Book AC · English

Index of statements

The 888 statements of Commutative Algebra, by chapter and §. Pick a kind, filter by a word, or put the most cited first.

Most cited first

I. Flat modules

§ 1. Diagrams and exact sequences

Definition 1 Let A be a ring, F, G, H three right (resp. left) A-modules, f a homomorphism of F to G and g a homomorphism of G to H. The ordered pair (f, g) is called an exact sequence if overl… 0421
Proposition 1 Consider a commutative diagram of commutative groups: beginarrayccccc A & xrightarrowu & B & xrightarrowv & C \\ a downarrow & & b downarrow & & c downarrow \\ A' & xrightarrowu' &… ↩ 3 0422
Lemma 1 Consider a commutative diagram of commutative groups: 0423
Proposition 2 Suppose that in the commutative diagram (10) the sequences (u, v) and (u', u') are exact. Then: ↩ 10 0424
Remark If the groups of the diagram (10) are all (for example, right) modules over a ring A and the homomorphisms are A-module homomorphisms, it is soon verified that the homomorphism d d… 0425
Corollary 1 Suppose that the diagram (10) is commutative and the two rows are exact. Then: ↩ 2 0426
Corollary 2 Suppose that the diagram (10) is commutative and the two rows are exact. Under these conditions: ↩ 5 0427

§ 2. Flat modules

Lemma 1 Let M' xrightarrowv M xrightarroww M'' → 0 be an exact sequence of left A-modules and E a right A-module. Then the sequence ↩ 11 042F
Lemma 2 If v : M' → M is injective and v(M') is a direct factor of M, the homomorphism 1, ⊗ v is injective and its image is a direct factor of E ⊗_A M. ↩ 3 042G
Definition 1 Let A be a ring, E a right A-module and M a left A-module. E is said to be flat for M (or M-flat) if, for every left A-module M' and every injective homomorphism v : M' → M, the ho… ↩ 3 042H
Lemma 3 For a tight A-module E to be M-flat, it is necessary and sufficient that for every finitely generated submodule M' of M the canonical homomorphism ↩ 2 042I
Lemma 4 Let E be a right A-module and M a left A-module such that E is M-flat. If N is either a submodule or a quotient module of M, then E is N-flat. ↩ 3 042J
Lemma 5 Let (Mᵢ)_i ∈ I be a family of left A-modules, M = ⨁_i ∈ I Mᵢ, their direct sum and E a right A-module. If, for all i ∈ I, E is flat for Mᵢ, then E is flat for M. ↩ 1 042K
Proposition 1 Let E be a right A-module. The following three properties are equivalent: ↩ 9 042L
Lemma 6 If M' xrightarrowv M xrightarroww M'' is an exact sequence of left A-modules and if E is an M''-flat right A-module, the sequence 042M
Definition 2 A right A-module E is called flat if it has the equivalent properties of Proposition 1. ↩ 1 042N
Proposition 2 (i) Let (Eₜ)_t ∈ I be a family of right A-modules. For E = ⨁_t ∈ I Eₜ to be flat, it is necessary and sufficient that each of the Eₜ be flat. ↩ 7 042O
Proposition 3 Let A be a ring and E a right A-module. ↩ 5 042P
Example Applying Proposition 3 to the ring Z, it is seen that Q is a flat Z-module, but that Z/nZ (for n ≥ 2) is not a flat Z-module. 042Q
Proposition 4 Let E be a right A-module. The three following properties are equivalent: ↩ 3 042R
Proposition 5 Let 0 → E' xrightarrowv E xrightarroww E'' → 0 be an exact sequence of right A-modules. Suppose E'' is flat. Then, for E to be flat it is necessary and sufficient that E' be flat. 042S
Lemma 7 Let E be a right A-module, F a left A-module and F', F'' two submodules of F such that F = F' + F''. Then the intersection of the canonical images of E \otimes F' and E \otimes F''… ↩ 1 042T
Proposition 6 Let E be a right A-module and F a left A-module such that E is F-flat. For every submodule F' of F, we denote by \phi(F') the image of E \otimes F' under the canonical mapping of E… ↩ 6 042U
Remark 1 With the hypotheses of Proposition 6, E ⊗_A F' is usually identified with φ(F') for every submodule F' of F, which gives the formula E ⊗_A (F' ∩ F'') = (E ⊗_A F') ∩ (E ⊗_A F''). 042V
Proposition 7 Let E be a right A-module, E' a submodule of E, F a left A-module and F' a submodule of F. Suppose that E/E' or F/F' is flat. Then the canonical image of E' ⊗ F' in E ⊗ F is the in… ↩ 5 042W
Corollary Let E be a right A-module and E' a submodule of E. ↩ 3 042X
Remark 2 The conclusion of Proposition 7 remains true if we assume only that E/E' is F-flat or that F/F' is E-flat. ↩ 2 042Y
Proposition 8 Let A, B be two rings, E a right A-module and F an (A,B)-bimodule. Suppose that E is flat and that F is flat as a B-module. Then the B-module E ⊗_A F is flat. ↩ 4 042Z
Corollary 1 Let C be a commutative ring, E, F two flat C-modules. Then E ⊗_C F is a flat C-module. ↩ 1 0430
Corollary 2 Let ρ be a homomorphism of a ring A to a ring B. If E is a flat right A-module, the right B-module ρ^*(E) = E_(B) obtained by extending the ring of scalars to B (Algebra, Chapter I… ↩ 4 0431
Corollary 3 Let R, S be two rings and φ : R → S a ring homomorphism. If M is a flat right S-module and φ_*(S_d) is a flat right R-module, then φ_*(M) is a flat right R-module. ↩ 4 0432
Proposition 9 Let (A_α, f_βα) be a direct system of rings, A = lim A_α its direct limit, (E_α, f_βα) a direct system of right A_α-modules with the same indexing set and E = lim E_α its direct li… ↩ 2 0433
Lemma 8 (i) Every module admitting a finite presentation is finitely generated. (ii) If A is a left Noetherian ring, every finitely generated left A-module admits a finite presentation. (i… ↩ 5 0434
Lemma 9 Let A be a ring and E a finitely presented A-module. For every exact sequence ↩ 5 0435
Proposition 10 Let A, B be two rings, E a right A-module, F a right B-module and G a (B, A)-bimodule. Suppose that F is flat. Then, if E is finitely generated (resp. finitely presented), the cano… ↩ 1 0436
Proposition 11 Let A be a commutative ring, B a ring, φ a homomorphism of A to the centre of B, and E and F two A-modules. Suppose that B is a flat A-module and E is finitely generated (resp. fin… ↩ 6 0437
Proposition 12 Let A, B be commutative rings, φ a homomorphism of A to B, E₁, E₂, E₃ three A-modules, f : E₁ × E → E₃ an A-bilinear mapping and ↩ 4 0438
Corollary 1 Let A, B be two commutative rings, p : A → B a homomorphism such that B is a flat A-module and E a finitely presented A-module. For every finitely generated submodule F of E, the s… 0439
Corollary 2 Let A, B be two commutative rings, p : A → B a homomorphism such that B is a flat A-module and E an A-module. Then, for every finitely generated submodule F of E, the annihilator o… ↩ 1 043A
Remark If there is no ambiguity over the modules E, nor the bilinear mapping f, F₃ : F₂ is sometimes used to denote the module denoted by T in Proposition 12 and it is called the transpor… 043B
Lemma 10 Let (f_λ)_λ ∈ L be a family of generators of F and (e_λ)_λ ∈ L a family of elements of E of finite support. For ∑_λ ∈ L e_λ ⊗ f_λ = 0, it is necessary and sufficient that there exi… ↩ 2 043C
Proposition 13 For E to be F-flat (no. 2, Definition 1), it is necessary and sufficient that the following condition be satisfied: ↩ 2 043D
Corollary 1 For a right A-module E to be flat, it is necessary and sufficient that the following condition hold: ↩ 1 043E
Corollary 2 For the ring B to be a flat right A-module, it is necessary and sufficient that it satisfy the following condition: (RP') Every solution (yₖ)_1 ≤ k ≤ n, consisting of elements of B… ↩ 1 043F

§ 3. Faithfully flat modules

Proposition 1 Let E be a right A-module. The following four properties are equivalent: ↩ 12 0444
Definition 1 A right A-module E is called faithfully flat if it satisfies the four equivalent conditions of Proposition 1. 0445
Remark If E is a faithfully flat A-module, E is a faithful A-module: for, if an element a ∈ A is such that xa = 0 for all x ∈ E, the homothety h : b ↦ ba in A is such that 1, ⊗ h = 0; whe… 0446
Proposition 2 Let E be a faithfully flat right A-module and u : N' → N a left A-module homomorphism. For u to be injective (resp. surjective, bijective), it is necessary and sufficient that 1_E… ↩ 6 0447
Proposition 3 Let 0 → E' → E → E'' → +0 be an exact sequence of right A-modules. Suppose that E' and E'' are flat and that one of them is faithfully flat. Then E is faithfully flat. ↩ 2 0448
Proposition 4 Let R, S be two rings, E a right R-module and F an (R, S)-bimodule. Suppose that E is faithfully flat. Then, for F to be a flat (resp. faithfully flat) S-module, it is necessary an… ↩ 4 0449
Corollary Let C be a commutative ring and E and F two faithfully flat C-modules. Then the C-module E ⊗_C F is faithfully flat. 044A
Proposition 5 Let ρ be a homomorphism from a ring A to a ring B. If E is a faithfully flat right A-module, the right B-module ρ^*(E) = E_(B) = E ⊗_A B is faithfully flat. ↩ 4 044B
Corollary If E is a faithfully flat right A-module and if a is a two-sided ideal of A, the right (A/a)-module E/Ea is faithfully flat. 044C
Proposition 6 Let A be a commutative ring, B an algebra over A and ρ : a ↦ a . 1 the canonical homomorphism of A to B. Suppose that B is a faithfully flat A-module. Then, for an A-module E to be… 044D
Proposition 7 Let A, B be two rings and ρ a homomorphism of A to B. Let E be a faithfully flat right B-module. For ρ^*(E) to be a flat (resp. faithfully flat) right A-module, it is necessary and… 044E
Proposition 8 Let A, B be two rings and ρ a homomorphism from A to B. Suppose that there exists a right B-module E such that ρ_*(E) is a faithfully flat A-module. Then: (i) For every left A-modu… ↩ 5 044F
Corollary Under the hypotheses of Proposition 8, if B is left Noetherian (resp. Artinian), so is A. 044G
Proposition 9 Let A, B be two rings and ρ a homomorphism of A to B. The following five properties are equivalent: (a) The right A-module B is faithfully flat. (b) The homomorphism ρ is injective… ↩ 3 044H
Proposition 10 Let B be a ring and A a subring of B such that B is a faithfully flat right A-module. Let F be a left A-module, F', F'' two submodules of F. Then: (i) The canonical mapping B ⊗_A F… ↩ 2 044I
Proposition 11 Let B be a ring and A a subring of B such that B is a faithfully flat right A-module. For a left A-module F to be definitely generated (resp. finitely presented), it is necessary a… ↩ 2 044J
Proposition 12 Let B be a ring and A a commutative subring of the centre of B such that B is a faithfully flat A-module. For an A-module F to be projective and finitely generated, it is necessary… ↩ 3 044K
Proposition 13 Let A be a subring of a ring B. For the ordered pair (A, B) to satisfy the linear extension property, it is necessary and sufficient that B be a faithfully flat A-module. 044L

§ 4. Flat modules and "Tor" functors

Proposition 1 Let E be a right A-module. The following four properties are equivalent: 044W
Proposition 2 Let R, S be two rings, ρ : R → S a homomorphism and F a left R-module. The following two properties are equivalent: (a) Tor₁ᴿ(ρ_*(E), F) = 0 for every right S-module E. (b) The lef… ↩ 3 044X
Remark Proposition 2 also follows from the existence of the exact sequence 044Y

II. Localization

§ 1. Prime ideals

Definition 1 An ideal p of a ring A is called prime if the ring A/p is an integral domain. ↩ 1 0450
Proposition 1 Let A be a ring, a₁, a₂, …, aₙ ideals of A and p a prime ideal of A. If p contains the product a₁ a₂ … aₙ, it contains at least one of the aᵢ. ↩ 7 0451
Corollary Let m be a maximal ideal of A; for every integer n > 0, the only prime ideal containing mⁿ is m. 0452
Proposition 2 Let A be a ring, a a non-empty set of A which is closed under addition and multiplication and (pᵢ)_i ∈ I a non-empty finite family of ideals of A. Suppose that a is contained in th… ↩ 5 0453
Proposition 3 Let a and b be two relatively prime ideals of a ring A. Let a' and b' be two ideals of A such that every element of a (resp. b) has a power in a' (resp. b'). Then a' and b' are rel… ↩ 3 0454
Proposition 4 Let a, b₁, …, bₙ be ideals of a ring A. If a is relatively prime to each of the bᵢ (1 ≤ i ≤ n), it is relatively prime to b₁ b₂ … bₙ. ↩ 2 0455
Proposition 5 Let (aᵢ)_i ∈ I be a non-empty finite family of ideals of a ring A. The following properties are equivalent = (a) For i ≠ j, aᵢ and aⱼ are relatively prime. (b) The canonical homomo… ↩ 8 0456
Proposition 6 Let (aᵢ)_i ∈ I be a non-empty finite family of ideals of a ring A which are relatively prime in pairs; let a be the intersection of the aᵢ. For every A-module M, the canonical mapp… ↩ 2 0457
Example Let K be a field, a, (1 ≤ i ≤ m) distinct elements of K and, for each i, let gᵢ be a polynomial in K[X]; the principal ideal (X - aᵢ) = mᵢ is maximal in K[X], hence, for every syst… 0458

§ 2. Rings and modules of fractions

Definition 1 Let A be a ring. A subset S of A is called multiplicative if every finite product of elements of S belongs to S. 045K
Proposition 1 Let A be a ring and S a subset of A. There exists a ring A' and a homomorphism h of A to A' with the following properties: (1) the elements of h(S) are invertible in A ; (2) for ev… ↩ 8 045L
Definition 2 Let A be a ring, S a subset of A and overlineS the multiplicative subset generated by S. The ring of fractions of A defined by S and denoted by A[S⁻¹] is the quotient set of A × ov… 045M
Proposition 2 Let A, B be two rings, S a subset of A, T a subset of B and f a homomorphism from A to B such that f(S) ⊂ T. There exists a unique homomorphism f' from A[S⁻¹] to B[T⁻¹] such that f… ↩ 12 045N
Corollary 1 Let A be a ring, S a subset of A and u an injective homomorphism of A to a ring B such that the elements of u(S) are invertible in B. The unique homomorphism u' of A[S⁻¹] to B such… ↩ 3 045O
Corollary 2 Let A be a ring and S and T two subsets of A such that S ⊂ T. There exists a unique homomorphism i_A^T,S from A[S⁻¹] to A[T⁻¹] such that i_Aᵀ = i_A^T,S ∘ i_A^S. ↩ 2 045P
Corollary 3 Let A, B, C be three rings, S (resp. T, U) a multiplicative subset of A (resp. B, C), f : A → B, g : B → C two homomorphisms and h : A → C the composite homomorphism g ∘ f; suppose… ↩ 1 045Q
Corollary 4 Let S be a subset of a ring A, B a subring of A[S⁻¹] containing i_A^S(A) and S' the set i_A^S(A). Let j be the canonical injection of B into A[S⁻¹]; the unique homomorphism g from… ↩ 1 045R
Proposition 3 Let A be a ring, S a subset of A, M an A-module, M' the A-module M ⊗_A A[S⁻¹] and f the canonical A-homomorphism x ↦ x ⊗ 1 of M to M'. Then: (1) For all s ∈ S, the homothety z ↦ sz… ↩ 5 045S
Definition 3 Let A be a ring, S a subset of A, overlineS the multiplicative subset of A generated by S and M an A-module. Then the module of fractions of M defined by S and denoted by M[S⁻¹] or… 045T
Proposition 4 Let S be a multiplicative subset of A and M an A-module. For ms = 0 (m \in M, s \in S), it is necessary and sufficient that there exist s' \in S such that s'm = 0. ↩ 3 045U
Corollary 1 For ms = m's' in S^{-1}M, it is necessary and sufficient that there exist t \in S such that t(s'm - sm') = 0. ↩ 1 045V
Corollary 2 Let M be a finitely generated A-module. For S^{-1}M = 0, it is necessary and sufficient that there exists s \in S such that sM = 0. ↩ 3 045W
Corollary 3 Let M be a finitely generated A-module. For an ideal a & A to be such that aM = M, it is necessary and sufficient that there exist a ∈ a such that (1 + a)M = 0. ↩ 2 045X
Lemma 1 For every ideal a of A, the set S of elements 1 + a, where a ∈ a, is a multiplicative subset of A and the set a' of elements of S⁻¹A of the form a/s, where a ∈ a and s ∈ S, is an i… ↩ 1 045Y
Proposition 5 Let A, B be two rings, S a multiplicative subset of A, T a multiplicative subset of B and f a homomorphism from A to B such that f(S) ⊂ T. Let M be an A-module, N a B-module and u… ↩ 3 045Z
Proposition 6 Let f be a homomorphism from a ring A to a ring B and S a multiplicative subset of A. There exists a unique mapping j from (f(S))⁻¹B to S⁻¹B (where B is considered as an A-module b… ↩ 2 0460
Proposition 7 Let A be a ring and S, T two multiplicative subsets of A. Set T' = i(TA^S). ↩ 7 0461
Proposition 8 Let A be a ring and S, T two multiplicative subsets of A such that S ⊂ T. The following properties are equivalent: ↩ 1 0462
Proposition 9 Let I be a right directed preordered set, (S_α)_α ∈ I an increasing family of multiplicative subsets of a ring A and S = ⋃_α ∈ I S_α. We write ρ_β α = i^S_β, S_α_A^S_β, S_α for α ≤… ↩ 1 0463
Corollary Under the hypotheses of Proposition 9, let M be an A-module. We write f_β α = i^S_β, S_α_M for α ≤ β, f_α = i^S, S_α_M for all α ∈ I and let f'_α be the canonical mapping from S_α⁻… 0464
Theorem 1 The ring S⁻¹A is a flat A-module (Chapter I, § 2, no. 1, Definition 2). ↩ 16 0465
Proposition 10 Let M be an A-module. For every submodule N' of the S⁻¹A-module S⁻¹M, let φ(N') be the inverse image of N' under i_M^S. Then: (i) Sφ(N') = N'. ↩ 3 0466
Corollary 1 Let N₁, N₂ be two submodules of an A-module M. For S⁻¹N₁ ⊂ S⁻¹N₂, it is necessary and sufficient that the saturation of N₁ with respect to S be contained in that of N₂. 0467
Corollary 2 If M is a Noetherian (resp. Artinian) A-module, S^{-1}M is a Noetherian (resp. Artinian) S^{-1}A-module. In particular, if the ring A is Noetherian (resp. Artinian), so is the ring… ↩ 3 0468
Proposition 11 Let A be a ring and S a multiplicative subset of A. For every ideal b' of S^{-1}A, let b = (iA^S)^{-1}(b') be such that b' = S^{-1}b. ↩ 16 0469
Remark For an ideal a of A to satisfy S⁻¹a = S⁻¹A (or, what amounts to the same thing by no. 4, Theorem I, S⁻¹(A/a) = 0), it is necessary and sufficient that a ∩ S ≠ ∅, as follows immedia… 046A
Corollary 1 Let A be a ring and S a multiplicative subset of A. Every ideal p of A which is maximal among those which do not meet S is prime. 046B
Corollary 2 Let A be a ring and S a multiplicative subset of A. For every ideal a of A not meeting S there exists a prime ideal containing a and not meeting S. ↩ 4 046C
Corollary 3 Let A, B be two rings, ρ a homomorphism from A to B and p a prime ideal of A. For there to exist a prime ideal p' of B such that overlineρ⁻¹(p') = p, it is necessary and sufficient… ↩ 1 046D
Corollary 4 Let A, B be two rings and ρ a homomorphism from A to B. (i) Suppose that there exists a B-module E such that ρ_*(E) is a faithfully flat A-module. Then, for every prime ideal p of… ↩ 3 046E
Corollary 5 Let A be a ring, S a multiplicative subset of A and B a ring such that i_A^S(A) ⊂ B ⊂ S⁻¹A. Let q be a prime ideal of B such that the prime ideal p = (i_A^S)⁻¹(q) of A does not mee… 046F
Definition 4 The ideal of nilpotent elements of a (commutative) ring A is called the nilradical of A. If a is an ideal of A, the inverse image, under the canonical mapping A → A/a, of the nilra… 046G
Proposition 12 Let p be a minimal prime ideal of a ring A. For all x ∈ p, there exists s ∈ A - p and an integer k > 0 such that sxᵏ = 0. ↩ 2 046H
Proposition 13 The nilradical of a ring A is the intersection of all the prime ideals of A and it is also the intersection of the minimal prime ideals of A. ↩ 1 046I
Lemma 2 Every prime ideal of a ring A contains a minimal prime ideal of A. 046J
Remark In § 4, no. 3, Corollary 3 to Proposition 14, we shall show that in a Noetherian ring the set of minimal prime ideals is finite; moreover we shall see later that every decreasing s… 046K
Corollary 1 The radical of an ideal a in a ring A is the intersection of the prime ideals containing a and it is also the intersection of the minimal elements of this set of prime ideals. ↩ 2 046L
Corollary 2 For an ideal a of a ring A we denote by r(a) the radical of a. Then, for two ideals a, b of A, r(a ∩ b) = r(ab) = r(a) ∩ r(b); in particular, if a ⊂ b, then r(a) ⊂ r(b). ↩ 1 046M
Proposition 14 For two ideals a, b of a ring to be relatively prime, it is necessary and sufficient that their radicals r(a) and r(b) be so. 046N
Proposition 15 In a ring A let a be an ideal and b a finitely generated ideal contained in the radical of a. Then there exists an integer k > 0 such that bᵏ ⊂ a. ↩ 3 046O
Corollary In a Noetherian ring the nilradical is a nilpotent ideal. 046P
Proposition 16 Let B be a ring and A a subring of B. For every minimal prime ideal p of A there exists a minimal prime ideal q of B such that q ∩ A = p. 046Q
Definition 5 A ring A is called reduced if its nilradical is reduced to 0, in other words if no element ≠ 0 of A is nilpotent. 046R
Proposition 17 Let A be a ring and 𝔑 its nilradical. For every multiplicative subset S of A, S⁻¹𝔑 is the nilradical of S⁻¹A. In particular, if A is reduced, then S⁻¹A is reduced. 046S
Proposition 18 Let A be a ring and S a multiplicative subset of A. ↩ 2 046T
Corollary Let M be an A-module and a an ideal of A. The sub-S⁻¹A-modules (S⁻¹a)(S⁻¹M), a(S⁻¹M), (S⁻¹a)j(M) (where j : M → S⁻¹M is the canonical mapping) and S⁻¹(aM) of S⁻¹M are identical. In… ↩ 2 046U
Remark Let M, N, P be three A-modules, f : M × N → P an A-bilinear mapping and S⁻¹f : (S⁻¹M) × (S⁻¹N) → (S⁻¹P) the S⁻¹A-bilinear mapping obtained from f by extending the ring of scalars t… 046V
Proposition 19 Let A be a ring and S a multiplicative subset of A. (i) If M and N are two A-modules and M is finitely generated (resp. finitely presented), the canonical homomorphism (Chapter I… ↩ 2 046W
Proposition 20 Let A, A' be two rings, p : A → A' a homomorphism, S a multiplicative subset of A, S' = ρ(S) and ρ' : S⁻¹A → S'⁻¹A' the homomorphism corresponding to p (no. 1, Proposition 2). 046X
Proposition 21 The ring S⁻¹A with the family ((S⁻¹A)ᵢ) is a graded ring and S⁻¹M with the family ((S⁻¹M)ᵢ) is a graded module over the graded ring S⁻¹A. The canonical mappings i^S_A and i^S_M are… 046Y
Proposition 22 Let A (resp. B) be a graded ring of type A, M (resp. N) a graded module over the graded ring A (resp. B), S (resp. T) a multiplicative subset of A (resp. B) all of whose elements a… 046Z

§ 3. Local rings. Passage from the local to the global

Proposition 1 Let A be a ring and I the set of non-invertible elements of A. The set I is the union of the ideals of A which are distinct from A. Moreover, the following conditions are equivalen… ↩ 2 047R
Definition 1 A ring A is called a local ring if it satisfies the equivalent conditions (a), (b) and (c) of Proposition 1. The quotient of A by its Jacobson radical (which is then the unique max… 047S
Definition 2 Let A, B be two local rings and m, n their respective maximal ideals. A homomorphism u : A → B is called local if u(m) ⊂ n. 047T
Proposition 2 Let A be a ring and p a prime ideal of A. The ring A, is local; its maximal ideal is the ideal pAₚ = pᵥ, generated by the canonical image of p in A; its residue field is canonicall… ↩ 2 047U
Definition 3 Let A be a ring and p a prime ideal of A. The ring A, is called the local ring of A at p, or the local ring of p, when there is no ambiguity. 047V
Proposition 3 Let A be a ring, p a prime ideal of A and S = A - p. For every ideal b' of Aₚ distinct from Aₚ, let b be the ideal (i_A^S)⁻¹(b') of A so that b' = bAₚ. ↩ 4 047W
Proposition 4 Let A be a ring which is not necessarily commutative, m a right ideal of A contained in the Jacobson radical of A and M a left A-module. Suppose that one of the following condition… ↩ 6 047X
Corollary 1 Let A be a ring which is not necessarily commutative, m a right ideal of A contained in the Jacobson radical of A, M and N two left A-modules and u : M → N an A-linear mapping. If… ↩ 5 047Y
Corollary 2 Let A be a ring which is not necessarily commutative, m a two-sided ideal of A contained in the Jacobson radical of A, M a left A-module and (xᵢ)_i ∈ I a family of elements of M. I… ↩ 5 047Z
Proposition 5 Let A be a ring which is not necessarily commutative, m a two-sided ideal of A contained in the Jacobson radical of A and M a left A-module. Suppose that one of the following condi… ↩ 7 0480
Corollary 1 Let A be a not necessarily commutative ring, m the Jacobson radical of A and M a left A-module. Suppose that A/m is a field, that the canonical homomorphism from m \otimesA M to M… ↩ 3 0481
Corollary 2 Let A be a not necessarily commutative ring, m the Jacobson radical of A and M a left A-module. Suppose that A/m is a field and that one of the following conditions holds: (i) M is… ↩ 3 0482
Proposition 6 Let A be a ring which is not necessarily commutative and m its Jacobson radical; suppose that A/m is a field. Let M and N be two finitely generated free A-modules and u: M → N a ho… ↩ 5 0483
Corollary Under the hypotheses of Proposition 6 the following properties are equivalent: ↩ 1 0484
Proposition 7 Let A be a reduced local ring, m its maximal ideal, (pᵢ)_i ∈ I the family of minimal prime ideals of A, Kᵢ the field of fractions of A/pᵢ and M a finitely generated A-module. For M… ↩ 1 0485
Proposition 8 Let A be a ring, m a maximal ideal of A and M an A-module. If there exists an ideal a of A such that m is the only maximal ideal of A containing a and aM = 0, then the canonical ho… ↩ 2 0486
Proposition 9 Let A be a ring, m a maximal ideal of A, M an A-module and k an integer geqslant 0. The canonical homomorphism M → Mₘ/mᵏ Mₘ is surjective, has kernel mᵏ M and defines an isomorphis… ↩ 5 0487
Corollary Let A be a ring, m₁, m₂, …, mₙ distinct maximal ideals of A, M an A-module and k₁, k₂, …, kₙ integers geqslant 0. The canonical homomorphism from M to ∏ⁿ Mₘ/m_i₁^k₁ Mₘ is surjectiv… ↩ 1 0488
Proposition 10 The A-module ⨁_m ∈ Ω Aₘ, the direct sum of the Aₘ where m ∈ Ω, is faithfully flat. ↩ 2 0489
Theorem 1 Let M, N be two A-modules, u : M → N an A-homomorphism and, for all m ∈ Ω, let uₘ : M, → Nₘ be the corresponding Aₘ-homomorphism (§ 2, no. 2, Remark 5). For u to be injective (resp… ↩ 11 048A
Corollary 1 Let M be an A-module, N a submodule of M and x an element of M. For x ∈ N, it is necessary and sufficient that, for all m ∈ R, the canonical image of x in M, belong to N. ↩ 5 048B
Corollary 2 Let M be an A-module and, for all m ∈ Ω, let fₘ be the canonical mapping M → Mₘ. The homomorphism x ↦ (fₘ(x)) of M to ∏_m ∈ Ω Mₘ is injective. ↩ 3 048C
Corollary 3 (i) Let b be an ideal of A and a an element of A. For a ∈ b, it is necessary and sufficient that, for all m ∈ Ω, the canonical image of a in Aₘ belong to bAₘ. 048D
Corollary 4 Let A be an integral domain, K its field of fractions and M a torsion-free A-module such that M is canonically identified with a sub-A-module of K ⊗_A M. Then, for all m ∈ Ω, Mₘ is… 048E
Corollary 5 Let A be a ring. Every system of generators of the A-module Aⁿ with n elements is a basis of Aⁿ. 048F
Proposition 11 Let M be an A-module, N a finitely generated A-module and u : M → N a homomorphism. For u to be surjective, it is necessary and sufficient that, for all m ∈ S₂, the homomorphism M/… ↩ 2 048G
Proposition 12 Let E, F, G be three A-modules and v : G → F, u : E → F homomorphisms. Suppose that E is finitely presented. For there to exist a homomorphism w : E → G such that u factors into E… ↩ 3 048H
Corollary 1 Let M be an A-module and N a submodule of M such that M/N is finitely presented. For N to be a direct factor of M, it is necessary and sufficient that, for all m ∈ S₂, N, be a dire… ↩ 1 048I
Corollary 2 Let M be a finitely generated free A-module and N a submodule of M which is a finitely generated free A-module. For N to be a direct factor of M, it is necessary and sufficient tha… 048J
Proposition 13 Let S be a multiplicative subset of a ring A and M an A-module. If M is jflat (resp. faithfully flat), S⁻¹M is ajflat (resp. faithfully flat) S⁻¹A-module and a flat A-module. ↩ 4 048K
Remark If N is an S⁻¹A-module, S⁻¹N is identified with N and this is consequently equivalent to saying that N is a flat S⁻¹A-module or a flat A-module. 048L
Proposition 14 Let A be a ring, B a commutative A-algebra and T a multiplicative subset of B. If N is a B-module which is flat as an A-module, T⁻¹N is a flat A-module. ↩ 1 048M
Proposition 15 Let A, B be two rings, φ : A → B a homomorphism and N a B-module. The following properties are equivalent: (a) N is a flat A-module. (b) For every maximal ideal n of B, Nₙ is a fla… ↩ 1 048N
Corollary For an A-module M to be flat (resp. faithfully flat), it is necessary and sufficient that, for every maximal ideal m of A, M, be a flat (resp. faithfully flat) Am-module. ↩ 2 048O
Proposition 16 Let A be a ring. The following properties are equivalent: (a) the set of maximal ideals of A is finite; (b) the quotient of A by its Jacobson radical is the direct composition of a… ↩ 1 048P
Definition 4 A ring is called semi-local if it satisfies the equivalent conditions (a), (b) of Proposition 16. 048Q
Example Every local ring is semi-local. Every quotient of a semi-local ring is semi-local. Every finite product of semi-local rings is semi-local. 048R
Proposition 17 Let A be a ring and 𝔭₁, …, 𝔭ₙ prime ideals of A. We write S = ⋂₁ⁿ (A - 𝔭ᵢ) = A - ⋃₁ⁿ 𝔭ᵢ. ↩ 2 048S
Corollary Let A be an integral domain and 𝔭₁, …, 𝔭ₙ prime ideals of A, no two of which are comparable with respect to inclusion. If A = ⋂₁ⁿ A_𝔭ᵢ in the field of fractions of A, the maximal i… ↩ 1 048T

§ 4. Spectra of rings and supports of modules

Definition 1 A topological space X is called irreducible if every finite intersection of non-empty open sets of X is non-empty. 0498
Proposition 1 Let X be a non-empty topological space. The following conditions are equivalent: (a) X is irreducible; (b) every non-empty open set of X is dense in X; (c) every open set of X is c… ↩ 2 0499
Proposition 2 In a topological space X, for a set E to be irreducible, it is necessary and sufficient that its closure overlineE be so. ↩ 5 049A
Proposition 3 (i) If X is an irreducible space, every non-empty open subset of X is irreducible. ↩ 1 049B
Proposition 4 Let X and Y be two topological spaces and f a continuous mapping from X to Y. For every irreducible subset E of X, f(E) is an irreducible subset of Y. 049C
Definition 2 Every maximal irreducible subset of a topological space X is called an irreducible component of X. 049D
Proposition 5 Let X be a topological space. Every irreducible subset of X is contained in an irreducible component of X and X is the union of its irreducible components. 049E
Corollary Every connected component of a topological space X is a union of irreducible components of X. ↩ 1 049F
Proposition 6 Let X be a topological space and (Pᵢ)_1 ≤ i ≤ n ajinite covering of X consisting & irreducible closed sets. Then the irreducible components of X are the maximal elements (with resp… ↩ 2 049G
Corollary Let X be a topological space and E a subspace with only ajinite number of distinct irreducible components Qᵢ (1 ≤ i ≤ n); then the irreducible components of the closure overlineE i… 049H
Remark Suppose that X has only ajinite number of distinct irreducible components Xᵢ (1 ≤ i ≤ n); then U₊ = C(⋃_j ≠ i Xⱼ) is open in X and dense in Xᵢ since Xᵢ ∉ ⋃_j ≠ i Xⱼ; the Uᵢ (1 ≤ i… 049I
Proposition 7 Let U be an open subset of a topological space X. The mapping V ↦ overlineV (closure in X) is a bijection of the set & irreducible subsets of U which are closed in U onto the set &… 049J
Definition 3 A topological space X is called Noetherian if every non-empty set of closed subsets of X, ordered by inclusion, has a minimal element. 049K
Proposition 8 (i) Every subspace of a Noetherian space is Noetherian. (ii) Let (Aᵢ)_i ∈ I be a finite covering of a topological space X. If the subspaces Aᵢ of X are Noetherian, X is Noetherian. ↩ 1 049L
Proposition 9 For a topological space X to be Noetherian, it is necessary and sufficient that every open set in X be quasi-compact. 049M
Lemma 1 ("Principle of Noetherian induction") Let E be an ordered set every non-empty subset of which admits a minimal element. Let F be a subset of E with the following property : if a ∈ E is such that the relation x < a impl… 049N
Proposition 10 If X is a Noetherian space, the set of irreducible components of X (and a fortiori the set of connected components of X) is finite. 049O
Definition 4 Let A be a ring. The set X of prime ideals of A, with the topology whose closed sets are the sets V(M), where M runs through 𝔓(A), is called the prime spectrum of A and denoted by… ↩ 1 049P
Proposition 11 Let A be a ring, a an ideal of A and Y a subset of X = Spec(A). (i) V(a) is closed in X and 𝔍(Y) is an ideal of A which is equal to its radical. (ii) 𝔍(V(a)) is the radical of a an… ↩ 8 049Q
Corollary 1 For every family (Yλ){λ ∈ L} of closed subsets of X, J(∩{λ ∈ L} Yλ) is the radical of the sum of the ideals J(Yλ). 049R
Corollary 2 Let r(a) denote the radical of an ideal a of A; if a and b are two ideals of A, the relation V(a) ⊂ V(b) is equivalent to b ⊂ r(a) and r(b) ⊂ r(a). ↩ 4 049S
Corollary 3 Let (fλ){λ ∈ L} be a family of elements of A. For an element g ∈ A to satisfy Xg ⊂ ∪{λ ∈ L} X{fλ}, it is necessary and sufficient that there exist an integer n 0 such that g^n belo… ↩ 2 049T
Corollary 5 For f ∈ A to satisfy X_f = ∅, it is necessary and sufficient that f be nilpotent. 049U
Corollary 6 The closure of a set consisting of a point p ∈ X = Spec(A) is the set V(p) of prime ideals containing p. For the set p to be closed in X (or, as we shall also say by an abuse of la… 049V
Corollary 7 If A is a Noetherian ring, X = Spec(A) is a Noetherian space. ↩ 1 049W
Proposition 12 For all f ∈ A, the open set X_f in X = Spec(A) is quasi-compact; in particular, the space X is quasi-compact. 049X
Proposition 13 Let A, A' be two rings, X = Spec(A), X' = Spec(A') and h a homomorphism from A to A'; the mapping ᵃ h : p' ↦ overlineh(p') from X' to X is continuous. ↩ 3 049Y
Remark If h is surjective and a is its kernel, it follows from the definition of the spectral topology that ᵃ h is a homeomorphism of X' onto the closed subspace V(a) of X; for a prime id… 049Z
Corollary Let S be a multiplicative subset of A, A' = S⁻¹A and h the canonical homomorphism i_A^S; then ᵃ h is a homeomorphism of X' = Spec(A') onto the subspace of X = Spec(A) consisting of… ↩ 1 04A0
Proposition 14 Let A be a ring. For a subset Y of X = Spec(A) to be irreducible, it is necessary and sufficient that the ideal 𝔍(Y) be prime. ↩ 1 04A1
Corollary 1 For a ring A to be such that X = Spec(A) is irreducible, it is necessary and sufficient that the quotient of A by its nilradical 𝔑 be an integral domain. 04A2
Corollary 2 The mapping p ↦ V(p) is a bijection of X = Spec(A) onto the set of irreducible closed subsets of X; in particular the irreducible components of a closed subset Y of X are the sets… ↩ 2 04A3
Corollary 3 The set of minimal prime ideals of a Noetherian ring A is finite. ↩ 1 04A4
Proposition 15 Let A be a ring, I a finite set and E the set of orthogonal families (eᵢ)_i ∈ I of idempotents eᵢ ≠ 0 of A such that ∑_i ∈ I eᵢ = 1. For all (eᵢ)_i ∈ I ∈ E, we set overlineω((eᵢ)_i… ↩ 2 04A5
Lemma 2 If e, f are two idempotents of A such that Ae and Af have the same radical then e = f. ↩ 1 04A6
Corollary 1 Let A be a ring, r a nil ideal of A and h : A → A/r the canonical homomorphism. For every finite orthogonal family (e'ᵢ)_i ∈ I of idempotents of A/r such that ∑_i ∈ I e'ᵢ = 1, ther… 04A7
Corollary 2 For the prime spectrum X = Spec(A) of a ring A to be connected, it is necessary and sufficient that A contain no idempotents other than 0 and 1. ↩ 1 04A8
Definition 5 Let A be a ring and M an A-module. The set of prime ideals p of A such that Mₚ ≠ 0 is called the support of M and is denoted by Supp(M). ↩ 1 04A9
Example Let a be an ideal of A; in the notation of no. 3, we have V(a) = Supp(A/a). If p is a prime of A such that a ∉ p, then (A/a)ₚ = 0 (§ 3, no. 1, Remark 3); if on the other hand a ⊂ p… 04AA
Proposition 16 Let A be a ring and M an A-module. (i) If N is a submodule of M, then Supp(M) = Supp(N) ∪ Supp(M/N). (ii) If M is the sum of a family (Nᵢ)_i ∈ I of submodules, then Supp(M) = ⋃_i ∈… ↩ 2 04AB
Corollary Let A be a ring, M an A-module, (mᵢ)_i ∈ I a system of generators of M and a, the annihilator of mᵢ. Then Supp(M) = ⋃_i ∈ I V(aᵢ). Supp(M) = ⋃_i ∈ I Supp(Amᵢ) by Proposition 16 (ii… ↩ 1 04AC
Proposition 17 Let A be a ring, M an A-module and a its annihilator; if M is finitely generated, then Supp(M) = V(a) and Supp(M) is therefore closed in Spec(A). Let (mᵢ)_1 ≤ i ≤ n be a system of… ↩ 6 04AD
Corollary 1 Let A be a ring, M a finitely generated A-module and a an element of A. For a to belong to every prime ideal of the support of M, it is necessary and sufficient that the homothety… 04AE
Corollary 2 Let A be a Noetherian ring, M a finitely generated A-module and a an ideal of A. For Supp(M) ⊂ V(a), it is necessary and sufficient that there exist an integer k such that aᵏ M = 0… 04AF
Proposition 18 Let M, M' be two finitely generated modules over a ring A; then 04AG
Lemma 3 Let B be a local ring and E and E' two finitely generated B-modules. If E ≠ 0 and E' ≠ 0, then E ⊗_B E' ≠ 0. 04AH
Corollary Let M be a finitely generated A-module and n its annihilator. For every ideal a of A, Supp(M/aM) = V(a) ∩ V(n) = V(a + n). 04AI
Proposition 19 Let A, B be two rings, φ : A → B a homomorphism and ᵃφ : Spec(B) → Spec(A) the continuous mapping associated with φ (Proposition 13). For every A-module M, Supp(M_(B)) ⊂ ᵃφ(Supp(M)… ↩ 1 04AJ
Lemma 4 Let A, B be two local rings, p : A → B a local homomorphism and E a finitely generated A-module. If E ≠ 0, then E_(B) ≠ 0. ↩ 1 04AK
Proposition 20 Let A be a ring and M a finitely generated A-module. For every prime ideal p ∈ Supp(M), there exists a non-zero A-homomorphism w : M → A/p. ↩ 1 04AL

§ 5. Finitely generated projective modules. Invertible fractional ideals

Proposition 1 Let f be an element of a ring A and φ : A → A_f, the canonical mapping. The mapping ᵃφ : Spec(A_f) → Spec(A) is a homeomorphism of Spec(A_f) onto the open subspace X_f of X = Spec(… 04BE
Proposition 2 Let A be a ring, u : M → N an A-module homomorphism and p a prime ideal of A. ↩ 2 04BF
Corollary If N is finitely presented and Nₚ is a free Aₚ-module of rank p, there exists f ∈ A - p such that N, is a free A_f-module of rank p. ↩ 1 04BG
Proposition 3 Let (fᵢ)_i ∈ I be a finite family of elements of a ring A, generating the ideal A of A. The ring B = ∏_i ∈ I Aᵢ, is then a faithfully flat A-module. ↩ 3 04BH
Corollary Under the hypotheses of Proposition 3, for an A-module M to be finitely generated (resp. finitely presented), it is necessary and sufficient that, for every index i, the A_fᵢ-modul… ↩ 1 04BI
Theorem 1 Let A be a ring and P an A-module. The following properties are equivalent: ↩ 15 04BJ
Corollary 1 Suppose that the equivalent properties of the statement of Theorem 1 hold. Let m be an integer 0 such that, for every family (xᵢ)_1 ≤ i ≤ m of elements of P, there exists a family… 04BK
Corollary 2 Every finitely presented flat module is projective. ↩ 1 04BL
Definition 1 Let P be a finitely generated projective A-module. For every prime ideal p of A, the rank of the free Aₚ-module Pₚ is called the rank of P at p and is denoted by rgₚ(P). ↩ 1 04BM
Definition 2 Let n be an integer geqslant 0. A projective A-module P is said to be of rank n if it is finitely generated and rgₚ(P) = n for every prime ideal p of A. ↩ 3 04BN
Theorem 2 Let P be an A-module and n an integer geqslant 0. The following properties are equivalent: ↩ 4 04BO
Remark If A is an integral domain, a projective A-module admits a well-defined rank (in the sense of Definition 2), as has been observed above; moreover, this rank coincides with the rank… 04BP
Corollary For a finitely generated projective A-module P to be of rank n, it is necessary and sufficient that Aⁿ P be of rank 1. 04BQ
Proposition 4 Let B be a commutative A-algebra and P a projective A-module of rank n. The B-module P_(B) = B ⊗_A P is then projective of rank n. ↩ 4 04BR
Proposition 5 Let A be a semi-local ring and P a finitely generated projective A-module. If the rank of P is defined, P is a free A-module. ↩ 1 04BS
Theorem 3 Let A be a ring and M ajnitely generated A-module. ↩ 2 04BT
Proposition 6 Let M and N be projective A-modules of rank 1. Then M ⊗_A N, Hom_A(M, N) and the dual M^* of M are projective of rank 1. 04BU
Proposition 7 The set P(A) of classes of projective A-modules of rank 1, with the law of composition (6), is a commutative group. If M is a projective A-module of rank 1 and M^* is its dual, the… 04BV
Definition 3 Let M be a sub-A-module of B. M is called non-degenerate if B.M = B. 04BW
Proposition 8 Let M be a sub-A-module of B. The following conditions are equivalent: (a) M is non-degenerate. (b) M meets S. (c) If j : M → B is the canonical injection, the homomorphism u = S⁻¹… ↩ 4 04BX
Corollary If M and N are two non-degenerate sub-A-modules of B, the A-modules M + N, M.N and M ∩ N are non-degenerate. 04BY
Proposition 9 Let M, N be two sub-A-modules of B. If M is non-degenerate, the canonical homomorphism from N : M to Hom_A(M, N) is bijective. ↩ 2 04BZ
Remark In particular, A : M is canonically identified with the dual M^* of M, the canonical bilinear form on M × M^* being identified with the restriction to M × (A : M) of the multiplica… 04C0
Definition 4 A sub-A-module M of B is called invertible if there exists a sub-A-module N of B such that M.N = A. 04C1
Example If b is invertible element of B, the A-module Ab is invertible, as is seen by taking N = Ab⁻¹. 04C2
Proposition 10 Let M be an invertible sub-A-module & B. Then: (i) There exists s ∈ S such that As ⊂ M ⊂ As⁻¹ (and in particular M is non-degenerate). (ii) A : M is the only sub-A-module N of B su… ↩ 3 04C3
Theorem 4 Let M be a non-degenerate sub-A-module of B. The following properties are equivalent: (a) M is invertible. (b) M is projective. (c) M is projective of rank 1. (d) M is a finitely g… ↩ 5 04C4
Corollary Every invertible sub-A-module of B is flat and finitely presented. This follows from Theorem 4 (c). ↩ 1 04C5
Proposition 11 Let M, N be two sub-A-modules of B. Suppose that M is invertible. Then: (i) The canonical homomorphism M ⊗_A N → M . N is bijective. (ii) N : M = N . (A : M) and N = (N : M) . M. ↩ 3 04C6
Remark The proof of (i) in Proposition 11 uses only the fact that M is flat and non-degenerate. 04C7
Proposition 12 The mapping cl : 𝔤 → P(A) defines, by taking quotients, an isomorphism from 𝔠 = 𝔤/θ(B^*) onto the kernel of the canonical homomorphism 04C8
Corollary 1 For two invertible sub-A-modules of B to have the same image in ℭ, it is necessary and sufficient that they be isomorphic. 04C9
Corollary 2 If the ring B is semi-local, the group ℭ of classes of invertible sub-A-modules of B is canonically identified with the group P(A) of classes of projective A-modules of rank 1. ↩ 2 04CA
Remark The hypothesis of Corollary 2 is fulfilled in the two following cases: 04CB

III. Graduations. Filtrations and topologies

§ 1. Finitely generated graded algebras

Definition 1 In a commutative algebra B over a commutative ring A, a family (xᵢ)_i ∈ I of elements of B is called algebraically free over A (or the xᵢ are called algebraically independent over… 04D1
Definition 2 A commutative algebra B over a commutative ring A is called finitely generated if it is generated by a finite family of elements. 04D2
Proposition 1 Let A = ⨁_i ∈ Z Aᵢ be a graded commutative ring with positive degrees, m the graded ideal ⨁_i ≥ 1 Aᵢ and (x_λ)_λ ∈ L a family of homogeneous elements of A of degrees ≥ 1. The follo… ↩ 1 04D3
Corollary Let A = ⨁_i ∈ Z Aᵢ, be a graded commutative ring with positive degrees and m the graded ideal, ⨁_i ≥ 1 Aᵢ. ↩ 3 04D4
Proposition 2 Let A = ⨁_i ∈ Z Aᵢ be a graded commutative ring with positive degrees and M = ⨁_i ∈ Z Mᵢ a graded A-module. Suppose that A is a finitely generated A₀-algebra and M a finitely gener… 04D5
Lemma 1 Let A be a graded commutative ring such that A = A₀[A₁], M a graded A-module and (y_λ)_λ ∈ L a system of homogeneous generators of M such that deg(y_λ) leqslant n₀ for all λ ∈ L. T… ↩ 2 04D6
Lemma 2 Let A be a graded commutative ring such that A = A₀[A₁] and let S = ⨁_i ≥ 0 Sᵢ, be a graded commutative A-algebra with positive degrees, which is a finitely generated A-module. The… ↩ 1 04D7
Proposition 3 Let R = ⨁_i ≥ 0 Rᵢ, be a graded commutative ring with positive degrees which is a finitely generated R₀-algebra. There exists an integer e ≥ 1 such that R⁽ᵐᵉ⁾ = R₀[Rₘₑ] for all m ≥… 04D8
Proposition 4 Let p = ⨁_i ≥ 0 pᵢ be a graded ideal of A; for p to be prime, it is necessary and sufficient that, if x ∈ A, \ y ∈ A, satisfy x ∉ p and y ∉ p, then xy ∉ p. ↩ 2 04D9
Proposition 5 Let a = ⨁_i ≥ 0 aᵢ be a graded ideal of A and n₀ an integer > 0. For there to exist a graded prime ideal p = ⨁_i ≥ 0 pᵢ such that p_n₀ = a, for n ≥ n₀, it is necessary and sufficie… ↩ 1 04DA
Proposition 6 Let d be an integer ≥ 1. 04DB
Proposition 7 Let p be a graded prime ideal of A, d an integer ≥ 1 and p' the graded prime ideal p ∩ A⁽ᵈ⁾ of A⁽ᵈ⁾; for every graded A-module M, the homomorphism (M⁽ᵈ⁾)_(p') → M₍ₚ₎ derived from t… ↩ 1 04DC
Proposition 8 Let m = ⨁_i ≥ 1 Aᵢ; let (p⁽ᵏ⁾)_1 ≤ k ≤ s be a finite family of graded prime ideals of A and a graded ideal 𝔞 ⊂ A such that 𝔞 ∩ m not⊂ p⁽ᵏ⁾ for all k; then there exists a homogeneou… 04DD

§ 2. General results on filtered rings and modules

Definition 1 An increasing (resp. decreasing) sequence (Gₙ)_n ∈ Z of subgroups of a group G is called an increasing (resp. decreasing) filtration on G. A group with a filtration is called a fil… 04DH
Definition 2 Given a ring A, a filtration (Aₙ)_n ∈ Z over the additive group A is called compatible with the ring structure on A if 04DI
Definition 3 Let A be a filtered ring, (Aₙ)_n ∈ Z its filtration and E an A-module. A filtration (Eₙ)_n ∈ Z on E is called compatible with its module structure over the filtered ring A if Aₘ Eₙ… 04DJ
Example 1 Let A be a commutative ring and t an element of A which is not a divisor of 0. Let us give A the (t)-adic filtration (no. 1, Example 3). Then the associated graded ring gr(A) is ca… ↩ 1 04DK
Example 2 Let K be a commutative ring, A the ring of formal power series 04DL
Example 3 More generally, let A be a commutative ring, b an ideal of A and A be given the b-adic filtration. If we write B=gr₀(A), F=gr₁(A)=b/b², we know (Algebra, Chapter III), that the ide… ↩ 9 04DM
Example 4 \ Let K be a commutative ring, mathfrak g a Lie algebra over K and U the enveloping algebra of mathfrak g. An increasing filtration (Uₙ)_n∈Z is defined on U by taking Uₙ=0 for n<0… ↩ 1 04DN
Example 5 Let A be a graded ring of type Z and E a graded A-module of type Z; let A₍ᵢ₎ (resp. E₍ᵢ₎) be the subgroup of homogeneous elements of degree i of A (resp. E). Let A and E be given t… 04DO
Proposition 1 Let A be a filtered ring, (Aₙ)_n ∈ Z its filtration and v its orderfunction. Suppose that gr(A) is a ring with no divisor of zero. Then, for every ordered pair of elements a, b of… ↩ 1 04DP
Corollary Let A be a filtered ring and (Aₙ)_n ∈ Z its filtration; let us set B = ⋃_n ∈ Z Aₙ, n = ⋂_n ∈ Z Aₙ. If the ring gr(A) has no divisors of zero, neither has the ring B/n. If a and b a… ↩ 2 04DQ
Remark Let G be a group which is not necessarily commutative with a filtration (Gₙ)_n ∈ Z such that Gₙ₊₁ is normal in G, for all n ∈ Z; again let grₙ(G) = Gₙ/Gₙ₊₁. The restricted product… 04DR
Proposition 2 Let G be a filtered commutative group and H a subgroup of G; let H be given the induced filtration and G/H the quotient filtration. If j : H → G is the canonical injection and p… ↩ 3 04DS
Proposition 3 Let A be a filtered ring, (An) its filtration, B the subring ⋃_n ∈ Z Aₙ of A, E a filtered B-module, (En) its filtration and F the sub-B-module ⋃_n ∈ Z Eₙ of E. Then the mapping (a… ↩ 2 04DT
Corollary The ring B is a topological ring and the B-module F a topological B-module. ↩ 1 04DU
Proposition 4 Let A be a commutative ring filtered by an exhaustive filtration (An) and p an ideal of A. Suppose that the ideal gr(p) = ⨁_n ∈ Z (p ∩ Aₙ)/(p ∩ Aₙ₊₁) of the ring gr(A) is prime. Th… 04DV
Proposition 5 Let G be a filtered group whose filtration (Gₙ) consists of invariant subgroups of G. The following conditions are equivalent: ↩ 1 04DW
Example 1 Let G be a complete filtered group. Every closed subgroup of G with the induced filtration is complete (General Topology, Chapter II, § 3, no. 4, Proposition 8). Every quotient gro… ↩ 1 04DX
Example 2 Let A be a filtered commutative ring whose filtration we denote by (aₙ)_n ∈ Z; let A' be the ring of formal power series A[[X₁, …, Xₛ]]. For all e = (e₁, …, eₛ) ∈ Nˢ, we write |e|… 04DY
Proposition 6 With the above notation, suppose that a₀ = A and let h denote the mapping P ↦ (α_e,P)_e ∈ Nˢ. Then h is an isomorphism of the additive topological group A onto the additive topolog… ↩ 3 04DZ
Corollary Let A be a commutative ring, ↩ 5 04E0
Proposition 7 Let A be a filtered ring, E a filtered A-module and (Eₙ) filtration of E; suppose that E₀ = E, that the Eₙ are submodules of E, that the A-modules E/Eₙ are Artinian and finally tha… ↩ 1 04E1
Proposition 8 Suppose that A and E satisfy the hypotheses of Proposition 7. Let (Fₚ) be a decreasing sequence of closed submodules of E such that ⋂ Fₚ = 0. Then, for every neighbourhood V of 0 i… ↩ 4 04E2
Theorem 1 Let X, Y be two filtered groups whose filtrations (Xₙ), (Yₙ) consist of normal subgroups; let u : X → Y be a homomorphism compatible with the filtrations. ↩ 4 04E3
Corollary 1 Suppose that X is Hausdorff and its filtration exhaustive. Then, if gr(u) is injective, u is injective. ↩ 4 04E4
Corollary 2 Suppose that one of the following hypotheses holds : (α) X is complete, Y is Hausdorff and its filtration is exhaustive; (β) Y is discrete and its filtration is exhaustive. Then, i… ↩ 2 04E5
Corollary 3 Suppose that X and Y are Hausdorff, the filtrations of X and Y exhaustive and X complete. Then, if gr(u) is bijective, u is bijective. 04E6
Proposition 9 Let A be a local ring, m its maximal ideal, M, N two A-modules and u : N \to M an A-homomorphism. M and N are given the m-adic filtrations and suppose that: (1) M satisfies propert… ↩ 1 04E7
Corollary Under the hypotheses of Proposition 9, if we assume also that N is Hausdorf with the m-adic filtration, then u is injective. 04E8
Proposition 10 Let A be a filtered commutative ring and (Aₙ) its filtration; suppose that there exists a subring C of A₀ such that C ∩ Aₙ = 0 and C is identified with a subring of gr₀(A). Let (xᵢ… ↩ 1 04E9
Proposition 11 Let A be a complete Hausdorff filtered commutative ring, C a subring of A₀ such that C ∩ Aₙ = 0 and (xᵢ)_1 ≤ i ≤ q a finite family of elements of A such that xᵢ ∈ A_nᵢ where nᵢ > 0… ↩ 3 04EA
Proposition 12 Let A be a filtered ring, E a filtered A-module and (Aₙ) and (Eₙ) the respective filtrations on A and E. Suppose that A is complete and the filtration (Eₙ) is exhaustive and separa… ↩ 5 04EB
Corollary 1 Let A be a complete filtered ring and E a filtered A-module whose filtration is exhaustive and separated. If gr(E) is a finitely generated (resp. Noetherian) gr(A)-module, then E i… ↩ 4 04EC
Corollary 2 Let A be a complete Hausdorff filtered ring whose filtration is exhaustive. If gr(A) is a left Noetherian ring, so is A. ↩ 2 04ED
Corollary 3 Let A be a complete filtered ring, (An) its filtration, E a Hausdorff filtered A-module, (En) its filtration and F a finitely generated submodule of E; suppose that A∞ = A and E0 =… ↩ 3 04EE
Proposition 13 Let A be a ring, m a two-sided ideal of A contained in the Jacobson radical of A and E an A-module. Let A and E be given the m-adic filtrations (no. 1, Example 3). Suppose that one… ↩ 2 04EF
Lemma 1 Let A be a graded ring of type Z, whose graduation (Aₙ) is such that Aₙ = 0 for all n < 0 or Aₙ = 0 for all n > 0. Let M be a graded A-module of type Z. For M to be a Noetherian A-… ↩ 1 04EG
Theorem 2 Let A be a graded ring of type N, M a graded A-module of type N and (Aₙ) and (Mₙ) their respective graduations. Suppose that there exists an element a ∈ A₀ such that Aₙ = A₀ aⁿ and… ↩ 2 04EH
Corollary 1 (Hilbert’s Theorem) For every commutative Noetherian ring C, the polynomial ring C[X] is Noetherian (cf. Exercise 10). ↩ 1 04EI
Corollary 2 For every commutative Noetherian ring C and every integer n > 0, the polynomial ring C[X1, ..., Xn] is Noetherian. ↩ 1 04EJ
Corollary 3 If C is a commutative Noetherian ring, every finitely generated commutative C-algebra is a Noetherian ring. ↩ 5 04EK
Corollary 4 Let A be a graded commutative ring of type N and let (Aₙ) be its graduation. For A to be Noetherian, it is necessary and sufficient that Aₙ be Noetherian and that A be ajinitely ge… ↩ 2 04EL
Corollary 5 Let A be a commutative ring and m an ideal of A such that A/m is Noetherian, m/m² is ajinitely generated (A/m)-module and A is Hausdorff and complete with the m-adic topology. Then… ↩ 4 04EM
Corollary 6 For every commutative Noetherian ring C and every integer n 0, the ring of formal power series C[[X₁, …, Xₙ]] is Noetherian. 04EN
Proposition 14 Let (Aᵢ, hᵢⱼ) be an inverse system of discrete commutative rings, whose indexing set is N and let (Mᵢ, uᵢⱼ) be an inverse system of modules over the inverse system of rings (Aᵢ, hᵢ… ↩ 4 04EO
Corollary 1 With the notation and hypotheses of Proposition 14 suppose further that M₀ is a finitely generated A-module and that the ideal n₁ of A, is finitely generated. Let m₁ be the kernel… ↩ 1 04EP
Corollary 2 Under the hypotheses of Corollary 1, for A to be Noetherian, it is necessary and sufficient that A, be so. ↩ 1 04EQ
Proposition 15 Let A be a filtered ring and E a filtered A-module, the respective filtrations (Aₙ), (Eₙ) of A and E being exhaustive. Then (hatAₙ) is a filtration compatible with the ring structu… ↩ 2 04ER
Lemma 2 Let X xrightarrowf Y xrightarrowg Z be an exact sequence of strict morphisms of topological groups (Algebra, Chapter II, § 1, no. 4, Remark). Suppose that X, Y, Z admit Hausdorff c… ↩ 3 04ES
Proposition 16 Let A be a filtered ring, (A,) its filtration, E an A-module and (Eₙ) the filtration on E derived from that on A consisting of the Eₙ = Aₙ E. Suppose that the filtration (A,) is ex… ↩ 5 04ET
Corollary 1 Under the conditions of Proposition 16, if A is complete, so is E. ↩ 1 04EU
Corollary 2 Let A be a commutative ring, m a finitely generated ideal of A and hatA the Hausdorff completion of A with respect to the m-adic topology. Then hatmⁿ = (hatm)ⁿ = mⁿ . hatA for ever… ↩ 2 04EV
Proposition 17 Let A be a commutative ring and (m_λ)_λ ∈ L a family of ideals of A, distinct from A, such that m_λ and m_μ are relatively prime for λ ≠ μ. For every family s = (s(λ))_λ ∈ L of int… ↩ 3 04EW
Corollary Let A be a principal ideal domain and P a representative system of extremal elements of A (Algebra, Chapter VII, § 1, no. 3). The topology on A with respect to which the ideals ≠ 0… 04EX
Proposition 18 Let A be a commutative ring, (mᵢ)_1 ≤ i ≤ q a finite family of distinct maximal ideals of A, t the product ideal m₁ m₂ … m_q = m₁ ∩ m₂ ∩ … ∩ m_q and S the multiplicative subset ⋂ᵢ₌… ↩ 1 04EY
Proposition 19 Let A be a commutative ring and m a maximal ideal of A. The Hausdorff completion hatA of A with respect to the m-adic topology is a local ring whose maximal ideal is m. 04EZ
Lemma 3 Let A be a complete Hausdorf topological ring, in which there exists a fundamental system G of neighbourhoods of 0 consisting of additive subgroups of A. (i) For all x ∈ A such tha… ↩ 3 04F0
Corollary Let A be a semi-local ring (Chapter II, § 3, no. 5), mᵢ its distinct maximal ideals (1 ≤ i ≤ q) and ↩ 3 04F1

§ 3. m-adic topologies on Noetherian rings

Definition 1 Let A be a commutative ring, m an ideal of A, E an A-module and (Eₙ) a filtration on the additive group E consisting of submodules of E. The filtrations (Eₙ) is called m-good if: (… 04FX
Theorem 1 Let A be a commutative ring, m an ideal of A, E an A-module and (Eₙ) a filtration of the additive group E consisting offinitely generated sub-A-modules. Suppose that m Eₙ ⊂ Eₙ₊₁ fo… ↩ 2 04FY
Lemma 1 Let A be a commutative Noetherian ring and m an ideal of A. Then the subring A' = ∑_n ≥ 0 mⁿ Xⁿ of A[X] is Noetherian. ↩ 1 04FZ
Proposition 1 Let A be a commutative Noetherian ring and m an ideal of A; let A be given the m-adic filtration. Let E, F be twofiltered A-modules and j : F → E an injective homomorphism compatib… ↩ 3 04G0
Corollary 1 (the Artin-Rees Lemma) Let A be a commutative Noetherian ring, m an ideal of A, E a finitely generated A-module and F a submodule of E. The filtration induced on F by the m-adic filtration on E is m-good… ↩ 2 04G1
Corollary 2 Let A be a commutative Noetherian ring and a, b two ideals of A. There exists an integer h 0 such that aʰ ∩ b ⊂ ab. 04G2
Corollary 3 Let A be a commutative Noetherian ring, m an ideal of A and x an element of A which is not a divisor of O. There exists an integer k > 0 such that, for all n ≥ k, the relation xy ∈… 04G3
Corollary 4 Let A be a commutative Noetherian ring, m an ideal of A, E ajnitely generated A-module and (Eₙ) and (E'ₙ) two filtrations consisting of submodules of E. Suppose that the filtration… 04G4
Lemma 2 Let A, B be two commutative Noetherian rings, φ : A → B a ring homomorphism, E a finitely generated A-module and F a finitely generated B-module. Then Hom_A(E, φ_*(F)) is a finitel… ↩ 1 04G5
Proposition 2 Let A be a commutative Noetherian ring, m an ideal of A and E, F two finitely generated A-modules. If (Fₙ) is an m-good filtration on F, the submodules Hom_A(E, Fₙ) form an m-good… ↩ 1 04G6
Proposition 3 Let A be a Noetherian ring and m an ideal of A such that A is Hausdorff and complete with respect to the m-adic topology. Let E be a filtered A-module over the filtered ring A, the… 04G7
Proposition 4 Let A be a commutative Noetherian ring, m an ideal of A and E a finitely generated A-module. All the m-good filtrations on E define the same topology (namely the m-adic topology). ↩ 1 04G8
Theorem 2 (Krull) Let A be a commutative Noetherian ring, m an ideal of A, E a finitely generated A-module and F a submodule of E. Then the m-adic topology on F is induced by the m-adic topology on… ↩ 6 04G9
Corollary Let A be a commutative Noetherian ring, m an ideal of A, E an A-module and F a finitely generated A-module. Every A-linear mapping u : E → F is a strict morphism (General Topology… ↩ 1 04GA
Proposition 5 Let A be a commutative Noetherian ring, m an ideal of A and E a finitely generated A-module. The closure ⋂ₙ₌₁ᵐ mⁿE of 0 in E with respect to the m-adic topology is the set of x ∈ E… ↩ 6 04GB
Corollary (Krull) Let A be a commutative Noetherian ring and m an ideal of A. The ideal ⋂ₙ₌₁ᵐ mⁿ is the set of elements x ∈ A for which there exists m ∈ m such that (1 - m)x = 0. In particular, for… ↩ 3 04GC
Remark The hypothesis that A is Noetherian is essential in this corollary. For example, let A be the ring of infinitely differentiable mappings from R to itself and let m be the (maximal)… 04GD
Definition 1 Let A be a topological ring. If a two-sided ideal m of A is such that the given topology on A is the m-adic topology, m is called a defining ideal & the topology on A. 04GE
Proposition 6 Let A be a commutative Noetherian ring and m an ideal of A. The following properties are equivalent: (a) m is contained in the Jacobson radical of A. (b) Every finitely generated A… ↩ 6 04GF
Definition 2 A topological ring A is called a Zariski ring if it is commutative and Noetherian and there exists a defining ideal m for the topology on A satisfying the equivalent conditions of… 04GG
Example 1 Let A be a commutative Noetherian ring and m an ideal of A. If A is Hausdorff and complete with the m-adic topology, A is a Zariski ring with this topology, by virtue of § 2, no. 1… ↩ 1 04GH
Example 2 Every quotient ring A/b of a Zariski ring is a Zariski ring, for it is Noetherian and, if m is a defining ideal of A, m(A/b) = (m + b)/b is contained in the Jacobson radical of A/b… 04GI
Example 3 Let A be a Noetherian semi-local ring and r its Jacobson radical. Then A with the r-adic topology is a Zariski ring. This will always be the topology in question (unless otherwise… ↩ 2 04GJ
Proposition 7 Let A, A' be two commutative rings and h : A → A' a ring homomorphism. Suppose that A is Noetherian and that A' is a finitely generated A-module (with the structure defined by h)… ↩ 2 04GK
Theorem 3 Let A be a commutative Noetherian ring, m an ideal of A and E, F, G threefinitely generated A-modules. Then: ↩ 7 04GL
Corollary 1 Let A be a commutative Noetherian ring, m an ideal of A, E a finitely generated A-module and F and G two submodules of E. Let A, E, F, G be given the m-adic topologies and let i be… ↩ 2 04GM
Corollary 2 Let A be a commutative Noetherian ring, m an ideal of A and hatA the Hausdorff completion of A with respect to the m-adic topology. If an element a ∈ A is not a divisor of 0 in A… ↩ 1 04GN
Corollary 3 If A is a commutative Noetherian ring, the ring of formal power series A[[X₁, …, Xₙ]] is a flat A-module. 04GO
Proposition 8 Let A be a commutative Noetherian ring, m an ideal of A, hatA the Hausdorff completion of A with respect to the m-adic topology and j the canonical mapping from A to hatA. The: (i)… ↩ 3 04GP
Corollary Let A be a Noetherian local (resp. semi-local) ring and m its Jacobson radical. Then hatA is a Noetherian local (resp. semi-local) ring whose Jacobson radical is hatm. 04GQ
Proposition 9 Let A be a commutative Noetherian ring and m an ideal of A; let A be given the m-adic topology. For hatA to be a faithfully flat A-module, it is necessary and sufficient that A be… ↩ 3 04GR
Corollary 1 Let A be a Zariski ring, E a finitely generated A-module and F a submodule of E. Then F = hatF ∩ E = (hatAF) ∩ E. 04GS
Corollary 2 Let A be a Zariski ring and E a finitely generated A-module. If hatE is a free A-module, E is a free A-module. ↩ 1 04GT
Corollary 3 Let A be a Zariski ring such that hatA is an integral domain and let a be an ideal of A. If the ideal ahatA of hatA is principal, a is principal. ↩ 1 04GU
Corollary 4 Let A be a Zariski ring such that hatA is an integral domain, L the field of fractions of hatA and K ⊂ L the field of fractions of A; then hatA ∩ K = A. ↩ 1 04GV
Corollary 5 Let A be a commutative Noetherian ring, E, F twofinitely generated A-modules and u : E → F an A-homomorphism. For every maximal ideal m of A, let A(m) (resp. E(m), F(m)) denote the… 04GW
Proposition 10 Let A, B be two Zariski rings, hatA, hatB their completions, f : A → B a continuous ring homomorphism and hatf : hatA → hatB the homomorphism obtained from f by passing to the comp… ↩ 1 04GX
Proposition 11 Let A be a Noetherian local ring, m its maximal ideal, hatA its m-adic completion and B a ring such that A ⊂ B ⊂ hatA. Suppose that B is a Noetherian local ring whose maximal ideal… 04GY
Remark A local ring B such that A ⊂ B ⊂ hatA, whose maximal ideal n is equal to mB and whose n-adic topology is induced by the m-adic topology on A, is not necessarily Noetherian (Exercis… 04GZ
Proposition 12 Let A be a commutative Noetherian ring, m an ideal of A, S the multiplicative subset 1 + m of A and E a finitely generated A-module. Under these conditions: 04H0

§ 4. Lifting in complete rings

Lemma 1 ("Euclid's Lemma") Let x, y be two strongly relatively prime elements of R; if z ∈ R is such that x divides yz, then x divides z. ↩ 1 04HS
Proposition 1 Let A be a commutative ring and P and P' two strongly relatively prime polynomials in A[X]. Suppose that P is monic and of degree s. Then every polynomial T in A[X] may be written… ↩ 3 04HT
Example For a polynomial P ∈ A[X] to be strongly relatively prime to X - a (where a ∈ A), it is necessary and sufficient that P(a) be invertible in A. For if P and X - a are strongly relat… 04HU
Proposition 2 Let A and B be two commutative rings, f a homomorphism from A to B and P, P' two polynomials in A[X]. If P and P' are strongly relatively prime in A[X], then barf(P) and barf(P') a… ↩ 2 04HV
Definition 1 A commutative topological ring A is said to be linearly topologized (and its topology is said to be linear) if there exists a fundamental system B of neighbourhoods 𝔬 consisting of… 04HW
Definition 2 Given a commutative topological ring A, a formal power series T = ∑_(nᵢ) c_n₁ n₂ … nₚ X₁^n₁ X₂^n₂ … Xₚ^nₚ in the ring A[[X₁, …, Xₚ]] is called restricted if, for every neighbourhoo… 04HX
Proposition 3 If the linearly topologized commutative ring A is Hausdorff and complete, the canonical homomorphism π is a topological ring isomorphism. ↩ 1 04HY
Proposition 4 Let A, B be two linearly topologized commutative rings, B being Hausdorff and complete, and u : A → B a continuous homomorphism. For every family b = (bᵢ)_1 ≤ i ≤ p of elements of… ↩ 3 04HZ
Remark 1 Proposition 4 proves that for every closed ideal a in a ring A which is assumed to be Hausdorff and complete, the relations bᵢ ∈ a for 1 ≤ i ≤ p imply f(b₁, …, bₚ) ∈ a for every re… ↩ 1 04I0
Remark 2 Suppose that A is linearly topologized; let r be an integer such that 1 ≤ r ≤ p and let the ring AX₁, …, Xᵣ be given the topology defined above. Then the topological ring AX₁, …, X… ↩ 1 04I1
Remark 3 With the notation of Remark 2, suppose further that A is Hausdorff and complete and let us write every restricted formal power series f ∈ AX₁, …, Xᵣ in the form ↩ 1 04I2
Proposition 5 Let A be a commutative ring and f = (f₁, …, fₙ) a system without constant term of n series in A[[X₁, …, Xₙ]]. Suppose that J_f(0) is invertible in A. Then there exists a system wit… ↩ 1 04I3
Proposition 6 Let A be a commutative ring, B a ring satisfying Hensel’s conditions and u : A → B a homomorphism. For every family x = (x₁, …, xₙ) of topologically nilpotent elements of B, there… ↩ 3 04I4
Corollary Let A be a ring satisfying Hensel's condition and x = (x₁, …, xₙ) a family of topologically nilpotent elements of A. Let g = (g₁, …, g_q) be a system without constant term of serie… ↩ 1 04I5
Proposition 7 Let A be a ring and m an ideal of A such that the ordered pair (A, m) satisfies Hensel's conditions. Let f = (f₁, …, fₙ) be a system without constant term of series in A[[X₁, …, Xₙ… ↩ 1 04I6
Corollary Let q be a closed ideal of A contained in m. Then the relation x ≡ x' pmodq^xⁿ is equivalent to f(x) ≡ f(x') pmodq^xⁿ. 04I7
Theorem 2 Let A be a ring and m an ideal of A such that the ordered pair (A, m) satisfies Hensel's conditions. Let f = (f₁, …, fₙ) be a system of n elements of ↩ 1 04I8
Corollary 1 Let A be a ring and m an ideal of A such that the ordered pair (A, m) satisfies Hensel's conditions. Let f ∈ AX, a ∈ A and write e = f'(a). If f(a) ≡ 0 (mod. e² m), then there exis… ↩ 2 04I9
Remark The corollary applies notably when e is invertible in A; we can then also deduce the existence of b from Hensel's Theorem, for the canonical image of f(X) in (A/𝔪)X is of the form… 04IA
Corollary 2 Let A be a ring and 𝔪 an ideal of A such that the ordered pair (A, m) satisfies Hensel's conditions. Let r, n be integers such that 0 ≤ r < n and f = (fᵣ₊₁, …, fₙ) is a system of n… ↩ 1 04IB
Corollary 3 With the notation & Corollary 2, let a ∈ Aⁿ; let us write e = Jᵣ^(n-r)(a) (not necessarily invertible in A) and suppose that f(a) ≡ 0 (mod. e^2m^×(n-r)). Then there exist n - r for… 04IC
Lemma 2 Let A be a ring and m an ideal of A such that the ordered pair (A, m) satisfies Hensel’s conditions. Let B be the quotient ring A/m and π : A → B be the canonical homomorphism. For… ↩ 1 04ID
Proposition 8 Let A be a ring and m an ideal of A such that the ordered pair (A, m) satisfies Hensel’s conditions. Let B be the quotient ring A/m and π : A → B the canonical homomorphism. If B i… ↩ 3 04IE
Remark Proposition 8 again gives the structure of a semi-local ring on A which is Hausdorff and complete with the c-adic topology (c the Jacobson radical of A), which has already been obt… 04IF

§ 5. Flatness properties of filtered modules

Definition 1 Let A be a commutative ring and 𝔤 an ideal of A. An A-module M is called ideally Hausdorff with respect to 3 (or simply ideally Hausdorff if there is no ambiguity) if, for every fi… 04IO
Example 1 If A is Noetherian and 𝔤 is contained in the Jacobson radical of A (in other words if A is a Zariski ring with the 3-adic topology), every finitely generated A-module is ideally Ha… ↩ 1 04IP
Example 2 Every direct sum of ideally Hausdorff modules is an ideally Hausdorff module, by virtue of the relations 𝔤ⁿ ( a ⊗_A ⨁_λ ∈ L M_λ ) = 𝔤ⁿ ⨁_λ ∈ L (a ⊗_A M_λ) = ⨁_λ ∈ L 𝔤ⁿ (a ⊗_A M_λ). 04IQ
Example 3 If an A-module M is flat and Hausdorff with the 𝔤-adic topology it is ideally Hausdorff, for a ⊗_A M is then identified with a submodule of M and the 𝔤-adic topology on a ⊗_A M is… 04IR
Theorem 1 Let A be a commutative ring, 𝔤 an ideal of A and M an A-module. Consider the following properties: (i) M is afat A-module. (ii) Tor₁ᴬ(N, M) = 0 for every A-module N annihilated by… ↩ 5 04IS
Remark If A / 𝔤 is a field (as often happens in applications) the condition "M/ 𝔤M is a flat (A / 𝔤)-module" holds automatically for every A-module M, which simplifies the statement of pr… 04IT
Proposition 1 Let A be a commutative ring, 3 an ideal of A and M an A-module. The following conditions are equivalent: (a) For all n ≥ 1, Tor₁ᴬ(A/𝔤ⁿ, M) = 0. (b) For all n ≥ 1, the canonical hom… 04IU
Proposition 2 Let A be a commutative ring, 𝔤 an ideal of A and B a commutative Noetherian A-algebra such that B is contained in the Jacobson radical of B. Then every finitely generated B-module… ↩ 2 04IV
Proposition 3 Let A be a commutative ring, B a commutative A-algebra, 𝔤 an ideal of A and M a B-module. Suppose that B is a Noetherian ring and a flat A-module and that M is ideally Hausdorff wi… 04IW
Corollary Let A be a commutative Noetherian ring, 𝔍 an ideal of A, hatA the Hausdorff completion of A with respect to the 𝔍-adic topology and M an ideally Hausdorff hatA-module with respect… ↩ 1 04IX
Proposition 4 Let A and B be two commutative Noetherian rings, h : A → B a ring homomorphism, 3 an ideal of A and 2 an ideal of B containing 𝔍B and contained in the Jacobson radical of B. Let ha… 04IY

IV. Associated prime ideals and primary decomposition

§ 1. Prime ideals associated with a module

Definition 1 Let M be a module over a ring A. A prime ideal p is said to be associated with M if there exists x ∈ M such that p is equal to the annihilator & x. The set of prime ideals associat… ↩ 1 04J7
Proposition 1 For every prime ideal p of a ring A and every submodule M ≠ 0 of A/p, Ass(M) = p. ↩ 2 04J8
Proposition 2 Let M be a module over a ring A. Every maximal element of the set of ideals Ann(x) of A, where x runs through the set of elements ≠ 0 of M, belongs to Ass(M). ↩ 6 04J9
Corollary 1 Let M be a module over a Noetherian ring A. Then the condition M ≠ 0 is equivalent to Ass(M) ≠ ∅. ↩ 8 04JA
Corollary 2 Let A be a Noetherian ring, M an A-module and a an element of A. For the homothety on M with ratio a to be injective, it is necessary and sufficient that a belong to no prime ideal… ↩ 1 04JB
Corollary 3 The set of divisors of zero in a Noetherian ring A is the union of the ideals p ∈ Ass(A). ↩ 2 04JC
Proposition 3 Let A be a ring, M an A-module and N a submodule of N. Then Ass(N) ⊂ Ass(M) ⊂ Ass(N) ∪ Ass(M/N). The inclusion Ass(N) ⊂ Ass(M) is obvious. Let p ∈ Ass(M), E be a submodule of M iso… ↩ 9 04JD
Corollary 1 If an A-module M is the direct sum of a family (Mᵢ)_i ∈ I of submodules, then Ass(M) = ⋃_i ∈ I Ass(Mᵢ). ↩ 2 04JE
Corollary 2 Let M be an A-module and (Qᵢ)_i ∈ I a finite family of submodules of M. If ⋂_i ∈ I Qᵢ = 0, then Ass(M) ⊂ ⋃_i ∈ I Ass(M/Qᵢ). The canonical mapping M → ⨁_i ∈ I (M/Qᵢ) is injective; t… ↩ 1 04JF
Proposition 4 Let M be an A-module and Φ a subset of Ass(M). Then there exists a submodule N of M such that Ass(N) = Ass(M) - Φ and Ass(M/N) = Φ. ↩ 3 04JG
Proposition 5 Let A be a ring, S a multiplicative subset of A, Φ the set of prime ideals of A which do not meet S and M an A-module. Then: (i) The mapping p ↦ S⁻¹p is a bijection of Ass_A(M) ∩ Φ… ↩ 2 04JH
Corollary If the ring A is Noetherian, the mapping 𝔭 ↦ S⁻¹𝔭 is a bijection of Ass_A(M) ∩ Φ onto Ass_S⁻¹A(S⁻¹M). ↩ 7 04JI
Proposition 6 Let A be a Noetherian ring, M an A-module, S a multiplicative subset of A and Y the set of elements of Ass_A(M) which do not meet S. Then the kernel N of the canonical mapping M →… ↩ 4 04JJ
Proposition 7 Let A be a ring and M an A-module. ↩ 3 04JK
Corollary 1 If M is a module over a Noetherian ring, then Ass(M) ⊂ Supp(M) and these two sets have the same minimal elements. ↩ 2 04JL
Corollary 2 The nilradical of a Noetherian ring A is the intersection of the ideals 𝔭 ∈ Ass(A). ↩ 1 04JM
Theorem 1 Let A be a Noetherian ring and M a finitely generated A-module. There exists a composition series (Mᵢ)_0 ≤ i ≤ n of M such that, for 0 ≤ i ≤ n - 1, Mᵢ / Mᵢ₊₁ is isomorphic to A / 𝔭… ↩ 3 04JN
Theorem 2 Let M be a finitely generated module over a Noetherian ring A and (Mᵢ)_0 ≤ i ≤ n a composition series of M such that, for 0 ≤ i ≤ n - 1, Mᵢ / Mᵢ₊₁ is isomorphic to A / 𝔭ᵢ where 𝔭ᵢ… ↩ 2 04JO
Corollary If M is a finitely generated module over a Noetherian ring A, Ass(M) is finite. ↩ 1 04JP
Proposition 8 Let A be a Noetherian ring, a an ideal of A and M a finitely generated A-module. The following conditions are equivalent: ↩ 1 04JQ
Proposition 9 Let A be a Noetherian ring, a an ideal of A and M a finitely generated A-module. For there to exist an integer n > 0 such that aⁿ M = 0, it is necessary and sufficient that a be co… ↩ 2 04JR
Definition 2 Given an A-module M, an endomorphism u of M is called almost nilpotent if, for all x ∈ M, there exists an integer n(x) > 0 such that uⁿ⁽ˣ⁾(x) = 0. 04JS
Corollary Let A be a Noetherian ring. M an A-module and a an element of A. For the homomorphism a_M : x ↦ ax of M to be almost nilpotent, it is necessary and sufficient that a belong to ever… 04JT
Proposition 10 Let A be a Noetherian ring, E a finitely generated A-module and F an A-module. Then ↩ 1 04JU

§ 2. Primary decomposition

Proposition 1 Let A be a Noetherian ring and M an A-module. The following conditions are equivalent: (a) Ass(M) is reduced to a single element. ↩ 5 04KE
Definition 1 Let A be a Noetherian ring, N an A-module and Q a submodule of N. If the module M = N/Q satisfies the conditions of Proposition 1, Q is called p-primary with respect to N (or in N)… ↩ 2 04KF
Remark Let q be a p-primary submodule of an A-module N. If N/Q is finitely generated, there exists an integer k > 0 such that pᵏ N ⊂ Q by § 1, no. 4, Proposition 9. 04KG
Example 1 If p is a prime ideal of A, p is p-primary (§ 1, no. 1, Proposition 1). 04KH
Example 2 Let q be an ideal of A such that there exists a single prime ideal m (necessarily maximal) containing q; then, if M is an A-module such that qM ≠ M, qM is m-primary with respect to… ↩ 1 04KI
Example 3 Let m be a maximal ideal of A; the m-primary ideals are then the ideals q of A for which there exists an integer n ≥ 1 such that mⁿ ⊂ q ⊂ m. For if mⁿ ⊂ q ⊂ m, m is the only prime… ↩ 1 04KJ
Example 4 In a principal ideal domain A, the primary ideals are (0) and the ideals of the form Apⁿ, where p is an extremal element and n ≥ 1; this follows immediately from Example 3. ↩ 2 04KK
Example 5 The powers of any prime ideal are not necessarily primary ideals (Exercise 1). On the other hand, there exist primary ideals which are not powers of prime ideals (Exercise 1). 04KL
Proposition 2 Let M be a module over a Noetherian ring A, p a prime ideal of A and (Qᵢ)_i ∈ I a non-empty finite family of submodules of M which are p-primary with respect to M. Then ⋂_i ∈ I Qᵢ… ↩ 4 04KM
Proposition 3 Let A be a Noetherian ring, S a multiplicative subset of A, p a prime ideal of A, M an A-module, N a submodule of M and i = i_A^S the canonical mapping of M to S⁻¹M. ↩ 5 04KN
Definition 2 Let A be a Noetherian ring, M an A-module and N a submodule of M. A finite family (Qᵢ)_i ∈ I of submodules of M which are primary with respect to M and such that N = ⋂_i ∈ I Qᵢ is… 04KO
Example Let us take A = Z, M = Z, N = nZ for some integer n > 0. If n = p₁^α₁ ⋯ pₖ^αₖ is the decomposition of n into prime factors, nZ = (p₁^α₁Z) ∩ … ∩ (pₖ^αₖZ) is a primary decomposition… 04KP
Theorem 1 Let M be a finitely generated module over a Noetherian ring and let N be a submodule of M. There exists a primary decomposition of N of the form N = ⋂_p ∈ Ass(M/N) Q(p) where, for… ↩ 4 04KQ
Definition 3 Let M be a module over a Noetherian ring and N a submodule of M. A primary decomposition N = ⋂_i ∈ I Qᵢ of N in M is called reduced if the following conditions are fulfilled: 04KR
Proposition 4 Let M be a module over a Noetherian ring, N a submodule of M, N = ⋂_i ∈ I Qᵢ a primary decomposition of N in M and, for all i ∈ I, let pᵢ = Ass(M/Qᵢ). For this decomposition to be… ↩ 7 04KS
Corollary Let A be a Noetherian ring, M an A-module, N a submodule of M and (Qᵢ)_i ∈ I a primary decomposition of N in M. Then Card(I) ≥ Card(Ass(M/N)); for (Qᵢ)_i ∈ I to be a reduced primar… ↩ 2 04KT
Proposition 5 Let A be a Noetherian ring, M an A-module, N a submodule of M, N = ⋂_i ∈ I Qᵢ a reduced primary decomposition of N in M and, for all i ∈ I, let pᵢ = Ass(M/Qᵢ). If pᵢ is a minimal e… ↩ 3 04KU
Remark The prime ideals pᵢ ∈ Ass(M/N) which are not minimal elements of this set are sometimes called the immersed prime ideals associated with M/N; if M/N is finitely generated, for p₀ ∈… 04KV
Proposition 6 Let A be a Noetherian ring, M an A-module, N a submodule of M and I = Ass(M/N). Let S be a multiplicative subset of A and J the subset of I consisting of the indices i such that S… ↩ 3 04KW
Corollary The mappings 04KX
Proposition 7 Let M be a finitely generated module over a Noetherian ring A. The following properties are equivalent: (a) M is of finite length. (b) Every ideal p ∈ Ass(M) is a maximal ideal of… ↩ 6 04KY
Corollary 1 For every module of finite length M over a Noetherian Zing A, Ass(M) = Supp(M). 04KZ
Corollary 2 Let M be a finitely generated module over a Noetherian ring A and p a prime ideal of A. For Mₚ to be a non-zero Aₚ-module of finite length, it is necessary and sufficient that p be… ↩ 1 04L0
Remark (1) Let M be a finitely generated module over a Noetherian ring A; let (M₁), … be a composition series of M such that, for 0 ≤ i ≤ n - 1, Mᵢ / Mᵢ₊₁ is isomorphic to A/pᵢ, where pᵢ is a… 04L1
Proposition 8 Let M be a module of finite length over a Noetherian ring A. ↩ 3 04L2
Corollary If M is a module of finite length over a Noetherian ring A, then 04L3
Proposition 9 Let A be a Noetherian ring. The following conditions are equivalent : (a) A is Artinian. (b) All the prime ideals 𝔞 in A are maximal ideals. (c) All the elements of Ass(A) are maxi… ↩ 8 04L4
Remark (2) The conditions of Proposition 9 for a Noetherian ring A imply that the spectrum of A is finite and discrete, every point of Spec(A) being therefore closed (Chapter 11, § 4, no. 3… 04L5
Corollary 1 Every Artinian ring A is isomorphic to the direct composition of a finite family of Artinian local rings. 04L6
Remark (3) This corollary can also be deduced from the fact that Spec(A) is finite and discrete and Chapter 11, § 4, no. 3, Proposition 15. 04L7
Corollary 2 Let A be a Noetherian ring and m an ideal of A. The following conditions are equivalent: ↩ 2 04L8
Corollary 3 Let A, A' be two rings and h a homomorphism from A to A'. Suppose that A is semi-local and Noetherian and that A' is a finitely generated A-module. Then the ring A' is semi-local a… ↩ 1 04L9
Corollary 4 Let A be a complete semi-local Noetherian ring, m a defining ideal of A, E a finitely generated A-module and (Fₙ) a decreasing sequence of submodules of E such that ⋂ₙ Fₙ = 0. Then… ↩ 1 04LA
Corollary 5 In a complete semi-local Noetherian ring every decreasing sequence φ ideals whose intersection is 0 is a filter base which converges to 0. 04LB
Proposition 10 Let A be a Noetherian ring and p₁, …, pₙ the prime ideals associated with the A-module A, where pᵢ ≠ pⱼ for i ≠ j. ↩ 1 04LC
Lemma 1 Let A be a Noetherian ring, p a prime ideal of A, E an A-module whose annihilator contains a power of p and such that Ass(E) = p and F a B-module such that ρ_*(F) is a flat A-modul… ↩ 2 04LD
Theorem 2 Let ρ : A → B be a ring homomorphism, E an A-module and F a B-module such that ρ_*(F) is a flat A-module. Then ↩ 6 04LE
Corollary 1 If A is Noetherian and 𝔓 ∈ Ass_B(E ⊗_A F), then ρ⁻¹(𝔓) ∈ Ass_A(E) and ρ⁻¹(𝔓) is the only prime ideal p of A such that 𝔓 ∈ Ass_B(F/pF). ↩ 1 04LF
Corollary 2 Suppose that A and B are Noetherian and that B is a flat A-module. Let p be a prime ideal of A, Q ⊂ E a p-primary submodule and 𝔓 a prime ideal of B. For Q ⊗_A B to be a q-primary… ↩ 1 04LG
Remark Suppose that A and B are Noetherian. Let 𝔓 be a prime ideal of B and let p = overlineρ⁻¹(𝔓); let us write S = A - p and let k(p) = S⁻¹(A/p) be the field of fractions of A/p. Since… 04LH
Proposition 11 Suppose that A and B are Noetherian and that B is a flat A-module. Let E be an A-module and E' a submodule of E such that, for every ideal p ∈ Ass_A(E/E'), pB is a prime ideal of B… ↩ 2 04LI
Corollary Suppose that pB is prime for all p ∈ Ass_A(E/E'). Then, if p₁, …, pₙ are the minimal elements of Ass_A(E/E'), the pᵢB are minimal elements of Ass_A((E ⊗_A B)/(E' ⊗_A B)). It follow… 04LJ

§ 3. Primary decomposition in graded modules

Proposition 1 Let A be a torsion-free commutative group, A a graded ring of type A and M a graded A-module of type A. Every prime ideal associated with M is graded and is the annihilator of a ho… ↩ 4 04MJ
Corollary For every (necessarily graded) prime ideal p associated with a graded A-module M, there exists an index k ∈ A such that the graded A-module (A/p)(k) obtained from the graded A-modu… ↩ 1 04MK
Proposition 2 Let A be a torsion-free commutative group, A a graded Noetherian ring of type A and M a graded finitely generated A-module of type A. There exists a composition series (Mᵢ)_0 ≤ i ≤… 04ML
Proposition 3 Let A be a torsion-free commutative group, A a graded Noetherian ring of type A, p a graded ideal of A and M a graded A-module of type A not reduced to 0. Suppose that for every ho… ↩ 1 04MM
Proposition 4 Let A be a torsion-free commutative group, A a graded Noetherian ring of type A and M a graded A-module of type A. Let p be a prime ideal of A and N, a submodule of M which is p-pr… ↩ 1 04MN
Proposition 5 Let A be a torsion-free commutative group, A a graded Noetherian ring of type A, M a graded A-module of type M, N a graded submodule of M and N = ⋂_i ∈ I Qᵢ a primary decomposition… 04MO

V. Integers

§ 1. Notion of an integral element

Theorem 1 Let A be a (commutative) ring, R an algebra over A (not necessarily commutative) and x an element of R. The following properties are equivalent: ↩ 4 04MS
Lemma 1 Let A be a ring, R an algebra (not necessarily commutative) over A and x an element of R. Let M be a faithful module over A[x] which is a finitely generated A-module. If q is an id… 04MT
Definition 1 Let A be a ring and R an A-algebra (not necessarily commutative). An element x ∈ R is called integral over A if it satisfies the equivalent properties (E₁), (E_II), (E_III) of Theo… 04MU
Example 1 Let K be a (commutative) field and R a K-algebra; to say that an element x ∈ R is integral over K is equivalent to saying that x is a root of a non-constant polynomial in the ring… ↩ 2 04MV
Example 2 Let R' be a sub-A-algebra of R; the elements of R' which are integral over A are just the elements of R which are integral over A and belong to R'; this often allows us not to spec… 04MW
Proposition 1 Let A be a ring, R an algebra over A (not necessarily commutative) and x an element of R. For x to be integral over A, it is necessary and sufficient that A[x] be contained in a su… ↩ 3 04MX
Corollary Let A be a Noetherian ring, R an A-algebra (not necessarily commutative) and x an element of R. For x to be integral over A, it is necessary and sufficient that there exist a finit… ↩ 1 04MY
Definition 2 Let A be a ring. An A-algebra R (not necessarily commutative) is called integral over A if every element of R is integral over A. R is called finite over A if R is a finitely gener… 04MZ
Proposition 2 Let A, A' be two rings, R an A-algebra, R an A'-algebra (not necessarily commutative) and f : A → A' and g : R → R' two ring homomorphisms such that the diagram ↩ 12 04N0
Corollary 1 Let A be a ring, B a (commutative) A-algebra and C a (not necessarily commutative) B-algebra. Then every element x ∈ C which is integral over A is integral over B. 04N1
Corollary 2 Let K be a field, L an extension of K and x, x' two elements of L which are conjugate over K (Algebra, Chapter V, § 6, no. 2). If A is a subring of K and x is integral over A, x' i… 04N2
Corollary 3 Let A be a ring, B a (commutative) A-algebra and C a (not necessarily commutative) B-algebra. If C is integral over A, C is integral over B. 04N3
Proposition 3 Let (Rᵢ)_1 ≤ i ≤ n be a finite family of (not necessarily commutative) A-algebras and let R = ∏ᵢ₌₁ⁿ Rᵢ be their product. For an element x = (xᵢ)_1 ≤ i ≤ n of R to be integral over… ↩ 1 04N4
Proposition 4 Let A be a ring, R an A-algebra (not necessarily commutative) and (xᵢ)_1 ≤ i ≤ n a finite family of elements of R which are pairwise permutable. If, for all i, xᵢ is integral over… ↩ 10 04N5
Corollary 1 Let A be a ring and R a (Commutative)A-algebra. The set of elements & R integral over A is a subalgebra of R. 04N6
Corollary 2 Let A be a ring, R an A-algebra (not necessarily commutative) and E a set & elements & R which are pairwise permutable and integral over A. Then the sub-A-algebra B & R generated b… 04N7
Proposition 5 Let A be a ring and A and R two (commutative)A-algebras. If R is integral over A, R ⊗_A A' is integral over A'. ↩ 6 04N8
Corollary Let R be a ring and A, B, C subrings & R such that A ⊂ B. If B is integral over A, C[B] is integral over C[A]. ↩ 1 04N9
Proposition 6 Let A be a ring, B a (commutative)A-algebra and C a (not necessarily commutative)B-algebra. If B is integral over A and C is integral over B, then C is integral over A. ↩ 6 04NA
Corollary Let A be a ring and R, R' two (commutative)A-algebras integral over A. Then R ⊗_A R' is integral over A. 04NB
Definition 3 Let A be a ring and R a (commutative)A-algebra. The sub-A-algebra A' of R consisting of the elements of R integral over A (no. 1, Corollary 1 to Proposition 4) is called the integr… ↩ 2 04NC
Definition 4 If A is an integral domain, the integral closure of A in its field of fractions is called the integral closure of A. An integral domain is called integrally closed if it is equal t… 04ND
Proposition 7 Let A be a ring and R an A-algebra. The integral closure A' of A in R is a subring integrally closed in R. ↩ 3 04NE
Corollary The integral closure of an integral domain A is an integrally closed domain. 04NF
Proposition 8 Let R be a ring, (B_λ)_λ ∈ L a family of subrings of R and for each λ ∈ L let A_λ be a subring of B_λ. If each A_λ is integrally closed in B_λ, then A = ⋂_λ ∈ L A_λ is integrally c… ↩ 1 04NG
Corollary Every intersection of a non-empty family of integrally closed subdomains of an integral domain A is an integrally closed domain. ↩ 3 04NH
Proposition 9 Let A be a ring, (Rᵢ)_1 ≤ i ≤ n a finite family of A-algebras and A'ᵢ the integral closure of A in Rᵢ, (1 ≤ i ≤ n). Then the integral closure of A in R = ∏ᵢ₌₁ⁿ Rᵢ, is equal to ∏ᵢ₌₁… 04NI
Corollary 1 Let A be a reduced Noetherian ring, pᵢ (1 ≤ i ≤ n) its distinct minimal prime ideals, K, the field of fractions of the integral domain A/pᵢ (canonically isomorphic to the local rin… 04NJ
Corollary 2 For a reduced Noetherian ring to be integrally closed in its ring of fractions it is necessary and sufficient that it be a direct composition of integrally closed (Noetherian) doma… 04NK
Proposition 10 Every principal ideal domain is integrally closed. ↩ 4 04NL
Lemma 2 Let R be a ring and P a monic polynomial in R[X]. There exists a ring R' containing R such that in the polynomial ring R'[X] the polynomial P is a product of monic polynomials of d… ↩ 1 04NM
Proposition 11 Let A be a ring, R an A-algebra and P and Q monic polynomials in R[X]. If the coefficients of PQ are integral over A, the coefficients of P and Q are integral over A. ↩ 4 04NN
Corollary Let A be an integral domain, K its field of fractions and x an element of a K-algebra K (not necessarily commutative). If x is integral over A, the coefficients of the minimal poly… ↩ 3 04NO
Proposition 12 Let A be a ring, R an A-algebra and P a polynomial in R[X₁, …, Xₙ]. For P to be integral over A[X₁, …, Xₙ], it is necessary and sufficient that the coefficients of P be integral ov… ↩ 3 04NP
Proposition 13 Let A be a ring, R an A-algebra and A' the integral closure of A in R. Then the integral closure of A[X₁, …, Xₙ] in R[X₁, …, Xₙ] is equal to A'[X₁, …, Xₙ]. ↩ 2 04NQ
Corollary 1 Let A be an integral domain and A' its integral closure. Then the integral closure of the polynomial ring A[X₁, …, Xₙ] is A'[X₁, …, Xₙ]. 04NR
Corollary 2 Let A be an integral domain. For the polynomial ring A[X₁, …, Xₙ] to be integrally closed, it is necessary and sufficient that A be integrally closed. ↩ 2 04NS
Corollary 3 If K is a field, every polynomial algebra K[X₁, …, Xₙ] is an integrally closed domain. 04NT
Definition 5 An integral domain A is called completely integrally closed if the following condition is satisfied: every element x of the field of fractions K of A such that all the powers xⁿ (n… ↩ 1 04NU
Proposition 14 Let A be a completely integrally closed domain. Then every polynomial ring A[X₁, …, Xₙ] (resp. every ring of formal power series A[[X₁, …, Xₙ]]) is completely integrally closed. ↩ 1 04NV
Proposition 15 Let A be a filtered ring whose filtration is exhaustive and such that every principal ideal of A is closed under the topology defined by the filtration. If the associated graded ri… 04NW
Proposition 16 Let A be a ring, R an A-algebra, A' the integral closure of A in R and S a multiplicative subset of A. Then the integral closure of S⁻¹A in S⁻¹R is S⁻¹A'. ↩ 6 04NX
Corollary 1 Let A be an integral domain, A' its integral closure and S a multiplicative subset of A such that 0 ∉ S. Then the integral closure of S⁻¹A is S⁻¹A'. ↩ 4 04NY
Corollary 2 Let A be an integral domain, K its field of fractions, R an algebra over K and B the integral closure of A in R. The elements of R which are algebraic over K (no. 1, Example 1) are… ↩ 1 04NZ
Corollary 3 Let A be an integral domain and Ω the set of maximal ideals of A. For A to be integrally closed, it is necessary and sufficient that, for all m ∈ Ω, Aₘ be integrally closed. 04O0
Corollary 4 Let A be an integral domain, K its field of fractions and S a multiplicative subset of A such that 0 ∉ S. 04O1
Corollary 5 Let A be an integral domain, A' its integral closure and f the annihilator of the A-module A'/A. Suppose that A' is afinitely generated A-module. The prime ideals p ∈ A such that A… 04O2
Proposition 17 Let A be a ring, B a (commutative) A-algebra and X a square matrix of order n over B; the following properties are equivalent: ↩ 3 04O3
Corollary 1 Let A be an integral domain, K its field of fractions and K' a finite-dimensional K-algebra (not necessarily commutative). If x \in K' is integral over A, the coefficients of the c… ↩ 1 04O4
Corollary 2 With the same hypotheses and notation as in Corollary 1, Tr{K'/K}(x) and N{K'/K}(x) are integral over A. ↩ 1 04O5
Remark (1) If K' is a simple central algebra over K and x \in K' is integral over A, the coefficients of the reduced characteristic polynomial of x (Algebra, Chapter VIII, § 12, no. 3) are in… 04O6
Proposition 18 Let A be an integrally closed domain, K its field of fractions, K' a finite-dimensional separable K-algebra (Algebra, Chapter VIII, § 7, no. 5) and A' the integral closure of A in… ↩ 1 04O7
Lemma 3 Under the hypotheses of Proposition 18, let (w1, ..., wn) be a basis of K' over K contained in A' (no. 5, Corollary 2 to Proposition 16); then there is a unique basis (w1^, ..., wn… ↩ 1 04O8
Corollary 1 If A is a Noetherian ring, the A-module A' is finitely generated and in particular the ring A' is Noetherian. ↩ 2 04O9
Corollary 2 If A is a principal ideal domain, A' is a free A-module of rank n. Every submodule of a free A-module is then free (Algebra, Chapter VII, § 3, Theorem 1). ↩ 2 04OA
Corollary 3 Let E be an extension of degree n of the field Q of rational numbers. The additive group of the integral closure in E of the ring Z of rational integers is a free commutative group… 04OB
Proposition 19 Let k be a field, L a separable extension of k and R an integrally closed k-algebra. If the ring L ⊗ₖ R is an integral domain, it is integrally closed. ↩ 1 04OC
Corollary Let k be a field and R and S two integrally closed k-algebras. Suppose that the ring R ⊗ₖ S is an integral domain and that the fields of fractions K and L of R and S respectively a… 04OD
Proposition 20 Let A = ⨁_i ∈ Z Aᵢ, be a graded ring and B a (commutative) graded A-algebra. The set A' of elements of B integral over A is a graded subalgebra of B. ↩ 2 04OE
Lemma 4 Let A = ⨁_i ∈ Z Aᵢ be a graded integral domain and S the set of homogeneous elements ≠ 0 of A. ↩ 2 04OF
Proposition 21 Let A = ⨁_i ∈ Z Aᵢ be a graded integral domain and S the set of homogeneous elements ≠ 0 of A. The integral closure A' of A is then a graded subring of S⁻¹A. If further Aᵢ = 0 for… ↩ 3 04OG
Corollary 1 With the hypotheses and notation of Proposition 21, if every homogeneous element of S⁻¹A which is integral over A belongs to A, then A is integrally closed. ↩ 1 04OH
Corollary 2 If A = ⨁_i ∈ Z Aᵢ is an integrally closed graded domain, the domain A₀ is integrally closed. 04OI
Corollary 3 Let A = ⨁_i ∈ Z Aᵢ be an integrally closed graded domain. Then, for every integer d > 0, the ring A⁽ᵈ⁾ (Chapter III, § 1, no. 3) is an integrally closed domain. 04OJ
Proposition 22 Let A be a (commutative) K-algebra and G a locally finite group of operators on A. Then A is integral over the subalgebra Aᴳ. ↩ 2 04OK
Theorem 2 Let A be a finitely generated K-algebra and G a locally finite group of operators on A. Then A is a finitely generated A-module; if further K is Noetherian, Aᴳ is affinitely genera… 04OL
Lemma 5 Let K be a Noetherian ring, B a finitely generated K-algebra and C a sub-K-algebra of B such that B is integral over C. Then C is a finitely generated A-algebra. 04OM
Remark The set of σ ∈ G such that σ aⱼ = aⱼ for 1 ≤ j ≤ m obviously leaves invariant every element of A. The normal subgroup H of G leaving invariant every element of A is therefore of fi… 04ON
Proposition 23 Let A be a K-algebra, G a locally finite group of operators on A, S a multiplicative subset of A stable under G and Sᴳ the set S ∩ A''. Then the canonical mapping of (Sᴳ)⁻¹A to S⁻¹… ↩ 5 04OO
Corollary Let A be an integral domain, K its field of fractions and G a locally finite group of operators on A. Then G operates on K and Kᴳ is the field of fractions of Aᴳ. 04OP

§ 2. The lift of prime ideals

Definition 1 Let A, A' be two rings and h : A → A' a ring homomorphism. An ideal a' ∈ A' is said to lie above an ideal a of A if a = overlineh¹(a'). 04PJ
Lemma 1 Let h : A → A' be a ring homomorphism, S a multiplicative subset of A, i = i_A^S : A → S⁻¹A, i' = i_A'^h(S) : A' → S⁻¹A' = (h(S))⁻¹A' the canonical homomorphisms and h₁ = S⁻¹h : S⁻… ↩ 3 04PK
Proposition 1 Let h : A → A' be a ring homomorphism such that A' is integral over A, 𝔭' a prime ideal of A' and 𝔭 = h⁻¹(𝔭'). For 𝔭 to be maximal, it is necessary and sufficient that 𝔭' be so. ↩ 12 04PL
Lemma 2 Let B be an integral domain and A a subring of B such that B is integral over A. For B to be a field, it is necessary and sufficient that A be a field. ↩ 1 04PM
Corollary 1 Let h : A → A' a ring homomorphism such that A' is integral over A, 𝔭 a prime ideal of A and 𝔭' and a' two ideals of A' lying above 𝔭 such that 𝔭' ⊂ a'. If 𝔭' is prime, then a' = 𝔭… ↩ 5 04PN
Corollary 2 Let A' be an integral domain, A a subring of A' such that A' is integral over A and f a homomorphism from A' to a ring B. If the restriction off to A is injective f is injective. ↩ 1 04PO
Corollary 3 Let h : A → A' be a ring homomorphism such that A' is integral over A and m a maximal ideal of A and suppose that there are in A' only a finite number of distinct maximal ideals m'… 04PP
Remark (1) If A' is Noetherian, it follows from (i) and (ii) that (q'ⱼ)_1 ≤ j ≤ n is the unique primary decomposition of mA' (Chapter IV, § 2, no. 3). 04PQ
Theorem 1 Let h : A → A' be an injective ring homomorphism such that A is integral over A and p a prime ideal of A. There exists a prime ideal p' of A' lying above p. ↩ 15 04PR
Corollary 1 With the hypotheses and notation of Theorem 1, barh⁻¹(pA') = p. 04PS
Corollary 2 Let h : A → A' be a ring homomorphism such that A' is integral over A, a and p two ideals of A such that a ⊂ p and a' an ideal of A' lying above a. Suppose that p is prime. Then th… ↩ 3 04PT
Corollary 3 Let A be a ring and A' a ring containing A and integral over A. If R' is the Jacobson radical of A', R' ∩ A is the Jacobson radical of A. 04PU
Corollary 4 Let A be a ring, A' a ring containing A and integral over A and f a homomorphism from A to an algebraically closed field L. Then f can be extended to a homomorphism from A' to L. ↩ 2 04PV
Remark (2) Let h : A → A' be a ring homomorphism such that A' is integral over A; then the associated continuous mapping ᵃ h : Spec(A') → Spec(A) is closed. For every ideal a' of A', A'/a' is… ↩ 2 04PW
Proposition 2 Let h : A → A' be a ring homomorphism such that A' is integral over A, 𝔭 a prime ideal of A, S = A - 𝔭, (p'ᵢ)_i ∈ I the family of all the prime ideals 𝔭 of A' lying above 𝔭 and S'… ↩ 3 04PX
Proposition 3 Let h : A → A' be a ring homomorphism such that A' is a finitely generated A-module; then, for every prime ideal 𝔭 of A, the set of prime ideals of A' lying above 𝔭 is finite. 04PY
Definition 2 Let A' be a ring and G a group operating on A' (§ 1, no. 9). Given a prime ideal p' ⊂ A' the subgroup of elements σ ∈ G such that σ.p' = p' is called the decomposition group of p'… ↩ 3 04PZ
Definition 3 With the notation of Definition 2, the subgroup of Gᶻ(p') consisting of those σ such that the endomorphism z ↦ σ.z of A'/p' is the identity is called the inertia group of p' (with… 04Q0
Lemma 3 Let A' be a ring, G a group operating on A', A the ring of invariants of G, p' a prime ideal of A' and S a multiplicative subset of A not meeting p'. Then G^Z(S⁻¹p') = G^Z(p'), G^Γ… ↩ 2 04Q1
Theorem 2 Let A' be a ring, G a finite group operating on A' and A the ring of invariants of G, so that A' is integral over A (§ 1, no. 9, Proposition 22). ↩ 9 04Q2
Corollary With the hypotheses and notation Of Theorem 2, let f₁, f₂ be two homomorphisms of A to a field L with the same restriction to A. Then there exists σ ∈ G such that ↩ 1 04Q3
Proposition 4 Let A' be a ring, G a finite group operating on A', H a subgroup of G, A and B the rings of invariants of G and H respectively and p' a prime ideal of A; let us write p = A ∩ p' an… ↩ 6 04Q4
Corollary 1 Under the hypotheses Of Theorem 2 the rings A/p and A^Z/(p' ∩ A^Z) have the same field Of fractions and the maximal ideal of the local ring (A^Z)'_p' ∩ A^Z is generated by p. 04Q5
Corollary 2 Let A' be an integral domain, G a finite group operating on A', A' the ring of invariants of G and p' a prime ideal of A'; let K, K^Z and K be the fields of fractions of A, A^Z and… ↩ 1 04Q6
Definition 4 With the hypotheses and notation of Corollary 2 to Proposition 4, a prime ideal 𝔭 of A is said to decompose completely in K' if the number of prime ideals of A' lying above 𝔭 is eq… 04Q7
Corollary 3 Let A' be an integral domain, G a finite commutative group operating on A', A the ring of invariants of G, 𝔭 a prime ideal of A and K and K' the fields of fractions of A and A' res… 04Q8
Proposition 5 With the hypotheses and notation of Theorem 2, the field of fractions kᵀ of Aᵀ/(𝔭' ∩ Aᵀ) is equal to the greatest separable extension k'ₛ of k contained in k'. ↩ 4 04Q9
Corollary If the order of the inertia group Gᵀ(p') is relatively prime to the characteristic exponent p of k, the field k' is a Galois extension of k. 04QA
Lemma 4 Let A be an integrally closed domain, K its field of fractions, p the characteristic exponent of K, K' a radicial extension of K and A' a subring of K' containing A and integral ov… ↩ 2 04QB
Remark (1) It follows from § 1, no. 3, Corollary to Proposition 11 that the integral closure of A in K' is the set of x ∈ K' such that there exists m ≥ 0 for which x^pᵐ ∈ A (Algebra, Chapter… 04QC
Proposition 6 Let A be an integrally closed domain, K its field of fractions, K' a quasi-Galois extension of K and A' the integral closure of A in K'. Then: ↩ 7 04QD
Corollary 1 Under the hypotheses of Proposition 6, let f be a homomorphism from A to a field L and g1, g2 two homomorphisms from A' to L which extend f. Then there exists a K-automorphism σ ∈… ↩ 1 04QE
Corollary 2 Let A be an integrally closed domain, K its field of fractions, K' a finite algebraic extension of K and A' a subring of K' containing A and integral over A. ↩ 1 04QF
Corollary 3 With the hypotheses and notation of Corollary 2, if A is semi-local, so is A'. 04QG
Corollary 4 Let A be an integrally closed domain, K its field of fractions, K a Galois extension of K, A' the integral closure of A in K', 𝔭' a prime ideal of A, 𝔭 = A ∩ 𝔭' and k and k' the fi… ↩ 1 04QH
Definition 5 With the hypotheses and notation of Proposition 6, the field of invariants K^Z(p') (resp. Kᵀ(p')) of the group G^Z(p') (resp. Gᵀ(p')) in the field K' is called the decomposition fi… 04QI
Proposition 7 Let A be an integrally closed domain, K its field of fractions, K' a Galois extension of K, G its Galois group, A' the integral closure of A in K', p' a prime ideal of A' and p = A… 04QJ
Theorem 3 Let A be an integrally closed domain and A' a ring containing A and integral over A. Suppose that 0 is the only element of A which is a divisor of 0 in A'. Let p, q be two prime id… ↩ 4 04QK
Corollary Under the hypotheses on A and A' of Theorem 3, let p be a prime ideal of A. The prime ideals of A' lying above p are the minimal elements of the set E of prime ideals of A' contain… 04QL

§ 3. Finitely generated algebras over a field

Theorem 1 (Normalization Lemma) Let A be a finitely generated k-algebra and let a₁ ⊂ a₂ ⊂ … ⊂ aₚ be an increasing finite sequence of ideals of A such that p ≥ 1 and aₚ ≠ A. There exists a finite sequence (xᵢ)_1 ≤… ↩ 4 04R8
Corollary 1 Let A be an integral domain and B a finitely generated A-algebra containing A as a subring. Then there exist an element s ≠ 0 of A and a subalgebra B' of B isomorphic to a polynomi… ↩ 4 04R9
Corollary 2 Let K be a field, A a subring of K and L the field of fractions of A. If K is a finitely generated A-algebra, [K:L] is finite and there exists a ≠ 0 in A such that L = A[a⁻¹]. ↩ 1 04RA
Corollary 3 Let A be an integral domain, B a finitely generated A-algebra and b an element of B such that zbⁿ ≠ 0 for all z ≠ 0 in A and every integer n > 0. Let φ : A → B be the canonical hom… ↩ 2 04RB
Theorem 2 Let A be a finitely generated integral k-algebra, K its field of fractions and A' the integral closure of A in a field K' which is a finite algebraic extension of K. Then A' is a f… ↩ 2 04RC
Proposition 1 Let A be a finitely generated algebra over a field k and L the algebraic closure of k. ↩ 2 04RD
Lemma 1 Let A be a finitely generated algebra over a field k, (aᵢ)_1 ≤ i ≤ n a system of generators of this algebra and c the ideal of algebraic relations between the aᵢ with coefficients… 04RE
Proposition 2 (Hilbert’s Nullstellensatz) Let k be a field and L an algebraic closure of k. 04RF
Definition 1 A ring A is called a Jacobson ring if every prime ideal of A is the intersection of a family of maximal ideals. 04RG
Proposition 3 For a ring A to be a Jacobson ring, it is necessary and sufficient that, for every ideal a of A, the Jacobson radical of A/a be equal to its nilradical (Chapter II, § 2, no. 6). ↩ 1 04RH
Corollary Let A be a Jacobson ring; for every ideal a of A, the radical of a is the intersection of the maximal ideals of A containing a. ↩ 1 04RI
Proposition 4 Let A be a principal ideal domain and (p_λ)_λ ∈ L a representative system of extremal elements of A (Algebra, Chapter VII, § 1, no. 3, Definition 2). For A to be a Jacobson ring, i… ↩ 2 04RJ
Proposition 5 Let A be a ring and B an A-algebra integral over A. If A is a Jacobson ring, so is B. ↩ 1 04RK
Theorem 3 Let A be a Jacobson ring, B a finitely generated A-algebra and φ : A → B the canonical homomorphism. Then: (i) B is a Jacobson ring. (ii) For every maximal ideal m' of B, m = φ⁻¹(m… ↩ 3 04RL
Corollary 1 Every finitely generated algebra A over Z is a Jacobson ring; for a prime ideal p ⊂ A to be maximal, it is necessary and sufficient that the ring A/p be finite. ↩ 1 04RM
Corollary 2 Let (P_λ)_λ ∈ L be a family of polynomials in Z[X₁, …, Xₙ] and let Q be a polynomial in Z[X₁, …, Xₙ] such that, for every system of elements (xᵢ)_1 ≤ i ≤ n belonging to a finite fi… 04RN
Corollary 3 Let A be a Jacobson ring. If there exists a finitely generated A-algebra B containing A and which is a field, then A is a field and B is an algebraic extension of A. 04RO

VI. Valuations

§ 1. Valuation rings

Definition 1 Let A and B be two local rings. B is said to dominate A if A is a subring of B and m(A) = A ∩ m(B). 04RZ
Proposition 1 Let A and B be local rings such that A is a subring of B. The following conditions are equivalent: (a) m(A) ⊂ m(B); (b) B dominates A; (c) the ideal Bm(A) generated by m(A) in B do… ↩ 1 04S0
Theorem 1 Let K be a field and V a subring of K. The following conditions are equivalent: (a) V is a maximal element of the set of local subrings of K, this set being ordered by the relation… ↩ 7 04S1
Lemma 1 Let A be a subring of a ring B and x an element of B not integral over A; then the ring of fractions B, (Chapter II, § 5, no. 1) is not reduced to O and there exists in the subring… ↩ 1 04S2
Definition 2 In the notation of Theorem 1, V is called a valuation ring for the field K if the equivalent conditions (a), (b), (c), (d), (e) hold. A ring is called a valuation ring if it is an… ↩ 1 04S3
Theorem 2 Let K be a field and h a homomorphism from a subring A of K to an algebraically closed field L. Then there exist a valuation ring V for K and a homomorphism h' from V to L such tha… ↩ 2 04S4
Corollary Every local subring A of a field K is dominated by at least one valuation ring of K. ↩ 3 04S5
Theorem 3 Let A be a subring of a field K. The integral closure A' of A in K is the intersection of the valuation rings of K which contain A; if A is local, A' is the intersection of the val… ↩ 5 04S6
Corollary 1 Every valuation ring is integrally closed. 04S7
Corollary 2 For an integral domain to be integrally closed, it is necessary and sufficient that it be the intersection of a family of valuation rings of its field of fractions. 04S8
Corollary 3 Let K be a field, K' an extension of K and A a valuation ring of K. The integral closure of A in K' is the intersection of the valuation rings V' of K such that V' ∩ K = A. ↩ 3 04S9
Proposition 2 Let A be a local ring whose maximal ideal is a principal ideal Ap. If ⋂ₙ₌₁^∞ Apⁿ = (0) (for example if A is Noetherian, cf. Chapter III, § 3, no. 2, Corollary to Proposition 5), th… ↩ 3 04SA
Proposition 3 Let A be a principal ideal domain and K its field of fractions. The valuation rings of K containing A and distinct from K are the rings of the form A_Ap, where p is an extremal ele… ↩ 3 04SB
Corollary 1 Every valuation ring of the field Q and distinct from Q is of the form Z₍ₚ₎ where p is a prime number. ↩ 1 04SC
Corollary 2 Let K be a field, K(X) the field of rational functions in one indeterminate over K and V a valuation ring of K(X) containing K and distinct from K(X). If X ∈ V, there exists an irr… ↩ 1 04SD

§ 2. Places

Definition 1 Let E and E' be two sets each with an internal law of composition denoted by (x, y) ↦ x * y, not necessarily everywhere defined. A mapping f : E → E' is a morphism if, for all x, y… 04SN
Definition 2 Let E and E' be two sets each with a family of internal laws of composition (x, y) ↦ x *_α y, α ∈ I. A mapping f : E → E' is a morphism if it is a morphism for each of the laws of… 04SO
Definition 3 Let K and L be two fields. Every morphism f of K to tildeL (for addition and multiplication) such that f(1) = 1 is called a place of K with values in L. 04SP
Proposition 1 Let K be a field, A a valuation ring of K and κ(A) the residue field of A. We extend the canonical mapping of A onto κ(A) to a mapping h_A : tildeK → (κ(A)) by the equation h_A(x)… ↩ 1 04SQ
Proposition 2 Let K and L be two fields and f a place of K with values in L. Then there exists a valuation ring A of K and an isomorphism j of κ(A) onto a subfield of L such that f = j ∘ h_A; th… 04SR
Proposition 3 Let K be a field, S a subring of K and f a homomorphism from S to an algebraically closed field L. Then there exists a place of K with values in L which extends f. ↩ 1 04SS
Proposition 4 Let K be a field, f a place of K with values in a field L and K' an extension of K. Then there exists an extension L' of L and a place f' of K' with values in L' which extends f. I… ↩ 1 04ST
Proposition 5 Let K be a field, S a subring of K, h a homomorphism from S to a field and p the kernel of h. For an element x of K to be integral over the local ring S, it is necessary and suffic… 04SU
Proposition 6 Let K be a field and S a subring of K. For an element x ∈ K to be integral over S, it is necessary and sufficient that every place of K which is finite on S be finite at x. ↩ 1 04SV

§ 3. Valuations

Definition 1 Let C be a (not necessarily commutative) ring and Γ a totally ordered commutative group written additively. A valuation on C with values in Γ is any mapping v : C → Γ_∞ which satis… 04T2
Proposition 1 Let v be a valuation on a (not necessarily commutative) ring C. For any elements xᵢ ∈ C (1 ≤ i ≤ n), ↩ 3 04T3
Corollary If a finite sequence of elements (xᵢ)_1 ≤ i ≤ n of C (for n ≥ 2) is such that ∑ᵢ₌₁ⁿ xᵢ = 0, there exist at least two distinct indices j, k such that 04T4
Proposition 2 Let K be a (not necessarily commutative) field and v a valuation on K with values in Γ. Then: (i) For x ≠ 0, v(x) ≠ +∞. (ii) The set A of x ∈ K such that v(x) ≥ O is a subring of K… ↩ 1 04T5
Proposition 3 For two valuations v, v' over a (not necessarily commutative) field K to be equivalent, it is necessary and sufficient that there exist an isomorphism λ of the ordered group v(K^*)… 04T6
Proposition 4 Let C be an integral domain, K its field of fractions, C^* = C - 0 and v : C → Γ_∞ a valuation on C. Then there exists a unique valuation w on K which extends v and w(K^*) is the s… ↩ 3 04T7
Proposition 5 Let K be a field, v a valuation on K and K' an extension of K. There exists a valuation v' on K' whose restriction to K is equivalent to v. ↩ 1 04T8
Proposition 6 Let K be a field, A a subring of K and x an element of K. For x to be integral over A, it is necessary and sufficient that every valuation on K which is positive on A be positive a… ↩ 1 04T9
Example (1) Every valuation on a finite field F is improper, since every element of F is a root unity. 04TA
Example (2) If K is a subfield of a field K', the restriction to K of a valuation on K' is a valuation on K. 04TB
Example (3) Let k be a field and K = k((T)). The mapping v which maps every non-zero formal power series to its order (Algebra, Chapter IV, § 5, no. 7) is a valuation on K whose order group is… ↩ 4 04TC
Example (4) Let A be a principal ideal domain, K its field of fractions and p an extremal element of A. For x ∈ K let vp(x) denote the exponent of p in the decomposition of x into extremal ele… ↩ 3 04TD
Example (5) The mapping P(X, Y) ↦ P(T, e^T) of C[X, Y] to C((T)) is injective (Functions of a real variable, Chapter IV, §2, Proposition 9) and therefore can be extended to an isomorphism of C… 04TE
Example (6) Let Γ be a totally ordered group and k a field. Let Γ+ be the monoid of positive elements of Γ and C the algebra of Γ+ over k. By definition, C has a basis (x\alpha){\alpha \in \Ga… ↩ 1 04TF
Definition 2 Let G be an ordered set. A subset of G is called major if the relations x ∈ M and y ≥ x imply y ∈ M. 04TG
Proposition 7 The mapping M ↦ a(M) is an increasing bijection of the set of major subsets of G onto the set of sub-A-modules of K. 04TH
Corollary Let G₊ be the set of positive elements in G. The mapping M ↦ a(M) is a bijection of the set of major subsets of G₊, onto the set of ideals of A. ↩ 1 04TI
Definition 3 Let K be a (not necessarily commutative) field, v a valuation on K and Γ the order group of v. v is called discrete if there exists a (necessarily unique) isomorphism of the ordere… ↩ 1 04TJ
Proposition 8 Let K be a (not necessarily commutative) field, v a discrete valuation on K, A the ring of v and u a uniformizer for v. The non-zero ideals of A are two-sided and of the form Auⁿ (… ↩ 1 04TK
Proposition 9 Let A be a local integral domain distinct from its field of fractions. The following conditions are equivalent: (a) A is the ring of a discrete valuation. (b) A is a principal idea… ↩ 6 04TL
Lemma 1 Let A be a valuation ring. Every finitely generated torsion-free A-module is free. Every finitely generated ideal 𝔞 of A is principal. Every torsion-free A-module is flat. ↩ 1 04TM

§ 4. The height of a valuation

Proposition 1 Let K be a field and A a valuation ring of K. Then: (a) Every ring B such that A ⊂ B ⊂ K is a valuation ring of K; (b) The maximal ideal m(B) of such a ring is contained in A and i… ↩ 3 04TV
Corollary The set & subrings of K containing A is totally ordered by inclusion. ↩ 2 04TW
Proposition 2 Let K be a field, B a valuation ring of K and h, the place & K associated with B (with values in κ(B)). Then the mapping A ↦ h_B(A) defines a bijection of the set A of valuation ri… ↩ 1 04TX
Corollary Let A and B be two valuation rings of K, where A ⊂ B; let A' = h_B(A), which is a valuation ring of κ(B). The residue field κ(A') of A' is canonically isomorphic to the residue fie… ↩ 1 04TY
Remark Conversely, let f be a place of K with values in K' and f' a place of K' with values in K''. Then f' ∘ f is a place of K whose ring is contained in the ring of the place f. 04TZ
Definition 1 A subgroup H of an ordered group G is called isolated if the relations 0 ≤ y ≤ x and x ∈ H imply y ∈ H. ↩ 1 04U0
Example (1) Let A and B be two ordered groups; let A × B be given the lexicographic order (i.e. "(a, b) ≤ (a', b')" is equivalent to "(a < a') or (a = a' and b ≤ b')"). The second factor B of… 04U1
Proposition 3 Let G be an ordered group and P the set of its positive elements. ↩ 3 04U2
Example (2) If we reconsider the example where G is a lexicographic product A × B and H = B, the ordered group G/H is canonically identified with A. ↩ 1 04U3
Proposition 4 The mapping B ↦ H_B is an increasing bijection of the set of subrings of K containing A onto the set of isolated subgroups of Γ_A. ↩ 1 04U4
Remark Under the above hypotheses, let f denote the canonical homomorphism of B onto κ(B) and A' = f(A); it is a valuation ring of κ(B) (Proposition 2, no. 1). Then f⁻¹(κ(B)^*) = U(B), f⁻… 04U5
Example Let k be a field, 04U6
Definition 2 Let G be a totally ordered group. If the number of isolated subgroups of G distinct from G is finite and equal to n, G is said to be of height n. If this number is infinite, G is s… 04U7
Definition 3 The height of the order group of a valuation is called the height of that valuation. 04U8
Proposition 5 The height of a valuation is equal to the number of non-zero prime ideals in its ring. ↩ 1 04U9
Proposition 6 Let K be a field and A a subring of K. Suppose that A is not a field. Then the following conditions are equivalent: (a) A is the ring of a valuation of height 1 on K; (b) A is a va… ↩ 3 04UA
Proposition 7 For a valuation on a field to be of height 1, it is necessary and sufficient that its order group be isomorphic to a non-zero ordered subgroup of \mathbf{R}. 04UB
Proposition 8 Let G be a totally ordered group not reduced to O. The following conditions are equivalent: (a) G is of height 1; (b) for all x 0 and y ≥ 0 in G, there exists an integer n ≥ 0 such… ↩ 2 04UC
Proposition 9 Let K be a field, v a non-improper valuation on K and A the ring of v. For A to be completely integrally closed (Chapter V, § 1, no. 4, Definition 5), it is necessary and sufficien… 04UD
Corollary Let K be a field, (v_α)_α ∈ I a family of valuations of height 1 on K and A the intersection of the rings of the v. Then A is a completely integrally closed domain. ↩ 1 04UE

§ 5. The topology defined by a valuation

Lemma 1 Let x ∈ K^*, y ∈ K^* and α ∈ G. If v(x - y) > sup(α + 2v(y), v(y)), then v(x⁻¹ - y⁻¹) > a. ↩ 2 04UQ
Proposition 1 The topology Tᵥ is Hausdorff and compatible with the field structure on K. The mapping v : K^* → G is continuous if G is given the discrete topology. ↩ 1 04UR
Proposition 2 Let K be a not necessarily commutative field, v a non-improper valuation on K, A the ring of v and m the ideal of v. For K with the topology Tᵥ to be locally compact, it is necessa… ↩ 4 04US
Remark Note that it is sufficient in this proof to suppose that A is complete. 04UT
Proposition 3 Let E be a left topological vector space over K which is Hausdorff and of dimension 1. Suppose that v is not improper. For all x₀ ≠ 0 in E, the mapping a ↦ ax₀ of Kₛ onto E is a to… ↩ 1 04UU
Corollary Let E be a left topological vectorspace over K, H a closed hyperplane of E and D a 1-dimensional vector subspace of E an algebraic supplement of H. Suppose that v is not improper… 04UV
Proposition 4 Suppose that v is not improper and K is complete. Let E be a left topological vector space over K, which is Hausdorff and of finite dimension n. For every basis (eᵢ)_1 ≤ i ≤ n of E… ↩ 1 04UW
Corollary Suppose that v is not improper and K is complete. Let E be a Hausdorff topological vector space over K and F a finite-dimensional vector subspace of E. Then F is closed. ↩ 2 04UX
Proposition 5 Let K be a not necessarily commutative field, v a valuation on K and G the group v(K^*) with the discrete topology. ↩ 2 04UY
Remark For all x ∈ hatK not belonging to hatA, there exists x₀ ∈ K such that hatv(x - x₀) > 0, hatv(x) = hatv(x₀) = v(x₀) < 0; then x₀⁻¹ x ∈ hatA and, as x₀⁻¹ ∈ A, it is seen that, if we… 04UZ

§ 6. Absolute values

Proposition 1 For a mapping f from K to R, satisfying (VA,) and (VAII) to belong to V(K), it is necessary and sufficient that f(1 + x) be bounded in the set of x ∈ K such that f(x) ≤ 1. ↩ 1 04VA
Proposition 2 Let f be a mapping of K to R₊ belonging to V(K); if there exists C > 0 such that f(n.1) ≤ C for every integer n > 0, f is an absolute value on K. ↩ 4 04VB
Corollary 1 For a mapping f from K to R₊ to be an absolute value, it is necessary and sufficient that it satisfy conditions (VA_I), (VA_II) and (U₂). ↩ 1 04VC
Corollary 2 For a mapping f of K to R₊ to belong to V(K), it is necessary and sufficient that it be of the form gᵗ, where t > 0 and g is an absolute value on K. ↩ 1 04VD
Proposition 3 Let f be a mapping of K to R₊. The following properties are equivalent: (a) f is an ultrametric absolute value. (b) There exists a valuation v on K with values in R and a real numb… ↩ 3 04VE
Corollary If K is a (not necessarily commutative) field of characteristic p > 0, every function on V(K) is an ultrametric absolute value. ↩ 1 04VF
Proposition 4 Let f be a mapping of Q to R₊ belonging to V(Q). Then: ↩ 2 04VG
Theorem 1 (Gelfand-Mazur) Let K be an algebra over the field R with the two following properties: (1) K is a (not necessarily commutative) field. (2) There exists on K a norm x ↦ |x| compatible with the alg… ↩ 3 04VH
Lemma 1 Let D be a field with centre L such that, for all x ∈ D, L(x) is an extension of L of degree ≤ m. Then the rank of D over L is ≤ m². ↩ 1 04VI
Theorem 2 (Ostrowski) Let K be a (not necessarily commutative) field and f an element V(K) which is not an ultrametric absolute value. Then there exist a unique real number s > 0 and an isomorphism j of… ↩ 2 04VJ
Lemma 2 Let L be a (not necessarily commutative) field and K a subfield of L such that L is ajinite-dimensional left vector space over K. Let g be an absolute value on L andf its restricti… 04VK

§ 7. Approximation theorem

Proposition 1 Let K be a field, (Aᵢ)_1 ≤ i ≤ n ajinitefamily of valuation rings of K and B = ⋂ᵢ₌₁ⁿ Aᵢ. We write pᵢ = B ∩ m(Aᵢ). Then Aᵢ = B_pᵢ for all i and the jield of ructions of B is K. ↩ 5 04VN
Lemma 1 Let vᵢ (1 ≤ i ≤ n) be ualuations on the jield K and x ∈ K^*. Then there exists a polynomialf (X) of theform ↩ 1 04VO
Proposition 2 With the hypotheses of Proposition 1 suppose further that Aᵢ not⊂ Aⱼ for i ≠ j. Then the pᵢ are distinct maximal ideals of B and every maximal ideal of B is equal to one of the pᵢ. ↩ 2 04VP
Corollary 1 Suppose that Aᵢ not⊂ Aⱼ for i ≠ j. For every family of elements aᵢ ∈ Aᵢ (1 ≤ i ≤ n), there exists x ∈ B such that x ≡ aᵢ pmodm(Aᵢ) for 1 ≤ i ≤ n. ↩ 1 04VQ
Corollary 2 Suppose that Aᵢ not⊂ Aⱼ for i ≠ j. There exist elements xᵢ (1 ≤ i ≤ n) of K such that vᵢ(xᵢ) = 0 and vⱼ(xᵢ) > 0 for i ≠ j. 04VR
Corollary 3 Every valuation ring of K containing B contains one of the Aᵢ. 04VS
Definition 1 Let A and A' be two valuation rings of the same field K. A and A' are called independent if K is the ring generated by A and A'. Two valuations on K are called independent if their… ↩ 1 04VT
Theorem 1 (Approximation Theorem for Valuations) Let vᵢ (1 ≤ i ≤ n) be valuations on a field K which are independent in pairs and Γᵢ the order group of vᵢ. Let aᵢ ∈ K and αᵢ ∈ Γᵢ (1 ≤ i ≤ n). Then there exists x ∈ K such that vᵢ(… ↩ 7 04VU
Lemma 2 Let A be a valuation ring and b an ideal of A distinct from A. Then the radical c of b is a prime ideal. 04VV
Corollary 1 For every family of elements γᵢ ∈ Γᵢ (1 ≤ i ≤ n), there exists x ∈ K such that vᵢ(x) = γᵢ (1 ≤ i ≤ n). ↩ 2 04VW
Corollary 2 Let Tᵢ be the topology defined on K by vᵢ; let Kⁿ be given the topology the product of the Tᵢ. If the vᵢ are not improper, the diagonal of Kⁿ is dense in Kⁿ. ↩ 1 04VX
Proposition 3 Let v and v' be two non-improper valuations on the same field K. For v and v' to define the same topology on K, it is necessary and sufficient that they be dependent. ↩ 1 04VY
Proposition 4 Let v₁, …, vₙ (n ≥ 2) be pairwise dependent valuations on the same field K. Then the rings A₁, …, Aₙ of v₁, …, vₙ generate a subring δ \ K distinct from K. ↩ 1 04VZ
Theorem 2 (Approximation Theorem for Absolute Values) Let fᵢ (1 ≤ i ≤ n) be absolute values on the same field K which are not improper and no two of which are equivalent. Let aᵢ \ (1 ≤ i ≤ n) be elements of K and ε a real number > 0… ↩ 1 04W0

§ 8. Extensions of a valuation to an algebraic extension

Definition 1 The index [𝔭ᵥ : Γ_v'] is called the ramification index of v' over v (or over K) and denoted by e(v'/v) (or e(A'/A), or sometimes e(L/K)). 04W4
Definition 2 The degree [κ(A') : κ(A)] is called the residue class degree of v' over v (or over K) and denoted by f(v'/v) (or f(A'/A), or sometimes f(L/K)). 04W5
Lemma 1 Let K, K', K'' be three fields such that K ⊂ K' ⊂ K'', v'' a valuation on K'' and v and v' its restriction to K and K'. Then there are the relations: (1) e(v''/v) = e(v''/v') e(v'/… ↩ 4 04W6
Lemma 2 Let K be a field, L a finite extension of K of degree n, v' a valuation on L and v its restriction to K. Then the inequality ↩ 8 04W7
Proposition 1 Let K be a field, L an algebraic extension of K, v' a valuation on L, v its restriction to K and A and A' the rings of v and v'. Then Γ_v'/Γᵥ is a torsion group and κ(A') is an alg… ↩ 4 04W8
Corollary 1 The height of v' is equal to that of v. 04W9
Lemma 3 Let G' be a totally ordered group, G a subgroup of G and G' (resp. G) the set of isolated subgroups of G' (resp. G). The mapping H' ↦ H' ∩ G maps G' onto G. This mapping is bijecti… ↩ 1 04WA
Corollary 2 For v' to be improper (resp. of height 1), it is necessary and sufficient that v be improper (resp. of height 1). ↩ 4 04WB
Corollary 3 Suppose that L is a finite extension of K. For v' to be discrete, it is necessary and sufficient that v be discrete. If v' is discrete, Γ_v' is isomorphic to a non-zero subgroup of… ↩ 3 04WC
Definition 3 Let K be afield, v a valuation on K and L an extension of K. A family (v'ᵢ)_i ∈ I of valuations on L which extend v and such that every valuation on L extending v is equivalent to… 04WD
Proposition 2 Let K be afield, v a valuation on K, hatK the completion of K with respect to v, hatv the continuous extension of v to hatK and L a finite extension of K of degree n. (a) Let v' be… ↩ 5 04WE
Corollary 1 If K is complete with respect to v and v is not improper, two valuations on L extending v are dependent. ↩ 1 04WF
Corollary 2 If hatK or L is separable over K, the canonical mapping 04WG
Remark Proposition 2 (b) shows that every composite extension of hatK and L over K (Algebra, Chapter VIII, § 8) is isomorphic to one of the completions hatLᵢ and that these are composite… 04WH
Theorem 1 Let K be a field, v a valuation on K and L a finite extension of K of degree n. Then: ↩ 2 04WI
Remark The intersection C of the rings Aᵢ' of the valuations vᵢ' (i ∈ I) is the integral closure of A in L (§ 1, no. 3, Corollary 3 to Theorem 3); it follows moreover from (c) and § 7, no… 04WJ
Definition 4 Let G be a totally ordered commutative group and H a subgroup of G of finite index. The number of major subsets of G consisting of strictly positive elements and containing all the… ↩ 1 04WK
Proposition 3 Under the hypotheses of Definition 4, if the set of strictly positive elements of G has no least element, then ε(G, H) = 1 for all H. If there exists a least element O of G and G'… ↩ 1 04WL
Corollary The initial index ε(G, H) divides the index (G : H) and is equal to it if G is isomorphic to Z. ↩ 1 04WM
Definition 5 Let K be a field, L a finite extension of K, w a valuation on L, v its restriction to K and Γ_w and Γᵥ their order groups. The initial index of Γᵥ in Γ_w is called the initial rami… ↩ 1 04WN
Proposition 4 Under the hypotheses of Definition 5, let A and m (resp. A' and m') be the ring and ideal of the valuation v (resp. w). Then ↩ 5 04WO
Proposition 5 Let K be a field, v a valuation on K, A its ring, m its ideal, L a finite extension of K of degree n, B the integral closure of A in L and (v'ᵢ)_1 ≤ i ≤ s a complete system of exte… ↩ 2 04WP
Corollary With the same hypotheses and notation, [B/mB : A/m] = ∑ᵢ₌₁ˢ ε(v'ᵢ/v) f(v'ᵢ/v) ≤ ∑ᵢ₌₁ˢ ε(v'ᵢ/v) f(v'ᵢ/v) ≤ n. We know that ε(v'ᵢ/v) ≤ ε(v'ᵢ/v) (no. 4, Corollary to Proposition 3) an… ↩ 1 04WQ
Theorem 2 With the hypotheses and notation of Proposition 5, the following conditions are equivalent : (a) B is a finitely generated A-module; (b) B is a free A-module; (c) [B/mB : A/m] = n… ↩ 4 04WR
Remark (1) If v is discrete, ε(v'ᵢ/v) = ε(v'ᵢ/v) (no. 4) and condition (d) reduces to ∑ᵢ₌₁ˢ ε(v'ᵢ/v) f(v'ᵢ/v) = n. 04WS
Corollary 1 With the same hypotheses and notation, suppose further that v is discrete and L separable. Then 04WT
Corollary 2 Let K be a field, v a discrete valuation on K with respect to which K is complete and L a finite extension of K of degree n. Then v admits a unique (upto equivalence) extension v'… ↩ 1 04WU
Corollary 4 Under the hypotheses of Corollary 3, if L is a separable extension of K, each of the hatLᵢ is a separable extension of hatK. If further L is a Galois extension of K with Galois gro… 04WV
Proposition 6 Let K be a field, v a valuation on K, A its ring, L an algebraic extension of K and A' the integral closure of A in L. Let 23 be the set of valuation rings on L which extend v and… ↩ 2 04WW
Proposition 7 Let K be a field, L a quasi-Galois extension of K and f and f' places of L with values in the same field F. Suppose that the restrictions off and f' to K coincide. Then there exist… ↩ 2 04WX
Corollary 1 Let K be a field, v a valuation on K, L a quasi-Galois extension of K and v' and v'' two extensions of v to L. Then there exists a K-automorphism s of L such that v'' is equivalent… ↩ 3 04WY
Corollary 2 Let K be a field, f a place of K (resp. v a valuation on K) and L a radicial extension of K. Then all the extensions of f (resp. v) to L are equivalent. 04WZ
Proposition 8 Let K be a field, v a valuation on K, L a finite quasi-Galois extension of K of degree n and (v'ᵢ)_1 ≤ i ≤ g a complete system of extensions of v to L. Then e(v'ᵢ/v) and f(v'ᵢ/v) h… 04X0
Proposition 9 Let K be a field, L an algebraic extension of K and 𝔣 an absolute value on K. Then 𝔣 can be extended to an absolute value on L. ↩ 1 04X1
Proposition 10 Let K be a field, f an absolute value on K such that K is complete and not discrete with respect to f and L an algebraic extension of K. Then f can be extended uniquely to an absol… 04X2
Proposition 11 Let K be a field, f a non-ultrametric absolute value on K, hatK the completion of K with respect to f, hatf the continuous extension off to hatK and L a finite extension of K of de… ↩ 1 04X3
Proposition 12 Let K be a field, f an absolute value on K, L a quasi-Galois extension of K and f' and f'' two extensions off to L. Then there exists a K-automorphism s of L such that f'' = f' ∘ s… 04X4

§ 9. Application: locally compact fields

Proposition 1 If K is a locally compactfield, thefunction mod, belongs to V(K) (§ 6, no. 1). Moreover: (i) If s > 0 is such that (mod_K)ˢ = g is an absolute value, then g defines the topology on… ↩ 1 04XQ
Proposition 2 Let K, K' be two (not necessarily commutative) locally compact fields such that K is a topological subfield of K' and K is not discrete. Then: (i) K' is a finite dimensional left (… ↩ 1 04XR
Corollary 1 Every locally compactfield whose centre is not discrete is of finite rank over its centre. 04XS
Corollary 2 Let K' be a locally compact field and K a closed subfield of K' (neither necessarily commutative). If K' is a left (resp. right) vector space of finite dimension n over K, then (2)… 04XT
Proposition 3 Let K be a (not necessarily commutative) locally compact field which is not discrete and whose topology is defined by a discrete valuation v; let A be the ring and m the ideal of v… ↩ 1 04XU
Lemma 1 Let x, y be two permutable elements of A such that x - y ∈ mʲ \ (j ≥ 1); then x^pⁿ - y^pⁿ ∈ mʲ⁺ⁿ for every integer n ≥ 0. 04XV
Theorem 1 Let K be a (not necessarily commutative) locally compact field which is not discrete. ↩ 1 04XW
Corollary 1 Every locally compact field which is not discrete is of finite rank over its centre. ↩ 1 04XX
Corollary 2 Every locally compact field is connected or totally disconnected; if it is connected, it is isomorphic to R, C or H. 04XY
Remark Let s be an integer > 0; the subfield F_q((Tˢ)) = L of K = F_q((T)) is closed in K and e(K/L) = s and f(K/L) = 1. It is therefore seen that there are closed subfields L of K which… 04XZ

§ 10. Extensions of a valuation to a transcendental extension

Lemma 1 Let K be a field, v a valuation on K, Γ its order group, Γ' a totally ordered group containing Γ and ξ an element of Γ'. There exists a unique valuation w on K(X) such that, for P… ↩ 5 04Y5
Proposition 1 Let K be a field, v a valuation on K, Γ its order group, Γ' a totally ordered group containing Γ and ξ an element of Γ' such that the relations nξ ∈ Γ, n ∈ Z imply n = 0. Then ther… ↩ 1 04Y6
Proposition 2 Let K be a field, v a valuation on K, Γ its order group and k its residue field. There exists a unique valuation w on K(X) extending v such that w(X) = 0 and the image t of X in th… ↩ 3 04Y7
Remark It should not be thought that the two types of extensions of v to K(X) which we have just met are the only ones; there may exist a third type of extension, where Γ'/Γ is a torsion… 04Y8
Definition 1 The rational rank of a commutative group G is the dimension of the vector Q-space G ⊗_Z Q. 04Y9
Proposition 3 Let G be a totally ordered Commutative group and H a subgroup of G. If h(G) and h(H) denote the heights of G and H (§ 4, no. 4), then the inequality ↩ 2 04YA
Corollary For every totally ordered commutative group G, h(G) ≤ r(G). We set H = (0) in Proposition 3. ↩ 1 04YB
Proposition 4 Let G be a totally ordered commutative group. Suppose that G is finitely generated and that h(G) = r(G). Then G is isomorphic to Zʳ⁽ᴳ⁾ ordered lexicographically. ↩ 1 04YC
Lemma 2 Let H be an isolated subgroup of a totally ordered commutative group G. If H is a direct factor of G, the ordered group G is isomorphic to the group (G/H) × H ordered lexicographic… ↩ 1 04YD
Theorem 1 Let x₁, …, x, be elements of the ring of v' whose canonical images overlinexᵢ in k' are algebraically independent over k and y₁, …, yᵣ elements of K' such that the canonical images… ↩ 1 04YE
Corollary 1 The inequality 04YF
Corollary 2 Let h and h' be the heights of v and v'. Then s(v'/v) + h' ≤ d(K'/K) + h. By Proposition 3, h' ≤ r(v'/v) + h. 04YG
Corollary 3 Suppose that K' is a finitely generated extension of K, that Γ is isomorphic to Z^h' (ordered lexicographically) and there is equality in formula (10). Then Γ' is isomorphic to Z^h… 04YH
Corollary 4 Suppose that v is improper (in which case k = K). Then h(Γ') + d(k'/K) ≤ r(Γ') + D(k'/K) ≤ d(K'/K). If, in particular, v' is of height 1, then d(k'/K) ≤ d(K'/K) - 1; further, if K'… 04YI

VII. Divisors

§ 1. Krull domains

Definition 1 Let A be an integral domain and K its field of fractions. Every sub-A-module a of K such that there exists an element d ≠ 0 in A for which da ⊂ A is called a fractional ideal 𝔡 of… ↩ 1 04YL
Definition 2 Every element a of I(A) such that a = tildea is called a divisorial fractional ideal of A. ↩ 1 04YM
Proposition 1 (a) If b is a divisorial ideal and a ∈ I(A), b : a is divisorial. ↩ 6 04YN
Remark In the course of the above proof it has been proved that A : a = A : (A : (A : a)) for every ideal a ∈ I(A), which is a special case of Set Theory, Chapter 111, § 1, no. 5, Proposi… 04YO
Proposition 2 (i) In D(A) every non-empty set bounded above admits a least upper bound. More precisely, if (aᵢ) is a non-empty family of elements of I(A) which is bounded above, then sup(div aᵢ)… ↩ 5 04YP
Corollary If x, y and x + y are in K^*, then div(x + y) ≥ inf(div(x), div(y)). 04YQ
Proposition 3 Let a, a', b, b' be elements of I(A). The relations a > a' and b > b' imply ab > a'b'. 04YR
Proposition 4 Let a, b be two divisorial fractional ideals of A. The following properties are equivalent: (a) div a and div b are equivalent divisors; (b) there exists x ∈ K^* such that b = xa. 04YS
Theorem 1 Let A be an integral domain. For the monoid D(A) of divisors of A to be a group, it is necessary and sufficient that A be completely integrally closed. ↩ 1 04YT
Corollary Let A be a completely integrally closed domain and a a divisorial fractional ideal of A. Then, for every fractional ideal b ≠ 0 of A, div(a : b) = div a - div b ↩ 1 04YU
Definition 3 A n integral domain A is called a Krull domain if there exists a family (vᵢ)_i ∈ I of valuations on the field of fractions K of A with the following properties: (AK_I) the valuatio… ↩ 5 04YV
Theorem 2 Let A be an integral domain. For A to be a Krull domain, it is necessary and sufficient that the twofollowing conditions be satisfied: (a) A is completely integrally closed; (b) ev… ↩ 3 04YW
Lemma 1 If A is a Krull domain and (vₜ)ₜ, is a family of normed valuations on K satisfying (AK_I), (AK_II) and (AK_III), the mapping a ↦ (vₜ(a))_t ∈ I is a decreasing injective mapping of… ↩ 1 04YX
Corollary For a Noetherian ring to be a Krull domain, it is necessary and sufficient that it be an integrally closed domain. ↩ 1 04YY
Proposition 5 Let A be a Krull domain, K its field of fractions and (v_P)_P ∈ P(A) the family of its essential valuations. Let (n_P)_P ∈ P(A) be a family of rational integers which are zero exce… ↩ 5 04YZ
Proposition 6 Let A be a Krull domain, K its field of fractions, (vₜ)_t ∈ I a family of valuations on K with the properties of Definition 3 and A, the ring of vₜ. Let S be a multiplicative subse… ↩ 8 04Z0
Corollary 1 Let P be an extremal divisor of A and p the corresponding divisorial ideal. Then p is prime, the ring of v_P is A, and the residue field of v_P is identified with the field of frac… ↩ 7 04Z1
Corollary 2 Let A be a Krull domain, K its field of fractions and (vₜ)_t ∈ I a family of valuations with the properties of Definition 3. Then every essential valuation of A is equivalent to on… ↩ 4 04Z2
Proposition 7 Let A be a Krull domain, (v_P)_P ∈ P(A) the family of its essential valuations and a ∈ I(A). Then the coefficient of P in div a is inf_y ∈ a (v_P(y)). If p is the divisorial prime… ↩ 1 04Z3
Proposition 8 Let A be an integrally closed Noetherian domain. (a) Let P be an extremal divisor of A and p the corresponding divisorial prime ideal; ↩ 3 04Z4
Proposition 9 Let v₁, …, vᵣ be distinct essential valuations of a Krull domain A and n₁, …, nᵣ rational integers. There exists an element x of the field of fractions K of A such vᵢ(x) = nᵢ for 1… ↩ 8 04Z5
Corollary 1 Let A be a Krull domain, K its field of fractions and a, b and c three divisorial fractional ideals of A such that a ⊂ b. There exists x ∈ K such that a = b ∩ xc. ↩ 1 04Z6
Corollary 2 Let A be a Krull domain. For a fractional ideal a of A to be divisorial, it is necessary and sufficient that it be the intersection of two fractional principal ideals. ↩ 2 04Z7
Definition 4 Let A be an integral domain. A prime ideal 𝔭 of A is said to be of height 1 if it is minimal among the non-zero prime ideals of A. 04Z8
Theorem 3 Let A be a Krull domain and 𝔭 an integral ideal of A. For 𝔭 to be the divisorial ideal corresponding to an extremal divisor, it is necessary and sufficient that 𝔭 be a prime ideal… ↩ 5 04Z9
Corollary 1 In a Krull domain every non-zero prime ideal m contains a prime ideal of height 1. If m is not of height 1, then div m = 0 and A : m = A. ↩ 1 04ZA
Corollary 2 Let A be a Krull domain, K its field of fractions, v a valuation on K which is positive on A and p the set of x ∈ A such that v(x) 0. If the prime ideal p is of height 1, v is equi… ↩ 2 04ZB
Theorem 4 Let A be an integral domain and M the set of its prime ideals of height 1. For A to be a Krull domain, it is necessary and sufficient that the following Properties are satisfied: (… ↩ 6 04ZC
Proposition 10 Let A be an integrallyclosed Noetherian domain and a an integral ideal of A. The following conditions are equivalent: (a) a is divisorial; (b) the prime ideals associated with A/a… 04ZD
Proposition 11 Let A be a local Krull domain (in particular an integrally closed local Noetherian domain) and m its maximal ideal. The following conditions are equivalent: (a) A is discrete valua… ↩ 1 04ZE
Proposition 12 Let A be a Krull domain, K its field of fractions, K' a finite extension of K and A' the integral closure of A in K'. Then A' is a Krull domain. The essential valuations of A' are… ↩ 3 04ZF
Corollary Let p (resp. p') be a prime ideal of A (resp. A') of height 1 and v (resp. u') the essential valuation of A (resp. A') corresponding to it. For p' to lie above p, it is necessary a… ↩ 1 04ZG
Proposition 13 Let A be a Krull domain and X₁, Xₙ, …, Xₙ indeterminates. The ring A[X₁, …, Xₙ] is a Krull domain. ↩ 2 04ZH
Remark The valuations v_f and barw introduced in the proof of Proposition 13 are the essential valuations of A[X]. It will be sufficient for us to show that, if V is the set of valuations… 04ZI
Proposition 14 Let A be a Krull domain and B a Krull domain containing A. Suppose that the following condition holds: ↩ 3 04ZJ
Proposition 15 Let A and B be Krull domains such that B contains A and is aflat A-module. Then: (a) condition (PDE) of Proposition 14 is fulfilled; (b) for every divisorial ideal a ∈ A, Ba is the… ↩ 1 04ZK
Corollary Let A be a local Krull domain and B a discrete valuation ring such that B dominates A and is aflat A-module. Then A is a field or a discrete valuation ring. 04ZL
Remark In the former of the two above cases, the mapping i : D(A) → D(B) is injective: as the elements of P(B) form a basis of D(B) and two distinct ideals of P(A) cannot be the traces on… 04ZM
Proposition 16 Let A be a Zariski ring such that its completion hatA is a Krull domain. Then A is a Krull domain and the canonical homomorphism tildei from C(A) to C(hatA) (which is defined since… ↩ 2 04ZN
Proposition 17 Let A be a Krull domain and S a multiplicative subset of A not containing 0. Then the canonical homomorphism bari from C(A) to C(S⁻¹A) is surjective. If further S is generated by a… ↩ 4 04ZO
Proposition 18 Let R be a Krull domain and A the polynomial ring R[X]. The canonical homomorphism of C(R) to C(R[X]) is bijective. ↩ 1 04ZP

§ 2. Dedekind domains

Definition 1 A Krull domain all of whose non-zero prime ideals are maximal is called a Dedekind domain. 050M
Theorem 1 Let A be an integral domain and K its field of fractions. The following conditions are equivalent: (a) A is a Dedekind domain; (b) A is a Krull domain and every non-improper valuat… ↩ 6 050N
Proposition 1 A semi-local Dedekind domain is a principal ideal domain. ↩ 3 050O
Proposition 2 Let A be a Dedekind domain, K its field of fractions and P the set of non-zero prime ideals of A; for p ∈ P let vₚ denote the corresponding essential valuation of A. Let p₁, …, p_q… ↩ 2 050P
Proposition 3 The ring A, (resp. A ) is identified with the completion of A (resp. K) with respect to the ring topology for which a fundamental system of neighbourhoods & 0 consists & all the in… ↩ 1 050Q
Proposition 4 The group SL(n, K) is dense in SL(n, A). 050R
Lemma 1 Let A be a Noetherian domain in which every non-zeroprime ideal is maximal and M afinitely generated torsion A-module. Then the length long_A(M) of M is infinite. ↩ 1 050S
Lemma 2 Let A be a ring, T an A-module and (Tᵢ) a right directed family of submodules of T with union T. Then long_A(T) = supᵢ (long_A(Tᵢ)). ↩ 1 050T
Remark This proof does not assume that A is commutative. 050U
Lemma 3 Let A be a Noetherian domain such that every non-zero prime ideal & A is maximal, M a torsion-free A-module of finite rank r and a non-zero element of A. Then A/Aa is an A-module o… ↩ 1 050V
Proposition 5 (Krull-Akizuki) Let A be a Noetherian domain each of whose non-zeroprime ideals is maximal, K its field of fractions, L a finite extension of K and B a subring of L containing A. Then B is Noether… ↩ 3 050W
Corollary 1 For every prime ideal p of A, the set of prime ideals of B lying above p is finite. ↩ 2 050X
Corollary 2 The integral closure of A in L is a Dedekind domain. 050Y
Corollary 3 The integral closure of a Dedekind domain in a finite extension of its field of fractions is a Dedekind domain. ↩ 2 050Z
Proposition 6 Let A be a Dedekind domain, K its field of fractions, L a finite extension of K and B the integral closure of A in L. Let p be a non-zero prime ideal of A, v the corresponding esse… 0510

§ 3. Factorial domains

Definition 1 A Krull domain all of whose divisorial ideals are principal is called a factorial (or unique factorization) domain. ↩ 4 051N
Theorem 1 Let A be an integral domain. The following conditions are equivalent: (a) A is factorial; (b) the ordered group of non-zero fractional principal ideals of A is a direct sum of grou… ↩ 4 051O
Proposition 1 Let A be a Krull domain. If every divisorial ideal of A is invertible, then, for every maximal ideal m of A, Aₘ is factorial. The converse is true if it is also assumed that every… ↩ 1 051P
Proposition 2 Let A be an integral domain. For A to be factorial, it is necessary and sufficient that there exist a subset P of A such that every a ∈ A - 0 may be written uniquely in the form a… 051Q
Proposition 3 Let A be a Krull domain and S a multiplicative subset of A not containing 0. ↩ 5 051R
Lemma 1 (Gauss) Let f, f' be non-zero elements of K[X] and c, c' contents off, f', f'. Then cc' is a content of ff'. ↩ 1 051S
Theorem 2 Let A be a factorial domain, K its field of fractions, (p_λ) a representative system of extremal elements of A and (P_λ) a representative system of irreducible polynomials of K[X]… ↩ 3 051T
Corollary If A is a factorial domain, the domain A[X₁, …, Xₙ] is factorial. 051U
Proposition 4 Let A be a Zariski ring and hatA its completion. If hatA is a factorial domain, A is a factorial domain. 051V
Corollary If the completion of a Noetherian local ring A is a factorial domain, A is a factorial domain. ↩ 1 051W
Lemma 2 Let f(X₁, X₂, …, Xₙ) be a formal power series ≠ 0 with coefficients in a ring E. There exist integers u(i) geqslant 1 (1 leqslant i leqslant n - 1) such that ↩ 1 051X
Lemma 3 Let f be a non-zero element of E[[X₁, …, Xₙ]]. There exist integers u(i) ≥ 1 (1 ≤ i ≤ n-1) such that the automorphism s of E defined by s(Xᵢ) = Xᵢ + Xₙᵘ⁽ⁱ⁾ (1 ≤ i ≤ n-1) and s(Xₙ)… ↩ 1 051Y
Proposition 5 Let f ∈ B be a series whose reduced series is non-zero. Let s denote its reduced order and M the sub-A-module of B with basis 1, X, …, Xˢ⁻¹. Then B is the direct sum of M and f is… ↩ 3 051Z
Corollary With the hypotheses and notation of Proposition 5, suppose that s ≥ 1, so that f ∈ Bm + BX. Then the A-homomorphism h of B' = A[[T]] to B = A[[X]] such that h(T) = f (Chapter III… 0520
Definition 2 A polynomial F ∈ A[X] is called distinguished if it is of the form F = Xˢ + aₛ₋₁ Xˢ⁻¹ + ⋯ + a₀, where aᵢ ∈ 𝔪 for 0 ≤ i ≤ s - 1. ↩ 1 0521
Proposition 6 (Preparation Theorem) Let f ∈ B be a series whose reduced series is not zero and s its reduced order. Then there exists a unique ordered pair (u, F) such that u is an invertible element of B, F a distin… ↩ 3 0522
Proposition 7 Let F be a distinguished polynomial and g, h two formal power series of B such that F = gh. Then there exists an invertible element u of B such that ug and u⁻¹h are distinguished p… ↩ 2 0523
Corollary Suppose further that A is an integral domain and F a distinguished polynomial of degree s. For F to be extremal in A[X], it is necessary and sufficient that it be extremal in B = A… ↩ 2 0524
Proposition 8 Let C be a ring which is either a field or a discrete valuation ring. Then the domain of formal power series C[[X₁, …, Xₙ]] is factorial. ↩ 3 0525

§ 4. Modules over integrally closed Noetherian domains

Definition 1 Let V be a finite-dimensional vector space over the field K. A lattice of V with respect to A (or simply a lattice of V) is defined to be any sub-A-module M of V satisfying the fol… 052X
Proposition 1 For a sub-A-module M of V to be a lattice of V, it is necessary and sufficient that KM = V and that M be contained in a finitely generated sub-A-module of V. ↩ 5 052Y
Corollary Suppose that A is Noetherian; for a sub-A-module M of V to be a lattice of V, it is necessary and sufficient that KM = V and M be finitely generated. ↩ 1 052Z
Remark (1) Recall that for every sub-A-module M of V the canonical mapping M ⊗_A K → V is injective and has image KM (Algebra, Chapter II, § 7, no. 10, Proposition 26); to say that KM = V mea… 0530
Proposition 2 Let M be a lattice of V and M, a sub-A-module of V. If there exist two elements x, y of K^* such that xM ⊂ M₁ ⊂ yM, M₁ is a lattice of V; conversely, if M₁ is a lattice of V, there… ↩ 3 0531
Proposition 3 (i) If M₁ and M₂ are lattices of V, so are M₁ ∩ M, and M₁ + M₂. ↩ 5 0532
Corollary Let U, V, W be three vector spaces of finite rank over K and f: U × V → W a left non-degenerate K-bilinear mapping (Algebra, Chapter IX, § 1, no. 1, Definition 3). If M is a lattic… 0533
Proposition 4 Let B be an integral domain, A a subring of B and K and L the respective fields & fractions of A and B. Let V be a finite-dimensional vector space over K. ↩ 3 0534
Corollary Suppose that A is a discrete valuation ring. Let hatA be its completion and let hatK be the field of fractions of hatA (Chapter VI, § 5, no. 3). The mapping φ, which maps every lat… 0535
Proposition 5 Let V, W vector spaces of finite rank over K, M a lattice of V and N a lattice of W. If M is finitely generated, then (in the notation of Proposition 3): ↩ 3 0536
Theorem 1 If M is a lattice of V, then M^* = ⋂_p ∈ P Mₚ^*. 0537
Remark (1) Let M be a finitely generated A-module; it is immediate that the dual M^* of M, identified with a sub-A-module of Hom_A(M, K) is a lattice of the vector K-space Hom_A(M, K); in par… ↩ 1 0538
Theorem 2 If M is a lattice of V, the following conditions are equivalent: (a) M is reflexive. (b) M = ⋂_p ∈ P Mₚ. (c) Ass(V/M) ⊂ P. ↩ 6 0539
Corollary Let M, N be two lattices of V with respect to A such that N is reflexive. In order that M ⊂ N, it is necessary and sufficient that, for all p ∈ P, Mp ⊂ Np. 053A
Proposition 6 (i) If M1 and M are reflexive lattices of V, so is M, n M. (ii) If W is a vector subspace of V and M is a reflexive lattice of V, M n W is a reflexive lattice of W. (iii) Let V, W… 053B
Proposition 7 Let 0 → M → N → Q → 0 be an exact sequence of A-modules. Suppose that N is finitely generated and torsion-free. (i) If M is reflexive, then Ass(Q) ⊂ P ∪ 0 (in other words, every id… ↩ 4 053C
Proposition 8 Let R and S be two commutative rings, p : R → S a ring homomorphism and M a finitely generated R-module. Suppose that R is Noetherian and that S is a flat R-module. Then, if M is r… ↩ 2 053D
Theorem 3 Let V be a vector space of finite rank over K and M a lattice of V with respect to A. ↩ 2 053E
Lemma 1 Let p and p' be two prime ideals of A such that (0) is the only prime ideal of A contained in p ∩ p'. Then, for every sub-A-module E of V, (Eₚ)_p' = K.E. 053F
Remark Let L be a free lattice of V with respect to A. Since Aₚ is a principal ideal domain for p ∈ P, N(p) is a free A-module of the same rank as L and there exists u(p) ∈ GL(V) such tha… 053G
Proposition 9 Let M be a finitely generated A-module. The following conditions are equivalent: ↩ 3 053H
Remark (1) The equivalent conditions of Proposition 9 mean also that Ass(M) contains no prime ideal of height ≤ 1. They may also be interpreted by saying that Supp(M) is of codimension ≥ 2 in… 053I
Definition 2 An A-module M is called pseudo-zero if it is finitely generated and it satisfies the equivalent conditions of Proposition 9. ↩ 3 053J
Definition 3 Let M and N be two A-modules and f : M → N a homomorphism. f is called pseudo-injective (resp. pseudo-surjective pseudo-zero) if Ker(f) (resp. Coker(f), Im(f)) is pseudo-zero; f is… 053K
Theorem 4 Let E be a finitely generated A-module, T the torsion submodule of E and M = E/T. There exists a pseudo-isomorphism ↩ 3 053L
Lemma 2 Let (pᵢ)_1 ≤ i ≤ k be a non-empty finite family of prime ideals of A of height 1 and let S = ⋂ᵢ (A - pᵢ); then the ring S⁻¹A is a principal ideal domain. ↩ 3 053M
Lemma 3 There exists a homomorphism g : E → T whose restriction to T is both a homothety and a pseudo-isomorphism. 053N
Theorem 5 Let T be a finitely generated torsion A-module. There exist two finite families (nᵢ)_i ∈ I and (pᵢ)_j ∈ I, where the nᵢ are integers ≥ 1 and the pᵢ are prime ideals of A of height… ↩ 3 053O
Remark (2) In the statement of Theorem 5, the modules A/pᵢ^nᵢ may be replaced by A/pᵢ^(nᵢ) (§ 1, no. 4, Proposition 8). For all p ∈ P, the canonical mapping g : A/pⁿ → A/p⁽ⁿ⁾ = A/(A ∩ pⁿAₚ) i… ↩ 1 053P
Definition 4 If T is a finitely generated torsion A-module, the divisor: ↩ 3 053Q
Proposition 10 (i) Let 0 → T₁ → T₂ → T₃ → 0 be an exact sequence of finitely generated torsion A-modules. Then ↩ 2 053R
Corollary Let 0 → Tₙ → Tₙ₋₁ → … → T₀ → 0 be an exact sequence of finitely generated torsion A-modules. Then ∑ᵢ₌₀ⁿ (-1)ⁱ χ(Tᵢ) = 0. ↩ 1 053S
Proposition 11 Let G be a commutative group written additively and φ : T(A) → G a mapping; for every finitely generated torsion A-module T, we also write, by an abuse of language, φ(T) = φ(cl(T))… ↩ 3 053T
Proposition 12 If a is an ideal ≠ 0 of A, ↩ 1 053U
Proposition 13 Let M be a lattice of V and u a K-automorphism of V. Then : (6) -χ(M, u(M)) = div(det(u)). For all p ∈ P, then bigwedgeⁿ (u(M)ₚ) = bigwedgeⁿ (u(Mₚ)); if (eᵢ)_1 ≤ i ≤ n is a basis o… 053V
Proposition 14 If M, M' are two lattices of V such that M' \subset M, M/M' is a finitely generated torsion A-module and: ↩ 1 053W
Corollary Let L1, L2 be twofree A-modules & the same rank n and let f: L1 \to L2 be a homomorphism. Let U be the matrix off with respect to bases of L1 and L2. For Coker(f) to be a torsion A… ↩ 1 053X
Example If A = Z, the divisor group of A is identified with the multiplicative group Q₊^* of rational numbers 0. For every finite commutative group T, χ(T) is the order of T; the above cor… ↩ 1 053Y
Proposition 15 Let M be a finitely generated A-module. There exists a free submodule L of M such that M/L is a torsion module and the element c(χ(M/L)) of C does not depend on the free submodule… ↩ 1 053Z
Proposition 16 (i) Let 0 → M₁ xrightarrowα M₂ xrightarrowg M₃ → 0 be an exact sequence of finitely generated A-modules. Then ↩ 7 0540
Corollary 1 Let 0 → M, xrightarrowu Mₙ₋₁ → ⋯ → M₀ → 0 be an exact sequence of finitely generated A-modules. Then ↩ 3 0541
Corollary 2 If a divisorial fractional ideal a ≠ 0 of A admits a finite free resolution, it is principal. ↩ 2 0542
Corollary 3 If every divisorial ideal ≠ 0 of A admits a finite free resolution, A is factorial. ↩ 1 0543
Proposition 17 Let G be a commutative group written additively and φ a mapping from the set F(A) of classes of finitely generated A-modules to G; for every finitely generated A-module M we also w… 0544
Proposition 18 (i) For R to be a pseudo-zero B-module, it is necessary and sufficient that the A-module R_[A] be pseudo-zero. ↩ 1 0545
Proposition 19 Let R be a finitely generated B-module. For R to be reflexive, it is necessary and sufficient that R_[A] be a reflexive A-module. ↩ 1 0546
Lemma 4 Let W be a vector space of finite rank over L and let R be a lattice of W with respect to B. Then, for all p ∈ P(A), (R_[A])ₚ = ⋂_𝔭/p R_𝔭. ↩ 1 0547
Corollary The ring B is a reflexive A-module. 0548
Proposition 20 (i) For a finitely generated A-module M to be pseudo-zero, it is necessary and sufficient that M ⊗_A B be a pseudo-zero B-module. (ii) If M is a finitely generated torsion A-module… 0549
Lemma 5 Let M₁, M, be two finitely generated A-modules, and f : M₁ → M₂ an injective homomorphism. Then the kernel of f ⊗ 1 ; M₁ ⊗_A B → M₂ ⊗_A B is pseudo-zero. 054A
Lemma 6 Let R be a commutative ring and pᵢ (1 ≤ i ≤ n) distinct prime ideals of R. ↩ 1 054B
Theorem 6 Let M be a torsion-free finitely generated A-module. There exists a free submodule L of M such that M/L is isomorphic to an ideal of A. ↩ 2 054C
Lemma 7 Let xᵢ (1 ≤ i ≤ m) be linearly independent elements of M (over A or K, which amounts to the same) and let L be the sub-A-module of M generated by the xᵢ. Then, for almost all p ∈ P… ↩ 1 054D
Lemma 8 Suppose that the rank n of M is geqslant 2; then there exists an element x ≠ 0 of M such that M/Ax is torsion-free. 054E
Remark If M is reflexive, the same is not necessarily true of M/L (Exercise 9). 054F
Proposition 21 Let A be a Dedekind domain. Every pseudo-zero A-module is zero. Every pseudo-injective (resp. pseudo-surjective, pseudo-bijective, pseudo-zero) A-module homomorphism is injective (… 054G
Proposition 22 Let A be a Dedekind domain and M a finitely generated A-module. The following properties are equivalent: (a) M is torsion-free. (b) M is reflexive. (c) M is projective. ↩ 2 054H
Corollary Let M be a finitely generated A-module and let T be its torsion submodule. Then T is a directfactor of M. ↩ 1 054I
Proposition 23 Let A be a Dedekind domain and T a finitely generated torsion A-module. There exist two finite families (nᵢ)_i ∈ I and (𝔭ᵢ)_i ∈ I, where the nᵢ are integers ≥ 1 and the 𝔭ᵢ elements… 054J
Proposition 24 Let A be a Dedekind domain and M a finitely generated torsion-free A-module & rank n ≥ 1. Then there exists an ideal b ≠ 0 of A such that M is isomorphic to the direct sum of the m… 054K