊗_A B → E' ⊗_A B rendering commutative the diag… 05YV

§ 2. Examples of algebras

Proposition 1 A quadratic algebra E is associative and commutative. ↩ 1 05YX
Proposition 2 Let E be a quadratic A-algebra and e its unit element. For all u ∈ E, let T(u) be the trace of the endomorphism mᵤ : x ↦ ux of the free A-module E (II, § 4, no. 3). Then the mappin… ↩ 1 05YY
Proposition 3 Let E be a quadratic A-algebra of type (α, β). 05YZ
Definition 1 A Cayley algebra over A is an ordered pair (E, s), where E is an algebra over A with a unit element e and s is an antiautomorphism of E such that ↩ 1 05Z0
Proposition 4 Let E be an A-algebra and s and s' antiautomorphisms of E such that (E, s) and (E, s') are Cayley algebras. If E admits a basis containing the unit element e, then s' = s. 05Z1
Proposition 5 (i) The ordered pair (F, t) is a Cayley algebra over A. ↩ 3 05Z2
Example (1) If we take E = A (and hence s = 1_A), the algebra F is a quadratic A-algebra with basis (e, j) where j² = γ e. 05Z3
Proposition 6 Let S be a magma, F an A-algebra and f a homomorphism of ↩ 3 05Z4
Corollary Let S, S' be two magmas and g: S → S' a homomorphism. Then there exists one and only one A-algebra homomorphism u: A^(S) → A^(S') rendering commutative the diagram 05Z5
Example Let V be an A-module and S a monoid which operates on V on the left; this means (I, § 5, no. 1) that there is given a mapping (s, x) ↦ s.x of S into V such that s.(x + y) = s.x + s… 05Z6
Definition 2 Let I be a set; let M(I) (resp. Mo(I), resp. N⁽ᴵ⁾) denote the free magma (resp. free monoid, resp. free commutative monoid) derived from I. The algebra of M(I) (resp. Mo(I), resp… ↩ 3 05Z7
Proposition 7 Let I be a set and F an algebra (resp. unital associative algebra, resp. unital commutative associative algebra) over A. For every mapping f : I → F, there exists one and only one… ↩ 6 05Z8
Remark (1) We shall later define an isomorphism of LibasA(I) onto the tensor algebra of the free module A⁽ᴵ⁾ (§ 5, no. 5) and also an isomorphism of LibascA(I) onto the symmetric algebra… 05Z9
Example (1) Take H = 1, 2, 3; the algebras which admit (X₁ X₂) X₃ - X₁ (X₂ X₃) as universal relator are the associative algebras. The algebras which admit X₁ X₂ - X₂ X₁ as universal relato… 05ZA
Example 3 Let L' be the universal unital associative algebra generated by I and subject to the identities uv - vu = 0 for every family of two elements of L'. It is seen, as in Example 2, tha… 05ZB
Proposition 8 Let A[(Xᵢ)_i ∈ I] be a polynomial algebra, J a subset of I and K the complement of J in I. Writing A' = A[(Xⱼ)_j ∈ J] and denoting by X'ₖ (k ∈ K) the indeterminates in the polynomi… ↩ 1 05ZC
Remark Let I and J be two sets and (Pj){j \in J} a family of elements of Z[(Xᵢ)_i ∈ I]; if Q is an element of Z[(Xⱼ)_j ∈ J] such that Q((Pⱼ)_j ∈ J) = 0, then, for every family (bᵢ)_i ∈ I… 05ZD

§ 3. Graded algebras

Definition 1 Let Δ be a commutative monoid, A a graded ring of type Δ (II, § 11, no. 2), (A_λ)_λ ∈ Δ its graduation and E an A-algebra. A graduation (E_λ)_λ ∈ Δ of type Δ on the additive group… ↩ 1 0603
Remark Definition 1 can also be interpreted by saying that E is a graded A-module and that the A-linear mapping 0604
Example (1) Let B be a graded ring of type Δ; if B is given its canonical Z-algebra structure (§ 1, no. 1, Example 2), B is a graded A-algebra (Z being given the trivial graduation). 0605
Proposition 1 Let A be a graded commutative ring of type Δ, E a graded A-algebra of type Δ and S a set of homogeneous elements of E. Then the sub-A-algebra (resp. left ideal, right ideal, two-si… ↩ 1 0606

§ 4. Tensor products of algebras

Definition 1 Given a family (Eᵢ)_i ∈ I of algebras over A, the tensor product of this family, denoted by ⨂_i ∈ I Eᵢ (or, when I is the interval 1, n of N, E₁ ⊗_A E₂ ⊗ ⋯ ⊗_A Eₙ, or simply E₁ ⊗ E… 0608
Proposition 1 Let E, F be two A-algebras and a (resp. b) a two-sided ideal of E (resp. F). Then the canonical A-module isomorphism 0609
Corollary 1 Let E be an A-algebra and a an ideal of A. Then the A-module aE is a two-sided ideal of E and the canonical (A/a)-module isomorphism ↩ 1 060A
Corollary 2 If a, b are two ideals of A, (A/a) ⊗_A (A/b) is canonically isomorphic to A/(a + b). 060B
Corollary 3 Let E, F be two A-algebras and a an ideal of A contained in the annihilator of F. Then the (A/a)-algebra E ⊗_A F is canonically isomorphic to (E/aE) ⊗_A/a F. 060C
Proposition 2 Let (E_λ)_λ ∈ L and (F_μ)_μ ∈ M be two families of A-algebras. The canonical mapping (II, § 3, no. 7) 060D
Proposition 3 Let A, B be two commutative rings, φ : A → B a ring homomorphism and E, F two A-algebras. Then the canonical B-module isomorphism ↩ 1 060E
Proposition 4 Let A, B be two commutative rings, ρ : A → B a ring homomorphism, E an A-algebra and F a B-algebra. Then the canonical A-module isomorphism 060F
Proposition 5 Let (Eᵢ)_i ∈ I be a finite family of A-algebras and, for each i ∈ I, let eᵢ be the unit element of Eᵢ. For each i ∈ I, let uᵢ : Eᵢ → E = ⨂_i ∈ I Eᵢ be the A-linear mapping defined… ↩ 7 060G
Corollary Let (Eᵢ)_i ∈ I, (Fᵢ)_i ∈ I be two finite families of A-algebras and, for all i ∈ I, let fᵢ : Eᵢ → Fᵢ be an algebra homomorphism. If uᵢ : Eᵢ → ⨂_j ∈ I Eⱼ, vᵢ : Fᵢ → ⨂_j ∈ I Fⱼ are t… 060H
Definition 2 Let E be a (unital) A-algebra. A left (resp. right) E-module is a left (resp. right) module over the underlying ring of E. 060I
Definition 3 Let E and F be two (unital) A-algebras and M a set with an E-module structure and an F-module structure. M is called a (left) bimodule over the algebras E and F if: ↩ 1 060J
Example (1) Let B be an A-algebra; the ring B has canonically a (B, B)-bimodule structure (II, § 1, no. 14, Example 1) and, if e is the unit element of B, then (α e)x = x(α e) = α x for al… 060K
Proposition 6 Let E, F be two non-zero algebras over a commutative field, K, C (resp. D) a subalgebra of E (resp. F) and C' (resp. D') the centralizer of C in E (resp. D in F). Then the centrali… ↩ 2 060L
Corollary If Z and Z' are the respective centres of E and F, the centre of E \otimesK F is Z \otimesK Z'. ↩ 1 060M
Definition 4 Given an algebra G over a commutative field K, two subalgebras E, F of G are said to be linearly disjoint over K if they satisfy the following conditions: (1) every element of E co… 060N
Proposition 7 Let G be an algebra over a commutative field K and E, F two subalgebras of G such that every element of E commutes with every element of F. For E and F to be linearly disjoint over… ↩ 2 060O
Corollary 1 For the canonical homomorphism of E ⊗_K F into G to be bijective, it is necessary and sufficient that there exist a basis of E over K which is a basis of the (right or left) F-modu… 060P
Corollary 2 Let E, F be two subalgebras of G, of finite rank over K and such that every element of E commutes with every element of F. For E and F to be linearly disjoint over K, it is necessa… 060Q
Definition 5 The direct limit E of the direct system (E_J, f_J'J) is called the tensor product of the family of A-algebras (Eᵢ)_i ∈ I. 060R
Proposition 8 (i) The homomorphisms fᵢ : Eᵢ → E = ⨂_k ∈ I Eₖ are such that for two indices i, j such that i ≠ j, fᵢ(xᵢ) and fⱼ(xⱼ) commute in E for all xᵢ ∈ Eᵢ and xⱼ ∈ Eⱼ; further, E is generat… ↩ 4 060S
Corollary Let (Eᵢ)_i ∈ I,\ (E'ᵢ)_i ∈ I be two families of A-algebras with the same indexing set and, for all i ∈ I, let uᵢ : Eᵢ → E'ᵢ be an algebra homomorphism. Then there exists one and on… ↩ 2 060T
Proposition 9 Let (Eᵢ)_i ∈ I be a family of A-algebras and for each i ∈ I let Bᵢ be a basis of Eᵢ such that the unit element eᵢ belongs to Bᵢ. Let B be the set of elements of the form ⨂_i ∈ I xᵢ… ↩ 1 060U
Lemma 1 Let A be a commutative ring, E an A-algebra, (xᵢ)_1 ≤ i ≤ n a finite sequence of elements of E, (λᵢ)_1 ≤ i ≤ n a finite sequence of elements of A and y an element of E; suppose tha… ↩ 3 060V
Lemma 2 Let A be a commutative ring, E an A-algebra and (xᵢ)_1 ≤ i ≤ n and (yᵢ)_1 ≤ i ≤ n two finite sequences of n elements of E; suppose that for 1 ≤ j ≤ i ≤ n, xᵢ yⱼ = λᵢⱼ yⱼ xᵢ with λᵢ… ↩ 1 060W
Lemma 3 Let A be a commutative ring, E an A-algebra, (xᵢ)_1 ≤ i ≤ n a finite sequence of elements of E and suppose that, for every ordered pair (i, j) of integers such that 1 ≤ j < i ≤ n, ↩ 3 060X
Definition 6 Let (Δᵢ)_i ∈ I be a finite family of commutative monoids written additively. A system of commutation factors over the Δᵢ with values in a commutative ring A is a system of mappings… ↩ 1 060Y
Example (1) The trivial system of commutation factors consists of the εᵢⱼ such that εᵢⱼ(αᵢ, βⱼ) = 1 for all i, j, αᵢ ∈ Δᵢ, βⱼ ∈ Δⱼ. 060Z
Proposition 10 Let A be a commutative ring and (Δᵢ)_i ∈ I a finite family of commutative monoids written additively; for each i ∈ I let Eᵢ be a graded A-algebra of type Δᵢ. Finally, let (εᵢⱼ) be… ↩ 8 0610
Remark (1) We recover the tensor product of algebras defined in no. 1 (with moreover the graduation the tensor product of those of its factors) taking εᵢⱼ(αᵢ, βⱼ) = 1 for all i, j, αᵢ and… 0611
Corollary 1 In the notation of Proposition 10, suppose further that F is a graded A-algebra of type Δ and that each fᵢ is a graded algebra homomorphism relative to φᵢ : Δᵢ → Δ; then f = ε ⨂ᵢ f… ↩ 1 0612
Corollary 2 Let (Eᵢ)_i ∈ I, (Fᵢ)_i ∈ I be two finite families of A-algebras, with Eᵢ and Fᵢ graded of type Δᵢ for all i ∈ I. For each i ∈ I, let gᵢ : Eᵢ → Fᵢ be a graded algebra homomorphism o… 0613
Remark In the proof of Proposition 10, a total ordering on I was used to define an algebra structure on the tensor product ⨂_i ∈ I Eᵢ of the A-modules ↩ 1 0614
Proposition 11 ("associativity" of the tensor ε-product). In the notation of Proposition 10, let (J_λ)_λ ∈ L be a partition of I and write Δ'_λ = ∏_i ∈ J_λ Δᵢ for all λ ∈ L. Let E'_λ be a graded… ↩ 1 0615
Proposition 12 ("associativity" of the tensor ε-product of graded algebras of the same types). With the notation of Proposition 10 of no. 7, suppose that all the Δᵢ are equal to the same monoid Δ… 0616
Remark (1) Let I be an infinite indexing set and Δ₀ a commutative monoid; let (Δᵢ)_i ∈ I denote the family such that Δᵢ = Δ₀ for all i and suppose given for every ordered pair of distinct… 0617
Definition 7 A graded A-algebra E of type Z is called anticommutative if for all non-zero homogeneous elements x, y of E ↩ 3 0618
Remark (1) Let E⁺ be the graded subalgebra of E the direct sum of the E₂ₙ (n ∈ Z); it follows from Definition 7 that if E is anticommutative, E⁺ is a subalgebra contained in the centre of… 0619
Lemma 4 Let E be a graded algebra of type Z and S a set of homogeneous elements ≠ 0; the set F of elements of E all of whose homogeneous components x ≠ 0 satisfy relation (36) for all y ∈… ↩ 1 061A
Proposition 13 Let E be a graded A-algebra of type Z and S a generating system of the algebra E consisting of homogeneous elements ≠ 0; for E to be anticommutative (resp. alternating), it is nece… ↩ 1 061B
Proposition 14 Let E and F be two graded A-algebras of type Z, both anticommutative (resp. alternating). Then the skew tensor product Eᵍ ⊗_A F (no. 7) is an anticommutative (resp. alternating) al… ↩ 3 061C
Corollary Let E be an anticommutative (resp. alternating) graded A-algebra of type Z. Then for every ring homomorphism ρ : A → B, the graded B-algebra ρ^*(E) (no. 8, Remark 2) is anticommuta… 061D
Remark Let E be an anticommutative graded A-algebra of type Z. Then the A-linear mapping of E ⊗_A E into E defined by multiplication of E (§ 1, no. 3) is a homomorphism of the graded A-al… 061E

§ 5. Tensor algebra. Tensors

Definition 1 For every module M over a commutative ring A, the tensor algebra of M, denoted by T(M), or Tens(M), or T_A(M), is the algebra ⨁_n ≥ 0 Tⁿ(M) with the multiplication defined in (1)… 061H
Proposition 1 Let E be a (unital) A-algebra and f : M → E an A-linear mapping. There exists one and only one A-algebra homomorphism g : T(M) → E such that f = g ∘ φ. ↩ 5 061I
Remark Suppose that E is a graded A-algebra of type Z, with graduation (Eₙ), and suppose also that f(M) ⊂ E₁. Then it follows from (2) that g(Tᵖ(M)) ⊂ Eₚ for all p ≥ 0 and hence g is a gr… 061J
Proposition 2 Let A be a commutative ring, M and N two A-modules and u : M → N ↩ 1 061K
Proposition 3 If u : M → N is a surjective A-linear mapping, the homomorphism T(u) : T(M) → T(N) is surjective and its kernel is the two-sided ideal of T(M) generated by the kernel P ⊂ M ⊂ T(M)… ↩ 2 061L
Proposition 4 Let N and P be two submodules of an A-module M such that their sum N + P is a direct factor in M and their intersection N ∩ P is a direct factor in N and in P. Then the homomorphis… ↩ 5 061M
Remark Note in particular that the hypotheses of Proposition 4 always hold for arbitrary submodules N, P of M when A is a field (II, § 7, no. 3, Proposition 4). Moreover, if N ⊂ P and N ≠… 061N
Corollary Let K be a commutative field and M a vector space over K. For every element z ∈ T(M), there exists a smallest vector space N of M such that z ∈ T(N) and N is of finite rank over K. ↩ 2 061O
Proposition 5 Let A, B be two commutative rings, ρ : A → B a ring homomorphism and M an A-module. The canonical extension ↩ 3 061P
Proposition 6 The A-homomorphism f' = lim → f'_α : lim → T_Aα(M_α) → T_A(M) is a graded algebra isomorphism. ↩ 2 061Q
Theorem 1 Let A be a commutative ring, M a free A-module and (e_λ)_λ ∈ L a basis of M. Then the elements eₛ = e_λ₁ ⊗ e_λ₂ ⊗ ⋯ ⊗ e_λₙ, where s = (λ₁, …, λₙ) runs through the set of all finite… ↩ 4 061R
Remark The above results can equally be obtained as a consequence of the universal properties of the free associative algebra of the tensor algebra, using II, § 3, no. 7, Corollary 2 to P… 061S
Corollary If M is a projective A-module, T(M) is a projective A-module. ↩ 2 061T
Proposition 7 Let Δ be a commutative monoid, M a graded A-module of type Δ and (M_α)_α ∈ Δ its graduation. For every ordered pair (α, n) ∈ Δ × N, let T^α, n(M) be the (direct) sum of the submodu… ↩ 3 061U
Remark 2 Let (e_λ)_λ ∈ L, (bare_λ)_λ ∈ L be two bases of M and P the matrix of passage from (e_λ) to (bare_λ). On account of what was seen in Example 1, the element of P appearing in the ro… 061V

§ 6. Symmetric algebras

Definition 1 Let A be a commutative ring and M an A-module. The symmetric algebra of M, denoted by S(M), or Sym(M), or SA(M), is the quotient algebra over A of the tensor algebra T(M) by the tw… ↩ 1 0626
Proposition 1 The algebra S(M) is commutative. 0627
Proposition 2 Let E be an A-algebra and f : M → E an A-linear mapping such that (1) f(x)f(y) = f(y)f(x) for all x, y in M. There exists one and only one A-algebra homomorphism g : S(M) → E such… ↩ 3 0628
Remark (1) Suppose that E is a graded A-algebra of type Z, with graduation (Eₙ), and suppose also that the linear mapping f (assumed to satisfy (1)) is such that (2) f(M) ⊂ E₁. Then the r… 0629
Proposition 3 Let A be a commutative ring, M and N two A-modules and u : M → N an A-linear mapping. There exists one and only one A-algebra homomorphism u' : S(M) → S(N) such that the diagram ↩ 1 062A
Proposition 4 If u: M → N is a surjective A-linear mapping, the homomorphism S(u): S(M) → S(N) is surjective and its kernel is the ideal of S(M) generated by the kernel P ⊂ M ⊂ S(M) of u. ↩ 1 062B
Proposition 5 Let N and P be two submodules of an A-module M such that their sum N + P is a direct factor in M and their intersection N ∩ P is a direct factor in N and in P. Then the homomorphis… 062C
Corollary Let K be a commutative field and M a vector space over K. For every element z ∈ S(M) there exists a smallest vector space N of M such that z ∈ S(N) and N is finite-dimensional. 062D
Proposition 6 Let A be a commutative ring and M and N two A-modules. If with every A-linear mapping g : Sⁿ(M) → N (n geqslant 1) is associated the n-linear mapping 062E
Remark Let σ be a permutation in 𝔖ₙ; as the mapping ↩ 1 062F
Proposition 7 Let A, B be two commutative rings, ρ : A → B a ring homomorphism and M an A-module. The canonical extension 062G
Proposition 8 The A-homomorphism f' = lim → f'_α : lim → S_A_α(M_α) → S_A(M) is a graded algebra isomorphism. 062H
Proposition 9 The canonical homomorphism g (formula (10)) is a graded algebra isomorphism (cf. § 4, no. 8, Remark 1). ↩ 4 062I
Corollary The A-module Sⁿ(M) is the image under isomorphism (10) of the submodule of ⨂_λ ∈ L S(M_λ) the direct sum of the submodules S^ν(M) for all the families ν = (n_λ) ∈ N⁽ᴸ⁾ such that ∑_… ↩ 1 062J
Theorem 1 Let A be a commutative ring and M a free A-module with basis (e_λ)_λ ∈ L. For every mapping α : L → N of finite support, we write ↩ 2 062K
Corollary If M is a projective A-module, S(M) is a projective A-module. 062L
Proposition 10 Let Δ be a commutative monoid, M a graded A-module of type Δ and (M_α)_α ∈ Δ its graduation. For every ordered pair (α, n) ∈ Δ × N, let S^α, n(M) be the submodule of Sⁿ(M) the dire… 062M

§ 7. Exterior algebras

Definition 1 Let A be a commutative ring and M an A-module. The exterior algebra of M, denoted by bigwedge(M) or Alt(M) or bigwedge_A(M), is the algebra over A the quotient of the tensor algebr… 062S
Proposition 1 Let E be an A-algebra and f : M → E an A-linear mapping such that (f(x))² = 0 for all x ∈ M. There exists one and only one A-algebra homomorphism g : bigwedge(M) → E such that f =… ↩ 4 062T
Remark (1) Suppose that E is a graded A-algebra of type Z, of graduation (Eₙ), and suppose also that the linear mapping f (assumed to satisfy (1)) is such that f(M) ⊂ E₁. Then the relatio… 062U
Proposition 2 Let A be a commutative ring, M and N two A-modules and u : M → N an A-linear mapping. There exists one and only one A-algebra homomorphism u'' : bigwedge(M) → bigwedge(N) such that… ↩ 1 062V
Proposition 3 If u: M → N is a surjective A-linear mapping, the homomorphism Λ(u): Λ(M) → Λ(N) is surjective and its kernel is the two-sided ideal of Λ(M) generated by the kernel P ⊂ M ⊂ Λ(M) of… ↩ 1 062W
Proposition 4 Let N and P be two submodules of an A-module M such that their sum N + P is a direct factor in M and their intersection N ∩ P is a direct factor in N and in P. Then the homomorphis… 062X
Corollary Let K be a commutative field and M a vector space over K. For every element z ∈ Λ(M), there exists a smallest vector subspace N of M such that z ∈ Λ(N) and N is finite-dimensional. ↩ 2 062Y
Proposition 5 (i) Let (xᵢ)_1 ≤ i ≤ n be a finite sequence of elements of the module M; for every permutation σ in the symmetric group Sₙ, ↩ 6 062Z
Corollary 1 Let H, K be two complementary subsets of the interval [1, n] of N and let (iₕ)_1 ≤ h ≤ p, (jₖ)_1 ≤ k ≤ n-p be the sequences of elements of H and K respectively, arranged in increas… ↩ 1 0630
Lemma 1 If σ ∈ Sₙ is the permutation such that σ(h) = iₕ for 1 ≤ h ≤ p, σ(h) = jₙ₋ₚ for p + 1 ≤ h ≤ n, then ε_σ = (-1)^ν. 0631
Corollary 2 The graded algebra bigwedge(M) is alternating (§ 4, no. 9). ↩ 1 0632
Proposition 6 If M is a finitely generated A-module, bigwedge(M) is a finitely generated A-module; also, if M admits a generating system with n elements, then bigwedgeᵖ(M) = 0 for p > n. ↩ 5 0633
Proposition 7 Let A be a commutative ring and M and N two A-modules. If with every A-linear mapping g : bigwedgeⁿ(M) → N (n ≥ 2) is associated the n-linear mapping (x₁, x₂, …, xₙ) ↦ g(x₁ ∧ x₂ ∧… ↩ 6 0634
Corollary 1 For every alternating n-linear mapping g : Mⁿ → N, for every permutation σ ∈ 𝔖ₙ, g(x_σ(1), x_σ(2), …, x_σ(n)) = ε_σ · g(x₁, x₂, …, xₙ) for all xᵢ ∈ M; moreover if xᵢ = xⱼ for two d… ↩ 1 0635
Corollary 2 Let (xᵢ)_1 ≤ i ≤ n be a sequence of n elements of M such that x₁ ∧ x₂ ∧ ⋯ ∧ xₙ = 0; then, for every alternating n-linear mapping g : Mⁿ → N, g(x₁, …, xₙ) = 0. 0636
Corollary 3 Let f : Mⁿ⁻¹ → A be an alternating (n-1)-linear form. If (xᵢ)_1 ≤ i ≤ n is a sequence of n elements of M such that x₁ ∧ x₂ ∧ ⋯ ∧ xₙ = 0, then ∑ᵢ₌₁ⁿ (-1)ⁱ f(x₁, …, hatxᵢ, …, xₙ) · x… 0637
Remark An element z of Tⁿ(M) is called a (contravariant) skew-symmetric tensor of order n if σ.z = ε_σ z for every permutation σ ∈ Sₙ (cf. § 6, no. 3, Remark); these elements form a sub-A… 0638
Proposition 8 Let A, B be two commutative rings, ρ : A → B a ring homomorphism and M an A-module. The canonical extension 0639
Proposition 9 The A-homomorphism f'' = varprojlim f''_α : varprojlim Λ_A_α(M_α) → Λ_A(M) is a graded algebra isomorphism. 063A
Proposition 10 The canonical homomorphism g (formula (14)) is a graded algebra isomorphism. ↩ 5 063B
Remark Let N = ⨁_λ ∈ L N_λ be the direct sum of another family of A-modules with L as indexing set and, for all λ ∈ L, let v_λ : M_λ → N_λ be an A-linear mapping, whence there is an A-lin… 063C
Corollary The A-module bigwedgeⁿ(M) is the image under isomorphism (14) of the submodule of ⨂_λ ∈ L bigwedge(M_λ) the direct sum of the submodules bigwedge^ν(M) for all families ν = (n_λ) ∈… ↩ 1 063D
Proposition 11 Let Δ be a commutative monoid, M a graded A-module of type Δ and (M_α)_α ∈ Δ its graduation. For every ordered pair (α, n) ∈ Δ × N, let bigwedge^α,n(M) be the submodule of bigwedge… ↩ 1 063E
Theorem 1 Let M be an A-module with a basis (e_λ)_λ ∈ L. Let L be given the structure of a totally ordered set (Set Theory, III, § 2, no. 3, Theorem 1) and for every finite subset J of L we… ↩ 6 063F
Corollary 1 Suppose that L = 1, m; then the basis (e_J)_J ∈ P(L) of bigwedge(M) has 2ᵐ elements. For p > m, bigwedgeᵖ(M) = 0; bigwedgeᵐ(M) has a basis consisting of a single element e_L; for 0… ↩ 3 063G
Corollary 2 If M is a projective A-module, bigwedge(M) is a projective A-module. ↩ 2 063H
Corollary 3 Let M be a projective A-module and (xᵢ)_1 ≤ i ≤ n a finite sequence of elements of M. For there to exist on M an alternating n-linear form f such that f(x₁, x₂, …, xₙ) ≠ 0, it is n… ↩ 2 063I
Proposition 12 Let M be a projective A-module. For elements x₁, x₂, …, xₙ of M to be linearly dependent, it is necessary and sufficient that there exist λ ≠ 0 in A such that λ x₁ ∧ x₂ ∧ ⋯ ∧ xₙ =… ↩ 3 063J
Corollary Let M and N be two projective A-modules and f : M → N an A-linear mapping. If f is injective, then Λ(f) : Λ(M) → Λ(N) is injective. ↩ 1 063K
Proposition 13 Let M be an A-module, N a direct factor submodule of M which is free of dimension p and u a basis of Λᵖ(N). For an element x ∈ M to belong to N, it is necessary and sufficient that… ↩ 3 063L
Theorem 2 Let M be an A-module with a finite basis (eᵢ)_1 ≤ i ≤ n. For a sequence (xᵢ)_1 ≤ i ≤ n of n elements of M to form a basis of M, it is necessary and sufficient that the element λ ∈… ↩ 1 063M

§ 8. Determinants

Definition 1 The determinant of an endomorphism u of a free A-module M of finite dimension n (II, § 7, no. 2, Corollary to Proposition 3 and Remark 1), denoted by det u, is the scalar λ such th… 0642
Proposition 1 (i) If u and v are two endomorphisms of a finite-dimensional free A-module M, then ↩ 1 0643
Proposition 2 Let M be a free A-module with a finite basis (eᵢ)_1 ≤ i ≤ n and v an endomorphism of M. For every sequence (xᵢ)_1 ≤ i ≤ n of n elements of M, 0644
Theorem 1 Let M be a finite-dimensional free A-module and u an endomorphism of M. The following conditions are equivalent: (a) u is bijective; (b) u is right invertible (II, § 1, no. 9, Coro… ↩ 5 0645
Proposition 3 Let M be a finite-dimensional free A-module. For every endomorphism u of M, the following conditions are equivalent: (f) u is injective; (g) det u is not a divisor of zero in A. ↩ 5 0646
Remark When A is a field, condition (e) of Theorem 2 is equivalent to condition (g) of Proposition 3, since they both mean that det u ≠ 0. In this case therefore all the conditions of The… 0647
Definition 2 Let I be a finite set, A a commutative ring and X a square matrix of type (I, I) over the ring A (II, § 10, no. 7). The determinant of the endomorphism u of the A-module Aᴵ, whose… 0648
Proposition 4 For n vectors x₁, …, xₙ of Aⁿ, let X(x₁, …, xₙ) denote the square matrix of order n whose i-th column is xᵢ for 1 ≤ i ≤ n. Then the mapping (x₁, …, xₙ) ↦ det(X(x₁, …, xₙ)) of (Aⁿ)ⁿ… 0649
Example (1) The determinant of the empty matrix is equal to 1; the determinant of a square matrix of order 1 is equal to the unique element of this matrix. For a matrix of order 2 064A
Proposition 5 If X and Y are two square matrices over a commutative ring A with the same finite indexing set, then ↩ 1 064B
Corollary Two similar square matrices have equal determinants. 064C
Proposition 6 For the columns of a square matrix X of finite order to be linearly independent, it is necessary and sufficient that det X be not a divisor of zero in A. ↩ 1 064D
Lemma 1 Let A be a commutative ring and M a free A-module with a basis (eⱼ)_j ∈ J, where the indexing set J is totally ordered. For every integer p ≤ Card(J), every alternating p-linear fu… 064E
Proposition 7 Let I be a finite set and X = (ξⱼᵢ)_(j, i) ∈ I × I a square matrix of type (I, I) over a commutative ring A. Then ↩ 1 064F
Proposition 8 For every square matrix X over a commutative ring, the determinant of the transpose matrix ᵗ X is equal to the determinant of X. ↩ 4 064G
Corollary 1 For n vectors x₁, …, xₙ of Aⁿ, let Y(x₁, …, xₙ) denote the square matrix of order n whose i-th row is xᵢ, for 1 ≤ i ≤ n. Then the mapping (x₁, …, xₙ) ↦ det(Y(x₁, …, xₙ)) from (Aⁿ)ⁿ… ↩ 2 064H
Corollary 2 For a square matrix X of finite order over a commutative ring A the following conditions are equivalent: (i) the rows of X are linearly independent; (ii) the columns of X are linea… 064I
Corollary 3 Let u be an endomorphism of a finite-dimensional free A-module M and ᵗ u the transpose endomorphism of the dual M^* (II, § 2, no. 5, Definition 5); then det(ᵗ u) = det(u). If X is… ↩ 1 064J
Proposition 9 Let M be an A-module with a basis (eᵢ)_i ∈ J (finite or otherwise) whose indexing set J is totally ordered. For every integer p > 0, let (e_H)_H ∈ 𝔖ₚ(J) be the corresponding basis… ↩ 2 064K
Proposition 10 Let M and N be two free A-modules of respective dimensions m and n, u : M → N a linear mapping and X the matrix of u with respect to a basis (eᵢ)_1 ≤ i ≤ m of M and a basis (fⱼ)_1… ↩ 3 064L
Proposition 11 Let M be a free A-module of finite dimension n; for every endomorphism u of M and every ordered pair of elements ξ, η of A, ↩ 2 064M
Corollary Under the same hypotheses as in Proposition 11, for the endomorphism bigwedge(u) of the A-module bigwedge(M) 064N
Proposition 12 For every invertible square matrix X of type (n, n), the inverse of X is given by the formula 064O
Example (1) Vandermonde determinant. Given a sequence (ζᵢ)_1 ≤ i ≤ n of n elements of A, the Vandermonde determinant of this sequence is the determinant 064P
Proposition 13 For a system of n linear equations in n unknowns over a commutative ring to admit at least one solution regardless of the right hand sides, it is necessary and sufficient that the… ↩ 1 064Q
Proposition 14 For a homogeneous linear system of n equations in n unknowns over a commutative ring to admit a non-zero solution, it is necessary and sufficient that the determinant of its matrix… 064R
Proposition 15 Let E be a vector space over a commutative field; for p vectors xᵢ ∈ E (1 ≤ i ≤ p) to be linearly independent, it is necessary and sufficient that x₁ ∧ x₂ ∧ ⋯ ∧ xₚ ≠ 0. ↩ 3 064S
Corollary Let X be a matrix of type (m, n) over a commutative field. The rank of X is equal to the greatest integer p such that there exists at least one minor of X of order p which is ≠ 0. ↩ 1 064T
Proposition 16 Let L = (λᵢⱼ) be the matrix (of type (m, n)) of system (41). Let M be the matrix of type (m, n+1) obtained by adding to L the (n+1)-th column (ηᵢ) (II, § 10, no. 1). Let p be the r… 064U
Proposition 17 Let K be a commutative field. The unimodular group SL(n, K) is generated by the matrices Bᵢⱼ(λ) with i ≠ j and λ ∈ K. ↩ 4 064V
Corollary The group SL(n, K) is the group of commutators of GL(n, K), except in the case where n = 2 and where K is a field with 2 elements. ↩ 1 064W
Remark (1) GL(2, F₂) = SL(2, F₂); this is a solvable group of order 6, whose commutator group is of index 2 (II, § 10, Exercise 14). 064X
Proposition 18 The sequence of A[X]-homomorphisms M[X] xrightarrowψ M[X] xrightarrowφ Mᵤ longrightarrow 0 is exact. ↩ 2 064Y
Proposition 19 For a mapping g of M into M' to be an A[X]-homomorphism of Mu into M'u, it is necessary and sufficient that g be an A-homomorphism of M into M' such that g ∘ u = u' ∘ g. When this… ↩ 2 064Z
Definition 3 Let M be a finite-dimensional free A-module and u an endomorphism of M. The determinant of the endomorphism X - baru of the free A[X]-module M[X] is called the characteristic polyn… ↩ 1 0650
Proposition 20 ("Cayley–Hamilton theorem") For every endomorphism u of a finite-dimensional free A-module, χᵤ(u) = 0. ↩ 2 0651

§ 9. Norms and traces

Definition 1 Let M be an A-module admitting a finite basis as a K-module. For all a ∈ A, let a_M be the endomorphism x ↦ ax of the K-module M. The trace, determinant and characteristic polynomi… 065S
Proposition 1 Let M = M₀ ⊃ M₁ ⊃ … ⊃ Mᵣ = 0 be a decreasing sequence of submodules of an A-module M such that each of the K-modules Pᵢ = Mᵢ₋₁/Mᵢ (1 ≤ i ≤ r) admits a finite basis. Then the K-modu… ↩ 4 065T
Proposition 2 Let A, A' be two K-algebras, M an A-module and M' an A'-module. Suppose that M and M' are free K-modules of respective dimensions n and n' and consider M ⊗_K M' as an (A ⊗_K A')-mo… ↩ 3 065U
Definition 2 Let A be a K-algebra which is a finite-dimensional free K-module. For every element a ∈ A, the trace (resp. norm,\footnote{† This notion should not be confused with the notion for… ↩ 2 065V
Example (1) Let A be a quadratic algebra over K of type (α, β) and (e₁, e₂) a basis of type (α, β) (§ 2, no. 3). For x = ξ e₁ + η e₂, Tr_A/K(x) = 2ξ + β η and N_A/K(x) = ξ² + β ξ η - α η²… 065W
Proposition 3 Let A be a K-algebra admitting a finite basis. For an element a ∈ A to be invertible, it is necessary and sufficient that its N_A/K(a) be invertible in K. ↩ 2 065X
Proposition 4 Let A be a K-algebra admitting a finite basis. For all a ∈ A, Pc_A/K(a; a) = 0. 065Y
Proposition 5 Let A be a K-algebra and m a two-sided ideal of A. Suppose that A₀ = A/m is a free K-module of finite dimension n, that there exists an integer r > 0 such that mʳ = 0 and that mⁱ⁻¹… 065Z
Proposition 6 Let A be a commutative K-algebra admitting a finite basis over K and V an A-module admitting a finite basis over A. Then V admits a finite basis over K and for every A-endomorphism… ↩ 3 0660
Lemma 1 Let Xᵢⱼ (1 ≤ i ≤ n,\ 1 ≤ j ≤ n) be n² indeterminates, X the square matrix (Xᵢⱼ) of order n and D(X₁₁, …, Xₙₙ) ∈ Z[X₁₁, …, Xₙₙ] the determinant det(X). On the other hand let A be a… 0661
Corollary Let A be a commutative K-algebra admitting a finite basis over K and B an A-algebra admitting a finite basis over A. Then B admits a finite basis over K, and, for all b ∈ B ("trans… 0662
Remark Suppose that the homomorphism λ ↦ λ . 1 of K into A is injective and let K be identified with its image in A; suppose that A admits a finite basis (eᵢ)_1 ≤ i ≤ n as a K-module. Let… 0663
Definition 3 Let A be a K-algebra admitting a finite basis of n elements. The discriminant of a sequence (x₁, …, xₙ) of n elements of A, with respect to K, denoted by D_A/K(x₁, …, xₙ), is the d… ↩ 1 0664
Example (1) If A is a quadratic algebra of type (α, β) over K, then (in the notation of § 2, no. 3) Tr(e₁) = 2, Tr(e₂) = β, 0665

§ 10. Derivations

Definition 1 Given the above and a commutation factor ε on Δ × Δ, an ε-derivation (or (K, ε)-derivation) of degree δ ∈ Δ of (A, A', A'') into (B, B', B'') is a triple of graded K-linear mapping… ↩ 3 0667
Remark (1) The notion of derivation can be defined for non-graded modules by agreeing to give these modules the trivial graduation. 0668
Example 1 Let A be an R-algebra of differentiable mappings of R into R and let x₀ be a point of R; R can be considered as an A-module with the external law (f, a) ↦ f(x₀)a. Then the mapping… ↩ 1 0669
Example 2 Let X be a differentiable manifold of class C^∞ and let A be the graded R-algebra of differential forms on X. The mapping which associates with every differential form ω on X its e… ↩ 1 066A
Example 3 Let A be an associative K-algebra. For all a ∈ A, the mapping x ↦ ax - xa is a derivation of the algebra A (cf. no. 6). 066B
Example 4 Let M be a K-module and A the exterior algebra bigwedge(M^*) with its usual graduation (§ 7, no. 1). It will be seen in § 11, no. 9 that, for all x ∈ M, the right interior product… 066C
Example 5 Returning to the general situation of Definition 1 of no. 2, let overlineK be another commutative ring and ρ : K → overlineK a ring homomorphism; let overlineA, overlineA', overlin… 066D
Example 6 Let A be a graded K-algebra of type Z; a graded linear K-mapping of degree 0, d : A → A, is defined by taking, for xₙ ∈ Aₙ (n ∈ Z), d(xₙ) = nxₙ. This mapping is a derivation of A… ↩ 1 066E
Proposition 1 Let d₁, d₂ be two ε-derivations of (A, A', A'') of respective degrees δ₁, δ₂. Then the ε-bracket [d₁, d₂]_ε = d₁ ∘ d₂ - ε_δ₁, δ₂ d₂ ∘ d₁ is an ε-derivation of degree δ₁ + δ₂. Moreo… ↩ 1 066F
Corollary Suppose that Δ = Z. Then: (i) The square of an antiderivation is a derivation. (ii) The bracket of two derivations is a derivation. (iii) The bracket of an antiderivation and a der… 066G
Proposition 2 With the above hypotheses and notation, consider 2n indeterminates T₁, …, Tₙ, T'₁, …, T'ₙ and for every polynomial F ∈ K[X₁, …, Xₙ] write F(T) = F(T₁, …, Tₙ), F(T') = F(T'₁, …, T'ₙ… ↩ 2 066H
Corollary (Leibniz's formula) Let dᵢ (1 ≤ i ≤ n) be n derivations of (A, A', A'') which are permutable with one another. For all α = (α₁, …, αₙ) ∈ Nⁿ, we write 066I
Proposition 3 Let d : A → E be an ε-derivation of degree δ. Then Ker(d) is a graded subalgebra of A; if A admits a unit element, it belongs to Ker(d). ↩ 3 066J
Corollary Let d₁ and d₂ be two ε-derivations from A to E of the same degree δ. If d₁ and d₂ take the same values at each element of a generating system of the algebra A, then d₁ = d₂. ↩ 1 066K
Proposition 4 Let d : A → E be an ε-derivation of degree δ. Suppose that A has a unit element 1 and let x be a homogeneous element of A with an inverse x⁻¹ in A. Then 066L
Proposition 5 Suppose that A is an integral domain and let L be its field of fractions. Every derivation of A into a vector space E over L (considered as an A-module) can be extended uniquely to… ↩ 2 066M
Proposition 6 Suppose that A is a unital associative graded K-algebra and E a graded (A, A)-bimodule. If d : A → E is an ε-derivation of degree δ, then, for every finite sequence (xᵢ)_1 ≤ i ≤ n… ↩ 1 066N
Corollary Suppose that A is a unital commutative associative algebra and E an A-module. If d : A → E is a derivation, then, for every integer n ≥ 0, d(xⁿ) = nxⁿ⁻¹(dx) for all x ∈ A. It suffi… 066O
Proposition 7 Suppose that A is a unital associative graded algebra and E a graded (A, A)-bimodule. Let d : A → E be an ε-derivation of degree δ and a a homogeneous element of Z_ε of degree α. T… ↩ 1 066P
Proposition 8 Let A be a graded K-algebra. 066Q
Corollary Let A be an associative graded algebra. For two homogeneous elements, a, b of A, [ad_ε(a), ad_ε(b)]_ε = ad_ε([a, b]_ε). (25) 066R
Proposition 9 Let A, B be two graded K-algebras, E an (A, A)-bimodule and F a graded (B, B)-bimodule; let ρ : A → B be a graded algebra homomorphism and θ : E → F a graded A-homomorphism of A-bi… ↩ 6 066S
Corollary Let S be a generating system of the algebra A. In order that d' ∘ ρ = θ ∘ d, it is necessary and sufficient that d'(ρ(x)) = θ(d(x)) for all x ∈ S. ↩ 1 066T
Proposition 10 Under the conditions of Proposition 9, for an ε-derivation d' : B → F to be A-linear for the left (resp. right) A-module structures on B and ρ^*(F), it is necessary and sufficient… ↩ 1 066U
Proposition 11 Under the above conditions, there is an exact sequence of K-homomorphisms 0 → D_B(C, G) xrightarrowu D_A(C, G) xrightarrowv D_A(B, G) where u is the canonical injection and v the h… ↩ 2 066V
Proposition 12 For a graded K-linear mapping f : A → E of degree δ to be an ε-derivation, it is necessary and sufficient that the mapping x ↦ (x, f(x)) of A into A ⊕ E(δ) be a graded K-algebra ho… ↩ 1 066W
Proposition 13 For the algebra A ⊕ E(δ) to be associative and unital, it is necessary and sufficient that A be associative and unital, and that the mappings (a, x) ↦ a.x and (a, x) ↦ x.a define o… ↩ 1 066X
Proposition 14 Let A be a commutative ring, M an A-module, B the A-algebra T(M) (resp. S(M), resp. Λ(M)) and E a (B, B)-bimodule. Let d₀ : A → E be a derivation of the ring A into the A-module E… ↩ 2 066Y
Corollary Let M be a graded K-module of type Δ; the K-algebras T(M), S(M) and bigwedge(M) are given the corresponding graduations of type Δ' = Δ × Z (§ 5, no. 5, Proposition 7, § 6, no. 6, P… ↩ 1 066Z
Proposition 15 If M is a free K-module of finite rank n, then, for every endomorphism s of M, the restriction to Λⁿ(M) of the derivation Dₛ is the homothety of ratio Tr(s). 0670
Proposition 16 Let A be a commutative K-algebra, Mᵢ (1 ≤ i ≤ n) and P A-modules and H the A-module of A-multilinear mappings of M₁ × M₂ × ⋯ × Mₙ into P. Suppose that there is given a K-derivation… ↩ 1 0671
Example (3) Applying Proposition 16 to the case n = 1, M₁ = M, P = A, then H = M^*, the dual of M, and it is seen that for a K-derivation (d₀, d, d') of (A, M, M) is derived a K-derivation… ↩ 1 0672
Lemma 1 The mapping δ_A : x ↦ x ⊗ 1 - 1 ⊗ x is a K-derivation of A into I and I is generated, as a left A-module, by the image of δ_A. ↩ 2 0673
Proposition 17 The derivation δ_A has the following universal property: for every (A, A)-bimodule E and every K-derivation d : A → E, there exists one and only one (A, A)-bimodule homomorphism f… ↩ 1 0674
Proposition 18 Let A be a commutative K-algebra and 𝔍 the ideal the kernel of the surjective canonical homomorphism m : A ⊗_K A → A, so that A is isomorphic to (A ⊗_K A)/𝔍 and 𝔍/𝔍² has canonicall… ↩ 2 0675
Example Let M be a K-module; it follows from Proposition 14 of no. 9 that for every S(M)-module E, the mapping D ↦ D M defines an S(M)-module isomorphism of DK(S(M), E) onto HomK(M, E); on… 0676
Proposition 19 Let 0677
Proposition 20 Suppose that A' = A ⊗_K K', with η': K' → A' and u: A → A' the canonical homomorphisms. Then the A'-linear mapping ↩ 1 0678
Proposition 21 The sequence of B-linear mappings ↩ 1 0679
Proposition 22 Let 𝔍 be an ideal of the commutative K-algebra A, B = A/𝔍 and u : A → B the canonical homomorphism. The sequence of B-linear mappings ↩ 2 067A
Proposition 23 Let A be an integral K-algebra, B its field of fractions and u: A → B the canonical injection. Then Ω0(u): ΩK(A) ⊗A B → ΩK(B) is an isomorphism. 067B

§ 11. Cogebras, products of multilinear forms, inner products and duality

Definition 1 A cogebr a over A (or A-cogebra, or simply cogebr a if no confusion can arise) is a set E with a structure defined by giving the following: (1) an A-module structure on E; (2) an A… 067H
Definition 2 Given two cogebras E, E', whose coproducts are denoted respectively by c and c', a morphism of E into E' is an A-linear mapping u: E → E' such that (1) (u ⊗ u) ∘ c = c' ∘ u, in oth… ↩ 1 067I
Example (1) The canonical isomorphism A → A ⊗_A A (II, § 3, no. 5) defines an A-cogebra structure on A. ↩ 1 067J
Example (9) It is immediate that the cogebras T(M), S(M) and Λ(M) defined above are graded cogebras. ↩ 1 067K
Example (1) Take E to be the graded cogebrà T(M) (no. 1) and suppose that N, N', N'' have the trivial graduation. A graded homomorphism of degree -p of T(M) into N (resp. N', N'') then cor… 067L
Remark Suppose that the Eₚ are finitely generated projective A-modules, so that the graded A-modules (E ⊗_A E)^*gr and E^*gr ⊗_A E^*gr can be canonically identified (II, § 4, no. 4, Corol… 067M
Proposition 1 Let E be a cogebra over A. In order that, for every associative A-algebra B, the A-algebra Hom_A(E, B) be associative, it is necessary and sufficient that the coproduct c : E → E ⊗… ↩ 1 067N
Example (4) It is immediately verified that the cogebra A (no. 1, Example (1)) the cogebra A^(X) (no. 1, Example 4) and the cogebra T(M) (no. 1, Example 5) are coassociative. If B is an as… ↩ 1 067O
Proposition 2 Let E be a cogebras over A. In order that, for every commutative A-algebra B, the A-algebra Hom_A(E, B) be commutative, it is necessary and sufficient that the coproduct c : E → E… ↩ 2 067P
Example (5) It is immediate that the cogebra A (no. 1, Example 1) and the cogebra A^(X) (II, § 11, no. 1, Example 4) are cocommutative. It follows from formula (5) of no. 1 that the cogebr… ↩ 5 067Q
Proposition 3 Let E be a cogebra over A. In order that, for every unital A-algebra B, the A-algebra Hom_A(E, B) be unital, it is necessary and sufficient that there exist a linear form γ on E re… ↩ 1 067R
Example (6) The identity mapping is the counit of the cogebra A; on the cogebra A^(X) (no. 1, Example 4) the linear form γ such that γ(eₓ) = 1 for all x ∈ X is the counit. On the cogebra T… ↩ 3 067S
Proposition 4 Let E be a cogebra admitting a counit γ and suppose that there exists in E an element e such that γ(e) = 1; then E is the direct sum of the sub-A-modules Ae and E_γ = Ker(γ) and ↩ 2 067T
Remark Let C be a counital coassociative A-coalgebra, B a unital associative A-algebra and M a left B-module. The A-bilinear mapping (b, m) ↦ bm of B × M into M defines an A-bilinear mapp… 067U
Proposition 5 (i) Let E be a graded cogebra admitting a counit γ; then γ is a homogeneous linear form of degree 0. ↩ 2 067V
Proposition 6 Let E be a graded cogebra over A. In order that, for every commutative A-algebra B, with the trivial graduation, the graded A-algebra of type Z Homgr_A(E, B) (no. 2) be anticommuta… ↩ 1 067W
Example It follows immediately from formula (8) of no. 1 that for every A-module M, the graded cogebra bigwedge(M) is anticocommutative. 067X
Definition 3 A graded bigebra (resp. skew graded bigebra) over a ring A is a set E with a graded A-algebra structure of type N and a graded A-cogebra structure of type N, with the same underlyi… ↩ 4 067Y
Example (1) Let S be a monoid with identity element u, so that the algebra E = A^(S) of the monoid S over A admits the unit element eᵤ (§ 2, no. 6); it has been seen on the other hand that… 067Z
Remark If M is an A-module such that M ⊗_A M ≠ 0, the graded A-algebra (§ 5, no. 1) and graded A-cogebra (no. 1, Example 5) structures on T(M) do not define a bigebra structure, for in ge… 0680
Lemma 1 (i) In the associative algebra E^* = Hom_A(E, A), the product of n elements u₁, u₂, …, uₙ of degree 1 is given by u₁ u₂ … uₙ = mₙ ∘ (u₁ ⊗ u₂ ⊗ ⋯ ⊗ uₙ) ∘ cₙ. (ii) Suppose also that… 0681
Proposition 7 Let M be a finitely generated projective A-module. Then the canonical homomorphisms θ_T : T(M^*) → T(M)^*gr and θ_Λ : bigwedge(M^*) → (bigwedge(M)^*gr)⁰ are bijective. Also the gra… ↩ 4 0682
Lemma 2 Let ↩ 1 0683
Proposition 8 Suppose that the ring A is an algebra over the field Q of rational numbers; then, for every finitely generated projective A-module M, the homomorphism ↩ 2 0684
Remark Let M be an A-module and ρ : A → B a commutative ring homomorphism. Then there is a commutative diagram of graded B-algebra homomorphisms 0685
Example In particular the above may be applied when E is one of the graded algebras T(M), S(M) or bigwedge(M) for an A-module M and P an A-module (with the trivial graduation). To find exp… 0686
Example In particular the above can be applied when E is one of the graded cogebras T(M), S(M) or Λ(M) for an A-module M (no. 1, Examples 5, 6 and 7). To find explicitly the bimodule struc… 0687
Proposition 9 Let E be a graded bigebra (resp. skew graded bigebra). For every element x of degree 1 in E, the left and right inner products by x are derivations (resp. antiderivations) (§ 10, n… ↩ 1 0688
Remark (1) Let E be a graded bigebra over A and N, N', N'' three graded A-modules. Let m be an A-bilinear mapping of N × N' into N''; for u ∈ Homgr_A(E, N) and v ∈ Homgr_A(E, N'), let u.v… 0689
Proposition 10 Let E be a graded bigebra (resp. skew graded bigebra). For every element f of degree 1 in E^*gr, the left and right inner products are derivations (resp. antiderivations) of the al… 068A
Example Propositions 9 and 10 apply in particular to the graded bigebra S(M) and the skew graded bigebra Λ(M). The inner products by elements of degree 1 in S(M) (resp. S(M)^*gr) are deriv… 068B
Proposition 11 The canonical homomorphism θ_T : T(M^*) → T(M)^*gr (resp. θ_S : S(M) → S(M)^*gr, resp. θ_Λ : Λ(M^*) → Λ(M)^*gr) is a right T(M)-module (resp. right S(M)-module, resp. left Λ(M)-mod… ↩ 2 068C
Remark Being given the basis (eᵢ)_1 ≤ i ≤ n allows us to identify the algebra S(M) with the polynomial algebra A[X₁, …, Xₙ] (§ 6, no. 6); formula (75) shows that the inner product by e^*α… 068D
Proposition 12 Let M be a free A-module of dimension n; let e ∈ ∧ⁿ(M) be an element forming a basis of ∧ⁿ(M) and let e^* be the element of ∧ⁿ(M^*) such that (-1)^n(n-1)/2 θ_∧(e^*) is the dual bas… ↩ 2 068E
Proposition 13 With the hypotheses and notation of Proposition 11, for every endomorphism g of the A-module M. ↩ 2 068F
Corollary For every automorphism g of E, (80) Λ^(tg⁻¹) = (det g)⁻¹φ ∘ (Λ(g)) ∘ φ⁻¹. ↩ 1 068G
Proposition 14 (i) The orthogonal of M_z in E^* is the set of x^* ∈ E^* such that x^* ⊥ z = 0. 068H
Proposition 15 Let E be a vector space of dimension n and let e be an element ≠ 0 of ∧ⁿ(E) (hence forming a basis of this vector space). Let φ : ∧(E) → ∧(E^*) be the vector space isomorphism asso… ↩ 1 068I
Corollary If E is of dimension n, every non-zero (n-1)-vector E is pure. 068J
Proposition 16 For an element z ≠ 0 of bigwedgeᵖ(E) to be pure, it is necessary and sufficient that, for all u^* ∈ bigwedgeᵖ⁻¹(E^*), 068K

App. 0. Alternative algebras. Octonions

Lemma 1 For all p, q, r, s in the algebra F, 069A
Proposition 1 For an A-algebra F, the following conditions are equivalent: (a) For every ordered pair of elements x, y of F, the subalgebra generated by x and y is associative. (b) The trilinear… ↩ 4 069B
Lemma 2 Let E be an A-algebra such that the trilinear mapping (x, y, z) ↦ a(x, y, z) is alternating, S a generating system of E and U a sub-A-module of E containing S and such that sU ⊂ U… ↩ 1 069C
Lemma 3 Suppose that the mapping a is alternating. If H is a strongly associative subset of F, the subalgebra of F generated by H is strongly associative. ↩ 1 069D
Lemma 4 Suppose that the mapping a is alternating. Then, for all x ∈ F, the subalgebra of F generated by x is strongly associative. ↩ 1 069E
Lemma 5 Suppose that the mapping a is alternating and let X, Y be two strongly associative subalgebras of F. Then the subalgebra of E generated by X ∪ Y is associative. 069F
Definition 1 An algebra F is called alternative if it satisfies the equivalent conditions of Proposition 1. 069G
Proposition 2 Let A be a ring, F a Cayley A-algebra, e its unit element, s : x ↦ overlinex its conjugation and N : F → A its Cayley norm (§ 3, no. 4). ↩ 2 069H
Proposition 3 Let E be a Cayley A-algebra, γ an element of A and F the Cayley extension of E defined by γ and the conjugation of E (§ 2, no. 5, Proposition 5). For F to be alternative, it is nec… ↩ 2 069I
Proposition 4 Let F be an octonion algebra of type (-1, 0, -1, -1) over A. There exists a vector space V of dimension 3 over the field with two elements Z/2Z and a bijection λ ↦ e'_λ of V onto t… 069J
Remark As bare'_λ = -e'_λ for λ ≠ 0, 069K

IV. Polynomials and rational fractions

§ 1. Polynomials

Proposition 1 Let u and v be two polynomials. 069O
Proposition 2 Let a = (aᵢ)_i ∈ I, be a family of elements of A and let u ∈ A[X]. If v is the polynomial obtained by substituting Xᵢ + aᵢ for Xᵢ for each i ∈ I, then the constant term of v is equ… ↩ 1 069P
Corollary 1 Let m be the ideal of polynomials u ∈ A[X] such that u(a) = 0. Then m is generated by the polynomials Xᵢ - aᵢ (for i ∈ I). ↩ 1 069Q
Corollary 2 Let X = (Xᵢ)_i ∈ I and Y = (Yᵢ)_i ∈ I be two families of indeterminates. The set of polynomials u ∈ A[X,Y] such that u(X,X) = 0 is the ideal of A[X,Y] generated by the polynomials… 069R
Proposition 3 Let u ∈ A[X] and let X.Z be the family (XᵢZ)_i ∈ I, of elements of the polynomial ring A[X][Z]. The coefficient of Zᵏ in u(X.Z) is the homogeneous component of degree k of u, for e… 069S
Corollary For a polynomial u ∈ A[X] to be homogeneous of degree k it is necessary and sufficient that : 069T
Remark — Let x ∈ Aᴵ and let f be the mapping u ↦ u(x) of A[X] into A. Given an A-module M, we consider the homomorphism 1 ⊗ f of M[X] = M ⊗_A A[X] into M ⊗_A A = M. For each v ∈ M[X] we h… 069U
Proposition 4 Let E be an associative, commutative and unital A-algebra, x = (xᵢ)_i ∈ I a family of elements of E, u an element of A[(Xᵢ)_i ∈ I] and y = u(x). Then for every derivation D of E in… ↩ 1 069V
Corollary — Let f ∈ A[X₁, ..., Xₚ] and gᵢ ∈ A[Y₁, ..., Y_q] for 1 ≤ i ≤ p, and write h = f(g₁, ..., gₚ), then for 1 ≤ j ≤ q we have 069W
Proposition 5 — For any u ∈ A[X] and ν ∈ N⁽ᴵ⁾ we have ↩ 1 069X
Proposition 6 (*« Euler's identity »*) — Let u ∈ A[X] be a homogeneous polynomial of degree r ; we have 069Y
Proposition 7 — Let f ∈ A[X] be a non-zero polynomial in an indeterminate and a its leading coefficient. If a is cancellable in A (in particular iff is monic) then we have for any non-zero eleme… ↩ 4 069Z
Proposition 8 — If A is an integral domain, then so is A[(Xᵢ)_i ∈ I]. ↩ 3 06A0
Corollary 1 — If A is an integral domain, and u, v are elements of A[(Xᵢ)_i ∈ I], then deg(uv) = deg u + deg v. ↩ 1 06A1
Corollary 2 — If A is an integral domain, the invertible elements of A[(Xᵢ)_i ∈ I] are the invertible elements of A. 06A2
Proposition 9 — Let u ∈ A[(Xᵢ)_i ∈ I]; then for u to be nilpotent in the ring A[(Xᵢ)_i ∈ I] it is necessary and sufficient for all its coefficients to be nilpotent in the ring A. 06A3
Remark Let u and v be elements of A[(Xᵢ)ᵢ ∈ ,], and suppose that A is an integral domain, v is a non-zero multiple of u and v is homogeneous; then u is also homogeneous. For let u' ∈ A[(X… 06A4
Proposition 10 Let f and g be non-zero elements of A[X] of degrees m and n respectively. Let α₀ be the leading coefficient of f and μ = sup(n - m + 1, 0). There exist u, v ∈ A[X] such that α₀^μ g… ↩ 1 06A5
Corollary (« Euclidean division of polynomials ») — Let f be a non-zero element of A[X] whose leading coefficient is invertible and m = deg f. (i) For every g ∈ A[X] there exist u,\ v ∈ A[X] such that g = uf + v , deg v < m . More… 06A6
Proposition 11 — Let K be a commutative field. (i) For every non-zero ideal a of K[X] there exists precisely one monic polynomial f in K[X] such that a = (f ). (ii) Let f₁ and f₂ be in K[X]; for… ↩ 1 06A7
Proposition 12 Let K be a commutative field and f, g two elements of K[X]. For every polynomial d in K[X] the following properties are equivalent: ↩ 3 06A8
Definition 1 With the notation of Prop. 12 we say that d is a greatest common divisor (gcd for short) of f and g. We say that f and g are relatively prime or that f is prime to g if 1 is a gcd… 06A9
Corollary 1 Let d be a gcd of f and g and K' a commutative field containing K as subfield. Then d is a gcd of f and g considered as elements of K'[X]. 06AA
Corollary 2 — Let d be a gcd off and g. (i) If u ∈ K[X], du is a gcd of fu and gu. (ii) If v ∈ K[X] is a divisor (≠ 0) of f and g, then d/v is a gcd of f/v and g/v. This follows from Prop. 12… ↩ 2 06AB
Corollary 3 — Let w be a common factor of f and g. For w to be a gcd off and g it is necessary and sufficient that f/w and g/w are relatively prime. This follows from Cor. 2. 06AC
Corollary 4 — Let f, g, h ∈ K[X]. If f divides gh and is prime to g, then f divides h. For f divides gh and fh, hence f divides every gcd of gh and fh, in particular h (Cor. 2, (i)). 06AD
Corollary 5 — Let f, g ∈ K[X]. For fand g to be relatively prime it is necessary and sufficient that the canonical image of g in K[X]/(f) should be invertible. For this condition means that th… 06AE
Corollary 6 — Let f, g₁, g₂, ..., gₙ ∈ K[X]. If f is prime to g₁, g₂, ..., gₙ, then f is prime to g₁g₂...gₙ. 06AF
Definition 2 — Let K be a commutative field. We say that f ∈ K[X] is irreducible if deg f ≥ 1 and f is not divisible by any polynomial g such that 0 < deg g < deg f. It comes to the same to say… 06AG
Proposition 13 — Let I be the set of irreducible monic polynomials in K[X]. Let f be a non-zero element of K[X] and a its leading coefficient; then there exists precisely one family of positive i… 06AH

§ 2. Zeros of polynomials

Proposition 1 — Let f ∈ A[X] and α ∈ A. The remainder of the division of f by X - α is f(α). For α to be a root of f it is necessary and sufficient that X - α should be a divisor of f in A[X]. ↩ 2 06AO
Proposition 2 — Let f ∈ A[X], a ∈ A, and let h be an integer ≥ 0. The following conditions are equivalent: (i) f is divisible by (X - α)ʰ but not by (X - α)ʰ⁻¹; (ii) there exists g ∈ A[X] such t… ↩ 2 06AP
Proposition 3 — Let f be a non-zero element of A[X] and a ∈ A. There exists just one integer h ≥ 0 satisfying the conditions (i) and (ii) of Prop. 2. This is clear for condition (i), bearing in… 06AQ
Definition 1 — With the above notation we say that a is of order h, or multiplicity h relative to f. If h > 0 we also say that a is a root of order h or multiplicity h off. A root of order 1 is… ↩ 1 06AR
Remark 1 If f = 0 we agree to say that a has order ≥ h relative to f, whatever a ∈ A and the integer h ≥ 0. For any f ∈ A[X] and a ∈ A, to say that a has order ≥ h relative to f means that… 06AS
Remark 2 Let B be a commutative ring containing A as subring. Let f ∈ A[X] be non-zero and a ∈ A. The order of a relative to f is the same, whether we consider f as element of B[X] or as el… 06AT
Proposition 4 — Let f and g be non-zero elements of A[X]. Let a ∈ A, and let the orders of a relative to f and g be p and q respectively. (i) The order of a relative to f + g is ≥ inf(p, q). It… ↩ 1 06AU
Proposition 5 Suppose that A is an integral domain. Let f be a non-zero element of A[X], and α₁, …, αₚ pairwise distinct roots of f in A, of orders k₁, …, kᵣ. We have ↩ 3 06AV
Theorem 1 Let A be an integral domain. Given a non-zero element f of A[X], of degree n, then the sum of the orders of all the roots off in A is ≤ n. ↩ 2 06AW
Corollary Assume that A is an integral domain and let f, g ∈ A[X], of degrees ≤ n. If there exist n + 1 pairwise distinct elements x₁, …, xₙ₊₁ of A such that f(xᵢ) = g(xᵢ) for 1 ≤ i ≤ n + 1… ↩ 1 06AX
Proposition 6 (Lagrange interpolation formula) — Let K be a commutative field, α₁, α₂, …, a, distinct elements of K and β₁, β₂, …, βₙ any elements of K. For i = 1, 2, …, n we put ↩ 2 06AY
Corollary — Suppose that A is an integral domain. Let f ∈ A[X], of degree < n, and let K be a subring of A which is a field. If there exist n distinct elements α₁, …, aₙ of A such that αᵢ ∈… 06AZ
Proposition 7 — Let f ∈ A[X] and let a ∈ A be a root off: For a to be a simple root off it is necessary and sufficient that a should not be a root of the derivative Df of f. ↩ 3 06B0
Proposition 8 — Let f ∈ A[X] and a ∈ A, and suppose that a has order k ≥ 1 relative to f: Then a has order ≥ k - 1 relative to Df. If k . 1 is cancellable in A, then a has order k - 1 relative t… ↩ 2 06B1
Corollary — Let f ∈ A[X], a ∈ A and p an integer ≥ 0, further, suppose that p! . 1 is cancellable in A. Then for a to be a root of order p of f, it is necessary and sufficient that a should… 06B2
Proposition 9 — Assume that A is an integral domain. Let I be a set, (Hᵢ)_i ∈ I a family of infinite subsets of A and H = ∏_i ∈ I Hᵢ ⊂ A'. Iff is a non-zero element of A[(Xᵢ)_i ∈ I], and H_f the… ↩ 3 06B3
Corollary 1 — We keep the hypothesis and notation of Prop. 9. If l is non-empty, then H_f is infinite. 06B4
Corollary 2 — Suppose that A is an infinite integral domain or that A is an algebra over an infinite field. For every f ∈ A[(Xᵢ)_i ∈ I], let tildef: Aᴵ → A be the polynomial function defined b… ↩ 4 06B5
Theorem 2 (Principle of extension of algebraic identities) — Suppose that A is an infinite integral domain. Let g₁, …, gₘ, f be elements of A[(Xᵢ)_i ∈ I] and assume the following hypotheses: ↩ 4 06B6
Scholium — Let A be an integral domain and f ∈ A[(Xᵢ)_i ∈ I]. Th. 2 provides a convenient means of proving that f = 0. It suffices to consider an infinite integral domain E containing A as… 06B7

§ 3. Rational fractions

Definition 1 Let K be a commutative field and I a set. The field of fractions (I, p. 116) of the integral domain K[(Xᵢ)_i ∈ I] is denoted by K((Xᵢ)_i ∈ I) or K(Xᵢ)_i ∈ I. Its elements are calle… 06BL
Proposition 1 — Let r, s be two rational fractions. 06BM
Proposition 2 Let K be a commutative field, E a unital associative K-algebra and x = (xᵢ)_i ∈ I a family of pairwise permutable elements of E. The set Sₓ⁻¹B of f ∈ K((Xᵢ)_i ∈ I) such that x is s… ↩ 2 06BN
Corollary Let L be a commutative field, K a subfield of L, x = (xᵢ)_i ∈ I a family of elements of L, M the set consisting of the xᵢ, U the set of all f ∈ K((Xᵢ)_i ∈ I) such that x is substit… 06BO
Proposition 3 — Let E be an algebra over K which is associative, commutative, unital and non-zero. Let f \in K((Xi){i \in I}) and for each i \in I, let gi \in K((Yl){l \in L}). Given a family y… 06BP
Proposition 4 — Let E be an associative, commutative and unital K-algebra, x = (xᵢ)_i, I a family of elements of E and f ∈ K((Xᵢ)_i, I). Suppose that x is substitutable in f and y = f(x). (i) Fo… 06BQ

§ 4. Formal power series

Lemma 1 Let L be an infinite set and (u_λ)_λ ∈ L a family of elements of A[[I]], and put u_λ = ∑ α_λ,ν X^ν for λ ∈ L. Then the following conditions are equivalent: (i) The family (u_λ)_λ ∈… 06BY
Remark — Suppose that I is finite. For every integer n ≥ 0 let bₙ be the set of formal power series u ∈ A[[I]] such that ω(u) ≥ n. The sequence (bₙ)_n ≥ 0 is a fundamental system of neigh… 06BZ
Proposition 1 Let (u_λ)_λ ∈ L and (v_μ)_μ ∈ M be two summable families of elements of A[[I]]. Then the family (u_λ v_μ)_(λ, μ) ∈ L × M is summable and we have 06C0
Proposition 2 Let (u_λ)_λ be a summable family of elements of A[[I]]. (i) The family (1 + u_λ)_λ is multipliable. (ii) Let 𝔗 be the set of all finite subsets of L. For any M ∈ S put u_M = ∏_λ ∈… ↩ 2 06C1
Proposition 3 Let u = ∑ a_ν X^ν ∈ A[[I]] and m an integer > 0. For every n ∈ N let (α_ν, n)_ν ∈ N⁽ᴵ⁾ be the family of coefficients of uⁿ. If α₀ᵐ = 0, then α_ν, n = 0 for n ≥ |ν| + m. ↩ 1 06C2
Corollary — Let u ∈ A[[I]]; then for lim_n → ∞ uⁿ = 0 to hold it is necessary and sufficient that the constant term of u should be nilpotent. ↩ 1 06C3
Proposition 4 — Let I be a set and E an associative, commutative, unital, linearly topologized separated complete A-algebra. ↩ 3 06C4
Proposition 5 — In the ring A[[T]] of formal power series in one indeterminate the polynomial 1 - T is invertible, and we have (1 - T)⁻¹ = ∑ₙ₌₀ᵐ Tⁿ. ↩ 3 06C5
Proposition 6 — Let u ∈ A[[I]]; then for u to be invertible in A[[T]] it is necessary and sufficient that its constant term should be invertible in A. ↩ 3 06C6
Remark — Let A be the set of all formal power series with constant term 1. By Prop. 6, M is a commutative group under multiplication; the multiplicative group of A[[I]] is thus the direct… 06C7
Remarks 1) We often say that Dᵢ u is the partial derivative of u with respect to X, ; we also use the notation D_xᵢ u, (∂ u)/(∂ xᵢ) and u'_xᵢ. For a single indeterminate X the unique parti… 0DHH
Proposition 7 — Let (yᵢ)_i ∈ I, be a family of elements of E. When I is infinite, we assume that yᵢ tends to 0 along the filter of complements of finite subsets of I. There exists then a unique… ↩ 1 06C8
Corollary 1 — Let A be a continuous derivation of the A-algebra E. For every formal power series u ∈ A[[I]] the family ((Dᵢu)(x) · Δ xᵢ)_i ∈ I is summable and we have ↩ 3 06C9
Corollary 2 — The derivation Dᵢ is the unique continuous derivation of the A-algebra A[[I]] such that 06CA
Corollary 3 — Let f ∈ A[[X₁, ..., Xₛ]] and gᵢ ∈ A[[Y₁, ..., Y_q]] for 1 ≤ i ≤ p. Suppose that for 1 ≤ i ≤ p the constant term of gᵢ is zero, and put h = f(g₁, ..., gₚ). Then for 1 ≤ j ≤ q we h… ↩ 1 06CB
Proposition 8 — Let X = (Xᵢ)ᵢ, be a finite family of indeterminates. (i) Every derivation of the ring of formal power series A[[X]] is continuous. (ii) Every derivation of the polynomial ring A[… 06CC
Proposition 9 — Let (u_λ)_λ ∈ L be a summable family of elements of A[[I]] without constant term and D a continuous derivation of the A-algebra A[[I]]. If f = ∏_λ ∈ L (1 + u_λ) (IV, p. 27, Prop… 06CD
Lemma 2 Let (gᵢ)ᵢ be a family of elements of order ≥ 2 in A[[I]]. When I is infinite, assume that gᵢ tends to 0 along the filter of complements of finite subsets of I. There exists one and… ↩ 2 06CE
Proposition 10 — Let f ∈ AI; then the following conditions are equivalent : (i) f is invertible in AI for the law ∘; (ii) T_f is invertible in the ring End(A⁽ᴵ⁾). ↩ 2 06CF
Corollary — Let fᵢ(Y₁, Y₂, …, Y_q, X₁, X₂, …, Xₚ) (1 ≤ i ≤ q) be q formal power series without constant term in A[[Y₁, …, Y_q, X₁, …, Xₚ]]. If the constant term of the formal power series D… 06CG
Proposition 11 — Suppose that A is an integral domain. 06CH
Proposition 12 — Every non-zero element u of K((X)) may be written in a unique way as u = Xᵏ v, where k ∈ Z and v is a formal power series in X of order 0. ↩ 1 06CI
Proposition 13 — In Q[[X, Y]] we have e^X+Y = e^X e^Y. ↩ 1 06CJ
Proposition 14 — We have l(e(X)) = e(l(X)) = X. 06CK

§ 5. Symmetric tensors and polynomial mappings

Definition 1 — If G is of finite index in H, we denote by Tr_H/G the mapping of Mᴳ into Mᴴ defined by 06CT
Proposition 1 — (i) Let m ∈ Mᴳ and h ∈ H. Then hm ∈ M^hGh⁻¹ and ↩ 1 06CU
Proposition 2 — (i) The A-algebra TS(M) is associative, commutative and unital. ↩ 2 06CV
Definition 2 If x ∈ M, the element x ⊗ x ⊗ … ⊗ x of TSᵏ(M) is denoted by γₖ(x). 06CW
Proposition 3 (i) If x ∈ M, the pth power of x, calculated in TS(M), is equal to p! γₚ(x). (ii) Let x₁, …, xₙ ∈ M; then γₚ(x₁ + x₂ + ⋯ + xₙ) = ∑_p₁ + p₂ + ⋯ + pₙ = p γ_p₁(x₁) γ_p₂(x₂) … γ_pₙ(xₙ)… ↩ 5 06CX
Remark 1 Let (xᵢ)_i ∈ I be a family of elements of M. For each ν ∈ N⁽ᴵ⁾ put x_ν = ∏_i ∈ I γ_νᵢ(xᵢ). If (λᵢ) ∈ A⁽¹⁾ and p ∈ N, then we have by Prop. 3 (ii), γₚ ( ∑_i ∈ I λᵢ xᵢ ) = ∑_ν ∈ N⁽ᴵ⁾… 06CY
Remark 2 Let M be the set of mappings of (1, p) into I. We define a mapping ρ ↦ ρ^* of M into N⁽ᴵ⁾ by putting ρ^*(i) = Card ρ⁻¹(i) ↩ 1 06CZ
Proposition 4 — Let M be free and (eᵢ)_i ∈ I a basis of M. ↩ 5 06D0
Lemma 1 Let H be a finite group and U a left A[H]-module. Suppose that the A-module U has a basis B which is stable under the operations of H in U, and put Ω = B/H. For each w ∈ Ω let u_ω… ↩ 1 06D1
Proposition 5 — Let M be a free A-module, k an integer ≥ 0, P the sub-A-module of TSᵏ(M) generated by γₖ(M). Assume A to be an infinite integral domain. Then for each z ∈ TSᵏ(M) there exists a ∈… ↩ 1 06D2
Proposition 6 — Let M be an A-module which is the direct sum of a family (M_λ)_λ, of submodules. If each M_λ is a free module, then the canonical homomorphism of ⊗_λ ∈ L TS(M_λ) into TS(M) is an… ↩ 1 06D3
Proposition 7 Let M be an A-module, u the mapping (x, y) ↦ x + y of M ⊕ M into M and f the composite mapping ↩ 2 06D4
Proposition 8 — Let M be a free A-module, then with its algebra and cogebra structures, TS (M) is a graded commutative and cocommutative bigebra. The counit is the A-linear mapping ε : TS(M) → T… 06D5
Proposition 9 — Let M and N be free A-modules and u an A-homomorphism of M into N ; then TS(u) is a bigebra homomorphism. ↩ 1 06D6
Proposition 10 — Let M be a free A-module, then the primitive elements (III, p. 602) of the bigebra TS(M) are the elements of M. ↩ 2 06D7
Proposition 11 — If M is free, then φ_M is a morphism of graded bigebras. 06D8
Proposition 12 — (i) If u ∈ Sⁿ(M), then ψ_M(φ_M(u)) = n!u. 06D9
Corollary 1 — If A is a Q-algebra, then the canonical homomorphism of S(M) into TS(M) is an algebra isomorphism. If moreover M is free, then it is an isomorphism of graded bigebras. ↩ 2 06DA
Corollary 2 — If A is a Q-algebra then the module TSⁿ(M) is generated by the n-th powers of elements of M in TS(M). This follows from Cor. 1 and the corresponding property of S(M) (III, p. 498… 06DB
Proposition 13 — Let M and N be A-modules, q an integer geqslant 0, and f a mapping of M into N. Suppose that M is free, then the following conditions are equivalent: (i) There exists a q-linear… ↩ 5 06DC
Definition 3 — Let M and N be A-modules and q an integer ≥ 0. Suppose that M is free, and denote by Pol_A^q(M, N) or simply Pol^q(M, N) the set of mappings of M into N satisfying the conditions… ↩ 1 06DD
Example 1 The homogeneous polynomial mappings of degree 1 of M into N are the A-linear mappings of M into N. 06DE
Example 2 Let (Nᵢ)_i ∈ I be a family of A-modules, fᵢ a mapping of M into Nᵢ, i ∈ I, and f : M → ∏_i ∈ I Nᵢ the mapping with components fᵢ. For f to be a homogeneous polynomial mapping of de… 06DF
Example 3 Let (Mⱼ)_j ∈ J be a finite family of free A-modules and u : ∏_j ∈ J Mⱼ → N a multilinear mapping ; then u is polynomial of degree Card(J). 06DG
Example 4 Let (Xᵢ)_i ∈ I be a family of indeterminates, N an A-module and u ∈ N[(Xᵢ)_i ∈ I] a homogeneous polynomial of degree q. The mapping (xᵢ)_i ∈ I ↦ u((xᵢ)_i ∈ I) of A⁽ᴵ⁾ into N is a h… 06DH
Example 5 The mapping (xᵢ)_i ∈ N ↦ x₀² + x₁² + … + xₙ² + … of A⁽ᴺ⁾ into A is a homogeneous polynomial mapping of degree 2. If A = 2/22, it coincides with the linear mapping (xᵢ)_i ∈ I ↦ x₀ +… 06DI
Example 6 Let f ∈ Pol_A^q(M, N), let B be a commutative ring, ρ a homomorphism of B into A and M' and N' the B-modules derived from M and N by means of ρ. Assume that M' is free; then f ∈ Po… 06DJ
Proposition 14 — Let M, N, P be A-modules, q and r integers geqslant 0, and assume that M and N are free. If f ∈ Pol^q(M, N), f' ∈ Polʳ(N, P), then f' ∘ f ∈ Pol^qr(M, P). 06DK
Proposition 15 — Let M be a free A-module, N an A-module and q an integer geqslant 0. We suppose that the mapping y ↦ q!y is an automorphism of N. Let f ∈ Pol^q(M, N), then there exists one and o… ↩ 1 06DL
Proposition 16 — Let M be a free A-module, N an A-module, q a positive integer and u the canonical homomorphism of Hom (TS^q(M), N) into Pol^Y(M, N). ↩ 2 06DM
Corollary — Let M be a free A-module, N an A-module, q a positive integer, h ∈ Pol^q(M, N) and (eᵢ)ᵢ, a basis of M. In the two cases of Prop. 16 there exists a unique family (u,, N⁽ᴵ⁾, |·|… ↩ 1 06DN
Definition 4 — Let M and N be A-modules and assume that M is free. Let Map(M, N) be the A-module of all mappings of M into N. The submodule ∑_q > 0 Pol_A^q(M, N) of Map(M, N) is denoted by Pol_… ↩ 2 06DO
Proposition 17 — Let M be a free A-module and B an associative, commutative and unital A-algebra. Then Pol_A(M, B) is a sub-B-algebra of the algebra Map(M, B). 06DP
Proposition 18 — Let M, N, P be A-modules, and assume that M and N are free. If f ∈ Pol(M, N), g ∈ Pol(N, P), then g ∘ f ∈ Pol(M, P). 06DQ
Lemma 2 — Let N be an A-module, n an integer ≥ 0 and ↩ 1 06DR
Proposition 19 — Let M be a free A-module, N an A-module, G an infinite additive subgroup of A, and suppose that the homotheties of N defined by the nonzero elements of G are injective. Then Pol(… ↩ 1 06DS
Corollary — Assume that A is an infinite integral domain; let M be a free A-module and N a torsion-free A-module. ↩ 2 06DT
Proposition 20 — If the A-module M is free and finitely generated, then 0 is an isomorphism of graded algebras. ↩ 1 06DU
Remark 1 — In the same way we see that if M is a finitely generated free A-module, then the graded algebra S(M)^*gr defined in III, p. 593, may be identified with TS(M^*). ↩ 1 06DV
Proposition 21 — The canonical homomorphism of A-modules (IV, p. 55) 06DW
Remark — 2) The composite homomorphism A, = u ∘ 0 : S(M^*) → Pol_A(M, A) is the unique unital homomorphism of algebras inducing the inclusion of ↩ 2 06DX

§ 6. Symmetric functions

Theorem 1 — Let us put E = A[X₁, …, Xₙ] and S = A[X₁, …, Xₙ]^sym. ↩ 2 06EE
Proposition 1 For every pair of positive integers k, n, let Sₖ⁽ⁿ⁾ be the A-module consisting of all symmetric polynomials in X₁, …, Xₙ, which are homogeneous of degree k. If the integer m satisf… 06EF
Example 3 By example 2 of IV, p. 64 and Prop. 1 above we thus have ∑ᵢ₌₁ⁿ Xᵢ³ = s_1,n³ - 3s_1,ns_2,n + 3s_3,n for every integer n ≥ 3. The commutativity of the diagram (10) moreover gives the… 06EG
Remark — Let n and k be two positive integers. We denote by Δ_k,n the set of elements of length k in Nⁿ and further equip A, , with the order relation, written a ≤ β, induced by the lexic… 06EH
Example 4 By Example 3 of IV, p. 65, we have 06EI
Proposition 2 — The field of symmetric rational fractions in X₁, …, Xₙ is the field of fractions of the ring of symmetric polynomials in X₁, …, Xₙ.. ↩ 2 06EJ
Corollary — Let s₁, s₂, …, sₙ be the elementary symmetric polynomials in X₁, …, Xₙ. For every rational fraction g ∈ K(S₁, S₂, …, Sₙ) the sequence (s₁, s₂, …, sₙ) is substitutable in g and th… 06EK
Theorem 2 a) If I is a finite set of n elements, then φ_I induces a bicontinuous isomorphism of A[[S₁, …, Sₙ]] onto A[[X]]^sym. ↩ 1 06EL
Lemma 1 Let J be a finite subset of I, r an integer such that Card J ≥ r and fa symmetric formal power series which is homogeneous of degree r in the Xᵢ (i ∈ I). Let barf be the formal pow… ↩ 2 06EM
Lemma 2 Let f be a symmetric formal power series of degree r in the Xᵢ (i ∈ I). There exists a unique polynomial P ∈ B ∩ A[S], isobaric of weight r, such that f = ψ₁(P). ↩ 1 06EN
Lemma 3 For every integer m ≥ 0 let cₘ be the ideal of the algebra A[[X]]ˢʸᵐ consisting of all symmetric formal power series of order ≥ m. The sequence (cₘ)_m ≥ 0 is a fundamental system o… 06EO
Lemma 4 (« Newton's relations ») — For every integer d ≥ 1 we have 06EP
Remark — With the notation of the above proof we have 06EQ
Proposition 3 — a) If I is a finite set with n elements and n! · 1 is invertible in A, then λ_I induces a bicontinuous isomorphism of A[[P₁, …, Pₙ]] onto A[[X]]^sym. b) If I is infinite and A is… ↩ 1 06ER
Corollary — Let ξ₁, …, ξₙ, η₁, …, ηₙ be elements of A and suppose that A is an integral domain. 06ES
Proposition 4 Let B be a commutative ring, ρ a homomorphism of A into B and ξ₁, ..., 5, elements of B. Suppose that the relation ^ρ f(X) = ∏ᵢ₌₁ⁿ (X - ξᵢ) holds in B[X]. Then there exists one and… 06ET
Proposition 5 In the universal decomposition algebra E_f the family of monomials x₁^ν(1) ... xₙ^ν(n) such that 0 ≤ ν(i) < i for 1 ≤ i ≤ n is a basis of the A-module E_f. In particular E_f is a f… 06EU
Corollary — The canonical homomorphism of A into the universal decomposition algebra of the monic polynomial f ∈ A[X] is injective. ↩ 1 06EV
Proposition 6 — Let f ∈ A[X] be a monic polynomial of degree n, and let P be a symmetric polynomial in X₁, ..., Xₙ with coefficients in A. Then there exists precisely one element a of A with the… 06EW
Examples 1) If P = sₖ, then P^*(f) = (-1)ᵏ aₖ. \ 2) Let g be a polynomial in A[X] and put P(X₁, ..., Xₙ) = g(X₁) ... g(Xₙ) . Then P^*(f) is just the resultant res(f, g), by Cor. 1 of IV, p… 0DHL
Definition 1 With the above notation the determinant of the matrix M(f, g, p, q) is called the resultant of the pair (f, g) for the degrees p and q, or simply the resultant of f and g if p = de… 06EX
Example 1 Given Λ, μ in Λ, we have the formulae res_p,0(f, λ) = λᵖ, res_0,q(μ, g) = μ^q res_p,1(f, λ) = λᵖ tₚ, res_1,q(μ, g) = (-1)^q μ^q u_q, whose proof is immediate. 06EY
Example 2 When p = q = 1, we have res_1,1(t₁ X + t₀, u₁ X + u₀) = t₁ u₀ - t₀ u₁ 06EZ
Remark 1 The matrix M(g, f, q, p) is obtained from M(f, g, p, q) by pq transpositions of columns, whence res_q,p(g, f) = (-1)^pq res_p,q(f, g) 06F0
Remark 2 Let ρ : A → B be a ring homomorphism. Def. / implies directly the formula res_p,q(ρ f, ρ g) = ρ(res_p,q(f, g)) . 06F1
Remark 3 Given λ, μ in A, we have res_p,q(λ f, μ g) = λ^q μᵖ res_p,q(f, g) . (28) ↩ 1 06F2
Remark 4 Suppose that p + q ≥ 1. By III, p. 532, formula (28), the image of φ contains the constant polynomial res_p,q(f, g). Hence there exist a pair of polynomials (u, v), with u ∈ S_q, v… 06F3
Remark 5 Suppose that p ≥ q, and let h ∈ A[X] be a polynomial of degree ≤ p - q. Let us show that res_p,q(f, g) = res_p,q(f + gh, g) . (29) For if we write ω(u, v) = (u, uh + v) for (u, v)… 06F4
Proposition 7 — Assume that f is monic of degree p; with the above notations we have¹ res_p,q(f, g) = N_E/A(g(x)) . (30) 06F5
Corollary 1 — Let f ∈ A[X] be a monic polynomial; for every polynomial g ∈ A[X] the following conditions are equivalent: (i) res(f, g) is invertible in A; (ii) there exist polynomials u, v in… ↩ 1 06F6
Corollary 2 — Suppose that A is a field and let f, g ∈ A[X], then the following conditions are equivalent when f, g are non-zero: (i) res(f, g) ≠ 0; (ii) the polynomials f and g are relatively… 06F7
Corollary 3 — For every λ ∈ A we have res_p,1(f, λ - X) = f(λ), res_1,q(X - λ, g) = g(λ). When f(X) = X - λ, the algebra E is equal to A and we have x = A; the second formula (32) now follows… 06F8
Proposition 8 — Suppose that f and g are monic of degrees p and q respectively. With the above notation, the A-module F is free with basis (xⁱ yʲ)_0 ≤ i < p,\ 0 ≤ j < q and we have ↩ 1 06F9
Proposition 9 — Let p₁ and q₁ be positive integers and f₁, g₁ polynomials in A[X] such that deg f₁ ≤ p₁, deg g₁ ≤ q₁; then we have ↩ 1 06FA
Lemma 5 Let t be an element of A. There exists a commutative ring C containing A as subring, a subring B of C containing A, an element τ of B invertible in C and a ring homomorphism p : B… 06FB
Corollary 1 — (i) Let λ, α₁, …, αₚ be elements of A and suppose that f(X) = λ (X - α₁) … (X - αₚ). We have res_p,q(f, g) = λ^q g(α₁) … g(αₚ) . (ii) Let μ, β₁, …, β_q be elements of A and suppo… ↩ 2 06FC
Corollary 2 — For every integer r ≥ 0 we have res_p,q+r(f, g) = tₚʳ · res_p,q(f, g) . Bearing in mind Example 1 (IV, p. 76) we need only take q₁ = r, g₁ = 1 in (35). 06FD
Example 3 Let us prove the formula res_2,2(aX² + bX + c, a'X² + b'X + c') = (ac' - ca')² + (bc' - cb')(ba' - ab') 06FE
Definition 2 Let f be a monic polynomial of A[X] of degree m, and denote by E the A-algebra A[X]/(f) and by x the canonical image of X in E. We define the discriminant of f, written dis(f), as… ↩ 1 06FF
Proposition 10 Let m ≥ 1. There exists a unique polynomial A ∈ Z[A₁, …, Aₘ] with the following property : for any commutative ring A and monic polynomial f = Xᵐ + ∑ᵢ₌₁ᵐ aᵢ Xᵐ⁻ⁱ in A[X] we have di… 06FG
Proposition 11 — Let f in A[X] be of degree ≤ m. ↩ 2 06FH
Corollary 1 — Let g ∈ A[X] and let n be a positive integer such that deg g ≤ n. We have ↩ 1 06FI
Corollary 2 — If a₀ is the coefficient of Xᵐ in f, we have ↩ 1 06FJ
Corollary 3 — Let A be a field, and let f be a non-constant polynomial in A[X]. For f and Df to be relatively prime, it is necessary and sufficient that dis(f) ≠ 0. This follows from Prop. 11… 06FK
Remark Adouble application of Cor. 2 above shows that we have disₘ(f) = 0 for every polynomial f of degree ≤ m - 2. 06FL
Example 3 Let m = 2. By Example2 (IV, p. 81) we have Δ(A₁, A₂) = A₁² - 4A₂, whence tildeΔ(A₀, A₁, A₂) = A₁² - 4A₀A₂. In other words, we have dis₂(a₀X² + a₁X + a₂) = a₁² - 4a₀a₂ . 06FM
Example 4 Consider the polynomial F = A₀X³ + 3A₁X² + 3A₂X + A, with coefficients in Q[A₀, A₁, A₂, A₃]. We have DF = 3(A₀X² + 2A₁X + A₂) , F - 1/3X · DF = A₁X² + 2A₂X + A₃ . By the formulae (… 06FN

V. Commutative fields

§ 1. Prime fields. Characteristic

Definition 1 A field is said to be prime if it is isomorphic either to Q or to one of the fields Fₚ. 06G3
Theorem 1 — Let A be a ring ; suppose that there exists a subfield of A. Then A has a unique subfield P which is a prime field. Moreover, P is contained in the centre of A and in every subfi… ↩ 1 06G4
Corollary 1 — Let K be a field. There exists a unique subfield of K which is a prime field, and this is the least subfield of K. 06G5
Corollary 2 — For a field to be prime it is necessary and sufficient that it should contain no subfield other than itself. ↩ 1 06G6
Proposition 1 — Let A be a ring not reduced to 0. a) For A to be of characteristic 0 it is necessary and sufficient that the mapping x ↦ n · x of A into itself should be bijective, for every int… ↩ 1 06G7
Remark 1 Let A and A' be two rings not reduced to 0. Suppose that the characteristic of A is defined and that there is a homomorphism u of A into A'. The image under u of the prime subfield… ↩ 2 06G8
Remark 2 Remark 1 shows that if A is a ring of characteristic 0 (resp. p ≠ 0), the same holds of any ring A' containing A as subring, or of any quotient of A by a two-sided ideal a ≠ A. In… 06G9
Remark 3 Let A be an algebra not reduced to 0 over a field K. Since the mapping λ ↦ λ · 1 of K into A is a ring homomorphism, Remark 1 shows that the characteristic of A is defined and equa… 06GA
Remark 4 Since the field Q is infinite, every ring of characteristic 0 is infinite; it follows that every finite field has non-zero characteristic. 06GB
Remark 5 Let A be a ring not reduced to 0, whose additive group is a torsion-free Z-module, and put B = Q ⊗_Z A. The mapping x ↦ 1 ⊗ x of A into B is injective (II, p. 314), hence A is isom… 06GC
Theorem 2 — Let A be a commutative ring of characteristic p. The mapping a ↦ aᵖ is an endomorphism of the ring A, that is, we have the relations ↩ 2 06GD
Lemma 1 Let p be a prime number and i an integer in the range from 1 to p - 1, then the binomial coefficient binompi is an integer divisible by p. 06GE
Proposition 2 — Let A be a commutative ring of characteristic p, K a subring of A, S a subset of A and f a positive integer. ↩ 1 06GF
Definition 2 — A ring A of characteristic p ≠ 0 is said to be perfect if it is commutative and the mapping a ↦ aᵖ is bijective. ↩ 1 06GG
Definition 3 — Let A be a commutative ring of characteristic p ≠ 0. By a perfect closure of A we understand a pair (A, u) where A is a perfect ring of characteristic p and u is a homomorphism o… 06GH
Theorem 3 — Let A be a commutative ring of characteristic p ≠ 0. There exists a perfect closure (A, u) of A. Moreover, the kernel of u is the set of all nilpotent elements of A and for each… ↩ 1 06GI
Proposition 3 — Let B be a perfect ring of characteristic p and A a subring of B. Write A^p^-∞ = ⋃_f ≥ 0 A^p⁻ᶠ and denote by j the canonical injection of A in A^p^-∞. Then A^p^-∞ is the smallest… ↩ 1 06GJ
Proposition 4 — Let K be a field of characteristic exponent q. For every integer f ≥ 0 the mapping x ↦ x^qᶠ is an isomorphism of K on one of its subfields (denoted by K^qᶠ). This follows from Th… ↩ 1 06GK
Definition 4 — A field K of characteristic exponent q is said to be perfect if we have K^q = K. When K^q ≠ K, K is called imperfect. By this definition a field is perfect if it is of characteri… 06GL
Proposition 5 — If K is a field of characteristic 0, or is finite, \ or algebraically closed \, it is perfect. In particular every prime field is perfect. Suppose that K has characteristic p ≠ 0… 06GM
Proposition 6 — Let K be a commutative ring, A the polynomial algebra K[Xᵢ]_i ∈ I, and S the set of elements F of A such that dF = 0. 06GN
Remark — We still have S = K when the additive group of K is torsion-free; this follows from the above proof or from Remark 5 of V, p. 3. 06GO
Corollary — Let K be a field and F(X) a polynomial with coefficients in K, whose derivative F'(X) is zero. a) If K is of characteristic 0, then F ∈ K. b) If K is of characteristic p ≠ 0, the… 06GP

§ 2. Extensions

Definition 1 — Let K be a field. By an extension of K we understand a K-algebra whose underlying ring is a field. By a subextension (or sub-K-extension) of the extension E we understand a sub-K… 06GV
Theorem 1 Let E be an extension of K and A an algebra over E. Then we have [A : K] = [A : E] · [E : K]. In particular, if F is an extension of E, we have ↩ 2 06GW
Corollary 1 Let K, E, F be three fields such that K ⊂ E ⊂ F and [F : K] is finite. Then the degrees [E : K] and [F : E] are divisors of [F : K]. 06GX
Corollary 2 Let K, E and F be three fields such that K ⊂ E ⊂ F. Suppose that [F : K] is finite, then the relation [E : K] = [F : K] is equivalent to E = F and the relation [F : E] = [F : K] is… 06GY
Proposition 1 Let A be an algebra of finite degree over a field K. If an element a ∈ A is not a left (resp. right) divisor of zero in A, then it is invertible in A. ↩ 3 06GZ
Corollary Let A be a commutative algebra of finite degree over a field K. If A is an integral domain, then it is a field. ↩ 1 06H0
Proposition 2 — If M and N are any two subsets of an extension of a field K, then K(M ∪ N) = K(M)(N) = K(N)(M). 06H1
Remark — Let P be the prime subfield of a field E (V, p. 2); for every subset A of E, P(A) is the least subfield of E containing A. In particular if K is a subfield of E, we have P(K ∪ A)… 06H2
Proposition 3 — Let F be a set of subfields of a field E, directed with respect to the relation ⊂. The union L of the fields of F is a field. ↩ 2 06H3
Corollary — Let E be an extension of a field K and A ⊂ E. The field K(A) is the union of the fields K(F) where F ranges over the set of all finite subsets of A. ↩ 1 06H4
Definition 2 — An extension E of a field K is said to be finitely generated if it has a finite generating family. It is called monogenous if there exists x in E such that E = K(x). ↩ 1 06H5
Proposition 4 — Let E, F be two extensions of K. 06H6
Remark — If p and p' are two distinct prime ideals of E ⊗_K F, the corresponding composite extensions of E and F (constructed by the procedure of the above proof) are not isomorphic. Howe… 06H7
Corollary There exist composite extensions of E and F. ↩ 1 06H8
Scholium Let K and E be two fields and u a homomorphism of K into E. If K' is a field containing K as subfield, there exists a field E' containing E as subfield and a homomorphism u' of K'… 06H9
Proposition 5 Let E and F be two extensions of K contained in Ω. ↩ 3 06HA
Proposition 6 Let E and F be two extensions of K contained in Ω, and A and B two subalgebras of Ω over K such that E is the field of fractions of A and F the field of fractions of B. Then for E… ↩ 3 06HB
Proposition 7 — Let E and F be two extensions of K contained in Ω; if E and F are linearly disjoint over K, then every subextension of E and every subextension of F are linearly disjoint over K… ↩ 1 06HC
Proposition 8 — Let E, F, G be three extensions of a field K contained in Ω, such that F ⊂ G. For E and G to be linearly disjoint over K it is necessary and sufficient that E and F should be lin… 06HD

§ 3. Algebraic extensions

Theorem 1 — Let A be an algebra over a field K, x an element of A algebraic over K, n the degree and f the minimal polynomial of x over K. ↩ 8 06HI
Corollary 1 — Let A be an algebra over a field K. For an element x of A to be algebraic over K it is necessary and sufficient that the subalgebra K[x] of A generated by x should be of finite d… ↩ 2 06HJ
Corollary 2 — Let E be an extension of K, A an algebra over E and x an element of A algebraic over K. Then x is algebraic over E, the minimal polynomial of x over E divides the minimal polynom… 06HK
Remark Let E be an extension of a field K and x an element of E which is root of a monic irreducible polynomial f ∈ K[X]. Th. 1, c) then shows that f is the minimal polynomial of x over K… 06HL
Example 1 In the field of complex numbers C, the number i is algebraic of degree 2 over the prime field Q; for if f(X) = X² + 1, then f(i) = 0, and x² + 1 ≠ 0 for all x ∈ Q, hence i ∉ Q. The… 06HM
Example 2 Let K be a field and F the field K(X) of rational functions in one indeterminate over K. Let E be the subfield K(X³) of F; we have F = E(X) and X is algebraic over E, since it is a… 06HN
Example 3 \ In the field R of real numbers it may be shown ¹ that the number π is transcendental over the prime field Q. \ 06HO
Definition 1 — An extension E of a field K is said to be algebraic (over K) if every element of E is algebraic over K. An extension E of K which is not algebraic is called transcendental (over… 06HP
Proposition 1 — For an extension E of K to be algebraic it is necessary and sufficient that every sub-K-algebra A of E should be a field. 06HQ
Proposition 2 — If an extension E of K is of finite degree n, then it is algebraic and the degree over K of each element of E divides n. ↩ 4 06HR
Theorem 2 — Let E be a finitely generated extension of K, generated by the elements a₁, ..., aₘ, which are algebraic over K; then E is an extension of finite degree of K. If the degree of aᵢ… ↩ 13 06HS
Corollary 1 — Let E be an extension of K and A a subset of E consisting of elements that are algebraic over K. Then K(A) is algebraic over K and we have K[A] = K(A). ↩ 10 06HT
Corollary 2 — Let L be an extension of K and E, F subextensions of L. If F is algebraic over K, the subring K[E, F] of L generated by E ∪ F is a field coinciding with E(F) and algebraic over E… ↩ 7 06HU
Remark 1 With the notation of Th. 2, E = K[a₁, a₂, ..., aₙ] and hence E is isomorphic to a quotient K[X₁, X₂, ..., Xₘ]/𝔞; since E is a field, a is a maximal ideal in K[X₁, ..., Xₘ]. 06HV
Remark 2 Let E be an algebraic extension of K, of infinite degree. By Th. 2 there exists an infinite sequence (aᵢ), of elements of E such that aᵢ ∉ K(a₁, a₂, ..., aᵢ₋₁); Th. 2 shows further… ↩ 2 06HW
Proposition 3 — Let E and F be two extension fields of a field K such that K ⊂ E ⊂ F. For F to be algebraic over K it is necessary and sufficient that E should be algebraic over K and F algebrai… ↩ 3 06HX
Definition 2 — A subfield K of a field E is said to be relatively algebraically closed in E if every element of E which is algebraic over K, belongs to K. 06HY
Proposition 4 — Let E be an extension of a field K; the set L of elements of E which are algebraic over K forms a subextension of E which is relatively algebraically closed in E. 06HZ

§ 4. Algebraically closed extensions

Proposition 1 — Let K be a field; then the following properties are equivalent: (AC) Every non-constant polynomial of K[X] splits in K[X] into a product of polynomials of degree 1 (distinct or n… 06ID
Definition 1 A field K is said to be algebraically closed if it possesses the (equivalent) properties (AC), (AC'), (AC), (AC). ↩ 1 06IE
Proposition 2 Let Ω be an algebraically closed field and K a subfield of Ω. Then the relative algebraic closure barK of K in Ω is an algebraically closed field. ↩ 1 06IF
Proposition 3 Every algebraically closed field is infinite. ↩ 1 06IG
Theorem 1 (Steinitz) Let K be a field, E an algebraic extension of K and Ω an algebraically closed extension of K; then there exists a K-homomorphism of E into R. ↩ 4 06IH
Definition 2 Let K be a field and (fᵢ)_i ∈ I a family of non-constant polynomials in K[X]. By a splitting extension of (fᵢ)ᵢ, we understand any extension E of K with the following properties : ↩ 1 06II
Remarks 1) For each i ∈ I let cᵢ be a non-zero element of K and let f'ᵢ = cᵢ fᵢ. Then it is clear that every splitting extension for the family (fᵢ)ᵢ, is also a splitting extension for the… ↩ 1 0EEN
Proposition 4 Let K be a field and (fᵢ)_i ∈ I a family of non-constant polynomials in K[X]; then there exists a splitting extension for the family (fᵢ)_i ∈ I. ↩ 4 06IJ
Proposition 5 — Let K be a field, (fᵢ)_i ∈ I a family of non-constant polynomials in K[X], E an extension of K, and F, F' subextensions of E which are each a splitting extension of (fᵢ)_i ∈ I. T… ↩ 2 06IK
Corollary — Let K be a field, (fᵢ)_i ∈ I, a family of non-constant polynomials in K[X] and F, F' splitting extensions of (fᵢ)_i ∈ I. Then there exists a K-isomorphism of F onto F. 06IL
Definition 3 — Let K be a field. By an algebraic closure of K we understand any extension of K which is algebraic and algebraically closed. 06IM
Example 1 The field C of complex numbers is an algebraic closure of the field R of real numbers (Gen. Top., VIII, p. 100) \ ↩ 1 06IN
Example 2 Let K be a field and Ω an algebraically closed extension of K. If overlineK is the relative algebraic closure of K in Ω, then by V, p. 20, Prop. 2, K is an algebraic closure of K… ↩ 2 06IO
Proposition 6 — Let Ω be an extension of a field K. For Ω to be an algebraic closure of K it is necessary and sufficient that it should be algebraic and that each non-constant polynomial in K[X]… ↩ 2 06IP
Remark 1 If Ω is algebraic over K and if every non-constant polynomial of K[X] has a root in Ω then Ω is an algebraic closure of K (V, p. 156, Ex. 20). 06IQ
Proposition 7 — Let Ω be an algebraic extension of a field K. a) If Ω is algebraically closed, then every algebraic extension of K is isomorphic to a subextension of R. b) Conversely suppose tha… 06IR
Theorem 2 (Steinitz) — Let K be a field; then there exists an algebraic closure of K. If Ω and Ω' are two algebraic closures of K, there exists a K-isomorphism of Ω onto Ω'. ↩ 9 06IS
Corollary — Let K and K' be two fields, Ω an algebraic closure of K and Ω' an algebraic closure of K'. For every isomorphism u of K onto K' there exists an isomorphism v of Ω onto Ω' extendi… 06IT
Remark 2 In the notation of the preceding Corollary there exist in general K-automorphisms of Ω distinct from the identity. Hence there is in general no uniqueness about the isomorphism v o… 06IU
Remark 3 Let K be a field and Ω an algebraic closure of K. Then the following construction may be given for a splitting extension for a family (fᵢ)_i ∈ I of non-constant polynomials in K[X]… 06IV
Remark 4 Let K be a finite field and Ω an algebraic closure of K. Then Ω is infinite (V, p. 20, Prop. 3); since every extension of finite degree of K is a finite field, Ω is an algebraic ex… 06IW

§ 5. p-radical extensions

Definition 1 Let K be a field and E an extension of K. A n element x of E is said to be p-radical over K if there exists an integer m ≥ 0 such that x^pᵐ ∈ K; the least of these integers is call… ↩ 1 06IZ
Proposition 1 Let E be an extension of a field K and x a p-radical element of height e over K; put a = x^pᵉ. Then a ∈ K and the minimal polynomial of x over K is X^pᵉ - a: Further we have [K(x)… 06J0
Lemma 1 Let K be a field and a an element of K such that a ∉ Kᵖ. Then for every integer e ≥ 0, the polynomial f(X) = X^pᵉ - a is irreducible in K[X]. 06J1
Definition 2 Let E be an extension of a field K. We shall say that E is p-radical ¹ (over K) if every element of E is p-radical over K. When this is so, E is said to be of finite height when th… 06J2
Proposition 2 Let E be an extension of a field K. For every integer n ≥ 0 let Eₙ be the set of elements of E which are p-radical of height ≤ n over K, and let Eᵣ be the set of all elements of E… ↩ 3 06J3
Corollary If an extension E of a field K is generated by a set of p-radical elements over K, then it is p-radical over K. 06J4
Proposition 3 — Let K be a field, E an extension field of K which is p-radical over K and u a homomorphism of K into a perfect field F. Then there exists a unique homomorphism v of E into F exte… ↩ 1 06J5
Corollary — For an extension E of a field K to be a perfect closure of K it is necessary and sufficient that it should be a p-radical extension of K and that the field E should be perfect. 06J6
Proposition 4 — Let E be a p-radical extension of finite degree of a field K. Then [E : K] is a power of the characteristic exponent p of K. 06J7

§ 6. Etale algebras

Theorem 1 — Let L be an extension of a field K and A an algebra over K; let H be the set of all K-algebra homomorphisms of A into L. Then H is a free subset of the vector space Hom_K(A, L) o… ↩ 5 06JB
Corollary 1 — Let Γ be a monoid, L a field and X a set of homomorphisms of Γ into the multiplicative monoid of L. Then X is a free subset of the vector L-space L^Γ of mappings of Γ into L. 06JC
Corollary 2 (Dedekind's theorem) Let E and L be two extensions of K. The set of K-homomorphisms of E into L is free over L. If E is of finite degree over K, the number of K-homomorphisms of E into L is at most equ… ↩ 5 06JD
Theorem 2 — Let K be an infinite field, L an extension of K and A an algebra over K. Let u₁, …, uₙ be distinct K-algebra homomorphisms of A into L and f a polynomial in L[X₁, …, Xⱼ]. If we h… 06JE
Definition 1 — Let A be an algebra over K; then A is said to be diagonalizable if there exists an integer n ≥ 0 such that A is isomorphic to the product algebra Kⁿ. We say that A is diagonalize… ↩ 7 06JF
Proposition 1 Let A be an algebra of finite degree n over the field K ; then the following conditions are equivalent : a) The algebra A is diagonalizable. b) There is a basis (e₁, …, eₙ) of A su… ↩ 4 06JG
Corollary Let L be an extension of K and H the set of algebra homomorphisms of A into L. We have Card H ≤ [A : K], with equality if and only if A is diagonalized by L. If A is diagonalized b… ↩ 1 06JH
Proposition 2 — Let A be an algebra over K and Ω an algebraically closed extension of K. The following assertions are equivalent: ↩ 6 06JI
Proposition 3 — Let A be an etale algebra over K. There exist only a finite number of subalgebras and ideals of A. Moreover, every extension of K which diagonalizes A also diagonalizes every sub… ↩ 6 06JJ
Corollary — Let A₁, …, Aᵣ be algebras over K and A = A₁ × ⋯ × Aᵣ. For A to be etale it is necessary and sufficient that A₁, …, Aᵣ are etale. 06JK
Lemma 1 — Let Ω be an algebraic closure of K; then we have h(L) ≤ h(Ω) for every extension L of K, with equality when L is algebraically closed. 06JL
Proposition 4 — Let A be a commutative algebra of finite degree over K; then [A: K], ≤ [A: K] with equality if and only if A is etale. ↩ 5 06JM
Corollary 1 — Let A, B be two commutative algebras over K, of finite non-zero degrees. Then for the algebra C = A ⊗_K B to be etale it is necessary and sufficient for A and B to be etale. ↩ 3 06JN
Corollary 2 — Let K' be an extension of K. ↩ 6 06JO
Theorem 3 — Let A be a commutative algebra of finite degree over K. For A to be etale it is necessary and sufficient that the module Ω_K(A) of K-differentials of A should be reduced to 0. ↩ 1 06JP
Lemma 2 Let A be a commutative algebra over K and e an idempotent of A; then we have de = 0 in Ω_K(A). 06JQ
Lemma 3 Let A be a commutative algebra of finite degree over the algebraically closed field K, such that Ω_K(A) = 0. Then we have m = m² for every maximal ideal m of A. 06JR
Lemma 4 Let A be a commutative ring and let a be a finitely generated ideal of A such that a = a². Then there exists an idempotent e in A such that a = Ae. ↩ 1 06JS
Definition 2 — Let A be a commutative ring; then A is said to be reduced if every nilpotent element (I, p. 98) of A is zero. ↩ 1 06JT
Proposition 5 — Let A be a commutative algebra of finite degree over K. For A to be reduced it is necessary and sufficient that there exist extensions L₁, …, Lₛ of finite degree over K such that… ↩ 2 06JU
Theorem 4 — Let A be a commutative algebra of finite degree over K. Then the following assertions are equivalent: a) The algebra A is etale. b) For every extension L of K, the ring L ⊗_K A i… ↩ 4 06JV
Lemma 5 Let B be a reduced algebra of finite degree over a perfect field P; then B is etale. 06JW
Corollary — Suppose that K has characteristic p ≠ 0. For A to be etale it is necessary and sufficient that A = K[Aᵖ]. For every basis (aᵢ)ᵢ, of A over K, the family (aᵢᵖ)ᵢ is then a basis of… 06JX

§ 7. Separable algebraic extensions

Definition 1 — Let E be an algebraic extension of K; then E is said to be separable (over K) if every subextension F of E of finite degree over K is an etale algebra over K (V, p. 28, Def. 1). ↩ 1 06K0
Proposition 1 — Let E be an algebraic extension of K. If E is separable, then every subextension E' of E is separable. Conversely, if every subextension of finite degree of E is separable then E… 06K1
Proposition 2 — For a field K to be perfect it is necessary and sufficient that every algebraic extension of K should be separable. ↩ 3 06K2
Corollary — Every algebraic extension of a field of characteristic 0, or of a finite field, is separable. This follows from V, p. 7, Prop. 5. 06K3
Proposition 3 — Let f be a non-zero polynomial in K[X] and let Ω be an algebraically closed extension of K. The following conditions are equivalent: a) The polynomial f is relatively prime to it… ↩ 6 06K4
Definition 2 — A polynomial f ∈ K[X] is said to be separable if it is non-zero and satisfies the equivalent conditions a), b), c), d) and e) of Prop. 3. 06K5
Remark — 1) Let L be an extension of K and f a non-constant polynomial in K[X]. By e) of Prop. 3 and V, p. 32, Cor. 2 it comes to the same to suppose that f is separable, whether consider… ↩ 1 06K6
Proposition 4 — Let f be an irreducible polynomial in K[X]. Then the following conditions are equivalent: ↩ 3 06K7
Corollary 1 — For a field K to be perfect it is necessary and sufficient that every irreducible polynomial of K[X] should be separable. ↩ 1 06K8
Corollary 2 — Let f ∈ K[X] be a non-zero polynomial. For f to be separable it is necessary and sufficient that there should exist an extension L of K which is a perfect field and such that f h… ↩ 1 06K9
Definition 3 — Let E be an extension of K. An element x of E which is algebraic over K is said to be separable over K if the algebraic extension K(x) of K is separable. 06KA
Proposition 5 — Let E be an extension of K, x an element of E algebraic over K and f the minimal polynomial of x over K. Then the following conditions are equivalent: a) x is separable over K; b… ↩ 4 06KB
Corollary 1 — If an element x of E is a simple root of a polynomial g of K[X], it is separable over K. For the minimal polynomial f of x over K divides g in K[X] (V, p. 16, Th. 1), so x is a s… ↩ 1 06KC
Corollary 2 — If an element x of E is algebraic and separable over K, it is algebraic and separable over every extension K' of K contained in E. Let f be the minimal polynomial of x over K. Th… ↩ 3 06KD
Corollary 3 — Suppose that K has characteristic p ≠ 0. For an element x of E to belong to K it is necessary and sufficient for it to be both separable algebraic and p-radical over K. The state… ↩ 1 06KE
Proposition 6 — Let E be an extension of K. a) If E is algebraic and separable over K, every element of E is algebraic and separable over K. b) Conversely, let A be a set of elements of E, algeb… ↩ 7 06KF
Corollary — For an algebraic extension E to be separable over K it is necessary and sufficient that every element of E should be a simple root of its minimal polynomial over K. It is enough… ↩ 1 06KG
Theorem 1 — Let E be an extension of K. Then the following conditions are equivalent: a) E possesses a primitive element; b) there exist only a finite number of subextensions of E. These con… ↩ 5 06KH
Proposition 7 — Suppose that K is infinite; let A be a commutative K-algebra possessing only a finite number of sub-algebras (for example an etale K-algebra, V, p. 30, Prop. 3) and let V be a ve… ↩ 2 06KI
Lemma 1 Let A be a vector K-space, V, A₁, …, Aₙ subspaces of A. If Card(K) ≥ n and if V is not contained in any of the Aᵢ, then V is not contained in A₁ ∪ … ∪ Aₙ. 06KJ
Proposition 8 — Let E be an extension of K and (Eᵢ)_i ∈ I a family of subextensions of E such that E = K ( ⋃_i ∈ I Eᵢ ). If each extension Eᵢ is algebraic and separable over K, the same is true… ↩ 1 06KK
Proposition 9 — Let F be an algebraic extension of K and E a subextension of F. For F to be separable over K it is necessary and sufficient that F should be separable over E and E separable over… ↩ 1 06KL
Proposition 10 — Let E and K' be two subextensions of the same extension of K and let E' = K'(E). Suppose that E is algebraic over K, hence E' algebraic over K' (V, p. 18, Cor. 2). ↩ 3 06KM
Proposition 11 — Suppose that K is of characteristic exponent p, and let E be an algebraic extension of K, generated by a set S. If E is separable over K, then E = K(S^pⁿ) for all integers n ≥ 0… ↩ 4 06KN
Corollary 1 — Every algebraic extension of a perfect field is a perfect field. ↩ 1 06KO
Corollary 2 (Mac Lane) — Let K be an algebraic closure of K and K^p^-∞ the perfect closure of K in overlineK. For a subextension E of overlineK to be separable over K it is necessary and sufficient that… ↩ 1 06KP
Remark — 1) When E is algebraic and of infinite degree over K, the condition E = K(Eᵖ) does not always ensure that E is separable over K. For example, if K is imperfect and E is a perfect… 06KQ
Proposition 12 — Let E be an extension of K and Eₛ, the set of elements of E which are algebraic and separable over K. Then Eₛ, is the largest subextension of E which is algebraic and separable o… ↩ 4 06KR
Proposition 13 — Let E be an algebraic extension of K and let Es be the relative separable algebraic closure of K in E. a) E is a p-radical extension of E,. b) If F is a subextension of E such th… ↩ 6 06KS
Corollary 1 — Let E and K' be two extensions of K contained in the same extension of K. Suppose that E is algebraic over K and denote by E, the relative separable algebraic closure of K in E… ↩ 2 06KT
Corollary 2 — If E has finite degree over K, then Eₛ = ⋂_n ≥ 0 K(E^pⁿ). 06KU
Remark Let E be an algebraic extension of K and E, the relative p-radical closure of E in K (V, p. 25). Then E, is the largest subextension of E which is p-radical over K (V, p. 25, Prop… 06KV
Definition 4 — A field K is said to be separably closed if every separable algebraic extension of K is trivial. ↩ 1 06KW
Definition 5 — Let K be a field. By a separable algebraic closure, or (by abuse of language) separable closure of K we understand any extension E of K which is algebraic and separable over K, a… 06KX
Proposition 14 — Let Ω be an algebraically closed extension of K. ↩ 1 06KY
Corollary — Let E be a separably closed extension of K and F a separable algebraic extension of K ; then there exists a K-homomorphism of F into E. 06KZ
Remark 1 Let E and E' be two separable closures of K. If K is not separably closed, there exist several K-isomorphisms of E onto E'. \ For E is then a non-trivial Galois extension of K, and… 06L0
Remark 2 \ Let E be an algebraic and separable extension of K. If every algebraic and separable extension of K is isomorphic to a subextension of E, then E is a separable closure of K. For… ↩ 1 06L1
Proposition 15 Let E, be the relative separable closure of K in E; then [E : K]ₛ = [Eₛ : K]. ↩ 1 06L2
Proposition 16 Let Ω be an extension of K and E, F two subextensions of Ω, of finite degree over K. 06L3

§ 8. Norms and traces

Proposition 1 — Let A be a commutative algebra of finite degree over K. Then the following conditions are equivalent: a) The algebra A is etale. b) There exists a basis of A whose discriminant i… ↩ 2 06L9
Lemma 1 Let V be a vector space of finite dimension over K and B a bilinear form on V x V. Let (v₁, …, vₙ) be a basis of V over K and Δ = det B(vᵢ, vⱼ). Then A ≠ 0 if and only if, for each… ↩ 1 06LA
Lemma 2 Let V be a vector space of finite dimension over K and u a nilpotent endomorphism of V, then Tr(u) = 0. 06LB
Corollary — Let E be an extension of finite degree of K. For E to be separable it is necessary and sufficient that there exist a in E such that Tr_E/K(a) ≠ 0. ↩ 2 06LC
Proposition 2 — Let F be an extension of finite degree of K and E a subextension of F. Then for every x ∈ F we have Tr_F/K(x) = Tr_E/K(Tr_F/E(x)) \\ N_F/K(x) = N_E/K(N_F/E(x)) . 06LD
Corollary — Put m = [F : E]; then for every x ∈ E we have Tr_F/K(x) = m · Tr_E/K(x) \\ N_F/K(x) = N_E/K(x)ᵐ . ↩ 1 06LE
Proposition 3 — Let E be an extension of finite degree n of K and x an element of E, of degree d over K. Write f(X) = Xᵈ + ∑ᵢ₌₁ᵈ aᵢ Xᵈ⁻ⁱ for the minimal polynomial of X over K. Then we have Tr_E… ↩ 1 06LF
Lemma 3 Let R be a commutative ring, f(X) = Xᵈ + ∑ᵢ₌₁ᵈ aᵢ Xᵈ⁻ⁱ a monic polynomial of R[X], A the R-algebra R[X]/(f) and x the residue class of X in A. Then Tr_A/R(x) = -a₁ and N_A/R(x) = (… 06LG
Proposition 4 Let p be the characteristic exponent of K and E an extension of finite degree of K. Let σ₁, …, σₙ be the distinct K-homomorphisms of E into an algebraic closure Ω of K, and let pᵉ… ↩ 1 06LH

§ 9. Conjugate elements and quasi-Galois extensions

Proposition 1 — Let E be an extension of K contained in Ω and u a K-homomorphism of E into Ω. ↩ 14 06LK
Definition 1 — Let E and F be two extensions of K contained in Ω. We shall say that E and F are conjugate (in Ω) if there exists a K-automorphism u of Ω such that u(E) = F. Two elements x and y… 06LL
Proposition 2 — Let x and y be two elements of Ω. Then the following conditions are equivalent: ↩ 1 06LM
Corollary 1 — Let x be an element of Ω of degree n over K, and let f be the minimal polynomial of x over K. The conjugates of x over K are-the roots off in Ω and their number is at most equal… 06LN
Corollary 2 — Let x be an element of Ω of degree n over K. For x to be separable over K it is necessary and sufficient that x should have n conjugates in Ω; when this is so, all the conjugates… ↩ 1 06LO
Corollary 3 — Let G be the group of K-automorphisms of Ω. The set of invariants of G in Ω is the relative p-radical closure of K in Ω (V, p. 25). 06LP
Definition 2 — Let E be an extension of K. Then E is said to be quasi-Galois or normal (over K), if it is algebraic and if every irreducible polynomial of K[X] which has at least one root in E… ↩ 1 06LQ
Proposition 3 — Let E be an extension of K contained in Ω. Then the following conditions are equivalent: a) E is a quasi-Galois extension of K. b) For each x ∈ E the conjugates of x over K in Ω… ↩ 3 06LR
Corollary 1 — For an extension E of K contained in Ω to be quasi-Galois it is necessary and sufficient that it should be identical to all its conjugates over K. ↩ 3 06LS
Corollary 2 — Let E and F be two algebraic extensions of K such that E ⊂ F. If F is quasi-Galois over K, then it is quasi-Galois over E. 06LT
Corollary 3 — Let N be a quasi-Galois extension of K contained in Ω, and E a subextension of N. Let u be a K-homomorphism of E into Ω; then u(E) ⊂ N and there exists a K-automorphism v of N wh… ↩ 1 06LU
Corollary 4 — Let E' be an extension of K and E, K' two subextensions of E'. Suppose that E is quasi-Galois over K and E' = K'(E); then E' is quasi-Galois over K'. ↩ 1 06LV
Remark 1 Let F be an extension of K and E a subextension of F, and suppose that E is quasi-Galois over K. Let us show that every K-automorphism u of F leaves E invariant. For let x ∈ E and… ↩ 3 06LW
Remark 2 Suppose that E is a quasi-Galois extension of K and F a quasi-Galois extension of E; it is not always the case that F is quasi-Galois over K (V, p. 153, Ex. 1). The reason is the f… 06LX
Remark 3 Let E be an algebraic extension of K and x, y two elements of E. If there exists a K-automorphism of E which maps x to y then x and y have the same minimal polynomial over K and so… ↩ 1 06LY
Proposition 4 — Let (Nᵢ)_i ∈ I be a family of quasi-Galois extensions of K contained in Ω. Let N = ⋂_i ∈ I Nᵢ and M = K( ⋃_i ∈ I Nᵢ ); then the extensions N and M of K are quasi-Galois. ↩ 1 06LZ
Proposition 5 — Let E be an extension of K contained in Ω and N the quasi-Galois extension generated by E. If A is a subset of Ω such that E = K(A) then N = K(B), where B is the set of elements… 06M0
Corollary 1 — If E is an extension of finite degree of K, then the quasi-Galois extension N of K generated by E is of finite degree over K. For we have E = K(A) where A is finite, hence the se… ↩ 1 06M1
Corollary 2 — Every quasi-Galois extension N of K is the union of quasi-Galois subextensions of N of finite degree over K. Let x ∈ N and let Nₓ be the quasi-Galois extension of K generated by… 06M2

§ 10. Galois extensions

Theorem 1 — Let N be an algebraic extension of K and Γ the group of K-automorphisms of N. Then the following assertions are equivalent: a) Every element of N invariant under Γ belongs to the… ↩ 4 06MB
Definition 1 — An extension N of K is said to be Galois if it is algebraic and satisfies the equivalent conditions a), b), c) of Th. 1. 06MC
Proposition 1 — Let N be an extension of K and (Nᵢ)_i ∈ I, a non-empty family of subextensions of N. We put E = ∩_i ∈ I Nᵢ and F = K ( ⋃_i ∈ I Nᵢ ). If all the extensions N, are Galois over K th… ↩ 3 06MD
Proposition 2 — Let N be a Galois extension of K and E a subextension of N, of finite degree over K. There exists a subextension F of N containing E, Galois and of finite degree over K. ↩ 5 06ME
Definition 2 Let N be a Galois extension of the field K. The group of all K-automorphisms of N will be called the Galois group of N over K and denoted by Gal(N/K). 06MF
Example 1 Suppose that the characteristic of K is different from 2 and let N be a quadratic extension of K. If x ∈ N − K, then we have N = K(x) and the minimal polynomial of x over K is of t… ↩ 1 06MG
Example 2 \ Let f = X³ + X² − 2X − 1 ∈ Q[X]. The polynomial f is irreducible, for if not, then it would have a root x ∈ Q; on writing x = a/b with a, b ∈ Z, a and b relatively prime, we shou… ↩ 1 06MH
Example 3 Suppose that K = Q and take f = X³ - 2. Using the decomposition of integers into products of prime factors (I, p. 51), we easily see that 2 is not the cube of an element of Q. Henc… ↩ 1 06MI
Example 4 Suppose that K is of characteristic p ≠ 0 and let K(T) be the field of rational fractions and U = Tᵖ - T. Put E = K[U] and F = K[T], then the polynomial f(X) = Xᵖ - X - U of E[X] h… 06MJ
Example 5 Let F = K(X₁, ..., Xₙ) be the field of rational fractions in n indeterminates X₁, ..., Xₙ with coefficients in K. Put sₖ = ∑_1 ≤ i₁ < ... < iₖ ≤ n X_i₁ ... X_iₖ for 1 ≤ k ≤ n and E… ↩ 4 06MK
Example 6 Suppose that f is monic of degree 0 and K is of characteristic ≠ 2. Define on A a total ordering, denoted by ≤, and put δ(f) = ∏_α < β (β - a), (α, β) ∈ A × A, and for each σ ∈ 𝔖_A… 06ML
Proposition 3 — The restriction homomorphism of Gal(N/K) into Gal(L/K) is surjective. ↩ 2 06MM
Proposition 4 — Let N be a Galois extension of K. Then the Galois group Γ = Gal(N/K) is compact and totally disconnected. 06MN
Proposition 5 — The homomorphism λ of Gal(N/K) into varprojlim Gal(Nᵢ/K) is an isomorphism of topological groups. 06MO
Proposition 6 — a) Let V be a vector space over N with a K-structure. For a vector x ∈ V to be rational over K it is necessary and sufficient that u_σ(x) = x for all σ ∈ Γ. For a vector sub-N-sp… 06MP
Corollary — Let V₀ be a vector space over K and let W be a vector sub-N-space of N ⊗_K V₀. Suppose that W is stable under the mappings σ ⊗ Id_V₀ for all σ ∈ Γ. Let W₀ be the set of x ∈ V₀ su… ↩ 1 06MQ
Proposition 7 — Let V be a vector space over N, (u_σ)_σ, r a family of mappings of V into itself satisfying (1) to (4) and V₀ the set of x ∈ V such that u_σ(x) = x for all σ ∈ Γ. ↩ 4 06MR
Proposition 8 — The N-linear mapping ψ of N ⊗_K M into V = ∏_σ ∈ Γ M^σ is injective, and it is bijective if Γ is finite. ↩ 3 06MS
Corollary — Let ψ be the K-linear mapping of N ⊗_K N into the product vector space N^Γ such that ψ(x ⊗ y) (a) = σ(x) y for x, y in N and σ ∈ Γ. Then ψ is injective and it is bijective when Γ… ↩ 3 06MT
Remark 1 Let F be an extension of K and N a subextension of F, and let Γ be a finite group of automorphisms of N, of which K is the field of invariants. Prop. 8 implies the existence of an… 06MU
Remark 2 The notation K, N and Γ has the same meaning as before. For each integer n geqslant 1, let Aₙ be the tensor product of n K-algebras identical with N; let Bₙ be the set of mappings… 06MV
Proposition 9 — Let (U_σ)_σ ∈ Γ be a family of elements of GL(n, N). For A to exist in GL(n, N) such that U_σ = A⁻¹ a(A) for all σ ∈ Γ it is necessary and sufficient that U_σ τ = U_σ · σ(U_τ) fo… ↩ 2 06MW
Corollary 1 Let (c_σ)_σ ∈ Γ be a family of non-zero elements of N. For a ≠ 0 to exist in N such that c_σ = σ(a) · a⁻¹ for all σ ∈ Γ it is necessary and sufficient that c_σ τ = c_σ · σ(c_τ) for… ↩ 2 06MX
Corollary 2 Let (a_σ)_σ ∈ Γ be a family of elements of N. For b to exist in N such that a_σ = σ(b) - b for all σ ∈ Γ it is necessary and sufficient that a_σ τ = a_σ + σ(a_τ) for σ, τ in Γ. ↩ 2 06MY
Theorem 2 (Artin) — Let N be a field, Γ a group of automorphisms of N and K the field of invariants of Γ. Let V be a vector sub-K-space of N of finite dimension over K. Then every K-linear mapping u… ↩ 1 06MZ
Theorem 3 — Let N be a field, Γ a finite group of automorphisms of N and K the field of invariants of Γ. Let n be the cardinal of Γ. a) We have [N : K] = n and N is a Galois extension of K w… ↩ 7 06N0
Theorem 4 — Let N be a Galois extension of K and Γ its Galois group. Let K be the set of subextensions of N and G the set of closed subgroups of Γ. For every subgroup A ∈ G we denote by k(Δ)… ↩ 2 06N1
Lemma 1 Let E be a subextension of N. Then N is a Galois extension of E and Gal(N/E) is a closed subgroup of Gal(N/K) with the induced topology. ↩ 2 06N2
Lemma 2 Let A be a subgroup of Γ. Let E be the field of invariants of A; then the Galois group of N over E is the closure of A in Γ. ↩ 1 06N3
Corollary 1 Let E and E' be two subfields of N containing K; then E ⊂ E' if and only if g(E) ⊃ g(E'). If A and A' are two closed subgroups of Γ, then A ⊂ A' if and only if k(Δ) ⊃ k(Δ'). ↩ 3 06N4
Corollary 2 Let (Eᵢ)_i ∈ I be a family of subfields of N containing K; put L = ⋂_i ∈ I Eᵢ and M = K( ⋃_i ∈ I Eᵢ ). Then g(L) is the smallest closed subgroup of containing U_i ∈ I g(Eᵢ) and we… ↩ 1 06N5
Corollary 3 For i = 1, 2 let Eᵢ be a subfield of N containing K and let Δᵢ = g(Eᵢ). For any σ ∈ Γ the relations σ(E₁) = E₂ and σ Δ₁ σ⁻¹ = Δ₂ are equivalent. ↩ 2 06N6
Corollary 4 Let E be a subfield of N containing K and let A = g(E). For E to be Galois over K it is necessary and sufficient that A should be a normal subgroup of Γ. When this is so, the restr… ↩ 5 06N7
Corollary 5 Let E be a subfield of N containing K. For E to have finite degree over K it is necessary and sufficient that g(E) should be open in Γ. When this is so, the index (Γ : g(E)) is fin… ↩ 2 06N8
Corollary 6 — For i = 1, 2 let Eᵢ be a subextension of N and Γᵢ the Galois group of N over Eᵢ. The following conditions are equivalent: a) The group Γ is the direct product of the subgroups Γ₁… ↩ 2 06N9
Remark With the notation of Cor. 6 suppose that conditions a) and b) hold. The restriction homomorphisms φᵢ : Γ → Gal(Eᵢ/K) for i = 1, 2 induce topological group isomorphisms Ψ₁ : Γ₂ → Ga… 06NA
Theorem 5 Let N' be an extension of K generated by two subextensions K' and N. Suppose that N is Galois over K, with Galois group Γ and that K' ∩ N = K. Then the extension N' of K' is Galois… ↩ 5 06NB
Corollary 1 Let E' be a subfield of N' containing K'. There exists a unique subfield E of N containing K and such that E' = K'(E). We have E = E' ∩ N. 06NC
Corollary 2 Let N be a Galois extension of K. Suppose that the Galois group Γ of N over K is the direct product of two closed subgroups Γ₁ and Γ₂ and denote by Eᵢ the field of invariants of Γᵢ… 06ND
Remark Let K and K' be two fields and u a homomorphism of K into K'. Let Kₛ (resp. K'ₛ) be a separable closure (V, p. 45, Prop. 14) of K (resp. K') and Π (resp. Π') the Galois group of Kₛ… 06NE
Theorem 6 Let N be a Galois extension of finite degree over K and let Γ be the Galois group of N over K. Then there exists a normal basis of N over K; in other words, the K^(Γ)-module N is f… ↩ 3 06NF
Proposition 10 Let x ∈ N, then for x to form a basis of the K^(Γ)-module N it is necessary and sufficient that det(σ τ(x))_σ, τ ∈ Γ ≠ 0. 06NG
Proposition 11 — Suppose that K is infinite and let P : N → K be a polynomial mapping which is non-zero on K. There exists x ∈ N such that P(x) ≠ 0 and {x} is a basis of N over K^(Γ). 06NH
Lemma 4 Let M be a torsion A-module generated by a finite number x₁, ..., xₕ of elements; then there exists an element x of M whose annihilator (II, p. 219) is equal to the annihilator of… ↩ 1 06NI
Lemma 5 Let X be a finite Γ-set and let x₁, …, xₙ be points of X such that the orbits Γ x₁, …, Γ xₙ form a partition of X. For 1 ≤ i ≤ n let Δᵢ be the stabilizer of xᵢ in Γ and let Lᵢ be t… ↩ 1 06NJ
Proposition 12 — For every finite Γ-set X the algebra Θ(X) is etale over K, of degree equal to the cardinal of X. Moreover, every etale algebra over K is isomorphic to an algebra of the form Θ(X)… ↩ 1 06NK
Remark a) It is easy to show that for every K-algebra homomorphism φ of Θ(X) into K, there exists a unique element x of X such that φ(f) = f(x) for all f ∈ Θ(X). 06NL
Proposition 13 — Let N be a quasi-Galois extension of K. We denote by Nᵣ, the field of invariants of the group of all K-automorphisms of N and by Nₛ, the relative separable algebraic closure of K… ↩ 2 06NM
Corollary — Let p the characteristic exponent of K, overlineK an algebraic closure of K, Kₛ, the relative separable closure of K in overlineK and K^p^-∞ the perfect closure of K. Then the ca… 06NN
Remark Let R (resp. S) be a p-radical (resp. separable algebraic) extension of K. Then the algebra R ⊗_K S is a field : for R (resp. S) is isomorphic to a subextension of K^p^∞ (resp. Kₛ)… 06NO

§ 11. Abelian extensions

Definition 1 — An extension E of K is said to be abelian if it is Galois and its Galois group is commutative. 06OC
Proposition 1 — Let E be a Galois extension of K and Γ its Galois group. Let A be the derived group of Γ (I, p. 10, Def. 4) and F the field of invariants of A. For a subextension L to E to be ab… ↩ 1 06OD
Corollary — Let E be an extension of K and let (Eᵢ)_i ∈ I be a family of subextensions of E such that E = K ( ⋃_i ∈ I Eᵢ ). Suppose that each extension Eᵢ is abelian, then the same is true o… 06OE
Definition 2 An element ζ of K is said to be a root of unity if there exists an integer n > 0 such that ζⁿ = 1; for every integer n > 0 such that ζⁿ = 1, ζ is called an n-th root of unity. 06OF
Theorem 1 Let p be the characteristic exponent of K and let n > 0 be an integer. Then the group μₙ(K) of n-th roots of unity in K is cyclic and its order divides n; when K is algebraically c… ↩ 4 06OG
Lemma 1 Let G be a finite subgroup of K^* of order m; then G is cyclic and we have G = μₘ(K). 06OH
Proposition 2 — Let K be algebraically closed and let p be its characteristic exponent. There exists an isomorphism of μ∞(K) onto the group Q/Z[1/p]. ↩ 1 06OI
Remark 1 When K is an algebraically closed field of characteristic 0, the group μ∞(K) is thus isomorphic (not canonically) to Q/Z. When K is the field C of complex numbers, we can write dow… 06OJ
Remark 2 The following result may be shown (cf. V, p. 165, Ex. 21): let G and H be two commutative groups all of whose elements are of finite order. Suppose that for every integer n ≥ 1, th… ↩ 1 06OK
Remark 3 For every prime number l put μ{l∞}(K) = U μ{ln}(K). When l is the characteristic p of K, we have μ{p∞}(K) = {1}. From I, p. 80, Th. 4, we deduce that μ∞(K) is a direct sum of the s… 06OL
Proposition 3 Let k and n ≥ 1 be two integers. Then the following assertions are equivalent: ↩ 1 06OM
Corollary 1 Let G be a cyclic group of order n and let d be a divisor of n. Then the number of elements of order d in G is equal to φ(d). In particular, φ(n) is the number of generators of G. ↩ 1 06ON
Corollary 2 For every integer n ≥ 1, we have ∑_d | n φ(d) = n, where the integer d runs over the set of divisors 0 of n ^1. 06OO
Proposition 4 — Let n ≥ 1 be an integer; suppose that there exist n n-th roots of unity in K (this is true for example if K is separably closed and n . 1, ≠ 0). Then the number of primitive n-th… ↩ 2 06OP
Proposition 5 — For every integer n ≥ 1 prime to p the extension Rₙ(K) of K is abelian of finite degree, its Galois group is isomorphic to a subgroup of (Z/nZ)^* and its degree divides the order… ↩ 1 06OQ
Lemma 2 Let f, g and h be monic polynomials in Q[X] such that f = gh. If f has integer coefficients, the same is true of g and h. ↩ 1 06OR
Remark \ For every integer n > 0 a function μ(n) is defined as follows: if n is divisible by the square of a prime number, we put μ(n) = 0, otherwise μ(n) = (-1)ʰ if n is the product of h… 06OS
Lemma 3 The roots of Φₙ in Kₛ are the primitive n-th roots of unity. ↩ 1 06OT
Proposition 6 — Let p be the characteristic exponent of K and let n ≥ 1 be an integer prime to p. For the polynomial Φₙ(X) to be irreducible in K[X] it is necessary and sufficient that the homom… ↩ 1 06OU
Theorem 2 (Gauss) — Let Q be an algebraic closure of Q and let n ≥ 1 be an integer. 06OV
Definition 3 — An extension E of K is said to be cyclic if it is Galois and its Galois group is cyclic. 06OW
Examples 1) Every Galois extension of prime degree is cyclic, because every finite group G of prime order p is cyclic (for every element x ≠ 1 of G is of order p, hence generates G). 0DI2
Theorem 3 (*Hilbert*) — Let E be a cyclic extension of K and let a be a generator of the Galois group Γ of E over K. 06OX
Lemma 4 Let Γ be a cyclic group of order n, σ a generator of Γ and M a commutative group on which Γ operates according to the rule γ . (m + m') = γ . m + γ . m' for all γ ∈ Γ and m, m' ∈ M… 06OY
Proposition 7 — Let G be a commutative group annihilated by n and H a subgroup of G. Then the restriction homomorphism Hom(G, T) → Hom(H, T) is surjective. ↩ 2 06OZ
Corollary 1 — If G is a commutative group annihilated by n, and G ≠ 1, then Hom(G, T) ≠ 1. ↩ 1 06P0
Corollary 2 — If G is a finite commutative group annihilated by n, then the groups G and Hom(G, T) have the same order. ↩ 1 06P1
Proposition 8 — Suppose that G and H are commutative and annihilated by n. If sf is bijective and H is finite, then df is bijective and we have Card(G) = Card(H). 06P2
Lemma 5 — K(A^1/n) is an abelian extension of K of exponent dividing n. 06P3
Lemma 6 — Let L be a Galois extension of K. There exists a unique mapping (σ, a) ↦ ⟨ σ, a ⟩ of Gal(L/K) × ((Lⁿ ∩ K^*)/K^*n) into μₙ(K) such that for every σ ∈ Gal(L/K) and every element 0… 06P4
Proposition 9 For every Galois extension of finite degree L of K, the homomorphism k_L is bijective. ↩ 1 06P5
Corollary If L is a Galois extension of K, the homomorphism k_L is injective and its image is the group Hom_c(Gal(L/K), μₙ(K)) of continuous homomorphisms of the topological group Gal(L/K) i… 06P6
Theorem 4 a) The mapping H ↦ K(H^1/n) is an inclusion preserving bijection of the set of subgroups of K^* containing K^*n onto the set of abelian subextensions of exponent dividing n of Ω. T… 06P7
Example 1 There exists a largest abelian extension of exponent dividing n of K, contained in Ω ; it is obtained by adjoining to K the n-th roots of all its elements ; its Galois group may be… 06P8
Example 2 \ Let us take K = Q and n = 2. Then Q^*/Q^*2 is a vector F₂-space having as a basis the union of -1 and the set of all prime numbers. The largest abelian extension of exponent 2 of… 06P9
Example 3 Let L be a cyclic extension of K of degree n; then the group (Lⁿ ∩ K^*)/K^*n is cyclic of order n. If a ∈ K^* is such that the residue class of a mod K^*n is a generator of this gr… 06PA
Example 4 Conversely let a ∈ K^* and let r be the least integer > 0 such that a' ∈ K^*n; then the subfield L of Ω generated by the roots of the polynomial Xⁿ - a is a cyclic extension of K o… ↩ 1 06PB
Remark Let a ∈ K^* and let r be the least integer > 0 such that a' ∈ Kⁿ. Let B be the set of roots in K of the polynomial X^n/r - a; then we have 06PC
Lemma 7 K(φ⁻¹(A)) is an abelian extension of K of exponent dividing p. 06PD
Lemma 8 Let L be a Galois extension of K. There exists a unique mapping (σ, a) ↦ [a, a') of Gal(L/K) × ((φ(L) ∩ K)/φ(K)) into Fₚ such that for all σ ∈ Gal(L/K) and every element x ∈ L such… 06PE
Proposition 10 For every Galois extension L of finite degree of K, the homomorphism a_L is bijective. 06PF
Corollary If L is a Galois extension of K, the homomorphism a_L is injective and its image is the group Hom_c(Gal(L/K), Fₚ) of continuous homomorphisms of the topological group Gal(L/K) into… 06PG
Theorem 5 a) The mapping A ↦ K(P⁻¹(A)) is a bijection of the set of subgroups of K containing P(K) onto the set of abelian subextensions of exponent dividing p in Ω. The inverse mapping is L… 06PH
Example 1 There exists a greatest abelian extension of K of exponent dividing p, contained in \Omega; it is K(B^1(K)), its Galois group may be identified with Horn(K/\mathfrak{p}(K),\mathbf{… 06PI
Example 2 Let L be a cyclic extension of K of degree p; then the group (\mathfrak{p}(L) \cap K)/\mathfrak{p}(K) is cyclic of order p. If a \in K is such that the residue class of a mod \math… 06PJ
Example 3 Conversely, if a \in K - \mathfrak{p}(K), then the polynomial X^p - X - a is irreducible and the subfield L of \Omega generated by its roots is a cyclic extension of K of degree p… 06PK

§ 12. Finite fields

Proposition 1 — Let K be a finite field with q elements. 06Q8
Proposition 2 — Let K be a finite field of q elements. The field K is a splitting field of the polynomial X^q - X of \mathbf{F}p[X] and it is the set of all roots of this polynomial. ↩ 2 06Q9
Corollary — Two finite fields of the same cardinal are isomorphic. ↩ 1 06QA
Proposition 3 — Let K be a finite field of q elements, Ω an algebraically closed extension of K and m an integer geqslant 1. a) There exists a unique subextension Kₘ of Ω which is of degree m ov… ↩ 2 06QB
Corollary — Let K be a finite field and Ω an algebraically closed extension of K. The relative algebraic closure overlineK of K in Ω consists of () and roots of unity and is an algebraic clo… 06QC
Proposition 4 — Let K be a finite field of q elements and Kₘ an extension of finite degree m of K. 06QD
Proposition 5 — There exists a unique isomorphism of topological groups π_K : Z → Gal(barK/K) such that π_K(1) = σ_q. 06QE
Proposition 6 — The image under φₙ of the Frobenius automorphism σ_q is the residue class of q mod n. 06QF
Corollary — The degree of Rₙ over K is the least integer f ≥ 1 such that qᶠ ≡ 1 pmodn. For the cyclotomic polynomial Φₙ to be irreducible over K it is necessary and sufficient that the group… 06QG
Example 1 The polynomial Φ₃(X) = X² + X + 1 is irreducible in F_q[X] if and only if q ≡ 2 pmod3. Likewise Φ₄(X) = X² + 1 is irreducible in F_q[X] if and only if q ≡ 3 pmod4 and for Φ₅ = X⁴ +… 06QH
Example 2 We have 5² ≡ 1 pmod12, hence the residue class of 5 (mod 12) does not generate (Z/12Z)^*. So the polynomial Φ₁₂(X) = X⁴ - X² + 1 is not irreducible in F₅[X]; in fact we have Φ₅(X)… 06QI

§ 13. p-radical extensions of height _ 1

Definition 1 Let K be a field and L a p-radical extension of height ≤ 1 of K. A family (xᵢ)_i ∈ I of elements of L is said to be p-free over K (resp. a p-basis of L over K) if xᵢ ∉ K for all i… 06QX
Proposition 1 Let K be a field, L a p-radical extension of height ≤ 1 of K and x = (xᵢ)_i ∈ I, a family of elements of L. Then the vector space K(xᵢ)_i ∈ I over K is generated by the products x'… ↩ 6 06QY
Corollary — Let L' be an extension of a field K, generated by two linearly disjoint subextensions K' and L. Suppose that L is p-radical of height ≤ 1 over K; then L' is p-radical of height ≤… 06QZ
Remark Let K be a field, L a p-radical extension of height ≤ 1 of K and (xᵢ)ᵢ, a family of elements of L. Let A be the polynomial algebra K[(Xᵢ)_i ∈ I], a the ideal of A generated by the… 06R0
Proposition 2 — Let K be a field, L a p-radical extension of height ≤ 1 of K and S a subset of L. For S to be p-free it is necessary and sufficient that K(T) ≠ K(S) for every subset T ≠ S of S. ↩ 1 06R1
Proposition 3 — Let K be a field, L a p-radical extension of height ≤ 1 of K and S, T two subsets of L. The following conditions are equivalent: a) The subset S is p-free over K and T is p-free… 06R2
Corollary — Let L be a p-radical extension of height ≤ 1 of K and M a subextension of L. Then L is p-radical of height ≤ 1 over M and M is p-radical of height ≤ 1 over K. Moreover, if B is a… 06R3
Proposition 4 — Let x = (xᵢ)_i ∈ I be a p-basis of K. For every integer n ≥ 1, the family (xᵘ)_u ∈ Λ(n) is a basis of K over K^pⁿ. 06R4
Proposition 5 Let K be a field and L an extension of K generated by an element x such that x ∉ K,\ xᵖ ∈ K. Let V be a vector space over L and A a derivation of K into V such that Δ(xᵖ) = 0. Then… ↩ 1 06R5
Corollary 1 Let K be a field, L a p-radical extension of height ≤ 1 of K and V a vector space over L. Every derivation A of K into V vanishing on Lᵖ extends to a derivation of L into V. ↩ 3 06R6
Corollary 2 —Let L be a p-radical extension of height ≤ l of a field K and let E be a subextension of L. Let U be the subspace of ΩK(L) generated by the differentials of elements of E. Then E… ↩ 3 06R7
Theorem 1 —Let L be a p-radical extension of height ≤ l of a field K and (ai){i ∈ I}, a family of elements of L. ↩ 3 06R8
Corollary —Let (xi){i ∈ I} be a p-basis of L over K. Let V be a vector space over L, A a derivation of K into V which vanishes on L^p and (ui){i ∈ I} a family of elements of V. Then there ex… 06R9
Theorem 2 — Let L be a p-radical extension of height ≤ 1 of a field K. a) There exist p-bases of L over K. More precisely, if S is a p-free subset of L over K and T a subset of L such that S… ↩ 4 06RA
Corollary — For each x ∈ L - K, there exists a p-basis of L over K containing X. 06RB
Proposition 6 — Let K be a field (of characteristic p) and L an extension of K. a) Let x ∈ L; the differential d_L/Kx vanishes if and only if x belongs to K(Lᵖ). b) Ω_K(L) = 0 if and only if L =… ↩ 1 06RC
Theorem 3 (Jacobson) — The mappings K ↦ g(K) and l ↦ I(h) are bijections of C onto L and of L onto C respectively, which are mutually inverse. If K ∈ C and h ∈ L correspond to each other, then [E : K]… 06RD
Lemma 1 — Let L be a field, V a vector space over L and u an endomorphism of V such that uᵖ = u. For each i ∈ Fₚ let Vᵢ be the kernel of u - i. Then we have ↩ 2 06RE
Lemma 2 Let D be a derivation of E such that Dᵖ = D and let K be the kernel of D. Suppose that there exists a non-zero x in E such that D(x) = x. Then K is a subfield of E containing Eᵖ an… 06RF
Lemma 3 Let 𝔥 ∈ L be of dimension s over E. Then I(𝔥) belongs to C and we have [E : I(𝔥)] = pˢ. 06RG
Remark — 1) The mutually inverse bijections K ↦ g(K) and 𝔥 ↦ I(𝔥) are inclusion reversing; therefore 𝔥 ↦ I(𝔥) is an isomorphism of the ordered set L onto the ordered set opposite to C. He… 06RH

§ 14. Transcendental extensions

Definition 1 Let E be an extension of a field K; a family x = (xᵢ)_i ∈ I of elements of E is said to be algebraically free over K if the monomials xᵃ = ∏_i ∈ I xᵢ^aᵢ with respect to the xᵢ (for… ↩ 1 06RO
Proposition 1 For a family (xᵢ)_i ∈ I of elements of an extension E of a field K to be algebraically free over K it is necessary and sufficient for the relation f((xᵢ)) = 0, where f is a polynom… 06RP
Definition 2 Let E be an extension of a field K. A family (xᵢ)_i ∈ I of elements of E is called a pure basis of E (over K) if it is algebraically free and E = K(xᵢ)_i ∈ I. The extension E of K… 06RQ
Proposition 2 Let E and E' be two fields and u an isomorphism of a subfield K of E onto a subfield K' of E'. Let x = (xᵢ)_i ∈ I (resp. x' = (x'ᵢ)_i ∈ I) be a family of elements of E (resp. E') w… ↩ 1 06RR
Corollary — For an extension E of a field K to be pure it is necessary and sufficient for E to be K-isomorphic to a field of rational fractions over K. More precisely, if the family (xᵢ)_i ∈… 06RS
Remark It is clear that in an extension E of K an algebraically free family over K consists of linearly independent elements over K (hence pairwise distinct); in other words, it is also a… 06RT
Proposition 3 — For a family (xᵢ)_i ∈ I, of elements of an extension E of a field K to be algebraically free over K it is necessary and sufficient that every finite subfamily of (xᵢ)_i ∈ I shoul… ↩ 4 06RU
Proposition 4 — Let E be an extension of a field K and S and T two subsets of E. Then the following properties are equivalent: a) S ∪ T is algebraically free over K and S ∩ T = ∅. ↩ 4 06RV
Corollary — Let E be an extension of a field K and S a subset of E which is algebraically free over K. If x \in E is transcendental over K(S), then S U {x} is algebraically free over K. ↩ 2 06RW
Proposition 5 — Let E be an extension of the field K. For a subset S of E to be algebraically free over K it is necessary and sufficient that for each x \in S, the element x should be transcende… ↩ 1 06RX
Proposition 6 — Let E be an extension of a field K and X an algebraically free subset of E over K. If K' \subset E is an algebraic extension of K then X is algebraically free over K'. ↩ 3 06RY
Definition 3 A subset B of an extension E of a field K is a transcendence basis of E (over K) if B is algebraically free over K and E is algebraic over K(B). 06RZ
Example — A pure basis is a transcendence basis. On the other hand, if E is a pure extension of K, a transcendence basis of E over K is not always a pure basis of E. For example, in K(X)… 06S0
Proposition 7 Let E be an extension of a field K. Every transcendence basis of E is a maximal element of the set (ordered by inclusion) of subsets of E algebraically free over K. Conversely, if… ↩ 2 06S1
Theorem 1 (Steinitz) — Every extension E of a field K admits a transcendence basis over K. In other words, every extension of a field K is an algebraic extension of a pure extension of K. ↩ 3 06S2
Theorem 2 Let E be an extension of a field K, S a subset of E such that E is algebraic over K(S) and T a subset of S which is algebraically free over K; then there exists a transcendence bas… ↩ 5 06S3
Corollary (« Exchange Theorem ») — Let E be an extension of K, S a subset of E such that E is algebraic over K(S), T a subset of E which is algebraically free over K; then there exists a subset S' of S such that T… 06S4
Theorem 3 — Let E be an extension of a field K. All transcendence bases of E over K have the same cardinal. ↩ 1 06S5
Definition 4 — Let E be an extension of a field K. The cardinal of every transcendence basis of E over K is called the transcendence degree of E over K and written tr. degK E. 06S6
Corollary 1 — Let S be a subset of E such that E is algebraic over K(S). Then Card(S) ≥ n; if the cardinal of S is equal to n, then S is algebraically free over K (hence it is then a transcend… ↩ 2 06S7
Corollary 2 — Suppose that E = K(x1, ..., xm), then m ≥ n; if moreover m = n, then (x1, ..., xm) is a pure basis of E over K, and E is then a pure extension of K. 06S8
Corollary 3 — Every subset of E which is algebraically free over K has at most n elements, and if it has exactly n elements, it is a transcendence basis of E over K. 06S9
Theorem 4 — Let K, E and F be three fields such that K ⊂ E ⊂ F. If S is a transcendence basis of E over K and T a transcendence basis of F over E, then S ∩ T is empty and S ∪ T is a transcen… 06SA
Corollary — Let K, E and F be three fields such that K ⊂ E ⊂ F. Then we have tr. deg_K F = tr. deg_K E + tr. deg_E F ↩ 1 06SB
Proposition 8 — Let Ω be an algebraically closed extension of a field K, E and F two subextensions of Ω and u a K-isomorphism of E onto F. For a K-automorphism v of O to exist which extends u, i… ↩ 1 06SC
Corollary 1 — Let O be an algebraically closed extension of a field K and E a subextension of Ω. Every K-automorphism of E extends to a K-automorphism of Ω. ↩ 1 06SD
Corollary 2 — Let Ω be an algebraically closed extension of a field K, E and F two subextensions of Ω and u a K-isomorphism of E onto F. If the transcendence degree of E over K is finite (in p… ↩ 1 06SE
Proposition 9 — Let K be a field and Ω an algebraically closed extension of K. Suppose that Ω is not algebraic over K; then the set of elements of Ω which are transcendental over K is infinite… ↩ 1 06SF
Proposition 10 — Let K be a field, Ω an algebraically closed extension of K and G the group of K-automorphisms of Ω. Let x ∈ Ω. ↩ 1 06SG
Definition 5 — Let L be an extension of a field K, and E and F two subextensions of L. Then E and F are said to be algebraically disjoint (over K) and E is said to be algebraically disjoint fro… ↩ 1 06SH
Remark 1 If E is a subextension of L which is algebraic over K, then it is algebraically disjoint from every subextension F of L. For an extension of K to be algebraic it is necessary and s… 06SI
Remark 2 It may happen that E is algebraically disjoint from F over K, but not over a subfield K_| of K. \ For example C is algebraically disjoint from itself over R but not over Q. \ 06SJ
Remark 3 It is clear that if E is algebraically disjoint from F over K, when E and F are considered as subextensions of L, the same is true when they are considered as subextensions of K(E… 06SK
Proposition 11 — If E and F are algebraically disjoint over K then E ∩ F is algebraic over K. 06SL
Proposition 12 — Let L be an extension of a field K and E, F subextensions of L. Then the following conditions are equivalent: ↩ 4 06SM
Corollary — Let E and F be algebraically disjoint over K. Let E' be the relative algebraic closure of E in L and F' that of F (V, p. 19). Then E' and F' are algebraically disjoint over K. 06SN
Proposition 13 — Let L be an extension of a field K and E, F two subextensions of L. ↩ 1 06SO
Proposition 14 — Let L be an extension of a field K, E and F two subextensions of L and B a transcendence basis of E over K. For E and F to be algebraically disjoint over K it is necessary and su… ↩ 2 06SP
Corollary 1 — If E and F are linearly disjoint, then E is algebraically disjoint from F over K. Conversely, if E is a pure extension of K and is algebraically disjoint from F over K, then E an… ↩ 1 06SQ
Corollary 2 — Every pure extension of K is linearly disjoint from every algebraic extension of K; in particular, K is relatively algebraically closed in every pure extension of K. ↩ 1 06SR
Definition 6 — Let L be an extension of a field K and (Eᵢ)_i ∈ I a family of subextensions of L. The family (Eᵢ)_i ∈ I is said to be algebraically free if the following condition is satisfied: 06SS
Remark By Prop. 3 (V, p. 107) it is enough to verify the condition (AF) for finite subsets Aᵢ. We thus obtain the following result : if (Eᵢ)_i ∈ I is an algebraically free family, the sam… 06ST
Proposition 15 — Let (Eᵢ)_i ∈ I be a family of subextensions of a given extension L of a field K. Then the following conditions are equivalent : ↩ 2 06SU
Proposition 16 — Let (Ei){i \in I} be a family of subextensions of an extension of a field K, and let E be the field generated by ⋃_i ∈ I Eᵢ. 06SV
Theorem 5 — Let (Ei){i \in I}, be a family of extensions of a field K and \Omega an algebraically closed extension of K. Suppose that the inequality ↩ 1 06SW
Corollary 1 — Let E and R be two extensions of a field K. Suppose that Ω is algebraically closed, of transcendence degree at least equal to that of E. Then E is K-isomorphic to a subextension… ↩ 2 06SX
Corollary 2 — Let R be an algebraically closed field of infinite transcendence degree over its prime subfield. Then every field of the same characteristic as Ω is the ascending directed union… ↩ 1 06SY
Proposition 17 — Let E be an extension of a field K and B a transcendence basis of E over K. For E to be finitely generated over K (V, p. 11, Def. 2) it is necessary and sufficient that B be fini… ↩ 1 06SZ
Corollary 1 — Suppose that E is a finitely generated extension of K, and let K' be the relative algebraic closure of K in E (V, p. 19). Then K' is of finite degree over K. ↩ 3 06T0
Corollary 2 — A field that is finitely generated over its prime subfield contains only finitely many roots of unity. 06T1
Corollary 3 — If E is a finitely generated extension of a field K, then every subextension E' of E is finitely generated. 06T2

§ 15. Separable extensions

Proposition 1 — Let A be a commutative ring and x an element of A. For x to be nilpotent it is necessary and sufficient that 1 - xT be invertible in the ring A[T]. ↩ 1 06TN
Proposition 2 — Let A be a commutative ring. The set of nilpotent elements of A is an ideal of A equal to the intersection of the set of all prime ideals of A. ↩ 3 06TO
Corollary — For a commutative ring to be reduced (V, p. 34, Def. 2) it is necessary and sufficient for it to be isomorphic to a subring of a product of fields. ↩ 2 06TP
Definition 1 Let A be a commutative algebra over a field K. Then A is said to be separable over K, or also a separable K-algebra if the ring L ⊗_K A is reduced for every extension L of K. ↩ 4 06TQ
Examples 1) Let A be a polynomial algebra K[Xᵢ]_i ∈ I. For every extension L of K the ring L ⊗_K A is isomorphic to L[Xᵢ]_i ∈ I (III, p. 449, Remark 2) and hence is an integral domain (IV… ↩ 1 0DI7
Proposition 3 Let K be a field. ↩ 2 06TR
Proposition 4 — Let A be a separable algebra over a field K and let B be the total ring of fractions of A; then the K-algebra B is separable. ↩ 2 06TS
Proposition 5 — Let K be a field and A, B commutative K-algebras. If A is reduced and B separable then A ⊗_K B is reduced. ↩ 7 06TT
Corollary 1 — Let K be a field, L a separable extension of K and f a polynomial in K[X]. If f has no multiple factors in K[X], it has also no multiple factors in L[X]. ↩ 1 06TU
Corollary 2 — Let A and B be two commutative K-algebras. If A and B are separable, the same is true of A ⊗K B. 06TV
Proposition 6 — Every pure extension of a field K is separable. This follows at once from Example 1 (V, p. 119) and Prop. 4 (V, p. 120). ↩ 2 06TW
Proposition 7 — Let E be a field, G a group of automorphisms of E and K the subfield of E consisting of the invariants of G. Then E is a separable extension of K. 06TX
Proposition 8 — Let L be an extension of a field K. If L is separable over K, every subextension of L is separable over K. Conversely, if every finitely generated subextension of L is separable… ↩ 1 06TY
Proposition 9 — Let L be an extension of a field K and M a commutative L-algebra (for example, an extension of L). If M is separable over L and L is separable over K, then M is separable over K. ↩ 2 06TZ
Theorem 1 — Let K be a field of characteristic 0. Every reduced K-algebra and in particular every extension of K is separable over K. ↩ 3 06U0
Theorem 2 — Let K be a field of characteristic p ≠ 0, K^p^-∞ a perfect closure of K and A a commutative K-algebra. The following properties are equivalent: a) A is separable. b) There exists… ↩ 6 06U1
Corollary 1 (Mac Lane) — Let K be a field of characteristic exponent p, Ω a perfect extension of K and L a subextension of Ω. Then the following conditions are equivalent: ↩ 5 06U2
Corollary 2 — Let K be a field of characteristic exponent p, K^p^-∞ a perfect closure of K and L a separable extension of K. Then the ring L ⊗_K K^p^-∞ is a field. If moreover L is algebraic o… ↩ 1 06U3
Corollary 3 — Let L be an algebraic extension of K and M a commutative L-algebra (for example an extension of L). If M is separable over K, it is separable over L. ↩ 1 06U4
Remark — Mac Lane's criterion may be formulated without introducing any extensions of K other than L. For by c) of Cor. 1, L is separable over K if and only if L and Kᴾ are linearly disjo… 06U5
Theorem 3 — Let K be a perfect field. a) Every algebraic extension of K is a perfect field. b) Every extension of K is separable. c) For a K-algebra to be separable it is necessary and suffi… ↩ 5 06U6
Theorem 4 — Let Ω be an algebraically closed extension of a field K and L a subextension of Ω. Then the following conditions are equivalent: a) L is separable over K. b) The intersection of… ↩ 1 06U7

§ 16. Differential criteria of separability

Proposition 1 — Suppose that the family x = (xᵢ)_i ∈ I, generates the K-algebra A. Let (mᵢ)_i ∈ I, be a family of elements of M and (f_λ)_λ, a family of polynomials generating the ideal a of all… ↩ 2 06UK
Proposition 2 — The following conditions are equivalent: a) The mapping α is injective. b) Every K-derivation of L into M extends to a K-derivation of M into M. c) Every K-derivation of L into a… ↩ 3 06UL
Proposition 3 — Let K be a field, L a pure extension of K and (xᵢ)_i ∈ I a pure basis of L (V, p. 106, Def. 2). ↩ 3 06UM
Corollary — Let P be a subfield of K. The canonical mapping α of Ω_P(K) ⊗_K L into Ω_P(L) is injective and the family (d_L/P xᵢ)_i ∈ I, is a basis (over L) of a subspace of Ω_P(L) supplement… 06UN
Proposition 4 — Let K be a field, L a separable algebraic extension of K and V a vector space over L. ↩ 7 06UO
Corollary 1 — We have Ω_K(L) = 0 if L is algebraic and separable over K. ↩ 2 06UP
Corollary 2 — If L is an algebraic and separable extension of a field K, the canonical mapping α : Ω_P(K) ⊗_K L → Ω_P(L) is an isomorphism for every subfield P of K. 06UQ
Corollary 3 — Let E be an extension of a field K and D a derivation of E into E, mapping K into K. If L is a subextension of E which is algebraic and separable over K, then D(L) ⊂ L. 06UR
Remark — 1) Later (V, p. 131, Cor. 3 and p. 135, Cor. 2) we shall prove a converse to Cor. 1 of Prop. 4. 06US
Theorem 1 — Let K be a field of characteristic 0, L an extension of K and V a vector space over L. Let A be a derivation of K into V, (xᵢ)_i ∈ I, a transcendence basis of L over K and (uᵢ)_i… ↩ 3 06UT
Corollary 1 — Every derivation of K into V extends to a derivation of L into V ↩ 1 06UU
Corollary 2 — Let E be a subextension of L and U the vector sub-L-space of Ω_K(L) generated by the differentials of elements of E. For an element x to be algebraic over E it is necessary and s… ↩ 3 06UV
Corollary 3 — For an element x of L to be algebraic over K it is necessary and sufficient that dx = 0 in Ω_K(L). In particular, for L to be an algebraic extension of K it is necessary and suff… 06UW
Corollary 4 — Let K, L and M be fields of characteristic 0 such that K ⊂ L ⊂ M: then the canonical mapping a : Ω_K(L) ⊗_L M → Ω_K(M) is injective. This follows at once from Cor. 1 and V, p. 12… 06UX
Theorem 2 — Let K be a field of characteristic 0, L an extension of K and (xᵢ)_i ∈ I, a family of elements of L. a) For (xᵢ)_i ∈ I, to be algebraically free over K it is necessary and suffic… 06UY
Corollary — We have [Ω_K(L) : L] = tr · deg_K L when K is of characteristic 0. 06UZ
Theorem 3 — Let K be a field and L an extension of K. For L to be separable over K it is necessary and sufficient that every derivation of K into a vector L-space should extend to a derivati… ↩ 2 06V0
Corollary — If L is a separable extension of K, the canonical mapping αₚ : Ωₚ(K) ⊗_K L → Ωₚ(L) is injective for every subfield P of K. Conversely, if there exists a perfect subfield P of K (… 06V1
Lemma 1 — Let U and V be two vector spaces of finite dimension over a field L. Every linear mapping f : U → V possesses an index equal to [V : L] - [U : L]. ↩ 1 06V2
Lemma 2 Let f : U → V and g : V → W be two L-linear mappings. Iff and g each have an index, the same is true of g ∘ f and we have ↩ 1 06V3
Theorem 4 Let P be a perfect field, L an extension of P and K a subextension of L; we suppose that L is a finitely generated extension of K. Then the canonical L-linear mapping a : Ωₚ(K) ⊗_K… ↩ 2 06V4
Corollary 1 — Let L be a finitely generated extension of a field K of transcendence degree s. The vector space Ω_K(L) is of dimension ≥ s over L, with equality if and only if L is separable ov… ↩ 4 06V5
Corollary 2 — Let L be a finitely generated extension of a field K. For L to be algebraic and separable over K it is necessary and sufficient that Ω_K(L) = 0. This follows immediately from Cor… ↩ 4 06V6
Corollary 3 — Let K be a field of characteristic p ≠ 0 and L a finitely generated extension of K of transcendence degree s. If [K : Kᵖ] is finite, the same is true of [L : Lᵖ] and we have [L… 06V7
Remark 1 Let K be a field of characteristic p ≠ 0 and L an extension of K. We have Ω_K(L) = 0 if and only L = K(Lᵖ) (V, p. 103, Prop. 6). Therefore if L is finitely generated over K, then i… 06V8
Remark 2 Let K be a field, F₁, …, Fₘ polynomials in K[X₁, …, Xₙ] and L an extension of K generated by the elements x₁, …, xₙ. Suppose that the polynomials F₁, …, Fₘ generate the ideal of K[… 06V9
Definition 1 Let L be an extension of a field K. A transcendence basis B of L over K is said to be separating if the algebraic extension L of K(B) is separable. 06VA
Theorem 5 Let K be a field, L an extension of K and (xᵢ)_i ∈ I, a family of elements of L. If the family (xᵢ)_i ∈ I, is a separating transcendence basis of L over K, then the family (dxᵢ)_i… ↩ 1 06VB
Corollary Let L be a separable finitely generated extension of K and let S be a subset of L such that L = K(S). Then there exists a separating transcendence basis B of L over K, contained in… ↩ 1 06VC
Remark — Let L be a separable finitely generated extension of a field K of characteristic p ≠ 0; there may exist transcendence bases of L which are not separating. It is enough to observe… 06VD
Proposition 5 — Let L and M be two extensions of a field K contained in a given extension and algebraically disjoint over K. If M is separable over K, then L(M) is separable over L. ↩ 2 06VE
Corollary — If L and M are separable extensions, algebraically disjoint over K, then the field K(L ∪ M) is separable over K. 06VF

§ 17. Regular extensions

Theorem 1 (Zariski) — Let K be a field, L an extension of K, K₁ the relative separable algebraic closure of K in L (V, p. 44), K₂ the relative algebraic closure of K in L (V, p. 19) and (Xᵢ)_i ∈ I a f… 06VP
Proposition 1 — Let Ω be an extension of a field K and let L, M be two subextensions of Ω which are algebraically disjoint over K. Suppose that the relative separable algebraic closure of K in L… 06VQ
Corollary Let L and M be two extensions of a field K. Suppose that the relative separable algebraic closure of K in L is equal to K. Then the set p of nilpotent elements of L ⊗_K M is a prim… 06VR
Definition 1 Let K be a field. An algebra A over K is said to be regular if L ⊗_K A is an integral domain for every extension L of K. 06VS
Proposition 2 — Let A and B be two algebras over a field K. If A is an integral domain and B is regular then A ⊗_K B is an integral domain. ↩ 1 06VT
Proposition 3 — Let K be a field. a) Every subalgebra of a regular K-algebra is itself regular. b) The tensor product of two regular K-algebras is again regular. c) Let A be a K-algebra and K' a… ↩ 2 06VU
Definition 2 — An extension of a field K is said to be regular if it is regular as K-algebra. ↩ 1 06VV
Proposition 4 — Let A be an algebra over field K which is an integral domain, and E its field of fractions. Let L be an extension of K; if the ring L ⊗_K A is an integral domain the same is true… ↩ 1 06VW
Corollary — For A to be a regular K-algebra it is necessary and sufficient that its field of fractions should be a regular extension of K. 06VX
Proposition 5 — Every pure extension of a field K is regular. 06VY
Proposition 6 — Let L be an extension of a field K. If L is regular, then every subextension of L is regular. Conversely, if every finitely generated subextension of L is regular, then L is regu… 06VZ
Proposition 7 — Let L be an extension of a field K and M an L-algebra (for example, an extension of L). If L is regular over K and M is a regular L-algebra, then M is regular as K-algebra. 06W0
Proposition 8 — Let L and M be two extensions of a field K. 06W1
Proposition 9 — Let K be a field, overlineK an algebraic closure of K and L an extension of K. Then the following conditions are equivalent: ↩ 1 06W2
Corollary 1 — Let A be an algebra over a field K. For A to be a regular K-algebra it is necessary and sufficient for the ring barK ⊗_K A to be an integral domain. ↩ 1 06W3
Corollary 2 — Let K be an algebraically closed field. Every K-algebra which is an integral domain is a regular K-algebra. In particular, every extension of K is regular. ↩ 1 06W4
Corollary 3 — Let K be an algebraically closed field. If A and B are two K-algebras which are integral domains, then the same is true of A ⊗_K B. 06W5
Proposition 10 — Let L and M be two extensions of a field K and (E, u, v) a composite extension of L and M (V, p. 12). Suppose that the ring L ⊗_K M is an integral domain and that the subextensio… ↩ 2 06W6
Corollary 1 Let Ω be an extension of a field K and L a subextension of Ω which is regular over K. Every subextension M of Ω which is algebraically disjoint from L over K is linearly disjoint. 06W7
Corollary 2 Let Ω be an extension of a field K and L, M two subextensions of Ω. Suppose that L is separable over K and that the relative separable closure of K in M is equal to K. If L and M a… 06W8

VI. Ordered groups and fields

§ 1. Ordered groups. Divisibility

Definition 1 — A commutative monoid structure (written additively) and an ordering (written ≤) on a set M are said to be compatible if they satisfy the following axiom: 06WE
Example 1 The additive group of the rational integers, and that of the rational numbers, are ordered groups under the ordering defined in I, pp. 21 and 117. The same is true for the additive… 06WF
Example 2 The additive group of finite numerical functions defined on a set E is an ordered group under the ordering « f(x) ≤ g(x) for all x ∈ E », written f ≤ g. This relation says that the… 06WG
Proposition 1 (« addition of inequalities ») — In an ordered monoid M, let (xᵢ) and (yᵢ) (1 ≤ i ≤ n) be two sequences of n elements such that xᵢ ≤ yᵢ for all i; then one has 06WH
Proposition 2 In an ordered group G the relations x ≤ y and x + z ≤ y + z are equivalent. ↩ 1 06WI
Corollary — In an ordered group G, the relations x ≤ y, 0 ≤ y - x, x - y ≤ 0 and -y ≤ -x are equivalent. 06WJ
Definition 2 — A pre-order relation (written ≤) and a commutative monoid structure (written additively) on a set M are said to be compatible if they satisfy the following axiom: 06WK
Proposition 3 — If P is a subset of an abelian group G, containing 0 and such that P + P ⊂ P, then the relation y - x ∈ P is a pre-order relation compatible with the group structure of G. For th… ↩ 2 06WL
Definition 3 In an ordered group G, any element x such that 0 ≤ x (resp. x ≤ 0) is called a positive (resp. negative) element. 06WM
Example — In the additive group Z × Z, let P be the set of elements (x, y) satisfying two inequalities ax + by ≥ 0 and cx + dy ≥ 0, where a, b, c, d are integers ( or real numbers ) such t… 06WN
Remark — By virtue of the condition P + P ⊂ P, the relation x ≥ 0 in an ordered group implies that nx ≥ 0 for every natural number n. If in addition the positive element x of G has finite… 06WO
Proposition 4 For an ordered group G to be filtered it is necessary and sufficient that it be generated by its positive elements, that is that every element be the difference of two positive ele… 06WP
Proposition 5 — If (xᵢ) is a finite family of elements of a filtered group G, then there exists z ∈ G such that xᵢ + z is positive for each i. 06WQ
Remark 1 The divisibility relation in K^* depends in an essential way on the particular ring A. If A = K we obtain the « trivial » relation, under which x | y for every pair (x, y) of eleme… 06WR
Remark 2 We will sometimes extend the definition of the relation x | y to pairs of elements of K (rather than only K^*), this relation being taken to mean « there exists z ∈ A such that y =… 06WS
Remark 1 In the commonest cases, the well ordered index set I will be a finite interval (1, n) in N. 06WT
Remark 2 The set of positive elements of the lexicographic product G consists of 0 and those nonzero elements whose nonzero component of least index is positive. 06WU
Proposition 6 — Let (xᵢ)_i ∈ A, (y_β)_β ∈ B be two families of elements of an ordered group G, each having a supremum. Then the family (x_α + y_β)_(α, β) ∈ A × B has a supremum, and sup_(α, β) ∈… 06WV
Proposition 7 — If x and y are two elements of an ordered group G, and if one of the elements inf(x, y), sup(x, y) exists, then so does the other, and x + y = inf(x, y) - sup(x, y). 06WW
Proposition 7 — If a, b ∈ K^*, and if d is a gcd of a and b, and m an lcm of a and b, then the product dm is an associate of ab. ↩ 1 06WX
Proposition 8 — Let P be the set of positive elements of an ordered group G. For G to be lattice ordered, it is necessary and sufficient that G = P - P, and that in addition P, equipped with the… 06WY
Theorem 1 (decomposition theorem) — Let (xᵢ)_1 ≤ i ≤ p and (yⱼ)_1 ≤ j ≤ q be two finite sequences of positive elements of a lattice ordered group G such that ∑ᵢ₌₁ᵖ xᵢ = ∑ⱼ₌₁^q yⱼ ; then there exists a double sequen… ↩ 1 06WZ
Corollary — Let y, x₁, x₂, …, xₙ be n + 1 positive elements of G such that y ≤ ∑ᵢ₌₁ⁿ xᵢ; then there exist n positive elements yᵢ (1 ≤ i ≤ n) such that yᵢ ≤ xᵢ and y = ∑ᵢ₌₁ⁿ yᵢ. 06X0
Definition 4 — In a lattice ordered group G the positive part (resp. negative part, absolute value) of an element x ∈ G is the element sup(x, 0) (resp. sup(-x, 0), sup(x, -x)), which is denoted… 06X1
Proposition 9 — a) For each element x of a lattice ordered group G we have x = x⁺ - x⁻ and inf(x⁺, x⁻) = 0. 06X2
Remark We deduce from d) that |x| = 0 implies x = 0 (for x⁺ and x⁻ are positive); thus x ≠ 0 implies |x| > 0. 06X3
Proposition 9 — If the group P^* of principal fractional ideals of K is lattice ordered, then every element x of K^* can be written in the form x = uv⁻¹, where u and v are elements of A such tha… ↩ 2 06X4
Definition 5 — In an ordered group two elements x and y are called coprime if inf(x, y) = 0. 06X5
Proposition 10 Let x, y and z be three elements of an ordered group ; for x – z and y – z to be coprime, it is necessary and sufficient that z = inf (x, y). ↩ 1 06X6
Proposition 11 If (xᵢ), (yⱼ) are two finite families of positive elements of a lattice ordered group, then ↩ 1 06X7
Corollary 1 If x and y are two coprime elements, and z a positive element, of a lattice ordered group, then inf(x, z) = inf(x, y + z). ↩ 1 06X8
Corollary 2 — In a lattice ordered group, if x and y are coprime and if z ≥ 0 and x ≤ y + z, then x ≤ z. 06X9
Corollary 3 — In a lattice ordered group, if x is coprime to y and to z, then it is also coprime to y + z. ↩ 1 06XA
Corollary 4 — If (xᵢ)_1 ≤ i ≤ n and (yⱼ)_1 ≤ j ≤ m are two finite families of elements of a lattice ordered group G, such that each xᵢ is coprime to each yⱼ, then x₁ + … + xₙ is coprime to y₁… ↩ 2 06XB
Corollary 5 — For any integer n ≥ 0 we have (nx)⁺ = nx⁺ and (nx)⁻ = nx⁻; for all n ∈ Z we have |nx| = |n| · |x|. Indeed nx = nx⁺ - nx⁻; since x⁺ and x are coprime, so are nx⁺ and nx⁻ if n ≥ 0… 06XC
Proposition 11 — Suppose the set P^* is a lattice, and let (aᵢ), (bⱼ) be two finite families of elements of A. Then every gcd of ∏ᵢ aᵢ and ∏ⱼ bⱼ divides the product ∏_i,j gcd(aᵢ, bⱼ). 06XD
Corollary 1 — If a, b, c are three elements of A such that a is coprime to b, then every gcd of a and c is also a gcd of a and bc. 06XE
Corollary 2 (Euclid’s lemma) — Let a, b, c be three elements of A. If a is coprime to b and divides bc, then it divides c. 06XF
Corollary 3 — If x is coprime to y and to z, then it is coprime to yz. 06XG
Corollary 4 — If (xᵢ) and (yⱼ) are two finite families of elements of A such that each xᵢ is coprime to each yⱼ, then the product of the xᵢ is coprime to the product of the yⱼ. ↩ 1 06XH
Corollary 5 — If d is a gcd of x and y, then dⁿ is a gcd of xⁿ and yⁿ for each positive integer n. Indeed xd¹ and yd¹ are coprime (Prop. 10 (DIV)), and hence so are xⁿ d⁻ⁿ and yⁿ d⁻ⁿ (Cor. 4). 06XI
Proposition 12 — Let xᵢ (1 ≤ i ≤ n) be n pairwise coprime elements in a lattice ordered group. Then sup(x₁, …, xₙ) = x₁ + ⋯ + xₙ . 06XJ
Remark Prop. 7 of VI, p. 10 also shows that a necessary and sufficient condition for x and y to be coprime is that x + y = sup(x, y). 06XK
Proposition 12 — Let aᵢ be a finite number n of pairwise coprime elements of A; then the product a₁ … aₙ, is an lcm of a₁, …, aₙ. 06XL
Proposition 13 — In a lattice ordered group G, let (x_α) be a family having an infimum (resp. supremum), and let z be an arbitrary element of G; then the family (sup(z, x_α)) (resp. (inf(z, x_α))… ↩ 1 06XM
Corollary — If an element z of a lattice ordered group G is coprime to each element x_α of a family having an infimum y, then z is coprime to y. 06XN
Remark Applying the formulae of Prop. 13 to a family of two elements (x, y), we obtain the following formulae, which express the fact that each of the laws of composition sup, inf in a la… 06XO
Definition 6 — An element x of an ordered group G is said to be irreducible if it is a minimal element in the set of strictly positive elements of G. 06XP
Examples (DIV) 1) An integer p > 0 is irreducible in Z if and only if it is prime (I, p. 50). 0EEO
Proposition 14 — For an element x > 0 of an ordered group G to be irreducible it is sufficient that it satisfy the following property : 06XQ
Proposition 14 — For a nonzero element p of A to be irreducible it is sufficient that it not be a unit, and that it cannot divide a product of two elements of A without dividing one or other of t… ↩ 2 06XR
Remark Proposition 14 (DIV) can also be expressed as follows: if p is a non zero element of A such that the ideal (p) is prime (I, p. 117, Def. 3) then p is irreducible; conversely, if P^… 06XS
Proposition 15 — In an ordered group G, let (pᵢ)_i ∈ I be a family of pairwise distinct positive elements of G satisfying condition (P) (and hence irreducible). Then the map ↩ 1 06XT
Theorem 2 — Let G be a filtered group. Then the following properties are equivalent : a) G is isomorphic to an ordered group of the form Z⁽ᴵ⁾. b) G is lattice ordered and satisfies the follo… ↩ 1 06XU

§ 2. Ordered fields

Definition 1 — Given a commutative ring A, we say that an ordering on A is compatible with the ring structure of A if it is compatible with the additive group structure of A, and if it satisfie… 06YV
Example 1 The rings Q and Z, with the usual orderings, are ordered rings. 06YW
Example 2 A product of ordered rings, equipped with the product ordering, is an ordered ring. In particular the ring Aᴱ of mappings from a set E to an ordered ring A is an ordered ring. 06YX
Example 3 A subring of an ordered ring, with the induced ordering, is an ordered ring. 06YY
Example There is only one totally ordered ring structure on Z: indeed 1 > 0, whence n > 0 for every natural number n ≠ 0, by induction. In contrast there exist ordered ring structures on Z… 06YZ
Example In Z, if we take P to be the set of positive (in the usual sense) even integers, we get a ring which is not totally ordered. 06Z0
Proposition 1 — A totally ordered ring is torsion free as a Z-module (II, p. 313). ↩ 1 06Z1
Definition 2 — A commutative field, equipped with a total ordering, is called an ordered field if its ordering and its ring structure are compatible. 06Z2
Examples 1) The field Q of rational numbers is an ordered field. 2) A subfield of an ordered field, with the induced ordering, is an ordered field. 3) \ The field of real numbers is an orde… 0DID
Proposition 2 — Let A be a totally ordered integral domain, and let K be its field of fractions. Then there exists one and only one ordering on K which restricts to the given ordering on A and m… ↩ 1 06Z3
Example — Since Z admits only one totally ordered ring structure (VI, p. 19, example), the field Q admits only one ordering which makes it an ordered field: this is the usual ordering. 06Z4
Definition 3 Let K be an ordered field. An ordered extension of K is a pair (E, u), where E is an ordered field and u is an increasing homomorphism from K to E. 06Z5
Example — 1) Every ordered field K is an ordered extension of Q. Indeed K is an extension of Q, since it is of characteristic zero, and on the other hand Q can only be ordered in one way… 06Z6
Theorem 1 For an extension E of K to admit the structure of an ordered extension of K, the following condition is necessary and sufficient: (OE) The relation p₁ x₁² + … + pₙ xₙ² = 0 implies… ↩ 3 06Z7
Lemma — Let P ∈ M and x ∉ P; then there exists P' ∈ M such that P ⊂ P' and -x ∈ P'. 06Z8
Corollary 1 (*« Artin-Schreier Theorem »*) — A necessary and sufficient condition for there to exist an ordering on a commutative field E which makes it an ordered field, is that the relation x₁² + … + xₙ² = 0 imply x₁ = …… 06Z9
Corollary 2 Let E be an extension of K admitting the structure of an ordered extension of K. For an element x ∈ E to be positive under every such structure on E, it is necessary and sufficient… 06ZA
Proposition 3 Let K be an ordered field and f an irreducible polynomial over K which changes sign between a and b in K. Then the extension E = K[X]/(f) of K admits the structure of an ordered ex… ↩ 2 06ZB
Remark There exist irreducible polynomials f over an ordered field K which do not change sign in K, but such that K[X]/(f) admits the structure of an ordered extension of K (cf. VI, p. 43… 06ZC
Proposition 4 — Let K be an ordered field and let f ∈ K[X]. There exists an interval in K, in the complement of which f takes the same sign as its highest degree term. ↩ 1 06ZD
Corollary 1 — Every extension of an ordered field of odd finite degree admits the structure of an ordered extension. 06ZE
Corollary 2 — If a is a positive element of an ordered field K, then every splitting field E of the polynomial X² - a admits the structure of an ordered extension of K. 06ZF
Remark When the ordered field K contains the « square roots » of a positive element a of K (roots of the polynomial X² - a) then the notation sqrta is generally reserved for the positive… 06ZG
Definition 4 — An ordered field K is maximal if every ordered algebraic extension of K is trivial. ↩ 1 06ZH
Example We will see later (Gen. Top., VIII, p. 1) that the field R of real numbers is a maximal ordered field. 06ZI
Theorem 2 — Every ordered field K admits an ordered algebraic extension which is a maximal ordered field. 06ZJ
Proposition 5 — Let K be a maximal ordered field, and let f be a polynomial in K[X] which changes sign between two elements a and b of K (with a < b). Then f has a root x in K such that a < x <… 06ZK
Proposition 6 — Every positive element of a maximal ordered field K has a square root in K. Every polynomial of odd degree in K[X] has at least one root in K. 06ZL
Corollary — On a maximal ordered field K there exists only one ordering compatible with the field structure. 06ZM
Theorem 3 (Euler-Lagrange) — Let K be an ordered field. Then the following three properties are equivalent: a) The field K(i) is algebraically closed (where i denotes a square root of -1). b) The ordered fie… ↩ 1 06ZN
Proposition 7 — Let K be an ordered field in which every positive element is a square. Then every element of K(i) is a square, and every polynomial of degree 2 over K(i) has a root in K(i). 06ZO
Proposition 8 — Let K be a commutative field (of arbitrary characteristic) and let K' be a splitting field for the polynomial X² + 1 ∈ K[X] (V, p. 21). Suppose: a) every polynomial in K[X] of od… 06ZP
Proposition 9 — If K is a maximal ordered field, then the only irreducible polynomials in K[X] are the first degree polynomials, and the second degree polynomials aX² + bX + c such that b² - 4ac… 06ZQ
Remark — Translating into the canonical trinomial form yields this stronger result: a necessary and sufficient condition for the polynomial aX² + bX + c over a given ordered field K to ha… 06ZR

VII. Modules over principal ideal domains

§ 1. Principal ideal domains

Definition 1 A principal ideal domain is an integral domain (I, p. 116) in which every ideal is principal. 070X
Example The ring Z of rational integers is a principal ideal domain (I, p. 111). If K is a commutative field, then the polynomial ring K[X] in one indeterminate over K is a principal ideal… 070Y
Proposition 1 Let K be the field of fractions of a principal ideal domain A, and let (xᵢ)_i ∈ I be a family of elements of K having a common denominator b ∈ K^* (in other words bxᵢ ∈ A for all i… ↩ 4 070Z
Remark Prop. 1 applies with no restrictions to an arbitrary family (xᵢ) of elements of A (take b = 1), and also to any finite family (xᵢ) of elements of K (if xᵢ = cᵢ bᵢ with cᵢ ∈ A and b… 0710
Corollary — Let (xᵢ) be an arbitrary family of elements of a principal ideal domain A contained as a subring in an integral domain B, and let d be a gcd of the family (xᵢ) in A. Then the fam… 0711
Theorem 1 (« Bezout's identity ») — Elements xᵢ \ (i ∈ I) of a principal ideal domain A are setwise coprime if and only if there exist elements aᵢ \ (i ∈ I) of A, all but finitely many of them zero, such that ∑ aᵢ… ↩ 1 0712
Proposition 2 — Let a, b, d, m and p be elements of the field of fractions K of a principal ideal domain A. a) « d is a gcd of a and b » is equivalent to « (d) = (a) + (b) ». b) « m is an lcm of… ↩ 2 0713
Proposition 3 Let a, b, c be elements of the field of fractions of a principal ideal domain A, and let d be a gcd of a and c; then the congruence ax ≡ b pmodc has a solution x₀ ∈ A if and only i… 0714
Proposition 4 Let (aᵢ), i ≤ n be a finite family of pairwise coprime elements of the principal ideal domain A. Then the canonical homomorphism (I, p. 110) from A / ( ∏ᵣ₌₁ⁿ aᵣ ) into the product… ↩ 1 0715
Definition 2 Let A be an integral domain. A system of representatives of irreducible elements of A is a family (p_α) of irreducible elements of A such that every irreducible element of A is an… ↩ 1 0716
Theorem 2 — Let A be a principal ideal domain and let (p_α) be a system of representatives of irreducible elements of A. Then every nonzero element x of the field of fractions of A can be un… ↩ 3 0717
Lemma 1 Let A be a ring such that every left ideal of A is finitely generated. Then every nonempty set Φ of left ideals of A, ordered by inclusion, has a maximal element. 0718
Remark The family (u, (n_α)) is called the decomposition of x into irreducible factors; by abuse of language, we also say that the formula (1) is the decomposition of x into irreducible f… 0719
Proposition 5 — The set of prime numbers is infinite. ↩ 1 071A
Proposition 6 For any field K, the set of monic irreducible polynomials in K[X] is infinite. 071B

§ 2. Torsion modules over a principal ideal domain

Proposition 1 The A-module M is the direct sum of its submodules Mᵢ. ↩ 2 0724
Remark 1 Conversely, let M'ᵢ be an (A/bᵢ)-module for each i, and consider the A-module M, the direct sum of the A-modules M'ᵢ; then the submodules Mᵢ constructed above coincide with the M'ᵢ… 0725
Remark 2 By the above proof, the projectors of M onto the components Mᵢ are homotheties. 0726
Remark 3 The A-module M is cyclic if and only if each Mᵢ is cyclic: if M = Am, then Mᵢ = Aᵢ eᵢ m; conversely if Mᵢ = Aᵢ mᵢ and m = ∑_i ∈ I mᵢ, then M = Am; indeed, if n ∈ M projects onto aᵢ… ↩ 1 0727
Remark 4 Let M and N be two A-modules, with components (Mᵢ) and (Nᵢ). Let u ∈ Hom_A(M, N) be an A-linear map from M to N; then for all i and for all m ∈ Mᵢ we have u(m) ∈ Nᵢ, so u induces a… 0728
Definition 1 Let π be an irreducible element of a principal ideal domain A; an A-module M is called rr-primary if, for all x ∈ M, there exists an integer n ≥ 1 such that πⁿ x = 0 (in other word… 0729
Lemma 1 Let M be a module over a principal ideal domain A; for all a ∈ A such that a ≠ 0, let α = ε ∑ᵢ₌₁ʳ πᵢⁿ⁽ⁱ⁾ be a decomposition of α into irreducible factors (VII, p. 4). The submodule… ↩ 2 072A
Lemma 2 Let M be a torsion module (II, p. 313) over an integral domain A. For every finite family (xᵢ)_1 ≤ i ≤ n of elements of M, there exists an element γ ≠ 0 in A such that the xᵢ all b… ↩ 2 072B
Theorem 1 Let M be a torsion module over a principal ideal domain A; for each irreducible element π of A, let M_π be the submodule of M consisting of elements annihilated by some power of π… ↩ 4 072C
Corollary 1 — Every submodule N of a torsion module M is the direct sum of its submodules N ∩ M. ↩ 3 072D
Corollary 2 — The submodule N of the torsion A-module M is a direct factor if and only if N, is a direct factor of M, for every irreducible element π of A. ↩ 2 072E
Corollary 3 — Let N be a submodule of the torsion A-module M. If, for every irreducible element π of A, either N, = 0 or (M/N)_π = 0, then N is a direct factor of M. ↩ 1 072F
Corollary 4 — Let A be a principal ideal domain which is not a field, and let M be an A-module. Then M is semi-simple if and only if M is a torsion module and M, = M(π) for every irreducible e… ↩ 1 072G
Remark 1 — Clearly the annihilator of every element ≠ 0 of a τ-primary module has the form Aπᵏ (k 0 an integer), since it is a principal ideal containing a power of τ. Let x be an element o… 072H
Proposition 2 — If M is a finitely generated torsion module over a principal ideal domain A, then the n-primary components of M are zero except for a finite number of them, and the projectors of… ↩ 2 072I
Remark 2 — By VII, p. 7, Remark 3, a finitely generated torsion A-module is cyclic if and only if each of its π-primary components is cyclic. ↩ 1 072J
Theorem 2 Let A be a principal ideal domain, let K be its field offructions and let P be a system of representatives of irreducible elements of A. Given an element x ∈ K, there exist a finit… ↩ 2 072K
Lemma 3 Let x, y, k be integers with k ≥ 0, and let p > 2 be a prime number. If x ≡ 1 + py mod p² then x^pᵏ ≡ 1 + pᵏ⁺¹y mod pᵏ⁺². If x ≡ 1 + 4y mod 8 then x^2ᵏ ≡ 1 + 2ᵏ⁺²y mod 2ᵏ⁺³. ↩ 1 072L
Proposition 3 Let p > 2 be a prime number and let n > 0 be an integer; then the group U(pⁿ) is cyclic of order pⁿ⁻¹; if n ≥ 2 then the residue class mod pⁿ of an integer x congruent to 1 mod p i… ↩ 1 072M
Lemma 4 Let A be a principal ideal domain and let 0 → N → M → P → 0 be an exact sequence of A-modules. Suppose that there exist coprime elements a, b ∈ A such that aN = 0 and bP = 0. Then… ↩ 1 072N
Theorem 3 If a = ∏ pᵢⁿ⁽ⁱ⁾ is the prime decomposition of the integer a > 1, then the group (Z/aZ)^* of invertible elements of the ring Z/aZ is isomorphic to the product of the groups (Z/pᵢⁿ⁽ⁱ… 072O
Remark Let p > 2 be a prime number and let x be an integer congruent to 1 mod p and not congruent to 1 mod p²; there is an exact sequence 072P

§ 3. Free modules over a principal ideal domain

Theorem 1 — Let A be a ring such that every left ideal of A is projective (II, p. 231, Def. 1) as an A-module. Then every submodule M of a free left A-module L is a direct sum of modules iso… ↩ 8 0735
Corollary 1 — If every left ideal of A is projective, then every submodule of a projective left A-module is projective. 0736
Corollary 2 — Every submodule of a free module over a principal ideal domain is free. This follows immediately from Th. 1, since every ideal of a principal ideal domain is free. ↩ 1 0737
Corollary 3 — Every projective module over a principal ideal domain is free. ↩ 1 0738
Remark The proof of Th. 1 shows that every submodule of A⁽ᴵ⁾ is isomorphic to a direct sum ⨁ aᵢ, where each aᵢ is an ideal of A. 0739
Lemma 1 — Let L be a module over a commutative ring A, generated by n elements, and let M be a free submodule of L; then M has rank ≤ n. Suppose first that L is free. Let i denote the cano… 073A
Corollary — Let E be a module over a principal ideal domain A, generated by n elements. Then any submodule F of E can be generated by at most n elements. Indeed there exists a homomorphism f… 073B

§ 4. Finitely generated modules over a principal ideal domain

Proposition 1 — Let E be a module over a commutative ring A; suppose E is a direct sum of n cyclic modules Λ/𝔞ₖ (1 ≤ k ≤ n), where the 𝔞ₖ are ideals of A; ↩ 1 073N
Proposition 2 — Let A be a commutative ring, and let E be a direct sum of n cyclic modules A / aₖ, where the aₖ satisfy a₁ ⊂ a₂ ⊂ … ⊂ aₙ. Then, for 1 ≤ p ≤ n, the ideal aₚ is the annihilator of… ↩ 1 073O
Corollary — In the notation of Prop. 2, if aₙ ≠ A, and if E is also isomorphic to the direct sum of m cyclic modules A / a'ⱼ with a'₁ ⊂ a'₂ ⊂ … ⊂ a'ₘ ≠ A, then m = n and aₖ = a'ₖ for 1 ≤ k ≤… ↩ 2 073P
Lemma 1 Let L be a free module over a principal ideal domain A and let x be an element of L. Then the following conditions are equivalent: 073Q
Remark — 1) If x is a nonzero element of L and c a content of x, then there exists a unique element y of L such that x = cy ; denote this element x/c ; then x/c is an indivisible element… 073R
Proposition 3 Let L be a free module over a principal ideal domain A and let M be a nonzero submodule of L. Let x be an element of M such that c_L(x) is maximal among the contents of elements of… ↩ 1 073S
Theorem 1 — Let L be a free module over a principal ideal domain A, and let M be a submodule of L of finite rank n. Then there exist a basis B of L, n elements eᵢ of B, and n nonzero element… ↩ 12 073T
Corollary — A submodule M of finite rank in a free module L over a principal ideal domain A has a complement in L if and only if L/M is torsion free. ↩ 1 073U
Remark It can happen that a submodule M of infinite rank in a free module L can be such that L/M is torsion free, but M has no complement in L (VII, p. 60, Ex. 6, b)). 073V
Definition 1 — In the notation and hypotheses of Th. 1, the ideals Aαᵢ of A are called the invariant factors of the submodule M with respect to the module L. ↩ 2 073W
Theorem 2 — Every finitely generated module E over a principal ideal domain A is isomorphic to a direct sum of a finite number m of cyclic modules A/aₖ, where the aₖ are ideals of A (some of… ↩ 7 073X
Corollary 1 — Every finitely generated module E over a principal ideal domain is the direct sum of the torsion submodule of E and a free module. ↩ 1 073Y
Corollary 2 — Every torsion free finitely generated module over a principal ideal domain is free of finite rank. This follows immediately from Cor. 1. ↩ 3 073Z
Definition 2 — In the notation and hypotheses of Th. 2, the ideals aₖ are called the invariant factors of the module E. As in Def. 1 (VII, p. 19), when A = Z or A = K[X], the canonical generato… ↩ 1 0740
Proposition 4 — Let A be a principal ideal domain, let L be a free A-module with finite basis (uⱼ) (1 ≤ j ≤ k), let M be a submodule of L, let (xᵢ) be a system of generators of M and let A αᵢ (1… ↩ 4 0741
Proposition 5 Let A be a principal ideal domain, and f a linear map of rank r from a free A-module L of rank m into a free A-module L' of rank n. Then there exist bases (eᵢ) (1 ≤ i ≤ m) of L and… ↩ 3 0742
Corollary 1 — Let X be a matrix of rank r, with n rows and m columns, over a principal ideal domain A; then there exists a matrix X₀ equivalent to X (II, p. 354) of the form 0743
Corollary 2 — Two matrices X and X' with n rows and m columns over a principal ideal domain A are equivalent if and only if they have the same invariant factors. 0744
Proposition 6 — Let X be a matrix of rank r over a principal ideal domain A, and let A αᵢ (1 ≤ i ≤ r) be its sequence of invariant factors. Then α₁ is a gcd of the elements of X; and the product… ↩ 1 0745
Theorem 3 — Every finitely generated abelian group G is a direct sum of its torsion subgroup F (subgroup of elements of finite order in G) und a free abelian group of finite rank p (isomorph… ↩ 4 0746
Remark While the orders n₁, …, n_q of the cyclic groups of which F is the direct sum are well defined by the divisibility condition of Th. 3, it is not the same for the groups themselves… 0747
Corollary 1 — In a finite abelian group G, there exists an element whose order is the lcm of all the orders of elements of G; this order n is the first invariant factor of G. ↩ 2 0748
Corollary 2 — Any finite abelian group G whose order is not divisible by the square of any integer > 1 is cyclic. Let us keep to the notation of Th. 3. Then p = 0 since G is finite, and q ≤ 1… 0749
Corollary 3 — Let L and M be two free Z-modules of rank n, let (eᵢ) be a basis for L and (fᵢ) a basis for M (1 ≤ i ≤ n), let u be a homomorphism from L into M, and let U be its matrix with res… ↩ 1 074A
Definition 3 — A left module M over a ring A is said to be decomposable if it is the direct sum of a family of proper nonzero submodules. Otherwise it is said to be indecomposable. The zero mod… ↩ 1 074B
Lemma 2 — The module A/𝔞 is indecomposable if and only if 𝔞 ≠ A and there is no pair (b, c) of ideals of A, distinct from A and 𝔞, such that A = b + c and b ∩ c = 𝔞. ↩ 1 074C
Proposition 7 — Let A be a commutative ring, let 𝔭 be a prime ideal of A (I, p. 117, Def. 3), and let q be an ideal of A contained in 𝔭. Suppose that for every x ∈ 𝔭 there exists an integer n >… ↩ 1 074D
Proposition 8 — Let A be a principal ideal domain and let M be a finitely generated A-module. ↩ 3 074E
Proposition 9 — Let A be a principal ideal domain, let P be a system of representatives of irreducible elements of A and let M be a finitely generated A-module. Then there exist positive integer… ↩ 4 074F
Definition 4 — In the notation of Prop. 9, those ideals (pⁿ) (p ∈ P, n ≥ 1 an integer) such that m(pⁿ) > 0 are called the elementary divisors of the module M, and the integers m(pⁿ) are called… 074G
Remark 1 If M is a finite abelian group, then its structure can be described by writing down its elementary divisors, each repeated as often as its multiplicity. We will say, for example, t… 074H
Remark 2 If a finitely generated torsion module M over a principal ideal domain A is given as a direct sum of cyclic modules isomorphic to A/(aᵢ) (in particular if the invariant factors of… 074I
Remark 3 To calculate the invariant factors of a torsion module whose elementary divisors are known, we again lean on the fact that, if the aᵢ are pairwise coprime elements of A, then the p… ↩ 1 074J
Remark 4 An A-module is called simple (I, p. 37) if it is nonzero and has no submodules other than itself and 0; it is then necessarily cyclic, so finitely generated, and indecomposable; si… ↩ 2 074K
Proposition 10 — If M is an A-module of finite length, then D(M) is (in general non-naturally) isomorphic to M, and the canonical map c_M : M → D(D(M)) is an isomorphism. Using VII, p. 19, Th. 2… ↩ 4 074L
Corollary — Let M, N be two A-modules of finite length, and let φ be an A-bilinear map from M × N into K/A, such that : 1) the relation φ(x, y) = 0 for all y ∈ N implies x = 0 ; and 2) the r… 074M
Proposition 11 — If M' xrightarrowu M xrightarrowv M'' is an exact sequence of A-modules of finite length, then the sequence D(M'') xrightarrowD(v) D(M) xrightarrowD(u) D(M') is exact [^1]. ↩ 2 074N
Proposition 12 Let M be an A-module of finite length. Then the map which sends every submodule N of M to N⁰ is a bijection from 𝔖(M) onto 𝔖(D(M)), and the inverse bijection sends every submodule… ↩ 1 074O
Example 1 For A = Z, the Z-modules of finite length are precisely the finite abelian groups; then K = Q, so K/A = Q/Z. Then to define D(M), we sometimes take, instead of Q/Z, a Z-module isom… ↩ 1 074P
Example 2 Let a be a nonzero element of A. The map x ↦ x/a from A into K induces an isomorphism on quotient modules from A/(a) onto the submodule (K/A)(a) of K/A consisting of elements annih… 074Q

§ 5. Endomorphisms of vector spaces

Definition 1 Let E be a vector space over a commutative field K, and u an endomorphism of E. An element x of E is said to be an eigenvector of u if there exists a ∈ K such that u . x = α x; if… 075G
Proposition 1 Suppose E is finite dimensional. Then an element a of K is an eigenvalue of the endomorphism u if and only if it is a root of the characteristic polynomial of u. 075H
Proposition 2 Let u be an endomorphism of a vector space E over a commutative field K; for each scalar a, let Vₐ be the eigenspace corresponding to a. Then the subspaces Vₐ are closed under u an… ↩ 1 075I
Proposition 3 Let E be a vector space of finite dimension n over a commutative field K, and let u be an endomorphism of E; for every monic irreducible polynomial p(X), let Mₚ be the vector subsp… ↩ 2 075J
Remark 1 Clearly the minimal polynomial of the restriction of u to Mₚ is the greatest power of p which divides the minimal polynomial of u. Moreover we have sₚ(u) · x = x for x ∈ Mₚ, from w… 075K
Proposition 4 — Let E be a vector space of finite dimension n over a commutative field K, and let u be an endomorphism of E. Then there exist n monic polynomials qᵢ(X) ∈ K[X] (1 ≤ i ≤ n) such th… ↩ 2 075L
Remark 2 By the above proposition, there exists a basis of E with respect to which the matrix U of u has the form 075M
Definition 2 — In the notation of Prop. 4, the n monic polynomials qᵢ(X) (1 ≤ i ≤ n) are called the similarity invariants of the endomorphism u. 075N
Corollary 1 — Let K be a field, let E and E' be two finite dimensional vector spaces over K, and let u (resp. u') be an endomorphism of E (resp. E'). Then u and u' are similar (VII, p. 29) if… ↩ 1 075O
Corollary 2 — Let u be an endomorphism of a finite dimensional vector space E over a field K, let (q₁, …, qₙ) be the family of similarity invariants of u, let L be an extension of K, let E_(L)… 075P
Corollary 3 — Let U and V be two square matrices of order n with coefficients in a commutative field K. If there exists an invertible square matrix P over some extension K' of K such that V =… 075Q
Proposition 5 — Let E be a vector space of finite dimension n over a commutative field K, let u be an endomorphism of E, and let U be its matrix with respect to some basis of E. Then for each in… ↩ 1 075R
Corollary 1 — Let u be an endomorphism of a vector space of finite dimension n over a commutative field K, with characteristic polynomial χᵤ(X) and similarity invariants qᵢ(X) (1 ≤ i ≤ n). The… ↩ 1 075S
Corollary 2 — In the notation of Cor. 1, let q(X) be the minimal polynomial of u; then q(X) divides χᵤ(X) and χᵤ(X) divides q(X)ⁿ. In particular the minimal polynomial and the characteristic p… ↩ 1 075T
Corollary 3 — An endomorphism u is nilpotent if and only if its characteristic polynomial has the form Xⁿ. ↩ 1 075U
Proposition 6 — Let E be a vector space of finite dimension n over a commutative field K, and let u be an endomorphism of E. Then E is the direct sum of subspaces Eₖ, closed under u and cyclic w… 075V
Proposition 7 — Let E be a finite dimensional vector space over a commutative field K, and let u be an endomorphism of E whose eigenvalues all belong to K. For each eigenvalue a of u, let Mₐ be… ↩ 2 075W
Definition 3 For every field K, every integer m ≥ 1, and every a ∈ K, the matrix U_m,α is called the Jordan matrix of order m and eigenvalue a. 075X
Proposition 8 Let E be a finite dimensional vector space over a commutative field K, and let u be an endomorphism of E. Then the following conditions are equivalent: ↩ 1 075Y
Definition 4 An endomorphism satisfying conditions (i), (ii) and (iii) of Prop. 8 is called triangularisable. 075Z
Corollary Let U be a square matrix over a commutative field K such that all the eigenvalues of U are in K; then there exists a matrix similar to U which is a diagonal block of Jordan matrice… ↩ 1 0760
Remark — 1) It follows from Prop. 6 of VII, p. 34, that, if U is similar to a diagonal block of Jordan matrices (Jₖ), then the number of Jₖ of the form U_m,α (for given m and α) is unique… 0761
Proposition 9 — If the square matrix U is similar to a diagonal block of Jordan matrices (U_mᵢ, αᵢ), then the minimal polynomial of U is the lcm of the (X - αᵢ)^mᵢ, and the characteristic polyno… ↩ 1 0762
Corollary — In the notation of Prop. 7, the dimension of the subspace M_α is the multiplicity of the eigenvalue α as a root of the characteristic polynomial of u. 0763
Proposition 10 — Let E be a vector space of finite dimension n over a commutative field K, let u be an endomorphism of E, and let χᵤ(X) = ∏ᵢ₌₁ⁿ (X - αᵢ) be a decomposition into linear factors of… ↩ 1 0764
Lemma 1 If B and C are lower triangular matrices of order n with diagonals (βᵢ) and (γᵢ), then the matrices B + C and BC are lower triangular with diagonals (βᵢ + γᵢ) and (βᵢ γᵢ). 0765
Corollary 1 For q(u) to be invertible, it is necessary and sufficient that q be coprime to χᵤ. ↩ 1 0766
Remark 1 — A polynomial is coprime to χᵤ if and only if it is coprime to the minimal polynomial of u (VII, p. 33, Cor. 2). 0767
Corollary 2 Let r ∈ K(X) be a rational function over K. Then u is substitutable (IV, p. 21) in r if and only if each of its eigenvalues is. In this case the following formulae hold: ↩ 1 0768
Corollary 3 For each integer s ≥ 0 we have Tr(uˢ) = ∑ᵢ₌₁ⁿ αᵢˢ. This formula is also valid for s < 0 provided u is invertible. ↩ 2 0769
Corollary 4 — Suppose the field K is of characteristic zero; then the endomorphism u is nilpotent if and only if Tr(uˢ) = 0 for 1 ≤ s ≤ n. 076A
Corollary 5 — Let Y be an indeterminate and let tildeu denote the endomorphism of the K(Y)-vector space K(Y) ⊗_K E induced from u by extension of scalars from K to the field K(Y) of rational f… ↩ 1 076B
Corollary 6 — Suppose the field K is of characteristic zero. Then, in the ring K[[T]] of formal power series, we have 076C
Remark 2 — By IV, p. 80, Cor. 1 and formula (8), we have, for each polynomial q ∈ K[X], that 076D
Corollary 7 We have det(Tr(uⁱ⁺ʲ)_0 ≤ i,j ≤ n) = dis(χᵤ). 076E
Proposition 11 Let E (resp. E') be a finite dimensional vector space over a commutative field K and let u (resp. u') be an endomorphism of E (resp. E'). Let 076F
Lemma 2 Let B and C be two lower triangular matrices of orders m and n respectively, with diagonals (βᵢ)_1 ≤ i ≤ n and (γⱼ)_1 ≤ j ≤ n. Let us identify the lexicographic product of the orde… 076G
Definition 5 — Let E be a finite dimensional vector space over a commutative field K and let 𝔉 be a set of endomorphisms of E. Then 𝔉 is said to be diagonal with respect to a basis (eᵢ) of E if… 076H
Proposition 12 — Let E be a finite dimensional vector space over a commutative field K, and let u be an endomorphism of E. Then the following conditions are equivalent: ↩ 2 076I
Corollary — If all the roots of the characteristic polynomial of u are in K, and they are all simple, then u is diagonalisable. 076J
Proposition 13 — Let E be a finite dimensional vector space over a commutative field K, let S be a set of endomorphisms of E, and let A be the subalgebra of End_K(E) generated by S and Id_E. Then… ↩ 4 076K
Lemma 3 Let g and h be two commuting endomorphisms of a vector space E. Then each eigenspace of g is closed under h. 076L
Corollary — The sum and the composite of two commuting diagonalisable endomorphisms of E are diagonalisable. 076M
Definition 6 — Let E be a finite dimensional vector space over a commutative field K. Then an endomorphism u of E is said to be semi-simple if every subspace of E which is closed under u has a… 076N
Proposition 14 — An endomorphism u of a finite dimensional vector space over a commutative field is semi-simple if and only if the minimal polynomial of u has no multiple factors. ↩ 4 076O
Corollary — Let u be an endomorphism of a finite dimensional vector space over a commutative field K, and let L be an extension of K. If u_(L) is semi-simple then u is semi-simple. If u is s… 076P
Proposition 15 — Let E be a finite dimensional vector space over a commutative field K, let u be an endomorphism of E and let q(X) be its minimal polynomial. Then the following conditions are equ… ↩ 1 076Q
Definition 7 — An endomorphism u satisfying conditions (i), (ii) and (iii) of Prop. 15 is said to be absolutely semi-simple. 076R
Corollary — A necessary and sufficient condition for u to be absolutely semi-simple is that there exist an extension L of K such that L is perfect and u_(L) is semi-simple. 076S
Proposition 16 — Let E be a finite dimensional vector space over a commutative field K, let 𝔉 be a set of endomorphisms of E and let A be the subalgebra of End_K(E) generated by 𝔉 and Id. Then th… ↩ 1 076T
Corollary The sum and product of two commuting, absolutely semi-simple endomorphisms are absolutely semi-simple. ↩ 1 076U
Remark — Suppose the conditions of Prop. 16 are satisfied and let L be an extension of K. By Prop. 13 the set 𝔉_(L) is diagonalisable if and only if the algebra L ⊗_K A is diagonalisable… 076V
Definition 8 Let E be a finite dimensional vector space over a commutative field and let u be an endomorphism of E. Then a Jordan decomposition of u is a pair (uₛ, uₙ), where uₛ is a semi-simpl… 076W
Theorem 1 Let E be a finite dimensional vector space over a commutative field K and let u be an endomorphism of E. Then u has a Jordan decomposition (uₛ, uₙ) if and only if the eigenvalues o… ↩ 1 076X
Lemma 4 Let E be a finite dimensional vector space over a commutative field K and let u be a triangularisable endomorphism of E. Then there exists a unique diagonalisable endomorphism v of… ↩ 1 076Y
Corollary 1 Suppose u has a Jordan decomposition, and let L be an extension of K. Then u_(L) has a Jordan decomposition, with (u_(L))ₛ = (uₛ)_(L) and (u_(L))ₙ = (uₙ)_(L). 076Z
Corollary 2 Suppose u has a Jordan decomposition. Then every endomorphism of E which commutes with u also commutes with uₛ and uₙ. ↩ 1 0770
Corollary 3 Let u and v be two commuting endomorphisms of E which have Jordan decompositions. a) The endomorphisms u, v, uₛ, vₛ, uₙ, vₙ, v, all commute. b) The endomorphisms u + v and uv have… 0771
Corollary 4 Suppose u has a Jordan decomposition, and let R be a polynomial in K[X]. Then the endomorphism R(u) has a Jordan decomposition with R(u)ₛ = R(uₛ). 0772
Remark 1 We have det(uₛ) = det(u) and Tr(uₛ) = Tr(u). ↩ 1 0773
Remark 2 A necessary and sufficient condition for u to be triangularisable is that u have a Jordan decomposition with uₛ diagonalisable. Then there exists a basis of E with respect to which… 0774
Remark 3 The notion of Jordan decomposition for a square matrix can be defined in an analogous manner. For example, for the Jordan matrix U_m,α we have (U_m,α)ₛ = α · Iₘ , (U_m,α)ₙ = U_m,0… 0775
Remark 4 If u is semi-simple but not absolutely semi-simple, then it has no Jordan decomposition. 0776
Proposition 17 — Let E be a finite dimensional vector space over a commutative field, and let f be an endomorphism of E. Then the following conditions are equivalent: (i) f has a Jordan decomposi… 0777
Proposition 18 — Let E be a finite dimensional vector space over a commutative field K, let u be an endomorphism of E and let E' be a subspace of E closed under u. Let u' (resp. u'') be the endom… 0778
Proposition 19 — Let E be a finite dimensional vector space over a commutative field K, and let S be a set of commuting triangularisable endomorphisms of E. Then there exists a basis of E with re… ↩ 1 0779

VIII. Semisimple Modules and Rings

§ 1. Artinian Modules and Noetherian Modules

Definition 1 Let A be a ring. We call an A-module M Artinian (resp. Noetherian) if it satisfies the following equivalent conditions: ↩ 1 0001
Example 1 A finite-dimensional vector space over a field is Artinian and Noetherian. 0002
Example 2 Let M be an A-module. If there exists an infinite family (Mᵢ)_i∈ I of nonzero submodules of M whose sum is direct, then M is neither Artinian nor Noetherian: indeed, for every stri… ↩ 3 0003
Example 3 We will see further on that the Z-module Z is Noetherian but not 0004
Example 4 Let p be a prime number and Mₚ the p-primary component of the torsion Z-module Q/Z (VII, §2, No. 2, p. 7). Every submodule of Mₚ is equal to either Mₚ or p⁻ⁿZ/Z for an integer n∈ N… 0005
Proposition 1 An A-module M has finite length (II, §1, No. 10, p. 212) if and only if it is both Artinian and Noetherian. ↩ 5 0006
Proposition 2 An A-module M is Noetherian if and only if every submodule of M is finitely generated. ↩ 2 0007
Lemma 1 Let M be a Noetherian A-module and E a subset of M. There exists a finite subset F of E generating the same submodule as E. 0008
Proposition 3 Let M be an A-module and N a submodule of M. Then M is Artinian (resp. Noetherian) if and only if N and M/N are. ↩ 5 0009
Lemma 2 Let M be an A-module and N, P, and Q submodules of M. Suppose that we have P⊂ Q, N∩ P = N∩ Q, and N + P = N + Q. We then have P = Q. ↩ 1 00QV
Corollary Let M be an A-module and (Mᵢ)_i∈ I a finite family of submodules of M. ↩ 1 000A
Example 5 Let (Mᵢ)_i∈ I be a finite family of A-modules. If the modules Mᵢ are Artinian (resp. Noetherian), then so is their direct sum ⨁_i∈ IMᵢ. ↩ 1 000B
Remark The definitions and results of this subsection extend to arbitrary abelian groups with operators by replacing the submodules in the statements with stable subgroups. 000C
Definition 2 A ring A is said to be left Artinian (resp. left Noetherian) if the left A-module Aₛ is Artinian (resp. Noetherian). Likewise, a ring A is said to be right Artinian (resp. right No… 000D
Example 1 A field is a ring that is both left and right Artinian and Noetherian. 000E
Example 2 Let A be a ring and D a subring of A. Suppose that D is a field and that A is a finite-dimensional left vector space over D. Then the ring A is left Artinian and left Noetherian be… 000F
Example 3 A principal ideal domain (VII, §1, No. 1, p. 1, Definition 1) is Noetherian. An integral domain A that is not a field is not an Artinian ring: for every nonzero noninvertible eleme… ↩ 2 000G
Example 4 Let M be an A-module that is the direct sum of an infinite family (Mᵢ)_i∈ I of nonzero submodules. Let E be the endomorphism ring of M. For every i∈ I, let 𝔞ᵢ (resp. 𝔟ᵢ) be the set… 000H
Theorem 1 Let A be a left Artinian ring. The A-module Aₛ has finite length. ↩ 7 00QW
Lemma 3 Let A be a ring and n a natural number. An Artinian A-module M that is the sum of a family of submodules of length leqslant n has finite length. ↩ 2 000I
Corollary Every left Artinian ring is left Noetherian. 000J
Example 5 The left and right lengths of a field are equal to 1. 000K
Proposition 4 a) Let A be a left Noetherian ring and M a finitely generated left A-module. The module M is Noetherian, and every submodule of M is finitely generated. ↩ 5 000L
Proposition 5 a) Let A be a left Artinian (resp. left Noetherian) ring, and let φ : A→ B be a ring homomorphism that makes B into a finitely generated left A-module. The ring B is left Artinian… ↩ 4 000M
Corollary The prime ideals of an Artinian commutative ring are its maximal ideals. 000N
Definition 3 Let A be a ring, M an A-module, and E = End_A(M) the endomorphism ring of M. The countermodule of M is the left E-module with the same underlying additive group as M and external l… ↩ 1 000O
Lemma 4 Let M be a left A-module with finitely generated countermodule. There exist a natural number m and an injective A_M-linear mapping from (A_M)ₛ to Mᵐ. ↩ 3 000P
Proposition 6 Let M be an Artinian (resp. Noetherian) left A-module with finitely generated countermodule. The ring of homotheties A_M of M is left Artinian (resp. left Noetherian). ↩ 4 000Q
Corollary Let A be a commutative ring. 000R
Remark Let A be a ring. An Artinian left A-module M with finitely generated countermodule has finite length: indeed, the ring of homotheties A_M of M is left Artinian (Proposition 6), and… 000S
Proposition 7 Let A,σ ,d be as above. There exists a unique ring structure on the group A[X] with the following properties: 00QX
Remark The ring A[X]_σ ,d has the following universal property: given a ring A', a ring homomorphism f: A→ A', and an element x of A' such that xf(a) =f(σ (a))x+f(d(a)) for every a∈ A, th… 000T
Theorem 2 Let A be a left Noetherian ring, and let σ be an automorphism of A and d an endomorphism of the additive group of A satisfying relation (1). The ring A[X]_σ ,d is left Noetherian. ↩ 1 000U
Corollary 1 (Hilbert) Let A be a Noetherian commutative ring. For every integer ngeqslant 0, the polynomial algebra A[X₁, . . . ,Xₙ] is a Noetherian ring. ↩ 1 00RW
Corollary 2 Let A be a Noetherian commutative ring. A commutative A-algebra generated by finitely many elements is a Noetherian ring. ↩ 1 000V
Corollary 3 Every commutative ring is the union of a right directed family of Noetherian subrings. ↩ 1 000W

§ 2. The Structure of Modules of Finite Length

Proposition 1 Let A be a nonzero ring, and let 𝔯 be the set of noninvertible elements of A. The following properties are equivalent: ↩ 2 001P
Definition 1 A local ring is a nonzero ring that has the equivalent properties of Proposition 1. ↩ 2 001Q
Example 1 Every field is a local ring. 001R
Example 2 Let A be a nonzero ring in which every element is invertible or nilpotent. Then A is a local ring. Indeed, if a∈ A is not invertible, then by assumption, there exists an integer ng… ↩ 1 001S
Example 3 Let X be a Cʳ manifold (VAR, R, 5.1.5) and x a point of X. Let Oₓ be the ring of germs at x of Cʳ functions with values in the field of scalars K. Then Oₓ is a commutative local ri… 001T
Example 4 Let A be a commutative local ring and B = A[[Xᵢ]]_i∈ I an algebra of formal power series with coefficients in A (III, §2, No. 11, p. 456). By Proposition 6 of IV, §4, No. 4, p. 30… ↩ 1 001U
Example 5 Let p be a prime number. We denote by Z₍ₚ₎ the subring of the field Q of rational numbers consisting of the fractions a/b with a∈ Z,b∈ Z, and b not divisible by p*(cf. Comm. Alg… ↩ 1 001V
Example 6 Let K be a commutative field of characteristic p >0 and G a p-group (I, §6, No. 5, p. 76, Definition 9). The algebra K[G] of the group G over K (III, §2, No. 6, p. 446) is a local… 001W
Proposition 2 a) Suppose that the sequence (Nₚ) is stationary. Then we have N_∞∩ I_∞= 0, the restriction of u to I_∞ is injective, and u induces a nilpotent endomorphism of N_∞. ↩ 3 001X
Remark 1 Let p be an integer such that Nₚ= Nₚ₊₁; the proof given above shows that N_∞∩ Iₚ= 0, and the restriction of u to Iₚ is injective. Likewise, let q be an integer such that I_q= I_q+1… 001Y
Remark 2 Suppose that M is a direct sum of two submodules N and I that are stable under u and that u induces a nilpotent endomorphism u_N of N and an automorphism of I. We then have N_∞= N… 001Z
Remark 3 The assumption of assertion a) is satisfied if the A-module M is Noetherian; the assumption of assertion b) is satisfied if M is Artinian; by Proposition 1 of VIII, p. 2, the assum… ↩ 1 0020
Corollary 1 Let A be a ring, and let M be an A-module. ↩ 3 0021
Corollary 2 In a left Noetherian ring, every left or right invertible element is invertible. 0022
Definition 2 An A-module M is called indecomposable if it is not the direct sum of a family of submodules distinct from 0 and M. 0023
Example Suppose that the ring A is a principal ideal domain. The indecomposable finitely generated A-modules are the A-modules isomorphic to either A or A/pⁿA, where p is an irreducible el… 0024
Proposition 3 A Noetherian or Artinian A-module M is the direct sum of a finite family of indecomposable submodules. ↩ 1 0025
Definition 3 A module is called primordial[^1] if its endomorphism ring is local. 0026
Proposition 4 a) A primordial module is indecomposable. ↩ 4 0027
Definition 4 A module is called semiprimordial if it is the direct sum of a family of primordial submodules. 0028
Example 1 Every simple module is primordial (VIII, p. 45); every semisimple module is therefore semiprimordial (VIII, p. 55, Definition 1). 00SE
Example 2 If A is a left Noetherian ring, then every injective A-module is semiprimordial (X, §1, nᵒ9, p. 21, proposition 14 and X, §1, nᵒ10, p. 22, théorème 3, b)). 01L7
Theorem 1 (Azumaya) Let A be a ring, L a primordial A-module, and M a semiprimordial A-module. There exists a unique cardinal, denoted by [M : L], with the following property: ↩ 5 00RX
Lemma 1 Let M be an A-module, M' a primordial submodule of M, and M'' a submodule of M supplementary to M'. Let u be an endomorphism of M. Then u or 1_M-u induces an isomorphism from M' to… ↩ 1 0029
Lemma 2 Let M be an A-module that is the direct sum of a family (Mᵢ)_i∈ I of primordial submodules, and let u be an endomorphism of M. Set v= 1_M-u and M_J=⨁_i∈ JMᵢ for every subset J of I… ↩ 4 002A
Lemma 3 Let M be an A-module that is the direct sum of a family (Mᵢ)_i∈ I of primordial submodules, and let p be a nonzero projector of M. There exists an index i∈ I such that p induces an… ↩ 3 002B
Lemma 4 Let M be an A-module that is the direct sum of a family (Mᵢ)_i∈ I of primordial submodules, let L be a primordial A-module, and let N be a direct factor submodule of M. Suppose tha… ↩ 1 002C
Corollary 1 Let M and N be semiprimordial modules. Then M and N are isomorphic if and only if we have [M : L] = [N : L] for every primordial module L. ↩ 2 002D
Corollary 2 Let M be a semiprimordial module. Let (Mᵢ)_i∈ I and (M'ⱼ)_j∈ J be families of primordial submodules of M such that ↩ 1 00QY
Corollary 3 Let M, N, and P be semiprimordial modules. Suppose that M⊕ P is isomorphic to N⊕ P and that [P : L] is finite for every primordial module L. Then M and N are isomorphic. 002E
Corollary 4 Let M and N be semiprimordial modules. Suppose that there exists an integer d >0 such that Mᵈ is isomorphic to Nᵈ. Then the modules M and N are isomorphic. ↩ 1 002F
Corollary 5 Let M be a semiprimordial module that is the direct sum of a finite family (Mᵢ)_i∈ I of primordial submodules. For any subset J of I, write M_J=⨁_i∈ JMᵢ. Let N be a direct factor s… ↩ 2 002G
Corollary 6 Every finitely generated projective module over a local ring is free.[^2] ↩ 3 002H
Remark Let M and M' be semiprimordial A-modules. It immediately follows from Lemma 4 of VIII, p. 33 that M' is isomorphic to a direct factor submodule of M if and only if we have [M': L]l… 002I
Theorem 2 (Krull–Remak–Schmidt) Let A be a ring and M an A-module of finite length. ↩ 5 00RY
Theorem 3 Let K be a commutative field, A a K-algebra, and M and N modules of finite length. Let K' be a nonzero commutative K-algebra such that the A_(K')-modules M_(K') and N_(K') are isom… ↩ 5 002J
Remark Let E and F be two finite-dimensional vector spaces over a commutative field K, and let K' be an extension of K. Let u be an endomorphism of E and v an endomorphism of F, and let u… 002K

§ 3. Simple Modules

Definition 1 Let A be a ring. An A-module M is called simple if it is nonzero and has no submodule distinct from 0 and M. ↩ 1 0035
Example 1 The module Aₛ is a simple A-module if and only if A is a field (I, §9, No. 1, p. 115, Theorem 1). The simple A-modules are then the vector spaces of dimension 1 over the field A. 0036
Example 2 Let A be a principal ideal domain (VII, §1, No. 1, p. 1, Definition 1) that is not a field. For every irreducible element π of A, the A-module Aₛ/(π ) is simple, and every simple A… 0037
Example 3 Let K be a field, V a nonzero right vector space over the field K, and A a subring of the ring End_K(V) containing the endomorphisms of V of finite rank (for example, A = End_K(V))… ↩ 2 0038
Proposition 1 Let A be a ring. ↩ 4 00QZ
Corollary 1 If the ring A is not reduced to 0, then there exist simple A-modules. 0039
Corollary 2 Let A be a local ring (VIII, p. 26, Definition 1) and 𝔯 its maximal ideal. The A-module Aₛ/𝔯 is simple, and every simple A-module is isomorphic to Aₛ/𝔯. 003A
Remark 1 Let A be a commutative ring and 𝔪 an ideal of A. Then 𝔪 is the annihilator (II, §1, No. 12, p. 219, Definition 11) of the A-module Aₛ/𝔪. Hence, if 𝔪 and 𝔪' are distinct ideals of A… ↩ 1 003B
Remark 2 We can give an example of a noncommutative ring A and two distinct maximal left ideals 𝔪 and 𝔪' of A such that the A-modules Aₛ/𝔪 and Aₛ/𝔪' are isomorphic (VIII, p. 52, Exercise 3)… 003C
Proposition 2 Let A be a ring, M and N two A-modules, and f a nonzero homomorphism from M to N. ↩ 5 003D
Corollary (Schur’s lemma) The endomorphism ring of a simple module is a field. ↩ 2 00RZ
Theorem 1 Let K be an algebraically closed commutative field, A a K-algebra, and M a simple A-module. Suppose that the dimension of M as a vector space over K is finite or, more generally, s… ↩ 16 003E
Lemma 1 Let E be a field and K a subfield of the center of E not equal to E. If the field K is algebraically closed, then we have [E : K]geqslant Card(K). ↩ 1 003F
Example Let A be a C-algebra generated by a countable family of elements; it has countable dimension over C. Let M be a simple A-module; it is monogenous, so admits a countable basis over… 003G
Corollary 1 Keep the assumptions of Theorem 1, and suppose, moreover, that the algebra A is commutative. Then M has dimension 1 over K. ↩ 1 003H
Corollary 2 Let K be a commutative field, A a K-algebra, and M an A-module. Suppose that for every extension L of K, the A_(L)-module M_(L) is simple. Then the endomorphism ring of M consists… 003I
Definition 2 Let A be a ring and M an A-module. A maximal submodule of M is a maximal element, for the inclusion, of the set of proper submodules of M. ↩ 2 003J
Proposition 3 Let M be a finitely generated A-module. Every proper submodule of M is contained in a maximal submodule. ↩ 3 003K
Corollary 1 Let M be a nonzero finitely generated A-module. There exists a two-sided ideal 𝔞 of A, annihilator of a simple A-module, such that 𝔞M is not equal to M. ↩ 1 003L
Corollary 2 Let A be a commutative ring and B an A-algebra. Let M be a simple B-module that is a finitely generated A-module, and let 𝔪 be the annihilator of the A-module M. Then 𝔪 is a maxima… 003M
Proposition 4 Let A be a ring that has a minimal left ideal 𝔞. Every faithful simple A-module is isomorphic to the A-module 𝔞. ↩ 2 003N
Proposition 5 Let A be a left Artinian ring and M a faithful A-module. ↩ 3 003O
Remark Proposition 5 applies, in particular, to the following two cases: 003P

§ 4. Semisimple Modules

Definition 1 A module is called semisimple if it is the direct sum of a family of simple modules.[^1]. ↩ 1 0040
Example 1 A module reduced to 0 and a simple module are semisimple modules. 0041
Example 2 If A is a field, then every A-module is semisimple by Theorem 1 of II, §7, No. 1, p. 292. This shows that, in general, a semisimple module decomposes in several ways into a direct… 0042
Example 3 Let A be a principal ideal domain, and let P be a system of representatives consisting of irreducible elements of A (VII, §1, No. 3, p. 3). Let M be an A-module, and, for every π ∈… 0043
Theorem 1 Let M be a module that is the (not necessarily direct) sum of a family (Sᵢ)_i∈ I of simple submodules, and let N be a submodule of M. There exists a subset J of I such that M is th… ↩ 6 0044
Corollary 1 Every module that is the sum of a family of simple modules is semisimple. ↩ 1 0045
Corollary 2 A module M is semisimple if and only if every submodule of M is a direct factor. ↩ 8 0046
Corollary 3 Let M be a semisimple module and N a submodule of M. The modules N and M/N are semisimple. More precisely, if M is the direct sum of a family (Sᵢ)_i∈ I of simple modules, then ther… ↩ 6 0047
Corollary 4 Let M be a semisimple module. Then M is simple if and only if the endomorphism ring E of M is a field. ↩ 1 0048
Remark Let K be an algebraically closed commutative field and A a K-algebra. Let M be a semisimple A-module that is a finite-dimensional vector space over the field K. Then M is simple if… 0049
Remark 1 Let V and V' be B-modules. The adjunction isomorphism ↩ 1 004A
Remark 2 Let M be a left A-module. The B-linear mappings ↩ 1 004B
Remark 3 Suppose that P is finitely generated as an A-module. Let M be the direct sum of a family (Mᵢ)_i∈ I of A-modules. The A-modules T(H(M)) and ⨁ᵢT(H(Mᵢ)) are canonically isomorphic. Wh… ↩ 1 004C
Proposition 1 Let M be an A-module. The following properties are equivalent: ↩ 2 004D
Definition 2 An A-module M is isotypical of type S if it has the equivalent properties of Proposition 1. The module M is called isotypical if there exists a simple A-module T such that M is iso… 004E
Proposition 2 If a module is the sum of isotypical submodules of type S, then it is isotypical of type S. The submodules and the quotient modules of an isotypical module of type S are isotypical… ↩ 5 004F
Remark Every nonzero isotypical module of type S has a quotient module and a submodule isomorphic to S; consequently, if M and M' are nonzero isotypical A-modules of type S, then the grou… 004G
Proposition 3 a) Let M be an isotypical A-module of type S. The A-linear mapping α_M: S⊗_DHom_A(S,M)→ M characterized by α_M(s⊗ f) =f(s) (VIII, p. 59) is bijective. ↩ 5 004H
Definition 3 Let M be an isotypical A-module of type S. A description of M with respect to S is a pair (V, α ), where V is a left vector space over the field D and α : S⊗_DV→ M is an isomorphis… 004I
Theorem 2 Let M be an isotypical A-module of type S and (V, α ) a description of M. Denote by D_D(V) the set, ordered by inclusion, of D-linear subspaces of V, and by D_A(M) that of A-submod… ↩ 4 004J
Remark 1 Let M be an isotypical A-module of type S and (V, α ) a description of M. Then M has finite length if and only if V is finite-dimensional, and in this case, we have the relation ↩ 1 004K
Corollary 1 Let M be an isotypical A-module of type S. For every A-submodule N of M, identify Hom_A(S,N) with the linear subspace of Hom_A(S,M) consisting of the mappings with image contained… 004L
Corollary 2 Let V be a left vector space over D and F a set of endomorphisms of V. An A-submodule of S⊗_DV is stable under all endomorphisms 1_S⊗ u, where u runs through F, if and only if it i… ↩ 6 004M
Theorem 3 Let M and M' be isotypical A-modules of type S. Let (V, α ) and (V', α ') be descriptions of M and M', respectively. For any D-linear mapping f: V→ V', denote by widetildef: M→ M'… ↩ 3 004N
Remark 2 Let M be an isotypical A-module of type S, and let (V, α ) be a description of M. Let B be a subring of the ring End_A(M)ᵒ. The ring isomorphism from End_D(V)ᵒ to End_A(M)ᵒ endows… ↩ 1 004O
Corollary Let M and M' be isotypical A-modules of type S. The mapping u↦ Hom(1_S, u) from Hom_A(M,M') to Hom_D(Hom_A(S,M), Hom_A(S,M')) is a group isomorphism. When M is equal to M', it is a… 004P
Definition 4 Let A be a ring, M an A-module, and S a simple A-module. The isotypical component of type S of M, denoted by M_S, is the sum of the submodules of M isomorphic to S. 004Q
Proposition 4 Let A be a ring. Denote the set of classes of simple A-modules by S. Let M be a semisimple A-module. ↩ 6 004R
Corollary Let A be a ring and S the set of classes of simple A-modules. Let M be a semisimple A-module and N a submodule of M. Then we have N =⨁_λ∈SN∩ M_λ and M/N =⨁_λ∈S(M_λ+ N)/N. 004S
Proposition 5 Let A be a ring, and let S be the set of classes of simple A-modules. Let M and N be A-modules. ↩ 8 004T
Remark Let A and B be rings. Let M be an (A, B)-bimodule. It follows from Proposition 5 that the isotypical components of the A-module M are sub-bimodules of M. This holds, in particular… 004U
Example Let us consider the case when the ring A is commutative. The mapping that sends a maximal ideal 𝔪 onto cl(A/𝔪) is a bijection from the set of maximal ideals of A to the set S of cl… 004V
Corollary 1 Let M be a semisimple A-module and N a submodule of M. The following properties are equivalent: ↩ 1 004W
Corollary 2 Let M be an A-module. The following two conditions are equivalent: ↩ 2 00R0
Proposition 6 Keep the notation above, and suppose that the A-module M is semisimple. The mapping u↦ h(u) from the right D-vector space Hom_A(M,S) to the dual of the left D-vector space Hom_A(S… ↩ 1 004X
Proposition 7 Let M be an A-module. The canonical mapping α_M is injective. For every λ ∈ S, the image under α_M of S_λ⊗_D_λ Hom_A(S_λ,M) is the isotypical component of M of type λ. The image of… ↩ 1 004Y
Definition 5 Let M be a semisimple A-module. A description of M (with respect to the family (S_λ)_λ∈S) is a pair ((V_λ)_λ∈S, α ), where V_λ is a left vector space over the field D_λ for each λ… ↩ 2 004Z
Proposition 8 Let M be a semisimple A-module and ((V_λ)_λ∈S, α ) a description of M. ↩ 3 0050
Corollary Let M be a semisimple A-module. For every submodule N of M and every element λ of S, identify Hom_A(S_λ,N) with the D_λ-linear subspace of Hom_A(S_λ,M) consisting of the mappings w… 0051
Proposition 9 Let M and M' be semisimple A-modules, and consider descriptions ((V_λ)_λ∈S, α ) and ((V'_λ)_λ∈S, α ') of M and M', respectively. For every family f= (f_λ)_λ∈S in ∏_λ∈S Hom_D_λ(V_λ… ↩ 2 0052
Corollary Let M be a semisimple A-module and M' an A-module. The mapping u↦ (Hom(1_S_λ, u))_λ∈S from Hom_A(M,M') to ↩ 2 0053
Proposition 10 Let M be a semisimple A-module. Let (Mᵢ)_i∈ I be a family of simple submodules with direct sum M. The following properties are equivalent: ↩ 4 0054
Proposition 11 Let M be a semisimple A-module that is the direct sum of a family (Mᵢ)_i∈ I of simple submodules. For every λ ∈ S, we denote by I(λ ) the set of indices i∈ I such that Mᵢ is of cla… 0055

§ 5. Commutation

Definition 1 The A-module M is balanced if we have A_M= A''_M. ↩ 1 005E
Proposition 1 Let (Aᵢ)_i∈ I be a family of rings, and for every i∈ I, let Mᵢ be an Aᵢ-module. Set A =∏ Aᵢ,M =∏ Mᵢ, and N =⨁ Mᵢ. Endow M with the structure of an A-module with law of action ((aᵢ)… ↩ 3 005F
Proposition 2 Let A be a ring, and let M be an A-module. Let I be a set. Then the bicommutant of the A-module M⁽ᴵ⁾ coincides with the ring of homotheties of the A''_M-module M⁽ᴵ⁾. ↩ 2 005G
Definition 2 An A-module M is called generating if every A-module N is generated by the images of the A-linear mappings from M to N. 005H
Theorem 1 Let M be a left A-module. The following properties are equivalent: ↩ 8 005I
Corollary A generating A-module is faithful. ↩ 1 00R1
Example 1 The A-module Aₛ is generating. 005J
Example 2 Every nonzero free A-module is generating. More generally, every module with a generating quotient is itself generating. ↩ 1 005K
Example 3 Let M be a semisimple A-module whose countermodule is finitely generated. Then M is a generating A_M-module. Indeed, by Lemma 4 of VIII, p. 8, there exists a natural number m such… ↩ 1 005L
Example 4 Let A be a principal ideal domain, and let P be a finitely generated A-module. There exist an integer ngeqslant 1 and an increasing sequence of ideals (𝔞ᵢ)_1leqslant ileqslant n of… ↩ 1 005M
Lemma 1 Let A be a commutative ring, M a finitely generated A-module, and Ann(M) its annihilator. Let 𝔞 be an ideal of A. The following properties are equivalent: ↩ 4 005N
Proposition 3 Let A be a commutative ring. Every finitely generated and faithful projective A-module is generating. More generally, a finitely generated projective A-module P is a generating A_P… ↩ 2 00R2
Lemma 2 Let A be a ring and M a projective A-module. The A_M-module M is projective. ↩ 1 005O
Theorem 2 A generating module is balanced. ↩ 4 005P
Corollary 1 A free module is balanced. ↩ 4 005Q
Corollary 2 Let A be a ring, and let n be an integer geqslant 0. The center of Mₙ(A) consists of the scalar matrices with entries in the center of A. We view Aⁿ as a left Mₙ(A)-module (II, §10… ↩ 3 00R3
Remark Let M be an A-module. If the A_M-module M is generating, then we have A_M= A''_M by Theorem 2 applied to the A_M-module M, so that M is a balanced A-module. 005R
Corollary 3 Every finitely generated projective module over a commutative ring is balanced. 005S
Corollary 4 Every finitely generated module over a principal ideal domain is balanced. ↩ 1 005T
Corollary 5 Let K be a commutative field, V a finite-dimensional vector space over K, and u and v endomorphisms of V. The following properties are equivalent: 005U
Proposition 4 A semisimple module with finitely generated countermodule is balanced. ↩ 2 005V
Corollary 1 Let (Sᵢ)_i∈ I be a finite family of pairwise nonisomorphic simple A-modules. For i∈ I, denote by Dᵢ the opposite ring of the field of endomorphisms of Sᵢ. Suppose that for every i∈… ↩ 4 005W
Corollary 2 (Burnside’s theorem) Let A be an algebra over an algebraically closed commutative field K, and let S be a simple A-module that is finite-dimensional as a K-vector space. Then the mapping a↦ a_S from A… 00S0
Remark 1 The canonical description α_M of the A-module M is an isomorphism of left (A, B)-bimodules. By the corollary of Proposition 9, VIII, p. 71, the mapping f↦ (Hom(1_S_λ, f))_λ∈S_M fro… ↩ 1 00R4
Proposition 5 a) The countermodule of M is semisimple. ↩ 4 005X
Remark 2 It follows from the proof that the mapping ↩ 1 00R5
Proposition 6 a) Viewed as an (A,Bᵒ)-bimodule, M is semisimple. ↩ 3 005Y
Proposition 7 a) For every element c of the bicommutant C of M and every λ ∈ S_M, there exists a unique element c_λ of End_D_λ(S_λ) such that for every φ ∈ Hom_A(S_λ,M) and every s∈ S_λ, we have… ↩ 3 005Z
Remark 3 Endow S_λ, for λ ∈ S_M, with the C-module structure given by Proposition 7, a). If we replace A with B and B with C in Proposition 5 (VIII, p. 85), then we see that for every λ ∈ S… 0060
Proposition 8 a) The mapping z↦ (z_S_λ)_λ∈S_M is an isomorphism from Z to the product of the fields Z_λ. ↩ 4 0061
Corollary Suppose that A is an algebra over an algebraically closed commutative field K and that M is a semisimple A-module that is finite-dimensional as a vector space over K. For every λ i… 0062
Theorem 3 (Jacobson) Let M be a semisimple A-module, and let c be an endomorphism of the additive group M. Then c belongs to the bicommutant A''_M of M if and only if it satisfies the following conditi… ↩ 4 00S1
Lemma 3 Let M be a semisimple A-module. Let B be the bicommutant of the A-module M. Then every A-submodule of M is a B-submodule of M. 0063
Remark Denote the endomorphism ring of the additive group of M by E. Endow M with the discrete topology; the ring E consists of mappings from M to M, and we can endow it with the topology… 0064
Proposition 9 Let L be a field and E a subring of End_Z(L) that contains the mapping γ ₐ:x↦ ax for every a∈ L. Denote by K the set of elements a of L such that u(xa) =u(x)a for every x in L and… ↩ 1 0065
Corollary Let L be a field. Let Γ be a subgroup of the automorphism group of the field L, and let K be the field of invariants of Γ. Let V be a right K-linear subspace of L of finite dimensi… 0066
Remark When the field L is commutative, this corollary reduces to Artin’s theorem (V, §10, No. 6, p. 65). 0067

§ 6. Morita Equivalence of Modules and Algebras

Proposition 1 Suppose that the mappings b↦ b_P from B to End_A(P) and a↦ a_P from A to End_B(P) are bijective. Then σ and widetildeσ are inverse isomorphisms, and we have the relations ↩ 2 006R
Remark 1 Suppose that the mapping b↦ b_P from B to End_A(P) is bijective. Then ↩ 2 006S
Remark 2 Under the assumptions of Proposition 1, the A-module P is balanced; since the rings A_P and A'_P have the same center (VIII, p. 77), there is an isomorphism φ from the center Z(A)… ↩ 1 006T
Proposition 2 Let A and B be algebras over k, and let P be an (A,B)ₖ-bimodule. Suppose that the mapping b↦ b_P from B to End_A(P) is bijective. The following assertions are equivalent: ↩ 2 006U
Proposition 3 Let A and B be algebras over k and P an (A,B)ₖ-bimodule. The following properties are equivalent: ↩ 2 006V
Proposition 4 Let A and B be k-algebras and P an (A,B)ₖ-bimodule. Suppose that the mapping b↦ b_P from B to End_A(P) is bijective. ↩ 2 006W
Corollary 1 The countermodule of a generating module is projective and finitely generated. The countermodule of a finitely generated projective module is generating. ↩ 1 006X
Corollary 2 Let A and B be k-algebras and P an (A,B)ₖ-bimodule. The following properties are equivalent: ↩ 1 006Y
Definition 1 Let A and B be k-algebras and P an (A,B)ₖ-bimodule. We say that P is invertible if there exists a (B,A)ₖ-bimodule Q such that P⊗_BQ is isomorphic to ₛA_d and Q⊗_AP to ₛB_d. Such a… 006Z
Definition 2 Two k-algebras A and B are called Morita equivalent if there exists an invertible (A,B)ₖ-bimodule. Rings A and B are called Morita equivalent if the Z-algebras A and B are Morita e… ↩ 1 0070
Lemma 1 — Let A and B be k-algebras, P be an invertible (A, B)ₖ-bimodule, M and N be B-modules, and u : M → N be a B-linear mapping. If the mapping 1_P ⊗ u : P ⊗_B M → P ⊗_B N is zero (res… ↩ 3 0071
Theorem 1 Let A and B be k-algebras and P an (A, B)ₖ-bimodule. Denote the (B, A)ₖ-bimodule Hom_A(P, Aₛ) by P^*. The following properties are equivalent: ↩ 14 0072
Corollary 1 Let A and B be Morita equivalent k-algebras, and let P be an invertible (A,B)ₖ-bimodule. There exists an isomorphism φ from the center Z(A) of A to the center of B determined by th… ↩ 3 077O
Corollary 2 Let A and B be Morita equivalent k-algebras, and let P be an invertible (A,B)ₖ-bimodule. Every (B,A)ₖ-bimodule inverse to P is isomorphic to the dual P^*= Hom_A(P,A) of P. More pre… 077P
Remark Under the assumptions of the corollary, let q be an element of Q such that we have λ (p⊗ q) = 0 for every p∈ P. We then have τ (q) = 0, that is, q= 0. Likewise, if p is an element… 0075
Example 1 Let B be a k-algebra, n an integer geqslant 1, and A the k-algebra Mₙ(B). The right B-module P = Bⁿ_d is projective, finitely generated, and generating, and A can be identified wit… ↩ 3 0076
Example 2 Let A be a commutative k-algebra and P an A-module. We view P as an (A,A)ₖ-bimodule whose two laws of action are equal. If the (A,A)ₖ-bimodule P is invertible, then the A-module P… 00SH
Theorem 2 (Morita) a) Let V and W be left B-modules. The mapping g↦ 1_P⊗ g is a bijection from Hom_B(V,W) to Hom_A(P⊗_BV,P⊗_BW). The inverse bijection sends an element h of Hom_A(P⊗_BV,P⊗_BW) to the… ↩ 3 00S2
Proposition 5 Let V be a left B-module. The mapping W↦ P⊗_BW is an isomorphism from the set of B-submodules of V, ordered by inclusion, to the set of A-submodules of P⊗_BV, ordered by inclusion… ↩ 1 0077
Example 1 Let us apply Proposition 5 of VIII, p. 104, to the specific case V = Bₛ. ↩ 2 0078
Example 2 Denote by B_A,B_B, and B_P the sets of two-sided ideals of A, two-sided ideals of B, and (A,B)ₖ-sub-bimodules of P, respectively. ↩ 1 0079
Proposition 6 Denote by B_A and B_B the sets of two-sided ideals of A and two-sided ideals of B. ↩ 4 007A
Example 3 Let V be a left B-module. Then the correspondence given above sends the annihilator of the B-module V to the annihilator of the A-module P⊗_BV. Indeed, denote the annihilator of th… ↩ 1 007B
Example 4 For any two-sided ideal 𝔞 of A, denote the subset of Mₙ(A) consisting of the matrices with entries in 𝔞 by Mₙ(𝔞). It is a two-sided ideal of Mₙ(A). We have Mₙ(𝔞)Aⁿ=𝔞ⁿ= Aⁿ𝔞. It foll… 007C
Remark We keep the assumptions and notation above and suppose that the (B,A)ₖ-bimodule Q is the dual P^* of the A-module P and that the isomorphisms λ and θ are the canonical mappings Λ… 007D
Proposition 7 a) We have φ (0) = 0 and φ (V) = M. ↩ 2 007E
Proposition 8 a) Let V' and V'' be submodules of V such that V'' contains V'. The B-module V''/V' is simple if and only if the A-module φ (V'')/φ (V') is simple. ↩ 2 007F
Lemma 2 Let H and H' be submodules of V such that H∩ H'= 0. The B-modules H and H' are isomorphic if and only if the A-modules φ (H) and φ (H') are isomorphic. ↩ 1 007G
Proposition 9 Let S be a simple submodule of V and T the simple submodule φ (S) of M. If V_S denotes the isotypical component of V of type S and M_T the isotypical component of M of type T, then… ↩ 2 00R6
Proposition 10 a) The B-module V is Artinian (resp. Noetherian, indecomposable, simple, finitely generated) if and only if M is. ↩ 2 007H
Proposition 11 Let ↩ 1 007I
Corollary Let f: V→ W be a B-linear mapping. Then f is injective (resp. surjective) if and only if 1_P⊗ f is. 007J
Proposition 12 Let V be a left B-module. The B-module V is projective (resp. generating, faithful, *injective, finitely presented ) if and only if the 007K
Corollary The ring A is left Artinian (resp. left Noetherian) if and only if the ring B is. 007L
Proposition 13 a) If two k-algebras are Morita equivalent, then their centers are isomorphic k-algebras. ↩ 3 007M
Proposition 14 Let A and B be k-algebras. Then A and B are Morita equivalent if and only if there exist an integer ngeqslant 1 and a square matrix e= (eᵢⱼ) in Mₙ(B) satisfying the following condi… ↩ 1 007N
Lemma 3 A right B-module P is projective, finitely generated, and generating if and only if there exist an integer ngeqslant 0 and an idempotent e= (eᵢⱼ) in Mₙ(B) with the following proper… 007O
Lemma 4 Let V be a B-module and E the endomorphism k-algebra of V. Let e be a projector of V and P the image of e. The mapping that sends v∈ eEe to the endomorphism x↦ v(x) of the B-module… ↩ 2 007P
Example 1 Let A be a k-algebra and e an idempotent in A such that AeA = A. The k-algebra eAe can be identified with the endomorphism k-algebra of the submodule eA_d of A_d (Lemma 4). Since A… 007Q
Example 2 Let A be a k-algebra and ngeqslant 1 an integer. We identify the matrix algebra Mₙ(A) with the endomorphism algebra of the right A-module Aⁿ_d. We have seen that A and Mₙ(A) are Mo… ↩ 4 007R
Example 3 Let A be a principal ideal domain and L a nonzero, finitely generated, free A-module. Let B be the endomorphism ring of L; then L is an invertible (A,B)_Z-bimodule, and the rings A… 007S

§ 7. Simple Rings

Proposition 1 Let A be a nonzero ring. The following conditions are equivalent: ↩ 6 0081
Definition 1 We say that a ring A is simple if it satisfies the equivalent conditions (i), (ii), and (iii) of Proposition 1. Let K be a commutative field; a K-algebra is called simple if its un… ↩ 2 0082
Remark 1 Recall that by Theorem 1 of II, §9, No. 1, p. 115, the following properties are equivalent: ↩ 1 0083
Remark 2 We sometimes say that a ring A is quasi-simple if it is not reduced to 0 and if its only two-sided ideals are 0 and A. We say that A is primitive if it admits a faithful simple mod… ↩ 3 0084
Theorem 1 (Wedderburn) A ring is simple if and only if it is isomorphic to a matrix ring Mᵣ(D), where rgeqslant 1 is an integer and D a field. ↩ 13 00S3
Lemma 1 Let A be a simple ring, S a simple left A-module, and D the opposite ring of the field End_A(S). Then S is an invertible (A,D)-bimodule. It is also a finite-dimensional right vecto… ↩ 5 0085
Lemma 2 Let D be a field and V a right vector space of finite dimension rgeqslant 1 over D. Then V is a simple module over the ring E = End_D(V), and its commutant is equal to D_V. The rin… ↩ 6 00R7
Remark 3 Let A be a simple ring, S a simple A-module, and D the opposite ring of the field End_A(S). Then the A-module Aₛ has finite length, and dim_D(S) is equal to long(A). Indeed, by Lem… 0086
Corollary 1 a) The center of a simple ring is a field. ↩ 6 0087
Corollary 2 Let r and r' be strictly positive integers, and let D and D' be fields. The rings Mᵣ(D) and M_r'(D') are isomorphic if and only if we have r=r' and the fields D and D' are isomorph… ↩ 2 0088
Corollary 3 Let K be a commutative field, and let A be a K-algebra of finite degree with simple underlying ring. There exist an integer r and a K-algebra D of finite degree over K that is a fi… ↩ 2 0089
Remark 4 Let K be an algebraically closed commutative field, and let A be an algebra of finite degree over K. The algebra A is simple if and only if there exists an integer ngeqslant 1 such… ↩ 1 008A
Lemma 3 Let A be a simple ring, and let S be a simple A-module. Denote the opposite field of the commutant of S by D. Every A-module is isomorphic to an A-module of the form S⊗_DV, where V… ↩ 2 008B
Proposition 2 Let A be a simple ring and S a simple A-module. ↩ 5 008C
Corollary Let k be a commutative field, and let A be a simple k-algebra of finite degree over k. Every simple A-module is finite-dimensional over k. Two A-modules are isomorphic if and only… 008D
Proposition 3 Let A be a simple ring. An A-module M is free if and only if its length is divisible by the length of A. If that is the case, then all bases of M have the same cardinal, denoted by… ↩ 1 008E
Proposition 4 Let A be a simple ring and M a nonzero A-module. Denote the endomorphism ring of the A-module M by B, and view M as a left B-module. ↩ 3 008F
Proposition 5 Let B be a ring, A a simple subring of B, and S a simple left A-module. Then B is a free left A-module of dimension long_A(B⊗_AS). ↩ 3 008G
Definition 2 Let B be a ring and A a simple subring of B. The dimension of the free left A-module B is called the (left) degree of B over A and is denoted by[^1] [B : A]ₛ. ↩ 1 008H
Proposition 6 Let C be a ring, B a simple subring of C, and A a simple subring of B. We then have [C : A]ₛ= [C : B]ₛ[B : A]ₛ. ↩ 2 008I
Remark 1 Suppose that A is a simple subring of a simple ring B and that the right degree [B : A]_d is finite. Let C be the endomorphism ring of B viewed as a right A-module; it is a simple… 008J
Remark 2 Let K be a commutative field. If A is a simple subalgebra of an algebra B of finite degree over K, then the left degree of B over A satisfies the relation [B : A]ₛ[A : K] = [B : K]… 008K
Proposition 7 a) The mapping W↦ 𝔞(W) is a bijection from the set of linear subspaces of V to the set of left ideals of A. 008L
Corollary a) The minimal left ideals of A are the ideals 𝔞(H), where H is a hyperplane in V. The maximal left ideals of A are the ideals 𝔞(L), where L is a line in V. 008M

§ 8. Semisimple Rings

Theorem 1 (Wedderburn) Let A be a ring. The following properties are equivalent: ↩ 4 00S4
Definition 1 We say that a ring A is semisimple if it has the equivalent properties (i) through (vi) of Theorem 1. An algebra A over a commutative ring k is a semisimple algebra if the ring und… 0093
Proposition 1 Let A be a semisimple ring. There exists a finite family (𝔪ᵢ)_i∈ I of minimal left ideals of A such that Aₛ=⊕_i∈ I𝔪ᵢ. If (𝔪ᵢ)_i∈ I is such a family, then every simple A-module is i… ↩ 5 0094
Example Let G be a finite group and K a commutative field. *We will see further on (VIII, p. 401, Corollary 1) that the algebra K[G] of the group G over the field K is a semisimple ring if… 0095
Remark 1 Let K be a commutative field, and let A be a semisimple algebra over K. Then there exist K-algebras D₁, . . . ,Dₛ that are fields and integers r₁geqslant 1, . . . , rₛgeqslant 1 su… ↩ 1 0096
Remark 2 Let K be an algebraically closed field, and let A be an algebra of finite degree over K. By Remark 4 of VIII, p. 122, the algebra A is semisimple if and only if there exist integer… 0097
Proposition 2 a) The center of a semisimple ring is semisimple. ↩ 3 0098
Proposition 3 Let A be a commutative ring. The following properties are equivalent: ↩ 4 0099
Proposition 4 Let A be a ring. The following properties are equivalent: ↩ 7 009A
Lemma 1 Let A be a left Artinian ring, and let M be a simple A-module. Then the ring A_M is simple. ↩ 2 009B
Proposition 5 Suppose that the ring A is semisimple. Let M be a left A-module. The countermodule of M is finitely generated, and we have A_M= A''_M. ↩ 3 009C
Proposition 6 Let M be a finitely generated semisimple A-module. Denote its support (VIII, p. 66) by S_M and its endomorphism ring by B. For every λ ∈ S_M, choose a simple A-module S_λ of class… ↩ 3 009D
Proposition 7 Let A be a semisimple ring. ↩ 1 009E
Proposition 8 a) For every λ ∈ S,S_λ is a finite-dimensional right vector space over the field D_λ. When passing to the quotient, the mapping a↦ a_S_λ defines a ring isomorphism from A/𝔟_λ to En… ↩ 4 009F
Example Let K be an algebraically closed commutative field, and let A be a semisimple algebra of finite degree over K. Let (Vᵢ)_i∈ I be a family of simple A-modules such that every simple… 009G
Proposition 9 a) Order the set 𝔓(S) of subsets of S and the set B_A of two-sided ideals of A by inclusion. The mapping Λ ↦ 𝔞_Λ is an isomorphism of ordered sets from 𝔓(S) to B_A. ↩ 3 009H
Corollary Let (Aᵢ)_i∈ I be a finite family of simple rings and f an isomorphism from A to ∏_i∈ IAᵢ. For every i∈ I, there exists a unique element φ (i) of S such that the kernel of prᵢ∘ f is… 009I
Proposition 10 Denote the center of A by Z. For λ ∈ S, let Z_λ be the center of the field D_λ. ↩ 1 009J
Corollary Let B be a ring. The following properties are equivalent: ↩ 1 009K
Proposition 11 Let λ ∈ S. The isotypical component 𝔞_λ of A is both the isotypical component of Aₛ of type S_λ and the isotypical component of A_d of type S^*_λ. Moreover, we have ↩ 2 009L
Scholium Let A be a semisimple ring and Z its center. There exist canonical bijections between the following sets: 009M
Proposition 12 Let M be a module over the semisimple ring A and S_M⊂ S the support of M. Then the annihilator Ann(M) of M is the two-sided ideal ∑_λ∈S-S_M𝔞_λ, and the trace ideal τ (M) of M is th… 009N
Corollary Let M be a module over the semisimple ring A. The following properties are equivalent: 009O
Proposition 13 a) A left ideal 𝔞 of A admits a supplement in Aₛ if and only if there exists an idempotent e in A such that 𝔞= Ae. The ideal 𝔞 then consists of the elements x of A such that x=xe. ↩ 1 009P
Remark 1 Let 𝔞 be a left ideal of A. By the proof above and the corollary of Proposition 12 of II, §1, No. 8, p. 209, the mapping e↦ A(1-e) defines a bijection from the set of idempotents e… 009Q
Remark 2 Let e and f be idempotents in A. By Proposition 13, b), we have Ae= Af if and and only if ef=e and f e=f. Consequently, if the ring A is commutative, then the relation Ae= Af is eq… 009R
Proposition 14 Let e be an idempotent in A. ↩ 1 009S
Remark 3 Applying the previous results to the opposite ring of A, we see, in particular, that a monogenous right A-module M is projective if and only if there exists an idempotent e in A su… 00R8
Proposition 15 Suppose that the ring A is semisimple. ↩ 2 009T
Remark 4 Suppose that the ring A is semisimple. Let (eᵢ)_i∈ I be a partition of 1 consisting of nonzero idempotents in the center Z of A. If Card(I) = Card(S), then the eᵢ are the indecompo… ↩ 1 009U

§ 9. Radical

Definition 1 Let A be a ring. The radical of an A-module M is the submodule defined as the intersection of the maximal submodules of M (VIII, p. 48, Definition 2) or, equivalently, the set of e… ↩ 1 00A1
Example 1 Let 𝔞 be a left ideal of A. The radical of the A-module Aₛ/𝔞 is equal to 𝔞'/𝔞, where 𝔞' is the intersection of the maximal left ideals of A containing 𝔞. In particular, the radical… ↩ 1 00A2
Example 2 Let A be a principal ideal domain that is not a field, and let K be its field of fractions. As a K-module, K is without radical. Let us prove that the radical of K, viewed as an A-… 00A3
Proposition 1 Let M and N be A-modules and f a homomorphism from M to N. We have f(ℜ(M))⊂ ℜ(N), and we even have equality if f is surjective and the kernel of f is contained in the radical of M. ↩ 3 00A4
Corollary 1 Let M be an A-module and N a submodule of M. ↩ 5 00A5
Corollary 2 Let (Mᵢ)_i∈ I be a family of A-modules, P its product, and S its direct sum. We have ℜ(P)⊂ ∏_i∈ Iℜ(Mᵢ) and ℜ(S) =⨁_i∈ Iℜ(Mᵢ). ↩ 3 00A6
Proposition 2 Let M be a finitely generated A-module. ↩ 4 00A7
Corollary Let M be an A-module, (xᵢ)_i∈ I a generating family of M, and x an element of M. The following properties are equivalent: ↩ 1 00A8
Proposition 3 a) A semisimple module is without radical. ↩ 5 00A9
Definition 2 The Jacobson radical (or simply radical) of a ring A, denoted by ℜ(A), is the radical of the A-module Aₛ, that is, the intersection of the maximal left ideals of A. ↩ 1 00AA
Proposition 4 A ring A is semisimple if and only if it is left Artinian and without radical. ↩ 4 00AB
Example 1 If A is a local ring, then it has a unique maximal left ideal 𝔯, consisting of the noninvertible elements of A (VIII, p. 25, Proposition 1); so 𝔯 is the radical of A. In particular… ↩ 2 00AC
Example 2 Let K be a commutative field and E the algebra K[[Xᵢ]]_i∈ I of formal power series in the variables Xᵢ with coefficients in K. By the previous example and Example 4 of VIII, p. 26… ↩ 2 00AD
Example 3 Suppose that A is a principal ideal domain, and let P be a system of representatives consisting of irreducible elements (VII, §1, No. 3, p. 3). If A is a field, then it is without… ↩ 1 00AE
Proposition 5 a) The radical of a ring A is the intersection of the annihilators of simple A-modules; it is also the smallest annihilator of a semisimple A-module. It is, in particular, a two-si… ↩ 7 00AF
Theorem 1 (Jacobson) The radical of a ring A consists of the elements x∈ A such that 1 +ax is left invertible (I, §2, No. 3, p. 15) for every a∈ A. It is also the largest two-sided ideal 𝔞 such that 1… ↩ 10 00S5
Corollary 1 The radical of A is equal to the radical of the opposite ring Aᵒ, that is, to the intersection of the maximal right ideals of A. ↩ 1 00AG
Corollary 2 An element of A is invertible if and only if its canonical image in the ring A/ℜ(A) is invertible. ↩ 1 00AH
Corollary 3 The radical of the product of a family (Aᵢ)_i∈ I of rings is the product of the ℜ(Aᵢ). ↩ 2 00AI
Corollary 4 The ring A is local if and only if the ring A/ℜ(A) is a field. If this is the case, then ℜ(A) is the set of noninvertible elements of A. ↩ 1 00AJ
Example 4 Let K be an integral domain, I a nonempty set, and A the polynomial ring K[Xᵢ]_i∈ I. Let us prove that the ring A is without radical. The only invertible elements of A are the inve… 00AK
Example 5 Let 𝔞 be a two-sided ideal of A. The 𝔞-adic topology on A is the topology, compatible with the ring structure of A, for which the ideals 𝔞ⁿ (for ngeqslant 1) form a fundamental sys… 00AL
Remark 1 By Theorem 1, every left nil ideal of a ring A is contained in ↩ 2 00AM
Remark 2 Let A be a commutative ring. The set of nilpotent elements a of A is an ideal 𝔑(A) of A, called the nilradical of A; it is the intersection of the prime ideals of A (V, §15, No. 1… ↩ 1 00AN
Proposition 6 For every A-module M, we have ℜ(A)M⊂ ℜ(M). We have equality if the A-module M is projective. ↩ 6 00AO
Theorem 2 (“Nakayama’s lemma”) Let M be an A-module and 𝔞 a two-sided ideal of A. Suppose that one of the following conditions is satisfied: ↩ 5 00S6
Corollary 1 We keep the assumptions of Theorem 2. Let (xᵢ)_i∈ I be a family of elements of M, and let overlinexᵢ be the canonical image of xᵢ in M/𝔞M. If the family (overlinexᵢ)_i∈ I generates… ↩ 1 00R9
Corollary 2 We keep the assumptions of Theorem 2. Furthermore, let M' be an A-module and u: M'→ M be a homomorphism. If the homomorphism overlineu from M'/𝔞M' to M/𝔞M deduced from u by passing… ↩ 1 00AP
Lemma 1 Let a be an element of a ring A such that a-a² is nilpotent. There exists a polynomial P in X + (X-X²)Z[X] such that P(a) is idempotent in A. ↩ 2 00AQ
Proposition 7 Let 𝔞 be a two-sided nil ideal of A, and let overlinee be an idempotent in the ring A/𝔞. There exists an idempotent e in A whose canonical image in A/𝔞 is equal to overlinee. ↩ 2 00AR
Corollary 1 Let M and P be A-modules and u a surjective A-linear mapping from P to M. Suppose that P is projective and that there exists a nilpotent two-sided ideal 𝔞 of A such that the kernel… ↩ 2 00AS
Corollary 2 Let A be a ring, and let 𝔞 be a nilpotent two-sided ideal of A. If P is a projective A-module, then P/𝔞P is a projective module over A/𝔞, and the A-module P is indecomposable if an… ↩ 1 00AT
Definition 3 Let A be a ring, and let M be an A-module. A projective cover of M is a pair (P, u), where P is a projective A-module and u a surjective homomorphism from P to M such that we have… 00AU
Remark 1 For every projective A-module M, the pair (M,1_M) is a projective cover of M. 00AV
Remark 2 Suppose that (P, u) is a projective cover of the A-module M. Let (xᵢ)_i∈ I be a family of elements of P, and let P' be the submodule of P that it generates; then u(P') is generated… ↩ 2 00AW
Proposition 8 Let M and M' be A-modules, (P, u) and (P', u') projective covers of M and M', respectively, and g: M→ M' an A-linear mapping. ↩ 2 00AX
Corollary 1 Let M be an A-module. Let (P, u) and (P', u') be projective covers of M. There exists an isomorphism f from P to P' such that u=u'∘ f. ↩ 1 00AY
Corollary 2 Let (P, u) be a projective cover of the A-module M. If Q is a projective A-module and g: Q→ M a surjective linear mapping, then there exists a surjective linear mapping f: Q→ P suc… 00AZ
Proposition 9 Let M be an A-module and (P, u) a projective cover of M. Denote the radical of the ring A by 𝔯. The homomorphism overlineu: P/𝔯P→ M/𝔯M deduced from u by passing to the quotients is… ↩ 3 00B0
Corollary If an A-module M has a projective cover, then the A/𝔯-module M/𝔯M is projective. 00B1
Remark 3 Suppose that the ring A is without radical. By Proposition 9, (P, u) is a projective cover of an A-module M if and only if u is an isomorphism. Hence, projective covers can only ex… 00B2
Remark 4 Suppose that every finitely generated A-module has a projective cover; then the quotient A' of A by its radical is semisimple. Indeed, every finitely generated module over the ring… 00B3
Proposition 10 Let M be an A-module, P a projective A-module, and u: P→ M a linear mapping. Let 𝔞 be a two-sided ideal of A. Suppose that the linear mapping overlineu: P/𝔞P→ M/𝔞M deduced from u b… ↩ 2 00B4
Corollary 1 Let P be a projective A-module. Suppose that P is finitely generated or that the radical 𝔯 of A is a nilpotent two-sided ideal. Denote the canonical mapping from P to P/𝔯P by u. Th… ↩ 1 00B5
Corollary 2 Let 𝔞 be a two-sided ideal of A and M an A-module such that the (A/𝔞)-module M/𝔞M is free. Suppose that one of the following conditions is satisfied: ↩ 2 00B6
Corollary 3 Every finitely generated module over a local ring has a projective cover. 00B7
Remark 5 Let A be a local ring and 𝔯 its radical. Let M be a finitely generated A-module, P a finitely generated projective A-module, and u: P→ M a homomorphism. By Corollary 6 of VIII, p… 00B8
Proposition 11 Let A be a ring and 𝔞 a nilpotent two-sided ideal of A. Let M be a projective A/𝔞-module. There exist a projective A-module P and a surjective A-linear mapping u: P→ M with kernel… ↩ 2 00B9
Corollary Let P and P' be projective A-modules. If the modules P/𝔞P and P'/𝔞P' are isomorphic, then P and P' are isomorphic. 00RA

§ 10. Modules over an Artinian Ring

Proposition 1 Let A be a left Artinian ring. The radical 𝔯 of A is the largest nilpotent two-sided ideal of A, and the ring A/𝔯 is semisimple. ↩ 11 00C5
Corollary 1 A ring is semisimple if and only if it is left Artinian and has no nilpotent two-sided ideals other than 0. 00C6
Corollary 2 In a left Artinian ring A, the radical consists of the elements x such that ax is nilpotent for every a in A. ↩ 2 00C7
Proposition 2 Let A be a left Artinian ring. For every A-module M, the following properties are equivalent: ↩ 4 00C8
Corollary Let A be a left Artinian ring. For every A-module M, the radical of M is equal to 𝔯M. ↩ 1 00C9
Proposition 3 Let A and B be rings and f a homomorphism from A to B such that B is generated by the union of f(A) and the commutant f(A)' of f(A) in B. Let M be a left B-module. Suppose that eit… ↩ 2 00CA
Corollary Let A be a commutative ring and B an algebra over A. Suppose that either the ring A is Artinian or B is a finitely generated A-module. We then have ℜ(A)B⊂ ℜ(B). 00CB
Proposition 4 Every module over a left Artinian ring has a projective cover. ↩ 3 00CC
Proposition 5 Let A be a left Artinian ring and 𝔯 its radical. ↩ 2 00CD
Corollary With each class P of projective A-modules, associate the class φ (P) of the module P/𝔯P over the ring A/𝔯. Every class of modules (resp. of finitely generated modules) over the rin… 00CE
Proposition 6 Let A be a left Artinian ring and 𝔯 its radical. Let P be a projective A-module. ↩ 1 00CF
Example Take P equal to Aₛ. For λ ∈ S, denote the opposite field of the commutant of the simple A-module S_λ by D_λ and the dimension of S_λ viewed as a right vector space over the field D… 00CG

§ 11. Grothendieck Groups

Definition 1 We say that the set C of classes of A-modules is additive if every zero module is of type C and the direct sum of two modules of type C is of type C. We say that C is hereditary if… ↩ 1 00CS
Example 1 The set of classes of A-modules of finite length is hereditary (II, §1, No. 10, p. 212, Proposition 16). 00CT
Example 2 The set of classes of finitely generated A-modules is additive. If the ring A is Noetherian, then this set is hereditary (VIII, p. 3, Proposition 3 and VIII, p. 7, Proposition 4). 00CU
Example 3 The set of classes of finitely generated projective A-modules is additive but generally not hereditary. 00CV
Definition 2 Let φ be a mapping from C to an abelian group G (written additively); set φ (E) =φ(cl(E)) for every A-module E of type C. We say that φ is an additive function of modules (resp. a… 00CW
Example 4 Let C be the set of classes of A-modules of finite length. The mapping long_A : C → Z that sends a class of A-modules of finite length to its length is an additive function of modu… 00CX
Proposition 1 Suppose that C is hereditary. Let E and F be A-modules and u : E → F a linear mapping. a) If E or F is of type C, then so is the image of u. b) If E is of type C, then so is the ke… ↩ 1 00CY
Corollary Let (Eᵢ)_0 ≤ i ≤ n be a finite sequence of modules of type C. If there exists an exact sequence 00CZ
Proposition 2 Suppose that the set C is hereditary. Let E be an A-module, and let M and N be submodules of E. 00D0
Proposition 3 Suppose that C is hereditary. Let E be a module of type C and (Eᵢ)_0leqslant ileqslant n a composition series of E (I, §4, No. 7, p. 41). For 1leqslant ileqslant n, the module Eᵢ₋₁… ↩ 2 00D1
Proposition 4 a) The mapping E↦ [E]_C from C to K(C) is additive. ↩ 3 00D2
Example Let A be a Noetherian commutative ring, and let Σ be its spectrum. For any integer ngeqslant 0, denote by C^geqslant n the set of classes of finitely generated A-modules with suppo… 00D3
Proposition 5 Let E and F be modules of type C. The equality [E]_C= [F]_C holds if and only if there exist exact sequences of modules of type C 00D4
Proposition 6 a) The mapping E↦ [E]'_C from C to K'(C) is a weakly additive function of modules. 00D5
Lemma 1 Let E, E', and E'' be A-modules of finite length and ↩ 3 00D6
Proposition 7 Let C be a hereditary set of classes of modules such that every module of type C has finite length. Let S be the set of classes of simple modules belonging to C. Then the family ([… ↩ 2 00D7
Corollary Let E and F be semisimple modules of type C. The module E is isomorphic to F if and only if we have [E]_C= [F]_C in K(C). ↩ 1 00D8
Remark The set K(C)⁺ is the submonoid of K(C) generated by the family ([S]_C)_S∈S. 00D9
Remark 1 Let SS(A) be the hereditary set of classes of semisimple A-modules of finite length. By Proposition 7 of VIII, p. 190, the Grothendieck group K(SS(A)) is a free Z-module whose elem… ↩ 1 00DA
Remark 2 Let K'(LF(A)) be the Grothendieck group of LF(A) for direct sums (VIII, p. 188). The Krull–Remak–Schmidt theorem (VIII, p. 37) implies that K'(LF(A)) is a free Z-module with basis… ↩ 2 00DB
Remark 3 Let E be an A-module of finite length. By c), there exists a semisimple A-module E' of finite length such that [E] = [E'], and by b), such a module is defined up to an isomorphism… 00DC
Proposition 8 Let A be a principal ideal domain that is not a field, and let L be its field of fractions. There exists an isomorphism φ : R(A)→ L^*/A^* such that we have ↩ 2 00DD
Remark 4 We keep the assumptions and notation of Proposition 8. Let E be an A-module of finite length, and let a₁A, . . . , aₙA be its invariant factors (VII, §4, No. 5, p. 20). Since E is… 00DE
Remark 5 Let A be a Dedekind ring that is not a field (Comm. Alg., VIII, §2, No. 1). By reasoning as in Proposition 8, we can prove the existence of an isomorphism φ from R(A) to the group… 00SI
Lemma 2 Let M and M' be A-modules that are finite-dimensional vector spaces over K. The supports (VIII, p. 190) of M and M' are disjoint if and only if there exists an a∈ A such that a_M=… ↩ 1 00DF
Theorem 1 The homomorphism u: R_K(A)→ R_L(A_(L)) defined above is injective. Let ξ be an element of R_K(A). Then ξ is effective if and only if u(ξ ) is. ↩ 6 00DG
Lemma 3 Let S and T be two nonisomorphic simple A-modules that are finite-dimensional over K. The supports of the A_(L)-modules S_(L) and T_(L) are disjoint. 00DH
Proposition 9 Let K be a commutative ring, A a bigebra over K with counit γ, and C an additive set of classes of A-modules with the following properties: ↩ 3 00DI
Remark Under only assumptions (i) and (ii) of Proposition 9, the Grothendieck group K'(C) for direct sums (VIII, p. 188) has a unique ring structure whose multiplication satisfies [E]'_C[… 00DJ
Corollary Let A be a bigebra with counit γ over a commutative field K. Then there exists a unique ring structure on the additive group R_K(A) whose multiplication satisfies 00DK
Example 1 Let K be a commutative field. Let G be a group, and let K[G] be the algebra of the group G. We identify G with its canonical image in K[G] (III, §2, No. 6, p. 446). ↩ 3 00DL
Example 2 Let 𝔤 be a Lie algebra over a commutative field K and U(𝔤) its enveloping algebra; we identify 𝔤 with its canonical image in U(𝔤) (Lie, I, §2, No. 7, p. 39, Corollary 2). We endow… 00DM
Proposition 10 Let A be a commutative ring, and let C be an additive set of classes of A-modules satisfying the following three conditions: ↩ 1 00DN
Remark Let A be a semisimple ring. Every A-module is then semisimple and projective (VIII, p. 138, Proposition 4); we therefore have the equality 00DO
Example If every finitely generated projective A-module is free, then the rank defines an isomorphism from K₀(A) to Z. This is, in particular, the case when A is a principal ideal domain (… 00DP
Proposition 11 a) The multiplicity [f_*(T) : S] of the simple A-module S in the semisimple A-module f_*(T) is equal to the dimension of H viewed as a right vector space over D. ↩ 1 00DQ
Corollary (Frobenius reciprocity) Suppose that A and B are semisimple algebras that are finite-dimensional over a commutative field K and that f is K-linear. Then the K-vector spaces S, T, D, E, and H are finite-di… 00S7
Proposition 12 Let B be a simple ring and A a simple subring of B. Suppose that B is a finitely generated left A-module. Let M be a nonzero finitely generated left B-module. Set A'= End_A(M) and… ↩ 2 00DR

§ 12. Tensor Products of Semisimple Modules

Proposition 1 Let M₁ be an A₁-module and M₂ an A₂-module, neither reduced to 0. If the module M = M₁⊗ M₂ over the ring A = A₁⊗ A₂ is simple (resp. isotypical, semisimple), then the A₁-module M₁… ↩ 2 00E6
Proposition 2 Let S be a simple module over the ring A = A₁⊗ A₂, finite-dimensional over K. For i∈ 1,2, there exists a simple Aᵢ-module Sᵢ such that the Aᵢ-module S is isotypical of type Sᵢ. The… ↩ 1 00E7
Theorem 1 Suppose that the field K is algebraically closed. ↩ 2 00E8
Remark 1 Assertion a) of Theorem 1 is no longer true when the field K is not assumed algebraically closed. We can give examples (VIII, p. 225, Exercise 4) where Mᵢ is a simple Aᵢ-module tha… 00E9
Remark 2 There exists a homomorphism φ from R_K(A₁)⊗_ZR_K(A₂) to R_K(A) characterized by the relation φ ([M₁]⊗ [M₂]) = [M₁⊗ M₂]. This can be proved in the same way as Proposition 9 of VIII… ↩ 1 00EA
Lemma 1 Let M₁ and N₁ be A₁-modules, and let M₂ and N₂ be A₂-modules. We make the following assumptions: ↩ 3 00RB
Theorem 2 Let A₁ and A₂ be algebras over the commutative field K; let S₁ be a simple A₁-module and S₂ a simple A₂-module. Let D₁ and D₂ be the respective commutants of S₁ and S₂. Set M = S₁⊗… ↩ 2 00EB
Corollary 1 The module S₁⊗ S₂ over the ring A₁⊗ A₂ is semisimple (resp. isotypical, simple) if and only if the ring D = D₁⊗ D₂ is semisimple (resp. simple, a field). In particular, S₁⊗ S₂ is s… ↩ 2 00EC
Corollary 2 We have ℜ_A(M) =ℜ(D)M. The A-module M is without radical if and only if the ring D is without radical. ↩ 2 00ED
Theorem 3 Let Z₁ and Z₂ be semisimple commutative algebras over K. The radical of the ring Z₁⊗ Z₂ is equal to the set of its nilpotent elements. ↩ 3 00EE
Proposition 3 Suppose that the algebras A₁ and A₂ are semisimple, with respective centers Z₁ and Z₂. Set Z = Z₁⊗ Z₂. ↩ 2 00EF
Corollary Suppose that the algebras A₁ and A₂ are simple and that the center Z₁ of A₁ is reduced to K. Then the ring A₁⊗ A₂ has no other two-sided ideals than 0 and itself. 00EG
Proposition 4 For i∈ 1,2, let Aᵢ be a K-algebra, Mᵢ a semisimple Aᵢ-module, and Zᵢ the center of the commutant of Mᵢ. Set A = A₁⊗ A₂, M = M₁⊗ M₂, and Z = Z₁⊗ Z₂. We have ℜ_A(M) =ℜ(Z)M. If one of… ↩ 2 00EH
Lemma 2 Let A₁ and A₂ be algebras over the commutative field K. Let M₁ be an A₁-module that is finite-dimensional over K and M₂ a A₂-module of finite length. Then the A₁⊗ A₂-module M₁⊗ M₂… ↩ 1 00EI
Proposition 5 Let A₁ and A₂ be algebras over the commutative field K. Let M₁ be a semisimple A₁-module that is finite-dimensional over K and M₂ a semisimple A₂-module. For i∈ 1,2, denote the com… ↩ 3 00RC
Corollary If Z₁ or Z₂ is a separable algebra over the field K (which is, for example, the case when K is perfect), then the A-module M₁⊗ M₂ is semisimple. 00EJ
Proposition 6 Let A₁ and A₂ be nonzero K-algebras. If the ring A₁⊗ A₂ is simple (resp. semisimple), then the rings A₁ and A₂ are simple (resp. semisimple). ↩ 1 00EK
Proposition 7 Let A₁ and A₂ be semisimple K-algebras, with respective centers Z₁ and Z₂. Suppose that A₁ has finite degree over K. Then the ring A₁⊗ A₂ is left Artinian, as is its center Z₁⊗ Z₂… ↩ 4 00EL
Corollary 1 Let A₁ and A₂ be semisimple K-algebras; suppose that A₁ is finite-dimensional over K. Suppose that the center of A₁ or A₂ is a separable algebra over K, which is, for example, the… ↩ 1 00EM
Corollary 2 Let A₁ and A₂ be simple K-algebras; suppose that A₁ is finite-dimensional over K. If the center of A₁ or A₂ is equal to K, then the algebra A₁⊗ A₂ is simple. This is, in particular… ↩ 8 00RD
Proposition 8 Let A be a K-algebra, M an A-module, and L an extension of the field K. Denote the commutant of M by D and the center of D by Z. ↩ 5 00EN
Corollary 1 a) Suppose that the A-module M is semisimple, that the extension L of K is separable, and that M or L is finite-dimensional over K. Then the A_(L)-module M_(L) is semisimple. ↩ 2 00EO
Corollary 2 Let L be an extension of the field K. Denote the center of the K-algebra A by Z. 00RE
Proposition 9 Let A be a K-algebra and L a separable extension of K. ↩ 1 00EP
Proposition 10 Let A be a K-algebra and L an extension of K. Let M be an A-module. ↩ 1 00EQ
Corollary Let L be a separable extension of K. We have ℜ(A_(L)) = ℜ(A)_(L) if L is algebraic over K or if the ring A is left Artinian. 00ER

§ 13. Absolutely Semisimple Algebras

Definition 1 Let K be a commutative field and A a K-algebra. We call an A-module M absolutely semisimple if the A_(L)-module M_(L) is semisimple for every extension L of K. 00F9
Proposition 1 Let K be a commutative field and A a K-algebra. ↩ 3 00FA
Proposition 2 Let K be a commutative field, A a K-algebra, and M an A-module that is finite-dimensional over K. The following properties are equivalent: ↩ 1 00FB
Lemma 1 Let D be a field and Z its center. Denote the characteristic exponent of Z by p. Suppose that for every element a of D, there exists an integer mgeqslant 0 such that a^pᵐ belongs t… ↩ 2 00FC
Proposition 3 Let K be a separably closed field (V, §7, No. 8, p. 45, Definition 4), and let D be an algebra of finite degree over K that is a field. Then D is commutative. ↩ 3 00FD
Corollary Let K be a separably closed field and A a semisimple algebra of finite degree over K. Then there exist an integer rgeqslant 0, strictly positive integers n₁, . . . , nᵣ, and extens… 00FE
Definition 2 Let K be a commutative field. We say that a K-algebra A is absolutely semisimple if the ring A_(L) is semisimple for every extension L of K. 00FF
Theorem 1 Let K be a commutative field and A a K-algebra. The following properties are equivalent: ↩ 5 00FG
Lemma 2 Let L be an algebraically closed field and D a field containing L in its center. If D is distinct from L, then there exists an extension L' of L such that the ring D⊗_LL' is not le… ↩ 1 00FH
Lemma 3 Let A be an absolutely semisimple K-algebra and L an algebraically closed extension of K. Then the algebra A_(L) is isomorphic to a product of finitely many matrix algebras over L. 00FI
Lemma 4 Let A and B be algebras over the field K that have finite generating sets, and let K' be an extension of K. If the K'-algebras A_(K') and B_(K') are isomorphic, then there exists a… ↩ 1 00FJ
Corollary 1 Let K be a commutative field, and let A₁ and A₂ be K-algebras. Suppose that A₁ is absolutely semisimple. ↩ 3 00FK
Corollary 2 Let K be a separably closed field, and let A be an absolutely semisimple K-algebra. Then there exist an integer rgeqslant 0 and strictly positive integers n₁, . . . , nᵣ such that… 00FL
Example A commutative K-algebra is absolutely semisimple if and only if it is étale: this follows from the definition (V, §6, No. 3, p. 28, Definition 1) and the equivalence of properties… 00FM
Proposition 4 Let K be a commutative field and A a K-algebra. ↩ 1 00FN
Corollary Let K be a commutative field, let A₁ and A₂ be K-algebras, and let M₁ be an absolutely semisimple A₁-module and M₂ a semisimple A₂-module. Then M₁⊗_KM₂ is a semisimple module over… 00RF
Proposition 5 The following properties are equivalent: ↩ 7 00FO
Remark 1 Let e=∑ʳᵢ₌₁aᵢ⊗ a'ᵢ be an element of B. The conditions in item (ii) of Proposition 5 translate to the formulas ↩ 1 00FP
Remark 2 Let K be a commutative field, let A be a K-algebra, and let M be an A-module. The group End_K(M) is endowed with an (A, A)-bimodule structure defined by ↩ 1 00FQ
Theorem 2 Let K be a commutative field and A a K-algebra. The following properties are equivalent: ↩ 3 00RG
Remark 3 We know (VIII, p. 234, Corollary 1) that the tensor product of two absolutely semisimple algebras over a commutative field is absolutely semisimple. Consequently, if the algebra A… 00FR
Proposition 6 Let K be a commutative ring and A a K-algebra. Let e=∑ʳᵢ₌₁aᵢ⊗ a'ᵢ be an element of B = A⊗_KAᵒ satisfying the conditions in item (ii) of Proposition 5 of VIII, p. 236. For any integ… ↩ 2 00FS
Remark 1 The morphisms ∂ⁿ: Cⁿ(A,P)→ Cⁿ⁺¹(A,P) define a complex (C(A,P), ∂ ) of K-modules (X, §2, nᵒ1, p. 24). The mapping γₙ therefore defines a homotopy from this complex to itself linking… 00SF
Lemma 5 We have the relation 00FT
Theorem 3 Let K be a commutative ring, A a K-algebra, and P an (A,A)-bimodule. Suppose that the (A⊗_KAᵒ)-module A is projective. We then have Hⁿ(A,P) = 0 for every integer ngeqslant 1. ↩ 3 00FU
Corollary Every K-derivation from A to P is inner. 00FV
Remark 2 The assumptions of Theorem 3 are, in particular, satisfied when K is a field and A an absolutely semisimple K-algebra (VIII, p. 238, Theorem 2). 00FW
Remark 3 Suppose that the K-module A is projective. Theorem 3 can also be proved as follows. The complex (⨁_ngeqslant 0Bₙ, d) and the homomorphism ε : B₀→ A define a projective resolution o… 00FX
Proposition 7 Suppose that the (A⊗_KoverlineAᵒ)-module overlineA is projective. ↩ 1 00FY
Corollary 1 (Wedderburn’s theorem) Let K be a commutative field, A a K-algebra, and 𝔯 the radical of A. Suppose that the K-algebra A/𝔯 is absolutely semisimple. ↩ 3 00S8
Corollary 2 Let A be a commutative algebra of finite degree over a perfect field K, and let 𝔯 be its radical. There exists a unique subalgebra S of A such that A = S⊕ 𝔯. Moreover, S is isomorp… 00FZ
Remark 1 The assumption that A/𝔯 is absolutely semisimple is essential in Corollary 1 (VIII, p. 246, Exercise 4). 00G0
Remark 2 Suppose that A is an Artinian algebra over the field K. If A is commutative, then we can show (VIII, p. 180, Exercise 9) that A is isomorphic to a product of algebras A₁× ⋯ × Aₙ su… 00G1

§ 14. Central Simple Algebras

Definition 1 We say that an algebra A over the field K is central if the mapping λ ↦ λ 1 is a bijection from K to the center of A. 00GE
Remark 1 Let A and B be K-algebras. If A⊗_KB is a central simple algebra, then so are A and B (III, §4, No. 4, p. 468, Corollary of Proposition 6 and VIII, p. 221, Proposition 6). Conversel… ↩ 2 00GF
Remark 2 Let A be a K-algebra and L an extension of the field K. If the L-algebra A_(L) is central simple, then the K-algebra A is central simple. Conversely, if one of the degrees [A : K]… ↩ 2 00GG
Remark 3 Let A and B be Morita-equivalent K-algebras. The algebra A is central simple if and only if B is (VIII, p. 105, Proposition 6; p. 111, Corollary; and p. 102, Corollary 1). 00GH
Remark 4 In particular, if A is a central simple K-algebra and ngeqslant 1, then Mₙ(A) is a central simple K-algebra (VIII, p. 102, Example 1). ↩ 1 00GI
Theorem 1 Let A be a K-algebra of finite degree. The following properties are equivalent: ↩ 14 00GJ
Corollary 1 Let A be a central simple algebra of finite degree over a separably closed field K. There exists an integer ngeqslant 1 such that A is isomorphic to the matrix algebra Mₙ(K). ↩ 1 00GK
Corollary 2 Let A be a central simple algebra of finite degree over K (for example, a field of finite degree over K with center K). There exists an integer ngeqslant 1 such that [A : K] =n². 00GL
Remark 5 Let A be a central simple algebra of finite degree over K whose reduced degree is a prime number ℓ. Then either A is a field, or A is isomorphic to M_ℓ(K). Indeed, A is isomorphic… 00GM
Lemma 1 Let f and g be homomorphisms from B to A. The following conditions are equivalent: ↩ 3 00GN
Lemma 2 Suppose that B is a semisimple ring that is a finitely generated module over its center Z. Let M and N be (B,A)-bimodules. Suppose that they have finite length (which is, in partic… ↩ 3 00GO
Remark It follows from the proof of Lemma 2 that M and N are (Z,A)-bimodules of finite length. Consequently, if B and A are two semisimple rings that are finitely generated modules over t… 00GP
Theorem 2 Let B be a semisimple ring and Z its center; suppose that B is a finitely generated Z-module. Let A be a right Artinian ring, and let f and g be ring homomorphisms from B to A; wri… ↩ 3 00GQ
Corollary Let A and B be algebras over the field K. Suppose that B is central simple and of finite degree and that A is right Artinian. Let f and g be K-algebra homomorphisms from B to A. Th… 00GR
Theorem 3 (Skolem–Noether) Let A and B be simple K-algebras and Z(A) and Z(B) their centers. Suppose that the algebra B has finite degree over K and that the algebra Z(A)⊗_KZ(B) is a field (which is, in part… ↩ 1 00S9
Corollary 1 Let A be a central simple algebra over K, and let L be an extension of K of finite degree. If f and g are K-algebra homomorphisms from L to A, then there exists an inner automorphi… ↩ 1 00GS
Corollary 2 Let A be a central simple algebra over K, and let L be a subalgebra of A that is a field. Every K-algebra homomorphism from L to A extends to an inner automorphism of A. ↩ 1 00GT
Corollary 3 Let D be a field of finite degree over K with center K. Every element of D is algebraic over K. Let x and y be elements of D; there exists an element a of D^* such that y=axa⁻¹ if… 00GU
Proposition 1 Let A be a central simple algebra of finite degree over K. Let B be a K-algebra, and let f and g be algebra homomorphisms from B to A. The following properties are equivalent: ↩ 2 00GV
Theorem 4 Let A be a semisimple ring, Z its center, and u an automorphism of A. Suppose that A is a finitely generated Z-module and that we have u(z) =z for every z in Z. Then u is an inner… ↩ 2 00GW
Example Theorem 4 applies in the following two specific cases: 00RH
Proposition 2 Let L be a commutative ring and V a free L-module of finite dimension m. Suppose that every L-module M such that Mᵐ is isomorphic to Lᵐ is isomorphic to L. Then every automorphism… 00GX
Theorem 5 Let A be a central simple K-algebra, and let B be a semisimple subalgebra of A of finite degree over K. ↩ 6 00GY
Theorem 6 Let A be a central simple K-algebra, B a subalgebra of A of finite degree, and B' its commutant in A. ↩ 2 00GZ
Corollary Suppose that the field K is algebraically closed and that A is a simple algebra of finite degree over K. Let B be a simple subalgebra of A, and let B' be the commutant of B in A. T… 00H0
Lemma 3 Let A be a K-algebra and L a subalgebra of A. ↩ 6 00H1
Proposition 3 Let A be a central simple K-algebra of finite degree, and let L be a semisimple commutative subalgebra of A. The following properties are equivalent: ↩ 10 00H2
Corollary Let A be a central simple algebra of finite degree over K, and let L be a semisimple commutative K-algebra such that [A : K] is equal to [L : K]². Let f and g be injective homomorp… 00H3
Lemma 4 Let A be a central simple algebra of finite degree over K, distinct from K. There exists an étale (V, §6, No. 3, p. 28, Definition 1) subalgebra of A distinct from K. ↩ 1 00H4
Proposition 4 Let A be a central simple algebra of finite degree over K. Let L be a subalgebra of A, and let L' be the commutant of L in A. ↩ 2 00H5
Corollary 1 Let A be a central simple K-algebra of finite degree. Every semisimple (resp. étale) commutative subalgebra of A is contained in a maximal commutative subalgebra of A that is semis… ↩ 2 00H6
Corollary 2 Let D be a field of finite degree over K with center K. ↩ 6 00H7
Proposition 5 Let A be a central simple algebra of finite degree over K. Let B be a semisimple subalgebra of A, and let B' be the commutant of B. 00H8
Proposition 6 Let L be the diagonalizable subalgebra of End_D(V) associated with a direct sum decomposition V =⊕_i∈ IVᵢ. ↩ 1 00H9
Proposition 7 Let L be a commutative algebra of finite degree over K. The following properties are equivalent: ↩ 1 00HA
Theorem 7 Let A be a central simple K-algebra of finite degree and L a subalgebra of A. The following properties are equivalent: 00HB

§ 15. Brauer Groups

Proposition 1 The set C_K, endowed with the law of composition given by (α , β )↦ cl(α ⊗_Kβ ), is a commutative monoid. The identity element of C_K is the class ε of the K-algebra K. Moreover, i… 00HV
Lemma 1 Let A be a K-algebra of finite degree and D a field of finite degree over K. We have [A] = [D] in M_K if and only if there exists an integer ngeqslant 1 such that A is isomorphic t… ↩ 4 00HW
Proposition 2 The set Br(K) is the set of invertible elements of M_K, where the inverse of an element [A] of Br(K) is [Aᵒ]. Consequently, the law of composition on the monoid M_K endows Br(K) wi… 00HX
Definition 1 The abelian group Br(K) is called the Brauer group of the field K. 00HY
Lemma 2 Let I and J be finite sets, k be a commutative ring, and A and B be k-algebras. Denote by M_I(A) the k-algebra of square matrices of type (I,I) with entries in A, and define the k-… ↩ 1 00HZ
Proposition 3 Let A and B be central simple K-algebras of finite degree. The following properties are equivalent: ↩ 2 00I0
Corollary Let A and B be central simple algebras of finite degree over K. Then A and B are isomorphic if and only if they are similar and have the same degree. 00I1
Proposition 4 Let K_K be the set of classes of central K-algebras of finite degree that are fields. The mapping D↦ [D] from K_K to Br(K) is bijective. 00I2
Proposition 5 Let A be a central simple K-algebra of finite degree, and let L be an extension of K of finite degree. The following properties are equivalent: ↩ 4 00I3
Lemma 3 Let A be a central simple K-algebra of finite degree, and let L be an extension of K of finite degree that is a splitting field for A. Let V be a simple A_(L)-module, so that the n… ↩ 2 00I4
Corollary 1 Let A be a central simple K-algebra of finite degree, and let L be an extension of K of finite degree. Suppose [A : K] = [L : K]². Then L is a splitting field for A if and only if… 00I5
Corollary 2 Let D be a field of finite degree over K with center K, and let L be an extension of K of finite degree that is a splitting field for D. The reduced degree of D divides [L : K]. ↩ 2 00I6

§ 16. Other Descriptions of the Brauer Group

Lemma 1 The relation 00I9
Example 1 The external semidirect product (F×_τG, i, p) (I, §6, No. 1, p. 66, Proposition 3) is a τ-extension; we denote it by I_τ. We say that a τ-extension is semitrivial if it is isomorph… 00IA
Example 2 For i∈ 1,2, take a group Gᵢ, an abelian group Fᵢ, and a group homomorphism τᵢ from Gᵢ to the automorphism group of Fᵢ. We denote by τ₁× τ₂: G₁× G₂→ Aut(F₁× F₂) the homomorphism def… 00IB
Proposition 1 The following diagram commutes: ↩ 1 00IC
Lemma 2 Let F'₁ be an abelian group, and let w: F'₁→ F and τ₁: G'→ Aut(F'₁) be group homomorphisms such that ↩ 2 00ID
Corollary 1 Let E₁ and E₂ be τ-extensions of G by F, and let ψ be a morphism of τ-extensions from E₁ to E₂. Denote by φ₁(resp. φ₂) the canonical homomorphism for u^*(E₁) (resp. u^*(E₂)). Then… ↩ 1 00IE
Corollary 2 Let u': G''→ G' be a group homomorphism, and let E be a τ-extension of G by F. Set τ ''=τ '∘ u', and denote by φ (resp. φ ',φ '') the canonical homomorphism associated with the τ '… ↩ 1 00IF
Example Let H be a subgroup of G and j: H→ G the canonical injection. Then for every τ-extension E= (Γ , ι , π ), the τ ∘ j-extension j^*(E) is isomorphic to (overset-1π(H), ι ', π '), whe… 00RI
Proposition 2 In the notation above, the following diagram commutes: ↩ 4 00IG
Lemma 3 Let G'₁ be a group, and let w: G→ G'₁ and τ₁: G'₁→ Aut(F') be group homomorphisms such that τ '=τ₁∘ w. Let E₁'= (Γ '₁, ι '₁, π '₁) be a τ₁-extension of G'₁ by F', and let φ₁: Γ → Γ… ↩ 1 00IH
Remark Denote by Σ the species of structures of τ '-extensions, and define α-mappings to be mappings from Γ to a group Γ ' underlying a τ '-extension that are group homomorphisms and that… 00II
Corollary 1 Let E₁ and E₂ be τ-extensions of G by F, and let ψ be a morphism of τ-extensions from E₁ to E₂. Denote by φ₁(resp. φ₂) the canonical homomorphism for v_*(E₁) (resp. v_*(E₂)). Then… ↩ 1 00IJ
Corollary 2 We keep the notation of the proposition. Let F'' be an abelian group, and let τ '': G→ Aut(F'') and v': F'→ F'' be group homomorphisms such that ↩ 1 00RJ
Example 1 Let j:1 → F be the canonical injection. The semitrivial extension I_τ is isomorphic to j_*((G, i, Id_G)), where i:e → G is the canonical injection. Let c: F→ F be the constant homo… ↩ 1 00IK
Example 2 Let E be a subgroup of F stable under the action of G. Denote by F' the quotient of F by E, by v: F→ F' the canonical homomorphism, and by τ ': G→ Aut(F') the action of G on F' cha… 00IL
Proposition 3 Let G and G' be groups, and let F and F' be abelian groups. Let τ : G→ Aut(F),τ ': G→ Aut(F'),u: G'→ G, and v: F→ F' be group homomorphisms such that ↩ 1 00IM
Remark Let E₁= (Γ₁, ι₁, π₁) and E₂= (Γ₂, ι₂, π₂) be τ-extensions of G by F. Let E₁E₂= (Γ , ι , π ) be the product of these extensions. Following the example of VIII, p. 289, the construct… 00IN
Proposition 4 The product of τ-extensions endows the set Ex_τ(G,F) with the structure of an abelian group. Its identity element is the class of the semitrivial extension I_τ. The inverse of the… 00IO
Proposition 5 a) Let G and G' be groups, and let F be an abelian group. Let τ : G→ Aut(F) and u: G'→ G be group homomorphisms. The mapping u^*: Ex_τ(G,F)→ Ex_τ∘ u(G',F) is a group homomorphism. 00IP
Proposition 6 The mapping c_σ is an element of the group Z²(G,F), and its class in the cohomology group H²(G,F) depends only on the class of the τ-extension E. 00IQ
Theorem 1 The mapping Θ is a group isomorphism from Ex_τ(G,F) to H²(G,F). 00IR
Remark Let G be a group, let F and F' be abelian groups, let τ (resp. τ ') be a group homomorphism from G to the automorphism group of F (resp. F'), and let v: F→ F' be a group morphism s… 00RK
Proposition 7 In the notation from above, the following diagram commutes: 00IS
Lemma 4 If c: H× H→ F is a 2-cocycle of H with values in F, then widetildecₛ is a 2-cocycle of G with values in F. 00IT
Lemma 5 If c is a 2-coboundary, then so is widetildecₛ. 00IU
Lemma 6 Let c be a 2-cocycle of H with values in F. The image of widetildecₛ in the group H²(G,F) does not depend on the choice of the section s. 00IV
Proposition 8 Let H be a subgroup of G of finite index. The endomorphism Corᴳ_H∘ Resᴳ_H of the group H²(G,F) coincides with multiplication by the index (G : H). ↩ 1 00IW
Lemma 7 a) Let S be a subset of G such that every element of G can be written uniquely as hs with h∈ H and s∈ S. Then the mapping that sends f to (f(s))_s∈ S is an isomorphism from the K-a… ↩ 3 00IX
Remark 1 Let G be a group, H a subgroup of G, and N a subgroup of H. Let E be a (K, N)-algebra. Let α be an element of F_H(G,F_N(H,E)). We have the relation 00IY
Lemma 8 a) For every g∈ G,σ ∈ O, and 𝔪∈ σ, we have ↩ 2 00RL
Lemma 9 The morphism u is surjective with kernel 𝔞_σ. The mapping deduced from u by passing to the quotient is an isomorphism of (K,G)-algebras from L_σ to F_𝔪. ↩ 1 00IZ
Lemma 10 Let L be a (K,G)-algebra that is étale over K. The mapping ψ from L_(Ω ) to F(H,Ω ) characterized by the relation ↩ 1 00J0
Theorem 2 Let G be a finite group, and let L be a commutative K-algebra of finite degree endowed with an action of G given by an injective homomorphism λ from G to Aut_K(L). The following pr… ↩ 9 00J1
Remark 2 In Theorem 2, we can replace the assumption that Ω is algebraically closed with the assumption that Ω is separably closed. Indeed, if L is étale, then the image of every K-algebra… 00J2
Remark 3 The normal basis theorem (V, §10, No. 9, p. 73, Theorem 6) is a specific case of the implication (i)⇒(iv) in Theorem 2. 00J3
Definition 1 Let G be a finite group, and let L be a nonzero commutative algebra of finite degree over K endowed with the structure of a (K,G)-algebra, where the action of G is given by an inje… 00J4
Remark 4 Suppose that L is an extension of K endowed with an action λ of G. Then L is a Galois algebra over K if and only if L is a Galois extension of K and λ is an isomorphism from G to t… 00J5
Remark 5 Suppose that the group G is abelian. If G acts faithfully and transitively on a set X, then the stabilizer of any point of X is reduced to the identity (because the stabilizers of… 00J6
Remark 6 By V, §10, No. 1, p. 75, Lemma 5; V, §10, No. 1, p. 76, Proposition 12; and VIII, p. 137, Proposition 3, a Galois algebra is commutative and semisimple. ↩ 1 00J7
Example 1 Let n be a strictly positive integer, prime to the characteristic exponent of K. Suppose that the group μₙ of n-th roots of unity in K has order n. Then for every divisor d of n, t… 00J8
Example 2 Now suppose that the field K has characteristic pnot= 0. Let c be an element of K. The polynomial f= Xᵖ-X-c is prime to its derivative f'=-1, so the algebra L = K[X]/(f) is étale (… 00J9
Proposition 9 Let G be a finite group, H be a subgroup of G, and E be a Galois algebra over the field K with group H. Then the (K,G)-algebra Coindᴳ_H(E) deduced from E by coinduction from H to G… 00JA
Proposition 10 Let G be a finite group. Let L be a Galois algebra over the field K with group G, and let K' be an extension of K. Then the (K',G)-algebra L_(K') is a Galois algebra over K'. ↩ 1 00JB
Proposition 11 Let G₁ and G₂ be groups. Let L₁ and L₂ be Galois algebras over K with respective actions λ₁: G₁→ Aut_K(L₁) and λ₂: G₂→ Aut_K(L₂). Set L = L₁⊗_KL₂,G = G₁× G₂, and λ (g₁, g₂) =λ₁(g₁)… ↩ 1 00JC
Lemma 11 Let G₁ and G₂ be groups, and let W₁ and W₂ be K-vector spaces. We endow W₁(resp. W₂) with an action of G₁(resp. G₂) given by a group homomorphism ρ₁: G₁→ Aut_K(W₁) (resp. ρ₂: G₂→ A… ↩ 1 00JD
Remark Let L be a Galois extension of the field K of finite degree, and let G be its Galois group; we denote the identity mapping on G by λ. Then L endowed with λ is a Galois algebra over… 00JE
Proposition 12 Let B be a K-algebra, u': L→ B a K-algebra homomorphism, and v': Γ → B^* a group homomorphism. We assume satisfied the relations ↩ 2 00JF
Remark Let σ : G→ Γ be a section of the mapping π. Set ε_g=v(σ (g)) for every g in G, and denote by c_σ the 2-cocycle associated with σ (VIII, p. 296). In particular, we have 00JG
Proposition 13 There exists a unique isomorphism of K'-algebras φ from A[E; L]_(K') to A[E'; L'] that satisfies the relations 00JH
Theorem 3 Let E= (Γ , ι , π ) be a τ-extension of G by L^*. The algebra A[E; L] is central simple and of degree n² over K. Moreover, the canonical homomorphism u from L to A[E; L] is an isom… ↩ 1 00JI
Lemma 12 a) There does not exist any ideal of L, other than 0 and L that is invariant under G. 00JJ
Theorem 4 Let A be a central simple algebra of finite degree over K, and let L be a maximal commutative subalgebra of A. Then there exists a τ-extension E of G by L^* such that A is isomorph… ↩ 1 00JK
Definition 2 Let S be the set of maximal ideals of L. We define 00JL
Theorem 5 There exists a group isomorphism 00JM
Remark 1 If L is an étale algebra over K and G is the automorphism group of L, then it is not always true that the algebra A[E; L] is central simple (for example, we can take L = Kⁿ and G =… 00JN
Remark 2 We can calculate a 2-cocycle c associated with an algebra A split by a finite Galois extension L with group G as follows. First, there exists a K-homomorphism φ : A→ Mₘ(L), where [… 00JO
Corollary The mapping Φ_L/K= Θ ∘ Ψ⁻¹ defines a group isomorphism from Br(L/K) to H²(G,L^*). 00JP
Proposition 14 The following diagram commutes: ↩ 1 00JQ
Theorem 6 Let K be a commutative field, and let A be a central simple A-algebra of finite degree over K. Let L be a separable extension of K of finite degree that is a splitting field for th… ↩ 2 00JR
Corollary 1 The exponent of a central simple algebra of finite degree over a commutative field divides its index. 00JS
Corollary 2 Let K be a commutative field, and let A be a central simple algebra of finite degree over K. Let p be a prime number. If p divides the index of A, then it divides the exponent of A… ↩ 1 00JT
Corollary 3 Let p be a prime number and let K be a perfect field of characteristic p. Let A be a central simple algebra of finite degree over K. Then p does not divide the index of A. ↩ 1 00JU
Remark By considering tensor products of quaternion algebras, it is possible to construct fields with center K, exponent 2, and arbitrarily large index (cf. VIII, p. 371, Exercises 7 and… 00JV

§ 17. Reduced Norms and Traces

Proposition 1 Let L be a commutative ring, M a free L-module of finite rank m geqslant 1, and u an endomorphism of M. There exists a unique endomorphism tildeu of M satisfying the relation ↩ 2 00KD
Lemma 1 Let p be an integer such that 0 leqslant p leqslant m. For any w in ∧ᵖ(M), let hₚ(w) be the linear mapping w' ↦ w ∧ w' from ∧ᵐ⁻ᵖ(M) to ∧ᵐ(M). The linear mapping hₚ : w ↦ hₚ(w) from… 00KE
Remark 1 The endomorphism widetildeu of M coincides with what we called the cotranspose of u in III, §8, No. 6, p. 532. 00KF
Remark 2 From formulas (1), (2), (4), and (6), we deduce the relation χᵤ(u) = 0 and therefore another proof of the Cayley–Hamilton theorem (III, §8, No. 11, p. 541). ↩ 1 00KG
Remark 3 Since the mapping cᵣ from End(M) to L is a homogeneous polynomial mapping of degree r for 0leqslant rleqslant m-1, it follows from formula (6) that the mapping u↦ widetildeu from E… 00KH
Proposition 2 Let a be an element of A and Pc(a; X) its characteristic polynomial. There exists a unique monic polynomial P in K[X] such that we have Pc(a; X) = P(X)ⁿ. 00KI
Lemma 2 Let P and Q be monic polynomials in K[X] and s is a strictly positive integer. If Pˢ= Qˢ, then we have P = Q. ↩ 5 00KJ
Definition 1 Let a be an element of the algebra A. The reduced characteristic polynomial of a (with respect to A) is the unique monic polynomial in K[X], denoted by Pcrd_A/K(a; X), that satisfi… ↩ 1 00RM
Definition 2 We call Trd_A/K(a) the reduced trace of A and Nrd_A/K(a) its reduced norm (with respect to the K-algebra A). ↩ 2 00KK
Proposition 3 An element a of A is invertible if and only if its reduced norm is nonzero. In particular, A is a field if and only if we have Nrd_A/K(a)not= 0 for every nonzero element a of A. ↩ 2 00KL
Remark Let L be the field K(X) of rational fractions in one variable X. The reduced characteristic polynomial of an element a of A is simply the reduced norm of the element X⊗ 1-1⊗ a of t… 00KM
Example 1 By Theorem 1 of VIII, p. 120, there exist an integer rgeqslant 1 and a field D such that A is isomorphic to Mᵣ(D). Let d be the reduced degree of D over K; we have r=n/d. Let M be… 00RN
Example 2 Consider the specific case when A is the algebra End_K(V) of endomorphisms of a finite-dimensional K-vector space V. Taking M to be the simple A-module V, we obtain the relations ↩ 1 00RO
Proposition 4 Let L be an extension of K and a an element of the central simple algebra A. We have the relations ↩ 2 00KN
Corollary 1 Let L be an extension of K. Let V be a vector space of dimension n over L and θ a K-algebra homomorphism from A to End_L(V). For every element a of A, we have ↩ 4 00KO
Corollary 2 Let a and a' be elements of A and λ an element of K. We have the relations 00KP
Corollary 3 Let Aᵒ be the opposite algebra of A. For every a in A, we have 00RP
Proposition 5 For any x in A, let tₓ be the linear form y↦ Trd_A/K(xy) on A. ↩ 2 00KQ
Corollary The linear subspace of A generated by the elements of the form xy-yx, where x and y run through A, is a hyperplane, the kernel of the nonzero linear form Trd_A/K. 00KR
Remark By formula (23) of VIII, p. 340, we have 00KS
Lemma 3 Let L be an extension of K, and let I be a set and T= (Tᵢ)_i∈ I a family of variables. We have K(T)∩ L[T] = K[T]. ↩ 1 00KT
Proposition 6 For every integer r such that 0leqslant rleqslant n, the mapping bᵣ from A to K is a homogeneous polynomial mapping of degree r. In particular, the reduced norm is a homogeneous po… ↩ 2 00KU
Lemma 4 Let u be the element ∑_i∈ ITᵢ⊗ eᵢ of the central simple K(T)-algebra A_(K(T)). Let P be the reduced characteristic polynomial of u. Then P belongs to the ring K[T][X]; viewed as an… ↩ 1 00KV
Lemma 5 For every extension K' of K and every element (tᵢ)_i∈ I of K'ᴵ, we have ↩ 1 00KW
Remark Let K' be a commutative K-algebra. Every element t of A_(K') can be written as ∑_i∈ Itᵢ⊗ eᵢ, where (tᵢ)∈ K'ᴵ. It follows from the proof of Lemma 5 that the characteristic polynomia… 00KX
Proposition 7 Let L be a maximal commutative semisimple subalgebra of A and a an element of L. We have ↩ 1 00KY
Corollary Let D be a field of finite degree over K with center K. Let a be an element of K and L a maximal commutative subfield of D containing a. We have 00KZ
Proposition 8 Let B be a simple subalgebra of A. Denote the center of B by L and the commutant of B in A by B'. Then B' is a central simple algebra over the field L; we denote its reduced degree… ↩ 2 00L0
Lemma 6 Let K' be a commutative algebra of finite degree d over K and 00L1
Proposition 9 Let V be a finite-dimensional right vector space over the field D. Let E be the algebra End_D(V) over the field K; it is central simple and of finite degree. For every invertible e… 00L2
Remark We have Nrd_E/K(u) = 0 if the element u of E is not invertible (VIII, p. 340, Proposition 3). 00L3

§ 18. Simple Algebras over a Finite Field

Theorem 1 (Chevalley–Warning) Let K be a finite commutative field of characteristic p. Let n be an integer geqslant 1 and (fᵢ)_i∈ I a finite family of nonzero elements of K[X₁, . . . ,Xₙ]. Denote by Z the set o… ↩ 2 00SA
Lemma 1 Let L be a field, G a finite group, and χ a nontrivial homomorphism from G to the multiplicative group L^*. We have ∑_x∈ Gχ (x) = 0. ↩ 1 00LB
Lemma 2 Let q be the cardinal of K. For any integer mgeqslant 0, set Sₘ= ∑_x∈ Kxᵐ. We have Sₘ=-1 if m is a nonzero multiple of q-1 and Sₘ= 0 in all other cases. 00LC
Corollary Let V be a vector space of finite dimension n over K and I a finite set, and for each i ∈ I, let Fᵢ : V → K be a homogeneous polynomial mapping of degree dᵢ > 0. If we have ∑_i ∈ I… 00LD
Theorem 2 (Wedderburn) Every finite field is commutative. ↩ 4 00SB
Corollary 1 Every finite simple ring is isomorphic to a matrix ring Mₙ(L), where n is a strictly positive integer and L is a finite commutative field. 00LE
Corollary 2 Let K be a finite commutative field. Every central simple algebra over K is isomorphic to a matrix algebra Mₙ(K), where n is a strictly positive integer. ↩ 2 00LF
Remark 1 Here is another proof of Theorem 2. Let D be a finite field, let K be its center, and let L be a maximal commutative subfield of D. Let x be an element of D^*. It belongs to a maxi… 00LG
Remark 2 Let L be a commutative field with the following property: (C₁) Let V be a vector space of finite dimension n over the field L, and let d be such that 0< d < n. For every homogeneou… ↩ 1 00LH
Remark 3 Suppose that the field K satisfies the following condition: 00LI

§ 19. Quaternion Algebras

Proposition 1 The characteristic polynomial of an element q of F is equal to (X²-T_F(q)X + N_F(q))². ↩ 1 00LQ
Remark 1 Suppose that the characteristic of K is different from 2, and set i'= 2i-β. Then (1, i') is a basis of E over K, and we have i'²= 4α +β². It follows that E is isomorphic to the qua… ↩ 1 00LR
Remark 2 The quaternion algebra of type (α ,0, γ ) is isomorphic to the quaternion algebra of type (γ ,0, α ) (III, §2, No. 5, p. 445). It is also isomorphic to the quaternion algebra of ty… ↩ 1 00LS
Remark 3 Let q be an element of F. Then q is nilpotent if and only if its characteristic polynomial is equal to X⁴, that is, T_F(q) = N_F(q) = 0; we then have q²= 0. ↩ 2 00LT
Example The matrix algebra M₂(K) is isomorphic to the quaternion algebra of type (0,1,1). Indeed, consider the quadratic algebra E = K× K (of type (0,1)) and the quaternion algebra F = E +… 00LU
Proposition 2 a) Suppose that the field K has characteristic different from 2. If γ or 4α +β² is nonzero, then the center of F is equal to K; otherwise, it has dimension 2 and is generated by 1… ↩ 3 00LV
Proposition 3 Let α ,β ,γ be elements of K, and let f be a quaternion algebra of type (α , β , γ ). Denote its Cayley trace and norm by T_F and N_F. The following properties are equivalent: 00LW
Proposition 4 Let A be a central simple algebra of degree 4 over K. Denote the characteristic of K by p. ↩ 2 00LX
Corollary 1 Let A be a central simple K-algebra of finite degree >1 whose elements are all algebraic of degree leqslant 2 over K. Then A is isomorphic to a quaternion algebra over K. ↩ 1 00LY
Corollary 2 Let (E, s) be a Cayley algebra over K such that the K-algebra E is central simple of finite degree >1 over K. Then E is isomorphic to a quaternion algebra over K. 00LZ
Proposition 5 The following properties are equivalent: 00M0
Remark Suppose that the quaternion algebra F is a field. It follows from the equality j²=γ that we have γ not= 0. By Proposition 2 of VIII, p. 363, the center of F is equal to K unless K… 00RQ
Theorem 1 Let D be an R-algebra of finite degree that is a field. Then D is isomorphic to R, C, or H. ↩ 1 00M1
Remark 1 Let O be the octonion algebra of type (-1,0,-1,-1) over R (III, Appendix, No. 3, p. 615). Let D be an alternative Cayley algebra over R such that every nonzero element of D has an… ↩ 1 00M2
Remark 2 The above applies to the field R of real numbers. Every R-algebra of finite degree that is a field is isomorphic to R,C, or H. 00M3
Remark 3 Let A be a normed algebra over the field R. Suppose that A is a field. Then A is isomorphic to R,C, or H (“Gelfand–Mazur theorem”) (cf. Comm. Alg., VI, §6, No. 4, p. 407, Theorem 1… 00M4

§ 20. Linear Representations of Algebras

Definition 1 Let M be a K-module. A linear representation of the algebra A in M is a K-homomorphism π from A to the algebra End_K(M). ↩ 1 00MD
Proposition 1 a) The mapping φ induces a bijection from the set of left (resp. right, two-sided) ideals of A of finite codimension to the set of right A-submodules (resp. left A-submodules, sub-… ↩ 1 00ME
Definition 2 The restricted dual ↩ 1 00MF
Remark 1 Suppose that M is finite-dimensional over K. Let (e₁, . . . , eₙ) be a basis of V, and let (e^*₁, . . . , e^*ₙ) be its dual basis. Denote by (πᵢⱼ(a)) the matrix of π (a) with respe… 00MG
Remark 2 Let E be an A-module, and let E' be a submodule of E. By the corollary of Theorem 5 of II, §7, No. 5, p. 299, we have Θ_E'(A)⊂ Θ_E(A) and Θ_E/E'(A)⊂ Θ_E(A). ↩ 1 00MH
Lemma 1 The mapping c''_E is bijective. If E is finite-dimensional, then so is c'_E. 00MI
Proposition 2 Let E be a left A-module. ↩ 4 00MJ
Remark 3 Let E be a left A-submodule of A^*. Let ε be the linear form y↦ y(1) on E. For every x in E, we have x=c_E(x, ε ), so E is an A-submodule of Θ_E(A). ↩ 1 00MK
Proposition 3 Let (V, π ) be a finite-dimensional linear representation of the algebra A. The kernel 𝔞 of π is a two-sided ideal of A of finite codimension, and Θ_π(A) is the orthogonal of 𝔞 in… ↩ 1 00ML
Corollary The restricted dual Θ(A) of A is the set of coefficients of the finite-dimensional linear representations of A. 00MM
Theorem 1 a) The set Θˢˢ(A) consists of the coefficients of the finite-dimensional semisimple representations of A. It is an (A,A)-sub-bimodule of A^*. ↩ 3 00MN
Corollary 1 Every simple A-module of finite dimension over K is isomorphic to an A-submodule of Θ(A). 00MO
Corollary 2 If two simple A-modules of finite dimension over K have a common nonzero coefficient, then they are isomorphic. 00MP
Proposition 4 a) There exists a group homomorphism λ_S from S⊗_DS^* to Θ_S(A), characterized by ↩ 1 00MQ
Remark When the field K is algebraically closed, we have D = K by Theorem 1 of VIII, p. 47, and λ_S is an isomorphism of (A, A)-bimodules from S⊗_KS^* to Θ_S(A). 00MR
Proposition 5 Let S be a simple A-module of finite dimension over K, let D be the opposite field of the commutant of S, and let Z be the center of D. The following properties are equivalent: 00MS
Corollary If the field K is perfect, then properties (i) through (iii) hold. 00MT
Proposition 6 Let S₀ be the set of classes of simple A-modules of finite dimension over K with nonzero trace. The family of linear forms (Tr_S)_S∈S₀ is free over K. ↩ 2 00MU
Remark Let S_K be the set of classes of simple A-modules of finite dimension over K. Proposition 6 also follows from the fact that the sum of the Θ_S(A), for S running through S_K, is dir… 00RR
Corollary a) Suppose that the field K has characteristic 0. The homomorphisms θ and θ_K are injective. Two semisimple A-modules of finite dimension over K are isomorphic if and only if their… ↩ 2 00MV
Theorem 2 Let A be a generating subset of the K-vector space A. The homomorphism χ_A: R_K(A)→ (1 + TK[[T]])^A defined by the relation χ_A(x) = (χₐ(x))_a∈A is injective. ↩ 1 00MW
Corollary 1 Let E and F be semisimple A-modules of finite dimension over K, and let A be a generating subset of the K-vector space A. Suppose that for every a ∈ A, the characteristic polynomia… ↩ 3 00MX
Corollary 2 Let A be a central simple algebra of finite degree over K. Let B be a semisimple K-algebra, let f and g be algebra homomorphisms from B to A, and let B be a generating subset of th… 00MY
Proposition 7 A left A-submodule of A^* of finite dimension over K is contained in Θ_C(A) if and only if it is of type C. ↩ 1 00MZ
Corollary Let E be an A-module of finite dimension over K. Then E is of type C if and only if Θ_E(A) is contained in Θ_C(A). 00N0
Proposition 8 On the vector space Θ(A), there exists a unique cogebra structure (III, §11, No. 1, p. 574) such that the mapping η : A→ Θ (A)^* is a homomorphism from A to the dual algebra (III… 00N1
Lemma 2 Let c : Θ(A) → Θ(A) ⊗ Θ(A) be a K-linear mapping. Then η is a homomorphism from A to the dual algebra of the cogebra (Θ(A), c) if and only if the following diagram commutes: ↩ 1 00RS
Remark 1 Let (V, π) be a finite-dimensional linear representation of the algebra A. Let us introduce a basis (e₁, …, eₙ) of V and the dual basis (e₁^*, …, eₙ^*) of V^*. By the proof of Lemm… 00N2
Definition 3 Let C be a cogebra over the field K and c its coproduct. A subcogebra of C is any linear subspace C₁ of C such that c(C₁) is contained in the canonical image of C₁⊗_KC₁ in C⊗_KC. 00N3
Proposition 9 Let Θ be a linear subspace of Θ(A). The following properties are equivalent: ↩ 1 00N4
Remark 2 Let Θ be a subcogebra of Θ(A). We endow K with the discrete topology and the algebra A with the coarsest topology for which the mappings f: A→ K for f running through Θ are continu… 00N5

§ 21. Linear Representations of Finite Groups

Definition 1 Let M be a K-module. A linear representation of G in M is a group homomorphism from G to the linear group GL(M) (II, §1, No. 2, p. 195). 00NC
Proposition 1 Suppose that K is a commutative field. Let π and π ' be finite-dimensional semisimple K-linear representations of G. 00ND
Example 1 The unit or trivial representation of G is the representation (K, ε ), where ε (g) = Id_K for every g∈ G. Its character is the constant function with value 1. ↩ 1 00NE
Example 2 The (left) regular representation of G is the representation γ of G in K[G] defined by γ (g)(x) =gx for g∈ G and x∈ K[G]. It corresponds to the left regular representation of the a… 00NF
Example 3 Given a linear representation (M, π ) of G, the contragredient or dual representation π^∨ of π is the representation of G in the K-module dual to M defined by the relation π^∨(g) =… 00NG
Example 4 Let (M, π ) and (M', π ') be linear representations of G. In Example 1 of VIII, p. 198, we defined a K[G]-module structure on M⊗_KM'. The corresponding linear representation is cal… 00NH
Example 5 Suppose that G is the product G' × G'' of two groups. The K-linear mapping from K[G'] ⊗_K K[G''] to K[G] that sends g' ⊗ g'' to (g', g'') for g' ∈ G' and g'' ∈ G'' is an algebra is… ↩ 1 00NI
Example 6 Let (V, π) be a linear representation of G such that V is a finite-dimensional free K-module. We define a representation ρ of G × G in End_K(V) by the formula ρ(g, g')(u) = π(g') ∘… ↩ 1 00NJ
Theorem 1 Suppose that the group G is finite. Let M be a K[G]-module, and let N be a K[G]-submodule of M. Suppose that N is a direct factor of the K-module M and that |G| is invertible in K… ↩ 3 00NK
Corollary 1 (Maschke) Suppose that the group G is finite and that K is a commutative field. The algebra K[G] is semisimple if and only if the element |G| of the field K is not zero. ↩ 2 00SC
Corollary 2 Suppose that the group G is finite and that |G| is invertible in K. An exact sequence of K[G]-modules splits if and only if it splits as an exact sequence of K-modules. ↩ 1 00NL
Corollary 3 Suppose that the group G is finite and that |G| is invertible in K. A K[G]-module is projective if and only if it is projective as a K-module. 00NM
Corollary 4 a) Suppose that the group G is finite and that K is a commutative field of characteristic 0. Two finite-dimensional linear representations of G are isomorphic if and only if they h… 00NN
Corollary 5 Suppose that the group G is finite and that K is a commutative field in which |G| not= 0. Let π and π ' be finite-dimensional linear representations of G. Then π and π ' are isomor… ↩ 1 00NO
Proposition 2 Let H be a subgroup of G, and let (M, σ ) be a linear representation of H. The canonical homomorphism ι : Indᴳ_H(σ )→ Coindᴳ_H(σ ) is injective. If the subgroup H has finite index… 00NP
Proposition 3 Let H be a subgroup of G of finite index. Let (M, σ ) be a linear representation of H. Suppose that M is a finite-dimensional free K-module. Denote by (V, π ) the representation of… ↩ 1 00NQ
Remark Let (V, π ) be a linear representation of G. By Maschke’s theorem, the K[G]-module V is semisimple. For any λ ∈ widehatG, denote by V^λ the isotypical component of type λ of the K[… 00NR
Remark We identify the algebra End_K(V_λ) with the matrix algebra M_d_λ(K) via the basis (e_λ ,j) of V_λ. The mapping F⁻¹ is an isomorphism from the algebra ∏_λM_d_λ(K) to the algebra K[G… 00NS
Proposition 4 (Orthogonality relation for characters) For λ and μ in widehatG, we have ⟨ χ_λ, χ_μ⟩_G=δ_λ μ. ↩ 3 00SG
Corollary Let π and π ' be finite-dimensional linear representations of G. In the field K, we have 00NT
Proposition 5 The family (χ_λ)_λ∈widehatG is a basis of the vector space of central functions. The number of classes of simple linear representations of G is equal to the number of conjugacy cla… ↩ 4 00NU
Proposition 6 Let f be a central function on G and u be a central function on H. We have ↩ 1 00NV
Proposition 7 Let f be a mapping from G to K. The following assertions are equivalent: 00NW
Remark Let n be the cardinal of the group G, and let μₙ(K) be the group of n-th roots of unity in K. For every g∈ G, we have gⁿ= 1; consequently, widehatG can be identified with the group… 00NX
Proposition 8 a) The homomorphism θ_G is an isomorphism from K⊗_ZR_K(G) to Z_K(G). ↩ 1 00NY
Proposition 9 Suppose that the field K has characteristic zero. Then the degree of every simple representation of G divides the cardinal of G. ↩ 4 00NZ
Lemma Let L be an extension of Q. Let A be a subring of L. Suppose that A is a finitely generated Z-module. We have A∩ Q=Z. 00O0
Proposition 10 There exists a unique mapping φ_G from S_K(G) to S_K'(G) such that λ and φ_G(λ ) are related through φ for every λ in S_K(G). Moreover, φ_G is bijective. ↩ 1 00O1
Remark 1 Suppose that K' is an extension of K and that the isomorphism φ is the mapping ξ ↦ ξ · 1; then the mapping φ_G is given by extension of scalars from K to K'. 00O2
Corollary 1 The mapping φ_G is a ring isomorphism from R_K(G) to R_K'(G). For every finite-dimensional representation π of G in a K-vector space, we have φ_G([π]) = [π'], where π' is a represe… 00O3
Corollary 2 The dimension of each simple representation of G divides the order of G. ↩ 2 00O4
Remark 2 Suppose that the group G is abelian. We saw in the remark of VIII, p. 414 that I_K(G) can be identified with the set Hom(G, μₙ(K)). Likewise, I_K'(G) can be identified with Hom(G… 00O5
Remark 3 Let π₁ and π₂ be linear representations of G in finite-dimensional vector spaces over K. For i = 1, 2, let πᵢ' be a representation related to πᵢ through φ. We have dim_K Hom_K(π₁… 00O6
Remark 4 Let H be a subgroup of G of cardinal m. The isomorphism φ restricts to an isomorphism from μₘ(K) to μₘ(K') and, consequently, a ring isomorphism φ_H from R_K(H) to R_K'(H). The fol… 00O7
Remark 5 Suppose that G is the product G'× G'' of two finite groups. We define isomorphisms φ_G' and φ_G'' as in the previous example. 00O8
Proposition 11 Let (M, π ) be a finite-dimensional linear representation of G. 00O9

App. 1. Algebras without Unit Element

Definition 1 Let A be a k-algebra and 𝔞 a left ideal of A. An element u of A such that au-a belongs to 𝔞 for every a∈ A is called a right unit modulo 𝔞. We say that the ideal 𝔞 is regular if th… 00P4
Proposition 1 Every regular proper left (resp. right) ideal of A is contained in a regular maximal left (resp. right) ideal. 00P5
Proposition 2 Let A be a commutative k-algebra. An ideal 𝔞 of A is a regular maximal ideal if and only if the pseudoring A/𝔞 (I, §8, No. 1, p. 98) is a field. 00P6
Example 1 Let V be an infinite-dimensional vector space over a commutative field K. Let Endᶠ_K(V) be the K-subalgebra of End_K(V) consisting of the endomorphisms of finite rank. Let W be a l… ↩ 1 00P7
Example 2 Let T be a locally compact space, and let C₀(T) be the commutative C-algebra of continuous mappings from T to C that tend to 0 at infinity (Gen. Top., X, §4, No. 4, p. 316); it is… 00P8
Example 3 Let L¹(R) be the convolution algebra of the locally compact group R. Recall (cf. Int., VIII, §4, No. 5, p. 38) that L¹(R) is the space of classes of functions on R that are integra… 00P9
Proposition 3 a) Let widetilde𝔞 be a left ideal of widetildeA such that widetildeA =widetilde𝔞+ A. We set 𝔞=widetilde𝔞∩ A. There exists an element u of A such that u-e belongs to widetilde𝔞. If… ↩ 2 00PA
Corollary The regular left ideals of A are the ideals of the form A∩ widetilde𝔞, where widetilde𝔞 is a left ideal of widetildeA such that widetildeA =widetilde𝔞+ A. 00PB
Proposition 4 a) Let 𝔞 be a regular maximal left ideal of A. There exists a unique left ideal widetilde𝔞 of widetildeA such that widetildeA =widetilde𝔞+ A and 𝔞=widetilde𝔞∩ A. This ideal is maxi… ↩ 1 00PC
Definition 2 We say that a pseudomodule M over A is simple if we have AM not= 0 and if 0 and M are the only sub-pseudomodules of M. ↩ 1 00PD
Proposition 5 a) Let 𝔪 be a regular maximal left ideal of A. Then the pseudomodule Aₛ/𝔪 is simple, and there exists a nonzero element of Aₛ/𝔪 with annihilator 𝔪. ↩ 1 00PE
Definition 3 The radical of the k-algebra A, denoted by Re (A), is the intersection of the regular maximal left ideals of A. 00RT
Example 1 The radical of the k-algebra Endᶠₖ(V) is reduced to 0 (VIII, p. 436, Example 1).* The same holds for the algebras C₀(T) and L¹(R). 00PF
Example 2 Let A be a pseudoring in which all elements are nilpotent. The radical of A is equal to A because A is the only regular left ideal. 00PG
Proposition 6 The radical of the algebra A is the intersection of the annihilators of the simple pseudomodules. It is, in particular, a two-sided ideal of A. 00PH
Proposition 7 The radical of A is the trace on A of the radical of widetildeA; it is also equal to the radical of the opposite algebra Aᵒ(that is, to the intersection of the regular maximal righ… ↩ 2 00PI
Remark Let x and y be elements of A; we say that x is a left adverse of y, or that y is a right adverse of x, if, in widetildeA,x-e is a left inverse of y-e, that is, if we have x+y=xy. B… 00PJ
Lemma 1 Let M be a semisimple left A-pseudomodule. Let B be the bicommutant of the widetildeA-module M. Then every A-sub-pseudomodule of M is a B-submodule of M. ↩ 1 00PK
Lemma 2 Let M be a semisimple left A-pseudomodule, and let x be an element of M. There exists an element a∈ A such that ax=x. ↩ 1 00PL
Theorem 1 (Jacobson’s density theorem) Let M be a semisimple left A-pseudomodule. Let b be an element of the bicommutant of the widetildeA-module M. Let F =x₁, . . . , xₙ be a finite subset of M. Then there exists an el… 00SD

App. 2. Determinants over a Noncommutative Field

Proposition 1 Let W be a hyperplane in V and e be a vector in V- W. Let φ be an element of Ω(W). There exists a unique element ω of Ω(V) such that we have ↩ 2 00Q1
Corollary 1 Let (e₁, . . . , eₙ) be a basis of V and t an element of D^*_ab. There exists a unique element ω of Ω(V) such that ω (e₁, . . . , eₙ) =t. ↩ 1 00Q2
Corollary 2 Let ω and ω ' be two elements of Ω(V). There exists a unique element t of D^*_ab such that ω '=tω. ↩ 2 00Q3
Remark Suppose that the field D is commutative. By definition, B(V) is a subset of Vⁿ. It is clear that the restriction to B(V) of a nonzero alternating n-linear form f: Vⁿ→ D belongs to… 00Q4
Proposition 2 a) Let u be an automorphism of V. There exists a unique element of D^*_ab, denoted by det u and called the determinant of u, such that we have ↩ 1 00Q5
Proposition 3 The mapping det is the unique homomorphism from GLₙ(D) to D^*_ab that satisfies the relations ↩ 3 00Q6
Example 1 The columns of the matrix Bᵢⱼ(λ ) are ε₁, . . . , εⱼ εᵢλ , εᵢ₊₁, . . . , εₙ. By formula (5) of VIII, p. 448, we have ↩ 2 00Q7
Example 2 Let σ be a permutation of the interval [1, n] in N, with signature ε (σ ). Let M(σ ) be the matrix of the permutation σ (II, §10, No. 7, p. 351). The columns of the matrix M(σ ) ar… ↩ 1 00Q8
Example 3 Suppose ngeqslant 1. For every invertible diagonal matrix of the form Δ = diag(d₁, . . . , dₙ), we have ΔBᵢⱼ(λ)Δ⁻¹=Bᵢⱼ(dᵢλ d⁻¹ⱼ). Let A be an element of GLₙ(D). By Corollary 1 of I… ↩ 2 00Q9
Example 4 Let D' be a field, and let u be a homomorphism from D to D'. When passing to the quotients, u defines a group homomorphism u_ab from D^*_ab to D^'*_ab. Let uₙ be the homomorphism f… 00QA
Example 5 We have GL₁(D) = D^*, and Proposition 3 shows that we have det(a) =π (a) for a in GL₁(D). 00QB
Example 6 Suppose n= 2. Let A = beginpmatrix a & b \\ c & d endpmatrix be an element of GL₂(D). The elements a and c of D are not both zero. We will give the determinant of A explicitly. ↩ 1 00QC
Example 7 Let Dᵒ be the opposite field of D. The mapping A↦ᵗA from Mₙ(D) to Mₙ(Dᵒ) satisfies ᵗ(AB) =ᵗBᵗA. Therefore, for every matrix A in GLₙ(D), the transposed matrix ᵗA is invertible in M… ↩ 1 00QD
Remark 1 Let V be a right vector space over the field D, of finite dimension n. We view the dual space V^*= Hom_D(V,D) as a right vector space over the opposite field Dᵒ of D. Let u be an a… 00QE
Remark 2 The results of Subsections 1 through 4 generalize to the case when D is a local ring (VIII, p. 459, Exercise 2). 00QF
Theorem 1 Suppose ngeqslant 2. ↩ 1 00QG
Remark If D is a field with two elements, then GL₂(D) is equal to SL₂(D), and it is a group of order 6 whose derived group has order 3. If D is a field with three elements, then the group… 00QH
Corollary Let V be a right vector space over the field D, of finite dimension ngeqslant 2. 00RU

App. 3. Hilbert’s Nullstellensatz

Theorem 1 Let A be an integral domain, K be its field of fractions, and L be a commutative K-algebra. Suppose that the A-algebra L is generated by finitely many elements and that L is a fiel… ↩ 1 00QM
Corollary 1 Let K be a commutative field, A be a commutative K-algebra generated by finitely many elements, and 𝔪 be a maximal ideal of A. ↩ 2 00RV
Corollary 2 Let K be a commutative field, n be a natural number, (Pᵢ)_i∈ I be a family of elements of K[X₁, . . . ,Xₙ], and Ω be an algebraically closed extension of K. The following condition… 00QN

App. 4. Trace of an Endomorphism of Finite Rank

Lemma 1 Suppose that the A-module F is projective. The group homomorphism θ is injective, and its image is Homᶠ_A(E,F). ↩ 1 00QO
Proposition 1 Let E be a free A-module. Let (eᵢ)_i∈ I be a basis of E, and let (e^*ᵢ)_i∈ I be its dual basis. Let u∈ Endᶠ_A(E). The family (⟨ u(eᵢ), e^*ᵢ⟩ )_i∈ I has finite support, and its sum… 00QP
Proposition 2 Let E, F be projective A-modules. Let u∈ Homᶠ_A(E,F) and v∈ Hom_A(F,E). We have the relation 00QQ
Corollary Let E be a projective A-module, u be an element of Endᶠ_A(E), and F be a projective A-submodule of E containing Im u. Denote by u_F the endomorphism of F induced by u. We have ↩ 1 00QR
Proposition 3 Let E be a projective A-module. 00QS