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IX. Sesquilinear forms and quadratic forms

§ 1. Sesquilinear Forms

Definition 1 A mapping Φ of the product E × F into G is said to be bilinear if it satisfies the following conditions: 0DIQ
Definition 2 Given a bilinear mapping Φ from E × F into G, the mapping d_Φ from F into L_A(E, G) (resp. the mapping s_Φ from E into L_B(F, G)) characterized by (5) is called the linear mapping… 0DIR
Definition 3 A bilinear mapping Φ of E × F into G is said to be degenerate on the right (resp. on the left) if there exists a nonzero element y₀ of F (resp. x₀ of E) such that Φ(x, y₀) = 0 for… 0DIS
Definition 4 We say that a mapping Φ of E × F into G is right sesquilinear for J if it satisfies conditions (1), (2), (3) (def. 1, No. 1) as well as (7) Φ(x, by) = Φ(x, y) · bᴶ for all x ∈ E,\… 0DIT
Definition 5 Let B be a ring, F a B-module to the left (resp. to the right) and J an antiautomorphism of B. We denote by Fᴶ the B-module to the right (resp. to the left) having the same underly… 0DIU
Remark Let A and B be two rings, J₁ an antiautomorphism of A, M an A-module to the right, N a B-module to the right and G a (A, B)-bimodule. One says that a mapping Φ from M × N into G is… 0DIV
Definition 6 Two elements x ∈ E and y ∈ F are said to be orthogonal with respect to Φ if Φ(x, y) = 0. Two subsets E' ⊂ E and F' ⊂ F are said to be orthogonal if, whatever x ∈ E' and y ∈ F', x a… 0DIW
Proposition 1 Let E be a left A-module and F a right B-module; put E' = A' ⊗_A E and F' = F ⊗_B B'. For every bilinear mapping Φ of E × F into G, there exists one and only one bilinear mapping Φ… 0DIX
Proposition 2 Suppose that B and B' are provided with anti-automorphisms J and I such that 0DIY
Proposition 3 The hypotheses and notations being those of prop. 1 (resp. prop. 2) suppose in addition that A, B, A', B' are fields and that the mappings α and β of A' ⊗_A G and G ⊗_B B' into G'… 0DIZ
Corollary The hypotheses and notations being those of Prop. 3, in order that Φ' be non-degenerate, it is necessary and sufficient that Φ be non-degenerate. 0DJ0
Remark Suppose that A, B, A' and B' are fields. Then, for the bimodules G' defined in the three examples above, the mappings α and β are injective, as follows immediately from Chap. III… 0DJ1
Proposition 4 Suppose that A is a field, and let Φ be a bilinear form (resp. sesquilinear form for J) on E × F; for E/F⁰ to be of finite dimension, it is necessary and sufficient that F/E⁰ be of… 0DJ2
Corollary 1 Suppose that A is a field and that Φ is non-degenerate. For a subspace M of E to be of finite dimension, it is necessary and sufficient that M⁰ be of finite codimension in F, and o… 0DJ3
Corollary 2 The hypotheses being those of cor. 1, let M, N be two subspaces of E; then one has (M + N)⁰ = M⁰ ∩ N⁰; if moreover M and N are of finite dimension, one has (M ∩ N)⁰ = M⁰ + N⁰. 0DJ4
Definition 7 Let A be a field (resp. a field endowed with an antiautomorphism J), E a left vector space over A, F a right (resp. left) vector space over A, and Φ a bilinear (resp. sesquilinear… 0DJ5
Proposition 5 The hypotheses and notations being those of def. 7, the linear mappings s_Φ and d_Φ associated with Φ have the same rank, and this rank is equal to the rank of the form Φ. 0DJ6
Proposition 6 The hypotheses and notations being those of def. 7, suppose in addition that E and F have the same finite dimension. Then the following conditions are equivalent: a) d_Φ is injecti… 0DJ7
Corollary The hypotheses and notations being those of Definition 7, we suppose in addition that E is of finite dimension and that Φ is non degenerate. Then we have dim E = dim F and, for eve… 0DJ8
Remark Let A be a ring endowed with an antiautomorphism J, M and N right A-modules, and Φ a left sesquilinear form for J on M × N (No. 2, Remark); it therefore satisfies the equality 0DJ9
Definition 8 Let Φ be a bilinear form on E × F whose associated mappings s and d are bijective. The bilinear form hatΦ on F^* × E^* defined by (27) is called the inverse form of Φ. 0DJA
Definition 9 Let Φ be a right sesquilinear form for J on E × F, whose associated mappings s and d are bijective. The left sesquilinear form widehatΦ for J on F^* × E^* is called the inverse for… 0DJB
Definition 10 The hypotheses and notations being as previously, one says that the homomorphism u^* from F'₁ into F satisfying (30) is the left adjoint of u, and that F'₁ is the defining submodul… 0DJC
Remark If the left adjoint u^* of u : E → E' is everywhere defined, and if s_Φ' and d_Φ' are injective, formula (30) shows that u is the right adjoint of u^*. 0DJD
Proposition 7 With the same notations as previously, suppose that d_Φ is bijective. Then every homomorphism u of E into E' admits an everywhere defined left adjoint, and one has u^* = (d_Φ)⁻¹ ∘… 0DJE
Remark When s_Φ is bijective, every homomorphism ν of F into F' admits an everywhere defined right adjoint, and one has 0DJF
Proposition 8 With the same notations as previously, suppose that s_Φ and d_Φ are bijective. Let u and ν be isomorphisms of E onto E' and of F onto F' respectively. Then, in order that Φ be the… 0DJG
Corollary Let A be a ring endowed with an antiautomorphism J, E a left A-module, Φ a sesquilinear form for J on E × E whose associated mappings are bijective, and u an automorphism of the A-… 0DJH
Remark Under the hypotheses of the cor. of prop. 8, suppose in addition that A be a field and that E be of finite dimension over A. Let ω be an endomorphism of E, ω1 and ω2 its right and… 0DJI
Definition 11 Given A-modules Eᵢ, Fᵢ (i = 1, …, m) and, for each i, a sesquilinear form Φᵢ for J on Eᵢ × Fᵢ, the sesquilinear form Φ for J on (⨂ᵢ Eᵢ) × (⨂ᵢ Fᵢ) characterized by (35) is called th… 0DJJ
Proposition 9 Let A be a commutative field endowed with an automorphism J, Eᵢ, Fᵢ finite-dimensional vector spaces over A, and Φᵢ a sesquilinear form for J on Eᵢ × Fᵢ (1 leqslant i leqslant m)… 0DJK
Definition 12 Given two A-modules E, F and a form Φ sesquilinear for J on E × F, the form Φ₍ₘ₎ sesquilinear for J on (∧ᵐ E) × (∧ᵐ F) characterized by (37) is called the extension of Φ to the m-t… 0DJL
Proposition 10 Let A be a commutative field endowed with an automorphism J, E and F two finite-dimensional vector spaces over A, and Φ a sesquilinear form for J on E × F. If Φ is nondegenerate, t… 0DJM
Remark Let E be a free A-module, and let θ be the canonical isomorphism of bigwedgeᵐ E onto the submodule of antisymmetrized tensors of order m (Chap. III, § 5, No. 6, Prop. 6). Let Φ be… 0DJN
Remark 1 When A is commutative, (45 G) reduces to (45 D), and (45 DG) to (45 GD), by means of the formula ᵗ(M'M'') = ᵗM'' · ᵗM' (cf. (42)), where the two members are here calculated in A. 2… 0DJO
Remark 2 When A is a field, the linear mappings s and d have the same rank. We see here that their matrices M(s) and M(d) have the same rank; indeed, a matrix over A and its transpose have… 0DJP
Remark 3 If M and N are right A-modules having finite bases (mᵢ) and (nₖ), Φ a left sesquilinear form for J on M × N (No. 6, Remark), s and d its associated mappings, and R = (Φ(mᵢ, nₖ)) it… 0DJQ
Remark 4 Give an example of a bilinear form Φ defined on a product E × F of two vector spaces, such that d_Φ is bijective, s_Φ injective but not bijective (take E of infinite dimension and… 0DJR
Remark 5 Let A be a ring endowed with an antiautomorphism J, E a left A-module, G an (A, A)-bimodule and Φ a mapping from E × E into G, right sesquilinear for J. Prove the identity (where Q… 0DJS
Remark 6 Soient K a commutative field of characteristic 2, A a separable quadratic extension of K; one has A = K(θ), where θ is root of an irreducible polynomial X² + X + β of K[X] and the… 0DJT
Remark 7 Soient A a field, E a vector space over A, Φ a sesquilinear form on E, u an endomorphism of E. 0DJU
Remark 8 Let E, E₁ be two A-modules, Φ (resp. Φ₁) a sesquilinear form on E (resp. E₁). Suppose that Φ₁ is non-degenerate and that there exist an element α ∈ A and a bijection u of E onto E₁… 0DJV
Remark 9 Let A be a field, E₁, E₂ two vector spaces not reduced to 0 over A, Φ₁ (resp. Φ₂) a non-degenerate sesquilinear form on E₁ (resp. E₂) for an antiautomorphism J₁ (resp. J₂) of A. Le… 0DJW
Remark 10 Let A be a field, E, F two left vector spaces over A, Φ a non-degenerate sesquilinear form on E × F for an antiautomorphism J of A. 0DJX
Remark 11 Let E, E' be two left vector spaces over fields A, A' respectively, of dimension geqslant 3; let 𝔉(E) (resp. 𝔉(E') be the lattice ordered set (for the relation of inclusion) formed… 0DJY

§ 2. Discriminant of a sesquilinear form

Definition 1 Given a sesquilinear form Φ for J on E and a system S = (x₁, …, xₙ) of n elements of E, the discriminant of Φ with respect to this system is called, and is denoted by D_Φ(x₁, …, xₙ… 0DK0
Example Let B be an algebra over the ring A, such that B is a free A-module of finite dimension n. Then the mapping (x, y) → Tr_B/A(xy) (Chapter VIII, § 12, no 2) is a bilinear form on B… 0DK1
Remark Soient (e₁, …, eₙ) a basis of B over A, and 0DK2
Proposition 1 Let Φ be a sesquilinear form for J on E, (tildex₁, …, xₙ) a system of n elements of E, and (aᵢⱼ)_(i,j = 1, …, n) a family of n² elements of A; put yᵢ = ∑ᵢ₌₁ⁿ aⱼᵢxᵢ. We then have 0DK3
Example Let B be an algebra over A which is a free A-module of finite dimension n, (x₁, …, xₙ) a basis of B over A, and m an ideal of A. If h denotes the canonical homomorphism of B onto B… 0DK4
Proposition 2 Suppose that A is an integral domain. Let Φ be a sesquilinear form for J on E and (e₁, …, eₙ) a basis of E, such that D_Φ(e₁, e₂, …, eₙ) ≠ 0. 0DK5
Proposition 3 Let Φ be a sesquilinear form for J on E, and let S be a basis of E. The following conditions are equivalent: 0DK6
Proposition 4 Suppose that A is integral. Let S be a basis of E. A necessary and sufficient condition for a sesquilinear form Φ on E to be nondegenerate is that one has D_Φ(S) ≠ 0. 0DK7
Proposition 5 Let A be a field, B a commutative algebra of finite dimension n over A, and S a basis of B. In order that B be separable (chap. VIII, § 7, No. 5, def. 1), it is necessary and suffi… 0DK8
Remark Suppose that B is an overfield of A. Let S = (x₁, …, xₙ) be a basis of B, and (s₁, …, sₙ) the A-isomorphisms of B into the algebraic closure A' of A (each sⱼ being repeated [B : A]… 0DK9
Proposition 6 Let Φ be an A-bilinear form on E, and K a subring of A such that A is a K-module free of finite dimension q. If (eᵢ)_i=1,…,n is a basis of E over A and (aⱼ)_j=1,…,q a basis of A ov… 0DKA
Corollary (« Transitivity Formula for Discriminants ») Let K be a commutative ring, A a commutative algebra admitting a finite basis (aⱼ)_j=1,…,q over K, and E an algebra over A admitting a finite basis (eᵢ)_i=1,…,n. Then (aⱼ eᵢ) is a… 0DKB

§ 3. Hermitian forms and quadratic forms

Definition 1 Let ε be an element of the center of A. A sesquilinear form Φ on E such that one has Φ(x, y) = ε overlineΦ(y, x) whatever x and y in E may be is called an ε-hermitian form. A 1-her… 0DKC
Remark 1 If Φ is an ε-hermitian form on E, one has Φ(x, y) = ε overlineΦ(x, y) overlineε whatever x, y in E may be. Hence, if Φ takes invertible values, one has ε overlineε = 1. 2) If there… 0DKD
Proposition 1 Let Φ₁ (resp. Φ₂) be a K-bilinear form on E₀ × E'₀, invariant under j and j'. The mapping which to Φ₁ (resp. Φ₂) associates the mapping Φ of E × E' into A defined by 0DKF
Remark 1 The formulas (7) and (8) show that, if x ∈ E, in order that Φ(x, x') = 0 for all x' ∈ E', it is necessary and sufficient that Φ₁(x, x') = 0 (resp. Φ₂(x, x') = 0) for all x' ∈ E'. 2… 0DKG
Definition 2 We suppose that the ring A is commutative. We say that a mapping Q of E into A is a quadratic form on E if 1) Q(α x) = α² Q(x) for α ∈ A and x ∈ E; 2) the mapping Φ : (x, y) → Q(x… 0DKI
Proposition 2 Suppose that A is commutative and that E admits a basis (eᵢ)_i ∈ I. Then, for every quadratic form Q on E, there exists a bilinear form f on E × E such that Q(x) = f(x, x) for ever… 0DKJ
Proposition 3 Let h be a homomorphism of A into a commutative ring A'. For every quadratic form Q on the A-module E, there exists one and only one quadratic form Q' on the A'-module A' ⊗_A E (ch… 0DKK

§ 4. Totally isotropic subspaces. Witt's Theorem

Definition 1 Given a module E over the ring A, an element x of E is said to be isotropic if Φ(x, x) = 0. A submodule F of E is said to be 1) isotropic if there exists an element x ≠ 0 of F orth… 0DKL
Proposition 1 Suppose that A is a field. Let F be a non-isotropic finite-dimensional subspace of E; then E is the direct sum of F and F⁰. 0DKM
Corollary If F is a finite-dimensional subspace of E, and if Φ is non-degenerate, the following conditions are equivalent: 0DKN
Definition 2 Let Q be a quadratic form on E. An element x of E is said to be singular (relative to Q) if Q(x) = 0. A submodule F of E is said to be: 1) singular if there exists an element x ≠ 0… 0DKO
Lemma 1 Let Φ be an ε-hermitian form satisfying (T) (resp. the bilinear form associated with a quadratic form Q) on E, and let F be a totally isotropic (resp. totally singular) subspace of… 0DKP
Proposition 2 Soit Φ une forme ε-hermitienne non dégénérée vérifiant (T) (resp. la forme bilinéaire associée à une forme quadratique non dégénérée Q), et soit F un sous-espace totalement isotrop… 0DKQ
Corollary 1 If F is a totally isotropic (resp. totally singular) subspace of dimension r, there exists a totally isotropic (resp. totally singular) subspace F' of dimension r such that F ∩ F'… 0DKR
Corollary 2 Two neutral ε-hermitian forms on spaces of the same dimension over A are equivalent. 0DKS
Remark Under the conditions of Corollary 1, E is the direct sum of F + F' and the orthogonal of F + F'. We therefore have a Witt decomposition of E. By prop. 2, a), there exist bases of F… 0DKT
Proposition 3 Let Φ be a non-degenerate ε-hermitian form satisfying (T) (resp. the bilinear form associated with a non-degenerate quadratic form Q). Let F₁ and F₂ be two maximal totally isotropi… 0DKU
Corollary 1 Under the hypotheses of Prop. 3, two maximal totally isotropic (resp. totally singular) subspaces of finite dimension have the same dimension. For every maximal totally isotropic (… 0DKV
Corollary 2 Let Q be a non-degenerate quadratic form on a vector space E of finite dimension n over an algebraically closed field A; there then exists a basis (ei){1 \leq i \leq n} of E such t… 0DKW
Definition 3 Suppose that E is of finite dimension and that Φ is a non-degenerate ε-hermitian form satisfying (T) (resp. the bilinear form associated with a non-degenerate quadratic form Q). We… 0DKX
Proposition 4 A non-degenerate quadratic form Q on E such that there exists a vector x ≠ 0 of E such that Q(x) = 0 being given, for every element a of A, there then exists y ∈ E such that Q(y) =… 0DKY
Theorem 1 (Witt) Let E and E' be two vector spaces of finite dimensions, provided respectively with two non-degenerate ε-hermitian forms Φ and Φ' satisfying condition (T) of No. 2 (resp. with two n… 0DKZ
Corollary 1 Let, for i = 1, 2, Ei be a finite-dimensional vector space, Φi a non-degenerate ε-hermitian form on Ei satisfying (T) (resp. Qi a non-degenerate quadratic form on Ei), E'i and E''i… 0DL0
Corollary 2 The hypotheses being those of Theorem 1, the group of metric automorphisms of E permutes transitively the totally isotropic (resp. totally singular) subspaces of given dimension of… 0DL1
Corollary 3 Let Q be a non-degenerate quadratic form on a finite-dimensional vector space E over an algebraically closed field A. The group of metric automorphisms of E permutes transitively t… 0DL2

§ 5. Special Properties of Alternating Bilinear Forms

Theorem 1 Let A be a (commutative) principal ring, E a free A-module of finite dimension n and Φ an alternating bilinear form on E. Then there exists a basis (eᵢ)_1 ≤ i ≤ n of E and an even… 0DL3
Corollary 1 Let A be a commutative field, E a finite-dimensional vector space of dimension n over A, and Φ an alternating bilinear form on E. There then exists a basis (eᵢ)_1 ≤ i ≤ n of E and… 0DL4
Remark It should be noted that this corollary is also an immediate consequence of Proposition 3 of § 4, No. 2 and of its Corollary 1, because with the notation of this proposition, one ne… 0DL5
Corollary 2 Let A be a commutative field, E a finite-dimensional vector space of dimension n over A. For every bivector z ∈ bigwedge² E, there exists a basis (eᵢ)_1 ≤ i ≤ n of E such that 0DL6
Corollary 3 Let A be a commutative field, R an alternating square matrix over A. The rank of R is an even number 2r, and there exists an invertible matrix P over A such that 0DL7
Remark If A is any commutative ring and R an alternating square matrix of odd order n over A, one has det R = 0. This results from Cor. 3 when A is a field. A direct proof in the case whe… 0DL8
Definition 1 Given an alternating matrix R = (αᵢⱼ) (i, j = 1, …, 2m) of even order 2m over an arbitrary commutative ring A, the pfaffian of R, denoted by Pf(R), is the element P((αₕₖ)) of A, wh… 0DL9
Example Suppose: 0DLA
Proposition 1 Let R be an alternating matrix of even order 2m over a commutative ring A, and let P be a square matrix of order 2m over A. We have 0DLB
Proposition 2 For every alternating matrix R of even order 2m over a commutative ring A, one has det R = (Pf(R))². Indeed, since the two members of (8) are polynomials with integer coefficients… 0DLC
Proposition 3 The determinant of a symplectic matrix is equal to 1. 0DLD

§ 6. Special Properties of Hermitian Forms

Definition 1 Let Φ be a Hermitian form on E. A basis (eᵢ) of E is said to be orthogonal for Φ if any two elements of this basis are orthogonal for Φ. 0DLE
Lemma 1 Suppose that A is a field and Φ a Hermitian form ≠ 0 on E. If all the vectors of E are isotropic, A is a commutative field of characteristic 2, the antiautomorphism J is the identi… 0DLF
Theorem 1 Suppose that A is a field and E a finite-dimensional vector space of dimension n over A. Then, for every Hermitian form Φ on E, E admits an orthogonal basis, unless the following c… 0DLG
Corollary 1 The notations being those of th. 1, suppose in addition that (C) is not satisfied, that Φ is non-degenerate and that, for every x ∈ E, there exists ρ ∈ A such that Φ(x, x) = ρ barρ… 0DLH
Remark The last assumption of the corollary is satisfied when J is the identity, and every element of A is the square of an element of A (for example when A is algebraically closed). 0DLI
Corollary 2 Let A be a field and R a hermitian matrix of order n and of rank r over A. Then, unless the following condition is satisfied: 0DLJ
Proposition 1 Suppose that A is a commutative field. Let Φ be a hermitian form on E, and let (xₙ) (n = 1, 2, …) be a sequence (finite or infinite) of linearly independent vectors of E such that… 0DLK
Proposition 2 Let Φ be a hermitian form on E, and let (eᵢ) (i = 1, …, n) be an orthogonal (resp. orthonormal) basis of E for Φ. Then, for every p geqslant 0, the basis of ⨂ᵖ E formed by the e_i₁… 0DLL
Proposition 3 Suppose that A is a commutative field and that E is of finite dimension > 0. Let Φ be a non-degenerate hermitian form on E. The mapping u → det u is a homomorphism of the unitary g… 0DLM
Lemma 2 For an endomorphism u of E to be such that u² = 1, it is necessary and sufficient that 1/2(1 - u) be a projector in E; then u is the difference of the two projectors 1/2(1 + u) and… 0DLN
Proposition 4 Let u ∈ GL(E) be an involution. The following properties are equivalent: a) u belongs to the unitary group associated with Φ ; b) the submodules U^+ = 1/2(1 + u)(E) and U^- = 1/2(1… 0DLO
Corollary 1 Suppose that A is a field and that E is finite-dimensional. The mapping u → 1/2(1 + u)(E) is a bijection from the set of involutions u belonging to the unitary group associated wit… 0DLP
Corollary 2 Suppose that A is a field and that E is finite-dimensional. In order that a projector ν in E be such that ν(E) and ν(0) are orthogonal (and consequently non-isotropic), it is neces… 0DLQ
Proposition 5 Suppose the vector space E to be of finite dimension n. The orthogonal group O(Q) associated with Q is then generated by the symmetries with respect to the non-isotropic hyperplane… 0DLR
Definition 2 Let A be a field, L an affine space over A and T the space of translations of L (Chapter II, 2nd ed., App. II). If T is endowed with a non-degenerate hermitian form Φ, L is said to… 0DLS
Definition 3 Let L be a hermitian space over a field A, T the space of translations of L. A displacement (resp. similitude) of L is any affine bijection u of L onto L such that the linear mappi… 0DLT
Proposition 6 Let L be a finite-dimensional hermitian space over A, whose metric form has index 0. Any similitude u of L, with multiplier μ ≠ 1, then admits one and only one fixed point. 0DLU

§ 7. Hermitian forms and ordered fields

Definition 1 A hermitian form Φ on E is said to be positive (resp. negative) if Φ(x, x) geqslant 0 (resp. Φ(x, x) leqslant 0) for every x ∈ E. 0DLV
Proposition 1 Suppose E is finite-dimensional, and A = K or A = K(i). If Φ is a positive non-degenerate hermitian form on E, then the extensions of Φ to ⨂ᵖ E and ∧ᵖ E (p > 0), as well as the inv… 0DLW
Proposition 2 Let Φ be a positive hermitian form on E. For x, y in E, one has Φ(x, y) overlineΦ(x, y) leqslant Φ(x, x) Φ(y, y). 0DLX
Corollary The set of isotropic vectors of E is the orthogonal subspace E⁰ of E for Φ. In order that Φ be non-degenerate, it is necessary and sufficient that Φ(x, x) > 0 for every x ≠ 0. 0DLY
Proposition 3 Suppose E finite-dimensional and A commutative (A = K or A = K(i)). Let Φ be a hermitian form on E, X its matrix with respect to a basis (xⱼ) (j = 1, …, n) of E; for every subset H… 0DLZ
Remark It follows from the existence of orthonormal bases that two non-degenerate positive hermitian forms on two vector spaces of the same finite dimension are equivalent (§ 1, no 6). Le… 0DM0
Theorem 1 (« law of inertia ») Suppose that A satisfies the hypotheses at the beginning of this paragraph, and that E is of finite dimension n. Let Φ be a hermitian form on E. Then: 0DM1
Definition 2 With the notations of Theorem 1, the pair (s, t) is called the signature of Φ. 0DM2
Remark 1 When J is the identity, we recover the notion of adjoint of a homomorphism defined in § 1, No. 8. 0DM3
Remark 2 Suppose still that A = K or A = K(i). Let Eᴶ be the right A-module defined in § 1, No. 2, Def. 5. The form Φ is a bilinear form on E × Eᴶ, Φᴶ is a bilinear form on Eᴶ × E, and u is… 0DM4
Theorem 2 Let S be a set of endomorphisms of E (resp. of semilinear mappings of E into E when A = K or A = K(i)) stable under the mapping u → u. Then, if V is a subspace of E stable under S… 0DM5
Corollary 1 Suppose that A = K or that A = K(i). Let B be a subalgebra of L_A(E), stable under the mapping u → u. Then E is a semisimple B-module, and is the direct sum of simple submodules pa… 0DM6
Corollary 2 The assumptions and notations being those of cor. 1, suppose in addition that the algebra B is commutative. Then all its elements are semisimple (normal) endomorphisms of E. When A… 0DM7
Proposition 4 Let u be a normal endomorphism of E. When A is equal to K(i) or to the quaternion field over K, there exists an orthonormal basis (for Φ) of E formed by eigenvectors of u. When A =… 0DM8
Remark It follows from th. 2 that the eigenspaces relative to two distinct eigenvalues of u are orthogonal. 0DM9
Proposition 5 Let u be a hermitian endomorphism of E. The eigenvalues of u belong to K, and there exists an orthonormal basis of E formed of eigenvectors of u. 0DMA
Proposition 6 a) Let Ψ be a hermitian form (resp. symmetric bilinear form if A = K or A = K(i)) on E. There exists a basis of E which is orthonormal for Φ and orthogonal for Ψ. 0DMB
Remark 3 One assumes that A = K or A = K(i) and that E is finite-dimensional. Let Φ₁, Φ₂ be two positive hermitian forms on E, and let V = (αᵢⱼ), W = (βᵢⱼ) be the matrices of these forms wi… 0DMC
Remark 4 One assumes that A = K or A = K(i). Let R be a hermitian matrix of order n over A; one says that R has signature (s, t) if the hermitian form on Aⁿ having R as matrix with respect… 0DMD
Remark 5 One assumes that A = K or A = K(i); let R, S be two hermitian matrices over A, of signatures (s, t) and (s', t') respectively (exerc. 4); show that the matrix R ⊗ S is a hermitian… 0DME
Remark 6 Let K be a maximal ordered field, L a simple algebra of finite rank over K. If (s, t) is the signature of the symmetric bilinear form (x, y) → Tr_L/K(xy) on L, now prove that one h… 0DMF
Remark 7 We suppose that A satisfies the conditions at the beginning of § 7 and that E is of finite dimension n. Let Φ be a non-degenerate positive hermitian form on E. a) Now prove that ev… 0DMG
Remark 8 On suppose A = K; soit L un espace hermitien de dimension finie sur A, dont la forme métrique est positive non dégénérée. Soient M, N deux parties de L, u une bijection de M sur N… 0DMH
Remark 9 On suppose remplies les conditions du n° 3, et en outre que A est le corps des quaternions sur K. Soit u un endomorphisme normal de E. Montrer que E est somme directe de sous-espac… 0DMI
Remark 10 On suppose remplies les conditions du n° 3, et en outre que A est égal à K(i) ou au corps des quaternions sur K. Soit u un endomorphisme normal de E. 0DMJ
Remark 11 On suppose remplies les conditions du n° 3, et en outre que A = K. Soit u un endomorphisme normal de E. 0DMK
Remark 12 On suppose remplies les conditions du n° 3, et en outre que A est égal à K(i) ou au corps des quaternions sur K. Montrer que pour tout endomorphisme u de E, il existe une base orth… 0DML
Remark 13 Soient A un corps vérifiant les conditions du début du § 7, E, F deux espaces vectoriels de dimension finie sur A, Φ (resp. Ψ) une forme hermitienne positive non dégénérée sur E (r… 0DMM
Remark 14 On suppose remplies les conditions du no 3. Soient u un endomorphisme de E, h₁, h₂ les deux endomorphismes hermitiens positifs de E, tels que l’on ait h₁² = u^*u, h₂² = uu^* (exerc… 0DMN
Remark 15 Suppose that the conditions of No. 3 are satisfied. Show that every positive Hermitian matrix H over A can be written in the form LL^*, where L = (λᵢⱼ) has only zeros below its dia… 0DMO
Remark 16 Suppose that the conditions of No. 3 are satisfied and in addition that A = K(i). Let u be an endomorphism of E. 0DMP
Remark 17 Suppose the conditions of No. 3 are satisfied; for every ξ ∈ A, denote by |ξ| the element ρ geqslant 0 of K such that ρ² = ξ overlineξ (absolute value of ξ). For every square matri… 0DMQ
Remark 18 The conditions of No. 3 are assumed to be satisfied. 0DMR
Remark 19 Suppose that the conditions of No. 3 are satisfied. Let M be a square matrix of order n over A; now prove that for every square submatrix N of M (obtained by deleting in M a certai… 0DMS
Remark 20 Suppose that the conditions of No. 3 are satisfied and in addition that A is equal to K or to K(i). Apply the results of Exercises 17 to 19 to the extension of Φ to the p-th exteri… 0DMT
Remark 21 Suppose that the conditions of No. 3 are satisfied. Let u be a hermitian endomorphism of E and let ∏ᵢ₌₁ⁿ (X-λᵢ) be the decomposition into linear factors of its characteristic polyn… 0DMU
Remark 22 Suppose that the conditions of No. 3 are fulfilled, and in addition that A is equal to K(i) or to the field of quaternions over K. Let u, v be two normal endomorphisms of E; let (E… 0DMV
Remark 23 Suppose that the conditions of No. 3 are fulfilled. Let u and v be two endomorphisms of E. 0DMW
Remark 24 Suppose that the conditions of No. 3 are fulfilled. Let Γ be a group of automorphisms of E such that every u ∈ Γ is normal. Now prove that there exists a decomposition of E into di… 0DMX
Remark 25 Suppose that A = K (maximal ordered field). Let L be an euclidean space of dimension n over A, whose metric form is positive non-degenerate. 0DMY
Remark 26 a) Let K be a commutative field, S a subset of K such that K is a maximal S-orderable field (chap. VI, § 2, exerc. 8). Let f be a polynomial of K[X], L its field of roots. Show tha… 0DMZ
Remark 27 Let K be a maximal ordered field, E a vector space of dimension n over K, Φ a non-degenerate symmetric bilinear form on E, of signature (s, t) distinct from (n, 0) and from (0, n)… 0DN0

§ 8. Types of quadratic forms

Proposition 1 Let Q and Q' be two quadratic forms over A, non-degenerate and of finite dimensions. In order that Q and Q' be equivalent, it is necessary and sufficient that they have the same di… 0DN1
Proposition 2 The relation « there exists a quadratic form Q over A, non-degenerate and of finite dimension, such that X = θ(Q) » is collectivizing in X (Ens., chap. II, § 1, n° 4). 0DN2
Proposition 3 Let Q be a non-degenerate quadratic form of finite dimension on a vector space V over A. Let – Q denote the quadratic form on V defined by (-Q)(x) = -Q(x) (x ∈ V). Then the form Q… 0DN3
Definition 1 The set of types of non-degenerate quadratic forms of finite dimensions over A, endowed with the addition defined by (2), is called the group of types of quadratic forms, or Witt g… 0DN4
Remarks 1) Every non-degenerate quadratic form Q of finite dimension whose type is zero (i.e. such that θ(Q) = T₀ with the above notation) is a neutral form. There indeed exist neutral for… 0EEP
Proposition 4 Let Q be a non-degenerate quadratic form on a finite-dimensional vector space E over A. Suppose that A is of characteristic ≠ 2, and let (x₁, …, xₙ) be an orthogonal basis of V. Le… 0DN5
Corollary The hypotheses and notation being those of prop. 3, the elements a.T₁ (a ∈ A^*) form a set of generators of the group of types of quadratic forms over A. 0DN6
Proposition 5 We suppose that A is a maximal ordered field. Let Q be a non-degenerate quadratic form of finite dimension over A, and (s, t) its signature (§ 7, no 2, def. 2). Then the type of Q… 0DN7
Remark 1 It is clear that, if a is an element of A^*, one has 0DN8
Remark 2 Since A is of characteristic ≠ 2, every element T of 𝔅 can be written in the form ∑ᵢ₌₁ⁿ aᵢ . T₁ where aᵢ ∈ A^* (No. 2, prop. 4). One has 0DN9
Remark 3 Suppose that A is a maximal ordered field. Then the ring 𝔅 is isomorphic to Z (prop. 5), the integer corresponding to the type of a form of signature (s,\ t) being s - t (ibid.). S… 0DNA

§ 9. Clifford Algebras

Definition 1 The Clifford algebra of Q, denoted by C(Q), is the quotient algebra of the tensor algebra T(E) of the module E by the two-sided ideal (denoted by I(Q)) generated by the elements of… 0DNB
Example If E admits a basis consisting of a single element e, T(E) is isomorphic to the polynomial algebra A[X], and C(Q) is a quadratic extension of A, having as basis (1,\ u), where u is… 0DNC
Proposition 1 Let f be a linear mapping of E into an algebra D over A such that f(x)² = Q(x).1 for all x ∈ E. There exists one and only one homomorphism barf of C(Q) into D such that f = barf ∘… 0DND
Proposition 2 Let A' be a commutative ring, φ a homomorphism of A into A', Q’ the quadratic form on E' = A' ⊗_A E deduced from Q by extension of scalars (§ 3, no 4, prop. 3). There exists one an… 0DNE
Lemma 1 Let f be an element of the dual E^* of E. There exists one and only one linear mapping i_f of T(E) into itself such that i_f(1) = 0 tag4 i_f ∘ eₓ + eₓ ∘ i_f = f(x).I tag5 for every… 0DNF
Lemma 2 There exists one and only one linear mapping λ_F of T(E) into itself such that 0DNG
Lemma 3 Let F and G be two bilinear forms on E. We have λ_F ∘ λ_G = λ_F+G. For every bilinear form F on E, λ_F is a bijection of T(E) onto itself. 0DNH
Proposition 3 Let Q and Q' be two quadratic forms and F a bilinear form on E such that Q'(x) = Q(x) + F(x, x) for every x ∈ E. The mapping λ_F maps the ideal I(Q') onto the ideal I(Q), and defin… 0DNI
Lemma 4 If the quadratic form Q is zero, C(Q) is none other than the exterior algebra of E. 0DNJ
Theorem 1 Suppose that the A-module E admits a basis (xᵢ)_i ∈ L, the set of indices L being endowed with a total ordering. For every finite subset H of L, put x_H = ρ(x_h₁) ρ(x_h₂) … ρ(x_h_q… 0DNK
Corollary 1 If E is a free module of dimension n, C(Q) is a free module of dimension 2ⁿ; moreover, if n > 0, C⁺ and C⁻ are free modules of dimension 2ⁿ⁻¹. 0DNL
Corollary 2 If E is a free module, the canonical mapping ρ of E into C(Q) and the mapping a → a.1 of A into C(Q) are injective. 0DNM
Corollary 3 Suppose that E is the direct sum of two free submodules E₁ and E₂. Let Qᵢ be the restriction of Q to Eᵢ and pᵢ the canonical mapping of C(Qᵢ) into C(Q) (i = 1, 2). Then the linear… 0DNN
Corollary 4 The hypotheses and notation being those of Cor. 3, suppose moreover that E₁ and E₂ are orthogonal, and transport to C(Q₁) ⊗ C(Q₂), by means of the bijection p, the algebra structur… 0DNO
Theorem 2 Suppose that the dimension of E is an even number m = 2r and that Q is neutral (§ 4, no. 2). Then the algebra C(Q) is separable (chap. VIII, § 7, no. 5, def. 1) and is isomorphic t… 0DNP
Corollary If m is even, but Q of arbitrary index, the algebra C(Q) is a central simple algebra of dimension 2ᵐ. Moreover, if m > 0, the subalgebra C⁺(Q) is separable, and its center Z is of… 0DNQ
Remark 1 Since the algebra C(Q) is simple, it has only one class of irreducible representations; they are called spinor representations; when one fixes one's attention on one of these repre… 0DNR
Remark 2 Suppose that A is of characteristic ≠ 2, and let (x₁, …, xₘ) (m = 2r) be an orthogonal basis of E. Put z = 2ʳ x₁ … xₘ ∈ C(Q); since xᵢ xⱼ + xⱼ xᵢ = 0 for i ≠ j, we have zxⱼ = -xⱼ z… 0DNS
Theorem 3 Suppose that the dimension of E is an odd number m = 2r + 1 (and hence that A is of characteristic ≠ 2). 0DNT
Remark The discriminant D of Φ with respect to the basis (xⱼ)_(j=0,…,2r) is equal to 2²ʳ⁺¹ Q(x₀) … Q(x₂ᵣ). Consequently Z is generated by 1 and by the odd element z' = 2ʳ⁺¹ z such that z'… 0DNU
Definition 2 The Clifford group of Q (resp. special Clifford group of Q) is called the multiplicative group of the invertible elements s of C(Q) (resp. C⁺(Q)) such that sEs⁻¹ = E. 0DNV
Theorem 4 Put, for s ∈ G and x ∈ E, φ(s).x = sx s⁻¹. 0DNW
Lemma 5 Every element s of G is of the form zs', where z is an invertible element of Z and s' belongs to G ∩ C⁺(Q) or to G ∩ C⁻(Q); the subgroup G⁺ is of index 2 in G when E ≠ 0. 0DNX
Proposition 4 Let β be the principal antiautomorphism of C(Q) (n° 1). For every s ∈ G⁺, β(s)s is a scalar. The mapping N : s → β(s)s is a homomorphism of G⁺ into the multiplicative group A^* of… 0DNY

§ 10. Angles

Proposition 1 Let A(Φ) be the subalgebra of L_A(E) generated by the direct similitudes of E. 0DNZ
Remark 1 Let ν be the similarity of matrix beginpmatrix 0 & δ \\ 1 & 0 endpmatrix with respect to (e₁, e₂); the similarity ω introduced in the proof of Prop. 1 is equal to -α₁ ν; one has ν²… 0DO0
Remark 2 The direct similitudes u = a + b ν (a, b in Λ, b ≠ 0) whose square is a homothety are the homotheties and the scalar multiples b ν of ν, since (a + b ν)² = (a² + δ b²) + 2ab ν. The… 0DO1
Remark 3 It is easily verified that, for x, y in E and u ∈ Λ(Φ), one has Φ(u(x), y) = Φ(x, overlineu(y)). Thus the adjoint endomorphism of a direct similitude u is the direct similitude ove… 0DO2
Remark 4 Since every inverse similitude of E is the product of a direct similitude and the symmetry with respect to Ae₁, the matrices No. 1 0DO3
Corollary 1 The group S⁺ of direct similitudes is commutative. Whatever the non-isotropic vectors x, y of E, there exists one and only one direct similitude u such that y = u(x). 0DO4
Corollary 2 The group O⁺ of rotations is commutative. Whatever the vectors x, y of E such that Φ(x, x) = Φ(y, y) ≠ 0, there exists one and only one rotation u such that y = u(x). 0DO5
Corollary 3 The group S⁺/H is commutative. It operates on the set of non-isotropic lines of E. Whatever the non-isotropic lines D, D' of E, there exists one and only one element φ of S⁺/H such… 0DO6
Proposition 2 The kernel of the canonical homomorphism of O⁺ into S⁺/H is 1, -1. 0DO7
Proposition 3 The homomorphism u → u/overlineu = u² / N(u) of S⁺ into itself has H as kernel, and defines, by passing to the quotient, an isomorphism of S⁺/H onto O⁺. 0DO8
Corollary Let us denote by i : O⁺ → S⁺/H and d : S⁺/H → O⁺ the homomorphisms defined in Propositions 2 and 3, and let us write the abelian groups O⁺ and S⁺/H additively; then d(i(θ)) = 2θ fo… 0DO9
Remark When –1 belongs to the class mod. (A^*)² in question, one generally chooses ω in such a way that ω² = -1, which determines it up to sign. When this class contains 1 but not –1, one… 0DOA
Proposition 4 a) When ω² = δ is not a square in A (that is to say, when E does not contain any isotropic right), the mapping t of S⁺/H into tildeA is a bijection. 0DOB
Proposition 5 Let us write O⁺ additively. For every pair of elements θ, θ' of O⁺ one has 0DOC
Proposition 6 Let d be the isomorphism of S⁺/H onto O⁺ defined in Prop. 3. For every element φ of S⁺/H such that t = t(φ) is finite, one has 0DOD
Corollary For every element θ of O⁺ such that t(θ) is finite, one has 0DOE
Remark Formulas (6) remain true for t = ∞ provided one replaces the rational functions occurring on the right-hand side by their canonical extensions to the projective field tildeA (Chapt… 0DOF
Proposition 7 Suppose that A is a maximal ordered field and that Φ is a positive form (§ 7). Then the canonical homomorphisms of O⁺ into S⁺/H⁺ and of O⁺/ 1, -1 into S⁺/H are bijective, and S⁺ is… 0DOG
Corollary Given two half-lines D, D' with origin 0, there exists one and only one rotation v such that v(D) = D'. 0DOH
Proposition 8 Suppose that A is a maximal ordered field, and that the form Φ is positive. Let D₁, D₂, D'₁, D'₂ be four lines (resp. half-lines) with origin 0 of E. A necessary and sufficient con… 0DOI
Remark 1 The set L of lines (resp. half-lines) with origin 0 of E is a homogeneous space for the abelian group S⁺/H (resp. O⁺) such that the identity element is the only operator leaving in… 0DOJ
Remark 2 In the definition of the group of angles of lines, one may, instead of the group S⁺/H, use the group O⁺/ -1, 1 which is canonically isomorphic to it (Prop. 7). The canonical homomo… 0DOK
Proposition 9 Suppose that the field A is maximal ordered, and that Φ is a positive form. For every integer n > 1, the number of elements θ of the group 𝔄₀ of angles of lines (resp. 𝔄 of angles… 0DOL
Corollary The right angle (resp. flat angle) is the only element of order 2 of the group 𝔄₀ (resp. 𝔄). 0DOM
Lemma 1 Let u be a direct similitude of E; all the bivectors of the form x ∧ u(x) belong to the same closed half-line of overset2∧ E. 0DON
Proposition 10 Suppose that the field A is maximal ordered, that the plane E is oriented, and that the form Φ is positive. For every pair of non-zero vectors x, y of E one has 0DOO
Remark 3 Given two lines (resp. half-lines) D, D' of an affine plane L attached to E, the angle of D and D' is defined to be, and is denoted by (widehatD, D'), the angle made by their direc… 0DOP
Remark 4 Let F be a vector space of arbitrary dimension over the maximal ordered field A, and let Ψ be a positive non-degenerate symmetric bilinear form on F. Given two linearly independent… 0DOQ
Proposition 11 Let E be an oriented plane over an ordered field A, D₀ a half-line of E, and G the set of half-lines of E distinct from D₀. The relation 0DOR
Corollary Let D₁ and D₂ be two distinct half-lines of E. For every half-line D₀ of E such that the sequence D₀, D₁, D₂ is direct, the set of half-lines D of E such that D₁ < D < D₂ (for the… 0DOS
Proposition 12 Let A be a maximal ordered field, E an oriented plane over A, D₀ a half-line of E, and G the set of half-lines of E distinct from D₀. The totally ordered sets A and G (prop. 11) ar… 0DOT

X. Homological algebra

§ 1. Complements of linear algebra

Proposition 1 Consider a commutative diagram of A-modules 0DOU
Lemma 1 Consider a commutative diagram of A-modules 0DOV
Proposition 2 Suppose that in the commutative diagram (4), the sequences (u, v) and (u', v') are exact. Then: (i) We have v₁ ∘ u₁ = 0; if u' is injective, the sequence (u₁, v₁) is exact. (ii) We… 0DOW
Corollary 1 Suppose that the diagram (4) is commutative and has exact rows. Then: 0DOX
Corollary 2 Suppose that the diagram (4) is commutative and has exact rows. Under these conditions: 0DOY
Definition 1 One says that the a-module E is flat if, for every exact sequence of right a-modules and homomorphisms 0DOZ
Proposition 3 In order that the a-module E be flat, it is necessary and sufficient that, for every injective A-homomorphism u : M' → M of right A-modules, the mapping u ⊗ 1 : M' ⊗_A E → M ⊗_A E… 0DP0
Proposition 4 (i) Let (Eᵢ)_i ∈ I be a family of a-modules, E = ⨁_i ∈ I Eᵢ their direct sum. In order that the a-module E be flat, it is necessary and sufficient that each of the Eᵢ be so. 0DP1
Example 1 It is clear that Aₛ is a flat A-module; it follows then from prop. 4 (i) that every free A-module, and more generally every projective A-module, is flat (see also II, p. 63, cor. 6… 0DP2
Example 2 According to prop. 4 (ii), every A-module which is an inductive limit of a filtered inductive system of free A-modules is flat. We shall prove a converse at No. 6. 0DP3
Example 3 If A is semisimple, every A-module is projective (VIII, § 5, No. 1, prop. 1), hence flat. 0DP4
Example 4 \ If A is an artinian local ring (not necessarily commutative), an A-module is flat if and only if it is free (AC II, § 3, No. 2, cor. 2 of prop. 5). \ 0DP5
Example 5 If A is integral, the field of fractions K of A is a flat A-module (II, p. 118, prop. 27). 0DP6
Example 6 In AC II and III, we shall study two important examples of A-modules flat when A is commutative: the rings of fractions S⁻¹ A, and, when A is noetherian, the separated completions… 0DP7
Example 7 Let a ∈ A be such that the mapping a_A : x ↦ ax of A into A is injective (“a is not a left divisor of 0”). If E is a flat A-module, then the homothety a_E is injective, since it is… 0DP8
Example 8 Let B be a ring and ρ : A → B a homomorphism. If E is a flat A-module, the B-module E_(B) = B ⊗_A E is flat. Indeed, let u : N' → N be an injective homomorphism of right B-modules… 0DP9
Example 9 Suppose that A = K[X, Y], where K is a field. Then the maximal ideal m generated by X and Y is a torsion-free A-module, but not flat. Consider indeed the ring B = A/(Y), which is i… 0DPA
Example 10 Suppose A is commutative. Let B be the algebra A[X₁, ..., Xₙ]/(P), where P is a nonzero polynomial. For every prime ideal p of A, denote by κ(p) the field of fractions of the integ… 0DPB
Example 11 \ Let S and X be two complex analytic spaces and let f : X → S be a morphism. We say that f is flat at a point x of X if O_X,x, considered as O_S,f(x)-module by means of the homomo… 0DPC
Proposition 5 (i) Every module admitting a finite presentation is finitely generated. (ii) If A is a left noetherian ring, every finitely generated a-module admits a finite presentation. (iii) E… 0DPD
Proposition 6 Let A be a ring, E an a-module of finite presentation. For every exact sequence 0DPE
Proposition 7 Let M be an a-module. There exists an ordered set I inductive to the right and an inductive system of a-modules of finite presentation (M_α, φ_βα) relative to I such that M is isom… 0DPF
Proposition 8 a) If the A-module E is of finite type (resp. of finite presentation), the canonical homomorphism (16) is injective (resp. bijective). 0DPG
Corollary Every flat module of finite presentation is projective. 0DPH
Lemma 2 Let I be a right filtered ordered set, (E_α, φ_βα) an inductive system of sets relative to I, E its inductive limit and φ_α : E_α → E, α ∈ I, the canonical mappings. Let f : I → I… 0DPI
Theorem 1 (D. Lazard) For every A-module E, the following conditions are equivalent: (i) E is flat. (ii) For every A-module P of finite presentation, the canonical homomorphism Hom_A(P, A) ⊗_A E → Hom_A… 0DPJ
Corollary Suppose A commutative. For every flat A-module E, the A-modules T(E), S(E), Λ(E), T^n(E), S^n(E), Λ^n(E) are flat. 0DPK
Remark Considering in (ii) a finite presentation As^J \xrightarrow{c} As^I \to P \to 0, one obtains the condition (ii’) still equivalent to the preceding ones: 0DPL
Definition 2 One says that the A-module E is injective if, for every exact sequence of A-modules and homomorphisms 0DPM
Lemma 3 For the a-module E to be injective, it is necessary and sufficient that, for every injective a-linear mapping u : M' → M, the mapping 0DPN
Remark Let E be an injective a-module, u : M' → M and f : M' → E homomorphisms of a-modules. If Keru⊂Kerf, there exists a homomorphism g : M → E such that g∘ u=f. This follows indeed from… 0DPO
Proposition 9 Let (Eᵢ)_i∈ I be a family of a-modules, E=∏_i∈ IEᵢ their product. For the a-module E to be injective, it is necessary and sufficient that each of the Eᵢ be so. 0DPP
Proposition 10 Let E be an a-module. For E to be injective, it is necessary and sufficient that, for every ideal a of A and every A-homomorphism f : a→ E, there exist e∈ E such that 0DPQ
Corollary 1 If the ring A is left Noetherian, every direct sum of injective A-modules is injective. 0DPR
Remark If every direct sum A-module of injective A-modules is injective, the ring A is left noetherian (X, p. 170, exercise 21). 0DPS
Corollary 2 Suppose A is integral. a) Every injective A-module is divisible. b) Every torsion-free (II, p. 115) and divisible A-module is injective. c) If A is principal, every divisible A-mod… 0DPT
Example 1 If A is integral, the field of fractions K of A is an A-module injective. If A is principal, K/A is an A-module injective. 2) For example, the Z-modules Q and Q/Z are injective. 3)… 0DPU
Proposition 11 Let B be a ring, F a B-module and P a (B, A)-bimodule. If F is an injective B-module and P a flat A-module, Hom_B(P, F) is an injective A-module. 0DPX
Definition 3 One says that the A-module E is a cogenerator if, for every A-module M and every nonzero element x of M, there exists an A-homomorphism u : M → E such that u(x) ≠ 0. 0DPY
Proposition 12 Let E be an injective A-module. In order that E be a cogenerator, it is necessary and sufficient that Hom_A(S, E) ≠ 0 for every simple A-module S. 0DPZ
Example The injective Z-module Q/Z (X, p. 18, example 2) is a cogenerator. Indeed, every Z-module simple module is isomorphic to a module Z/pZ, p≠ 0, and Hom_Z(Z/pZ,Q/Z) is nonzero (it con… 0DQ0
Proposition 13 Let B be a ring, F an injective cogenerator B-module, P a (B,A)-bimodule. Suppose that P is flat over A and such that P⊗_A S≠ 0 for every simple A-module S ( that is, faithfully fl… 0DQ1
Corollary 1 The A-module E_A=Hom_Z(A,Q/Z) is injective and a cogenerator. 0DQ2
Corollary 2 The A-module I⁰(M) is injective and the A-homomorphism e_M:M→ I⁰(M) is injective. 0DQ3
Corollary 3 Every A-module is isomorphic to a submodule of an injective A-module. 0DQ4
Corollary 4 For the A-module E to be injective, it is necessary and sufficient that every injective A-homomorphism f:E→ F possess an A-linear retraction. 0DQ5
Definition 4 Let M be an A-module. An injective envelope of M is a pair (I,i), where I is an injective A-module and i:M→ I is a homomorphism possessing the following property: 0DQ6
Example 1 Suppose that A is integral and M is torsion-free. Let K be the field of fractions of A and let i : M → K ⊗A M be the canonical homomorphism. Then (K ⊗A M, i) is an injective envelo… 0DQ7
Theorem 2 Let M be an A-module. a) M possesses injective envelopes. b) If (I, i) and (J, j) are two injective envelopes of M, there exists an isomorphism f : I → J such that f ∘ i = j. 0DQ8
Remark 1 Let (I, i) be an injective envelope of M and j : M → J an injective homomorphism of M into an injective A-module J. By the proof above, there exists an injective homomorphism f : I… 0DQ9
Remark 2 Let (I, i) be an injective envelope of M. Identify M with the submodule i(M) of I. For every submodule N of M, there exists an injective submodule of I which is an injective envelo… 0DQA
Proposition 14 Let I be a nonzero injective A-module. The following conditions are equivalent: (i) I is indecomposable (VII, § 4, no 8, def. 3); (ii) 0 is not the intersection of two nonzero subm… 0DQB
Corollary 1 The relation “I is a class of indecomposable injective A-modules” is collectivizing. 0DQC
Corollary 2 Let M be an A-module, I an injective envelope of M. In order that I be indecomposable, it is necessary and sufficient that 0 not be the intersection of two nonzero submodules of M. 0DQD
Example 2 If A is commutative and noetherian, the indecomposable injective A-modules are exactly the injective envelopes of the modules A/p where p is a prime ideal (X, p. 171, exercise 27). 0DQE
Lemma 4 Let M be a nonzero noetherian A-module, I an injective envelope of M. Then I possesses an indecomposable injective submodule. 0DQF
Theorem 3 Let I be an injective A-module. 0DQG

§ 2. Complexes of a-modules

Definition 1 A differential complex of A-modules is a pair (C,d) formed by a graded A-module C and an endomorphism d:C→ C graded of descending degree -1 and such that d∘ d=0. 0DRD
Example If C has zero differential, one has Z(C) = C, B(C) = 0 and H(C) identifies canonically with C. 0DRE
Definition 2 Let (C, d) and (C', d') be two complexes. A morphism ¹ from (C, d) to (C', d') is a graded A-homomorphism u of degree 0 from C to C' such that d' ∘ u = u ∘ d . 0DRF
Definition 3 Let (C, d) and (C', d') be two complexes. A homological isomorphism (or quasi-isomorphism) from (C, d) to (C', d') is a morphism u from (C, d) to (C', d') such that H(u) is bijecti… 0DRG
Proposition 1 For any family of complexes (Cᵢ)_i ∈ I, the canonical homomorphisms ⨁_i ∈ I H(Cᵢ) → H(⨁_i ∈ I Cᵢ), H(∏_i ∈ I Cᵢ) → ∏_i ∈ I H(Cᵢ) are bijective. 0DRH
Lemma 1 The image φ(Γ) of Γ in H(C'') × H(C') is the graph of a graded A-homomorphism of degree -1 from H(C'') into H(C'). 0DRI
Theorem 1 The unlimited sequence of homomorphisms of A-modules 0DRJ
Corollary 1 If two of the complexes C, C', C'' have zero homology, the third one does also. For u (resp. v) to be a homomorphism, it is necessary and sufficient that C'' (resp. C') have zero h… 0DRK
Corollary 2 Let u be a morphism of complexes. If Ker u and Coker u have zero homology, then u is a homologism. 0DRL
Proposition 2 Consider a commutative diagram of complexes with exact rows 0DRM
Example Let C be a complex. Consider the canonical exact sequence 0DRN
Example Consider the product of spheres S₂ × S₂ and the complex projective space P₂(C). Let b ∈ S₂, we define a cellular decomposition (Yₙ) of Y = S₂ × S₂ by setting 0DRO
Definition 4 Let (C, d) and (C', d') be two complexes, f and g two morphisms from C to C'. A homotopy connecting f to g is any graded A-homomorphism s of degree 1 from C to C' such that g - f =… 0DRP
Proposition 3 If f and g are two homotopic morphisms from C into C', then H(f) = H(g). Let s be a homotopy connecting f to g. One has (g - f)(Z(C)) = (d' ∘ s + s ∘ d)(Z(C)) = (d' ∘ s)(Z(C)) ⊂ B(… 0DRQ
Corollaire A morphism homotopic to a homology isomorphism is a homology isomorphism. 0DRR
Proposition 4 Let C, C', D, D' be four complexes, f : C → C', g : C → C', u : D → C, v : C' → D' four morphisms. If s is a homotopy connecting f to g, then v ∘ s ∘ u is a homotopy connecting v ∘… 0DRS
Corollaire Let C, C', C'' be three complexes, f and g two morphisms of C into C', f₁ and g₁ two morphisms of C' into C''. If s and s₁ are homotopies relating f to g and f₁ to g₁ respectively… 0DRT
Definition 5 A morphism of complexes f : C → C' is called a homotopism if there exists a morphism f' : C' → C such that f' ∘ f and f ∘ f' are homotopic to 1_C and 1_C' respectively. 0DRU
Proposition 5 A homotopism is a homologism; a morphism composed of homotopisms is a homotopism. A morphism homotopic to a homotopism is a homotopism. 0DRV
Corollary Let C, C', D, D' be four complexes, f : C → C' a morphism, u : D → C and v : C' → D' homotopy equivalences. For v ∘ f ∘ u to be a homotopy equivalence (resp. a homology isomorphism… 0DRW
Example Let u : M → N and v : N → P be homomorphisms of A-modules such that v ∘ u = 0; let C be the complex such that C₂ = M, C₁ = N, C₀ = P, Cᵢ = 0 for i ≠ 0, 1, 2, d₂ = u, d₁ = v, dᵢ = 0… 0DRX
Proposition 6 Let (C, d) be a complex. The following conditions are equivalent: (i) there exists a homotopy equivalence from (C, d) to (H(C), 0); (ii) there exists an A-endomorphism s of C, grad… 0DRY
Definition 6 A complex C is said to be split if it satisfies the equivalent conditions of Prop. 6. 0DRZ
Example 1 A complex with zero differential is split. 0DS0
Example 2 The complexes homotopic to zero are the split complexes of zero homology, i.e. the complexes C such that H(C)=0 and that Z(C) is a direct factor in C. 0DS1
Example 3 Let f:M→ N be a homomorphism of A-modules and C the complex such that C₁=M, C₀=N, Cᵢ=0 for i≠0,1, d₁=f, dᵢ=0 for i≠1. Then C is split if and only if Kerf is a direct factor in M an… 0DS2
Example 4 The complex C is split as soon as B(C) and H(C) are projective (resp. as soon as Bₙ(C) and Hₙ(C) are injective for each n). Indeed, according to the exact sequences (Iₙ) to (IVₙ) o… 0DS3
Example 5 In particular, if A is principal, a free complex C is split if and only if H(C) is free (that is to say Hₙ(C) free for every n∈Z). 0DS4
Remark a) Suppose that the canonical exact sequence of graded A-modules 0DS5
Lemma 2 (Cyl (u), overlineD) and (Con (u), D) are complexes of A-modules. 0DS6
Definition 7 The complexes Cyl (u) and Con (u) are called respectively the cylinder and the cone of the morphism u. 0DS7
Example Let u : M → N be a homomorphism of A-modules; then the only non-zero homogeneous components of Cyl (u) and Con (u) are Cyl₁ (u) = M , Cyl₀ (u) = M ⊕ N , Con₁ (u) = M , Con₀ (u) = N… 0DS8
Proposition 7 a) Les applications tildeu, α, β, π, tildeπ, δ sont des morphismes de complexes : on a u = β ∘ tildeu, π = tildeπ ∘ α, β ∘ α = 1_C. 0DS9
Corollaire Pour tout morphisme de complexes u : C' → C, il existe un morphisme injectif de complexes tildeu : C' → C₁ et un homotopisme β : C₁ → C tel que u = β ∘ tildeu. 0DSA
Lemme 3 a) L'homomorphisme de liaison ∂ₙ₊₁(π, δ) : Hₙ(C') → Hₙ(C) relatif à la suite exacte (7) est égal à -Hₙ(u). 0DSB
Proposition 8 On a la suite exacte illimitée (8) ⋯ longrightarrow Hₙ(C') xrightarrowHₙ(u) Hₙ(C) xrightarrowHₙ(π) Hₙ(Con (u)) xrightarrowHₙ(δ) Hₙ₋₁(C') longrightarrow ⋯ . 0DSC
Corollaire Pour que u soit un homologisme, il faut et il suffit que Con (u) soit d'homologie nulle. 0DSD
Remarque Considérons le diagramme 0DSE
Proposition 9 Les applications β et φ sont des homologismes de complexes. Pour β, cela résulte de la prop. 7, c). On a beginalign* φ ∘ D(y', x) &= φ(-d'y', dx - u(y')) = v(dx - u(y')) \\ &= v(dx… 0DSF
Remarque L'homologisme β est un homotopisme, mais φ n'est pas en général un homotopisme (cf. X, p. 173, exercice 8). 0DSG
Corollaire Le diagramme de A-modules gradués 0DSH
Example 1 If A is a field, we may take as C the set of classes of finite-dimensional vector spaces and as φ the isomorphism of K( C) onto Z defined by φ([M]) = dim(M). 0DSI
Example 2 We may take as C the set of classes of modules of finite length and as φ: K( C) → Z the homomorphism defined by φ([M]) = long_A(M). 0DSJ
Definition 8 Let M be a bounded graded A-module of type C and (Mₙ) its grading. We call the φ-characteristic of M and denote χ_φ(M) or simply χ(M) the element ∑ (-1)ⁿ φ([Mₙ]) of G. 0DSK
Example 3 If M is bounded of type C, it is analogously the case for M(p) for every p ∈ Z, and one has χ(M(p)) = (-1)ᵖ χ(M). 0DSL
Remark One calls the Poincaré polynomial of M relative to φ the element P_M(t) = ∑ φ([Mₙ]) tⁿ ∈ G ⊗ Z[t, t⁻¹]. One has P_M(1) = φ([M]) and P_M(-1) = χ(M). 0DSP
Lemma 4 Let C be a bounded complex of type C. If H(C) = 0, one has χ(C) = 0. This follows from VIII, § 6, No. 1, cor. of Prop. 1. 0DSQ
Proposition 10 Let C and C' be two bounded complexes of type C. If there exists a homologism u : C' → C, one has χ(C) = χ(C'). Indeed, Con (u) is bounded of type C and one has χ(Con(u)) = χ(C) -… 0DSR
Proposition 11 Let C be a bounded complex of type C. a) If C is stable, H(C) is of type C. b) If H(C) is of type C, it is analogously the case for B(C) and Z(C), and one has χ(H(C)) = χ(C). 0DSS
Corollary If C is stable and C is bounded of type C, the graded module H(C) is bounded of type C and one has χ(H(C)) = χ(C). 0DST
Proposition 12 Let 0 → C' → C → C'' → 0 be an exact sequence of complexes. 0DSU
Lemma 5 Let M → N → P → Q → R be an exact sequence of A-modules. If C is stable, and if M, N, Q and R are of type C, the module P is of type C. Set N' = Coker(M → N) and Q' = Ker(Q → R). T… 0DSV
Corollary Assume C stable, and let u : C' → C be a morphism of complexes such that H(C) and H(C') are bounded of type C. Then H(Con(u)) is bounded of type C, and one has χ(H(Con(u))) = χ(H(C… 0DSW
Remark Let E be a complex, h : E → C and h' : E → C' be homotopy equivalences with C and C' bounded of type C. Then χ(C) = χ(C'). Indeed, if h₁ is an inverse of h up to homotopy, h' ∘ h₁… 0DSX
Proposition 13 There exists a unique k-antiderivation d : Ω_A/k → Ω_A/k of degree 1, square zero, which extends the derivation d⁰ : A → Ω¹_A/k. 0DSY
Example 1 Take A to be the ring k[X₁, ..., Xₙ]. Then Ω_A/k¹ is a free A-module with basis dX₁, ..., dXₙ (III, p. 134, example). Consequently, if, for every subset I = i₁, ..., iₚ of 1, n , o… 0DSZ
Example 2 Suppose that k = C and A = C [X₁, ..., Xₙ]/(P₁, ..., Pᵣ), where the Pᵢ are polynomials in X₁, ..., Xₙ, such that the set of points of Cⁿ where all the Pᵢ vanish is an analytic subv… 0DT0
Proposition 14 (i) There exists a unique mapping k-linéaire ∇ from the right Ω_A/k-module M ⊗_A Ω_A/k into itself, graded of degree 1, which extends ∇⁰ in degree 0 and satisfies the identity : 0DT1

§ 3. Resolutions

Lemma 1 Consider a diagram of A-modules and homomorphisms 0DTO
Lemma 2 If in the commutative diagram of A-modules and homomorphisms 0DTP
Proposition 1 Let (P, d_P) and (E, d_E) be two complexes of A-modules and r an integer. 0DTQ
Definition 1 Let M be an A-module. A left resolution of M is a pair (P,p) where P is a complex zero on the right and p:P→ M is a homomorphism. A right resolution of M is a pair (e,E) where E is… 0DTR
Proposition 2 Let P be a complex zero on the right and p:P→ M a morphism. For (P,p) to be a left resolution of M, it is necessary and sufficient that the sequence 0DTS
Definition 2 A projective resolution (resp. free, resp. flat) of the A-module M is a left resolution (P,p) of M such that the complex P is projective (resp. free, resp. flat) (X, p. 25). An inj… 0DTT
Example 1 Suppose that the ring A is principal; let M be an A-module and (xᵢ)_i∈ I a generating family of M. Denote by L₀ the free module A⁽ᴵ⁾, by (eᵢ) its canonical basis, and define p:L₀→… 0DTU
Example 2 Suppose A commutative; let E be an A-module and u an endomorphism of E. Denote by Eᵤ the A[X]-module obtained by endowing E with the structure defined by 0DTV
Example 3 If A is principal, the exact sequence 0DTW
Proposition 3 Let f : M' → M be a homomorphism of A-modules, p' : P' → M' a morphism into M' of a complex null on the right and projective P', and p : P → M a left resolution of M. There exists… 0DTX
Corollary Let (P, p) and (P', p') be two projective resolutions of M. There exists a morphism, and only one up to homotopy, α : P' → P such that p ∘ α = p'. 0DTY
Corollary Let (e, E) and (e', E') be two injective resolutions of N; there exists a morphism α : E → E' and only one up to homotopy such that α ∘ e = e'. 0DTZ
Proposition 4 L(f) : L(M) → L(N) is a morphism of complexes of a-modules; one has p_M ∘ L(f) = f ∘ p_N. 0DU0
Remark If f, g ∈ Hom_A(M, N), one does not have L(f + g) = L(f) + L(g). However these two morphisms are homotopic according to X, p. 49, prop. 3. 0DU1
Proposition 5 I(f) : I(M) → I(N) is a morphism of complexes of A-modules; one has I(f) ◦ eM = eN ◦ f. This is proved in a manner analogous to Prop. 4. 0DU2
Remark If f, g ∈ Hom_A(M, N), one does not have I(f + g) = I(f) + I(g). However, these two morphisms are homotopic by X, p. 49, prop. 3 bis. 0DU3
Proposition 6 When A is left noetherian, every finitely generated A-module M possesses a free resolution p : L → M such that Lₙ is finitely generated for every integer n. More generally: 0DU4
Proposition 7 Let C be an A-complex and let a ∈ Z be such that Hₙ(C) = 0 for n < a. a) There exists a free A-complex L such that Lₙ = 0 for n < a and a homologism f : L → C. 0DU5
Lemma 3 Let C be a complex and let r∈ Z. There exists a complex C' and a homomorphism f:C'→ C such that fₙ:C'ₙ→ Cₙ is an isomorphism for n<r and such that C'ᵣ is a free A-module. If A is n… 0DU6
Proposition 8 Let M be an A-module, P and P' two minimal projective resolutions of M, and f:P→ P' a morphism of resolutions. Then f is an isomorphism. In particular, two minimal projective resol… 0DU7
Corollary Let M be an A-module, P and P' two projective resolutions of M; suppose that P is minimal. Let f:P→ P' and g:P'→ P be two morphisms of resolutions. Then f is injective, g is surjec… 0DU8
Proposition 9 Let M be an A-module and (P,p) a projective resolution of M. Let r denote the radical of A. Suppose either that Pₙ is a finitely generated A-module for every n, or that r is nilpot… 0DU9
Proposition 10 Suppose that A is a left noetherian local ring, and let M be a finitely generated A-module. Then M possesses a minimal resolution (P, p); for every n geqslant 0, Pₙ is a finitely g… 0DUA
Proposition 11 If M is a graded A-module bounded below (resp. if M is a graded A-module of finite type and if A is left noetherian), there exists an exact sequence of graded A-modules unbounded t… 0DUB
Proposition 12 The mapping overlineε_M : B(A, M) → M is a homotopism of complexes of k-modules. In particular, the complex B(A, M) is split over k, and (B(A, M), overlineε_M) is a left resolution… 0DUC
Definition 3 The left resolution (B(A, M), overlineε_M) of M is called the standard resolution of the A-module M. 0DUD
Theorem 1 Let C₀ and C be two additive and left exact sets of class of A-modules such that C₀ ⊂ C and every A-module of type C possesses a bounded left resolution of type C₀. Then the homomo… 0DUE
Lemma 4 Let f : M' → M be a homomorphism of A-modules of type C, and p : P → M a bounded left resolution of P of type C₀. There exist a bounded left resolution p' : P' → M' of type C₀ and… 0DUF
Lemma 5 Consider a commutative diagram 0DUG

§ 4. Torsion product

Remark 1 When C' is reduced to C'₀=M, then (C⊗_A C')ₙ=Cₙ⊗_A M and D=d⊗ 1_M; for example C⊗_A Aₛ is naturally identified with C. Similarly, when C is reduced to C₀=P, then (C⊗_A C')ₙ=P⊗_A C'… 0DV2
Remark 2 For every integer r, we have (C⊗_A C')(r)=C(r)⊗_A C', but (C⊗_A C')(r) and C⊗_A C'(r) do not in general have the same differential. 0DV3
Proposition 1 If the complexes C and C' are zero on the right, C ⊗_A C' is zero on the right and the canonical k-linear mapping γ_0,0(C, C') : H₀(C) ⊗_A H₀(C') → H₀(C ⊗_A C') is bijective. 0DV4
Proposition 2 The mapping σ(C, C') : C ⊗_A C' → C'^∘ ⊗_A C^∘ is an isomorphism of complexes of k-modules, whose reciprocal isomorphism is σ(C'^∘, C^∘). 0DV5
Remark 3 One can define the tensor product of a finite family of complexes without endowing the index set with a total order (X, p. 185, Exercise 3). 0DV6
Remark 4 Suppose that each C⁽ⁱ⁾ is endowed with a graded algebra structure compatible with its grading and such that the d⁽ⁱ⁾ are antiderivations (III, p. 117). Let us then endow ⨂_i ∈ I C⁽… 0DV7
Proposition 3 Let C,\ C₁ be two complexes of right A-modules, C',\ C'₁ two complexes of left A-modules, and u : C → C₁,\ v : C → C₁,\ u' : C' → C'₁,\ v' : C' → C'₁ be morphisms of complexes. 0DV8
Corollary 1 Let C' be a split complex of left A-modules such that H(C') is flat. For every complex C of right A-modules, the canonical mapping 0DV9
Corollary 2 Let N be a flat left A-module. For every complex C of right A-modules, the canonical homomorphisms 0DVA
Corollary 3 Let C' be a complex of left A-modules such that B(C') and H(C') are projective. For every complex C of right A-modules, the mapping γ(C,C') is bijective. 0DVB
Remark 1 Using the commutation isomorphisms, one deduces from corollaries 1, 2 and 3 the analogous statements obtained by exchanging the roles of the two arguments of the tensor products. 0DVC
Remark 2 We shall see below (X, p. 79, Cor. 4) that the conclusion of Cor. 1 is also true when one assumes C' and H(C') flat and C' bounded on the right. 0DVD
Lemma 1 Let C be a complex of right A-modules and E a complex of left A-modules. We assume that H(C)=0 and that E is flat and bounded on the right. Then H(C⊗_A E)=0. 0DVE
Lemma 2 If u:C→ C' is a morphism of complexes of right A-modules and E a complex of left A-modules, then the complexes Con(u)⊗_A E and Con(u⊗ 1_E) are isomorphic. 0DVF
Proposition 4 Let u:C→ C' be a homology isomorphism of complexes of right A-modules and E a complex of left A-modules, flat and bounded on the right. Then 0DVG
Remark Using the commutation isomorphisms, one deduces from the preceding statements the analogous statements obtained by exchanging the roles of the two arguments of the tensor products. 0DVH
Definition 1 Let M be a right A-module and N a left A-module. We call the torsion product of M and N the graded k-module 0DVI
Remark 1 We will see below (X, p. 107, Proposition 6) finiteness properties of the modules Torᴬ(M, N). For example, if A is commutative noetherian and if M and N are A-modules of finite typ… 0DVJ
Remark 2 The complex morphism p_M ⊗ p_N : L(M) ⊗_A L(N) → M ⊗_A N induces on the degree 0 homology the isomorphism 0DVK
Proposition 5 The k-homomorphisms 0DVL
Corollary If M or N is flat, Torᵢᴬ(M, N) = 0 for i geqslant 0. 0DVM
Remark 3 If g : N → N' is a homomorphism of left A-modules, then 0DVN
Proposition 6 The mapping (f, g) ↦ Torᴬ(f, g) : 0DVO
Corollary Let λ ∈ k. If λ annihilates M or N, it annihilates Torᴬ(M, N). 0DVP
Proposition 7 Let I and J be two sets, (M_α)_α ∈ I a family of right A-modules, (N_β)_β ∈ J a family of left A-modules. The homomorphism 0DVQ
Proposition 8 Let I (resp. J) be a right filtered preordered set, ((M_α, u_α', α) (resp. ((N_β, v_β', β))) an inductive system of right A-modules (resp. left) relative to I (resp. J). The homomo… 0DVR
Corollary Let I be a right filtered preordered set, (Mᵢ, uⱼᵢ) (resp. (Nᵢ, vⱼᵢ)) an inductive system of right A-modules (resp. left) relative to I. The homomorphism of graded k-modules 0DVS
Definition 2 The linking homomorphism of torsion products, relative to the module M and the exact sequence E, is the composite homomorphism 0DVT
Theorem 1 The left-infinite sequence of k-module homomorphisms 0DVU
Corollary 1 If Tor₁ᴬ(M, N'') = 0, the sequence 0DVV
Corollary 2 Let 0 → C' xrightarrowu C xrightarrowv C'' → 0 be an exact sequence of complexes of left A-modules, and let E be a complex of right A-modules. If C'' or E is flat, the sequence 0DVW
Example Let a be an ideal of A. The exact sequence 0DVX
Proposition 9 Let f : M → M₁ be a homomorphism of right a-modules and 0DVY
Proposition 10 Let us denote by ( F^∘) the exact sequence of left A-modules 0DVZ
Theorem 2 Let E be a right A-module. The following conditions are equi- valent : 0DW0
Corollary 1 Let 0 → E' → E → E'' → 0 be an exact sequence of right A-modules. Assume that E'' is flat. Then E is flat if and only if E' is flat. 0DW1
Corollary 2 Let 0longrightarrow Eₙlongrightarrow Eₙ₋₁longrightarrow⋯longrightarrow E₁longrightarrow0 be an exact sequence of right A-modules. If Eᵢ is flat for i=1,…,n-1, then Eₙ is flat. 0DW2
Lemma 3 The connecting homomorphism H(B ⊗_A C') → H(Z ⊗_A C') associated with the exact sequence (12) is equal to H(i ⊗ 1). 0DW4
Corollary 1 If B(C) and Z(C) are flat, then for every left A-module N and every integer n, there is an exact sequence 0DW5
Corollary 2 Suppose B(C) and B(C') are projective and Z(C) is flat. Then the sequences of k-modules (11) and (13) are exact and split. This follows from the theorem and the following lemma: 0DW6
Lemma 4 If B(C) and B(C') are projective, then the canonical homomorphism γ(C, C') : H(C) ⊗_A H(C') → H(C ⊗_A C') has a k-linear retraction. Indeed by X, p. 65, remark b), there exist homo… 0DW7
Corollary 3 (universal coefficient formula) Suppose the ring A is principal. If the complexes C and C' are free, the sequences of A-modules (11) are exact and split; if the complex C is free, the sequences of A-modules (13)… 0DW8
Corollary 4 (« Künneth formula ») Suppose C bounded on the right, C and H(C) flat; then the canonical homomorphism 0DW9
Corollary 5 Let u : C \to C' be a homologism of complexes of right a-modules, flat and bounded on the right. For every complex E of left a-modules, the morphism u ⊗ 1_E : C ⊗_A E → C' ⊗_A E is… 0DWA
Proposition 11 Suppose A left noetherian, and let C be a bounded and flat complex of left a-modules such that H(C) is a finitely generated a-module. Let a and b be two integers such that a \leq b… 0DWB

§ 5. Modules of extensions

Proposition 1 If C is zero on the right and C' zero on the left, then HomgrA (C, C') is zero on the left, and the canonical k-linear mapping 0DWL
Proposition 2 a) Let C' xrightarrowu C xrightarrowv C'' be an exact sequence of A-complexes, P a projective complex, E an injective complex (X, p. 25). Then the sequences 0DWM
Proposition 3 Let C, tildeC, C', tildeC' be four A-complexes, u : tildeC → C, v : tildeC → C, u' : C' → tildeC' and v' : C' → tildeC' four morphisms of complexes. 0DWN
Corollary 1 If C is split and H(C) projective (resp. if C' is split and Hₙ(C') injective for each n), then the canonical homomorphism 0DWO
Corollary 2 Assume that B(C) and H(C) are projective (resp. Bₙ(C') and Hₙ(C') are injective for every n). Then λ(C, C') is bijective. 0DWP
Corollary 3 Let M be a projective A-module (resp. an injective A-module). For every complex C of A-modules and every integer n, the canonical homomorphism 0DWQ
Lemma 1 a) If C or C' is bounded on the right, if C is projective and if H(C') = 0, then H(Homgr_A(C, C')) = 0. b) If C or C' is bounded on the left, if C' is injective and if H(C) = 0, th… 0DWR
Proposition 4 Let u : C' → C be a morphism of complexes, P a projective complex, E an injective complex. 0DWS
Definition 1 Let M and N be two A-modules. The module of extensions of N by M is the graded k-module 0DWT
Remark 1 We will see below (X, p. 107, Prop. 6) some finiteness properties of the modules Extⁿ_A(M,N). For example, if A is commutative and noetherian, and if M and N are A-modules of finit… 0DWU
Remark 2 The morphism of complexes 0DWV
Proposition 5 The k-homomorphisms beginaligned &φ_M(N) : H(Homgr_A(L(M), N)) → Ext_A(M, N) \\ &and overlineφ_N(M) : H(Homgr_A(M, I(N))) → Ext_A(M, N) are bijective. endaligned 0DWW
Corollaire If M is projective (resp. if N is injective), then Ext_Aⁱ(M, N) = 0 for i > 0. Indeed, Homgr_A(1, e_N) : Hom_A(M, N) → Homgr_A(M, I(N)) (resp. Homgr_A(p_M, 1) : Hom_A(M, N) → Homgr… 0DWX
Proposition 6 The mapping (f,g) ↦ Ext_A(f,g) : Hom_A(M',M)×Hom_A(N,N')longrightarrow Homgrₖbigl(Ext_A(M,N),Ext_A(M',N')bigr) is k-bilinear. 0DWY
Corollaire Let λ ∈ k. If λ annihilates M or N, then it annihilates Ext_A(M,N). Indeed, λ · 1_Ext_A(M,N) = Ext_A(λ 1_M, 1_N) = Ext_A(1_M, λ 1_N). 0DWZ
Proposition 7 Let I and J be two sets, (M_α)_α∈ I and (N_β)_β∈ J two families of A-modules; the homomorphism Ext_A(⨁_α∈ IM_α,∏_β∈ JN_β) longrightarrow ∏_substackβ∈ J,\ α∈ I Ext_A(M_α,N_β) induce… 0DX0
Remarque 5 Let P^∘ and Q^∘ be two right A-modules. We define Ext_A(P,Q) by 0DX1
Definition 2 We call the composite homomorphism the connecting homomorphism of the extension modules relative to the module M and to the exact sequence E 0DX2
Theorem 1 The right-unbounded sequence of homomorphisms of k-modules 0DX3
Corollary If Ext_A¹(M, N') = 0, the sequence 0DX4
Proposition 8 Let f : M₁ → M be a homomorphism of A-modules and 0DX5
Definition 3 One calls connecting homomorphism of extension modules relative to the exact sequence (F) and to the module N, the composite homomorphism δ(F, N) : overlineφ_N(M'') ∘ ∂(F_N) ∘ over… 0DX6
Theorem 2 The right-unbounded sequence of homomorphisms of k-modules 0 longrightarrow Hom_A(M'', N) xrightarrowHom(s, 1) Hom_A(M, N) xrightarrowHom(r, 1) Hom_A(M', N) xrightarrowδ⁰(F, N) Ext… 0DX7
Corollary If Ext_A¹(M'', N) = 0, the sequence 0DX8
Proposition 9 Let g : N → N₁ be a homomorphism of A-modules and 0DX9
Proposition 10 Let M be an A-module. The following conditions are equivalent: 0DXA
Proposition 11 Let N be an A-module. The following conditions are equivalent: 0DXB
Theorem 3 Suppose the A-modules B(C) and Z(C) are projective. There exists, for each n, a unique homomorphism of k-modules 0DXC
Remark One can prove an analogous statement, assuming Bₙ(C') and C'/Bₙ(C') are injective for each n. 0DXD
Lemma 2 The connecting homomorphism 0DXE
Corollary 1 Assume B(C) and Z(C) projective and B^n(C') injective for each n. Then the exact sequences (11) are split. This follows from the theorem and the following lemma: 0DXF
Lemma 3 If B(C) is projective and Bn(C') injective for each n, the canonical homomorphism λ(C, C') : H(Homgr_A(C, C')) → Homgr_A(H(C), H(C')) has a linear section k. 0DXG
Corollary 2 If B(C) and Z(C) are projective, one has, for every A-module N and every integer n, a split exact sequence: 0DXH
Corollary 3 (« universal coefficient formula ») Assume A principal and C free. One has for every A-module N and every integer n a split exact sequence (13). Indeed B(C) and Z(C) are free as submodules of the free module C (VII… 0DXI
Corollary 4 If C is bounded on the right and if C and H(C) are projective, then 0DXJ
Proposition 12 The homomorphisms (14), (15), (16) are morphisms of complexes. 0DXK

§ 6. Use of noncanonical resolutions

Theorem 1 a) If one of the resolutions R or S is flat, then ψ(S, R) is an isomorphism of graded k-modules. 0DXT
Corollary If R is a flat resolution of M, the homomorphism 0DXU
Remark The diagram of k-modules 0DXV
Proposition 1 If a₁, b₁, c₁ are homologisms, ψ'(S₁, R₁) and φ'(R₁, E₁) are the reciprocal bijections of the bijections ψ(S₁, R₁) and φ(R₁, E₁) respectively. 0DXW
Example 1 Let a be an element of A such that the mapping φ : x ↦ xa of A into itself is injective (" a is not a right zero divisor "). Using the resolution 0DXX
Example 2 Suppose that A is an integral domain; let K be the field of fractions of A and M an A-module. Using the flat resolution 0DXY
Example 3 Suppose that A is a local noetherian ring, let m denote its maximal ideal, and put κ = A/m. Let M be a finitely generated A-module and P a minimal projective resolution of M (X, p… 0DXZ
Proposition 2 The following two diagrams are commutative: 0DY0
Remark Similarly, consider commutative diagrams of morphisms of complexes 0DY1
Proposition 3 a) Let P be an A-module, b : S → P a left resolution of P; suppose that the sequence of complexes of k-modules 0DY2
Remark 1 Using the commutation isomorphisms, one deduces from a) the analogous statement obtained by exchanging the roles of the two arguments of the tensor product. 0DY3
Remark 2 With the notation of a), suppose either that S is flat, or that R, R', R'' are flat; then on the one hand sequence (3) is exact (X, p. 72, cor. 2) and one may apply prop. 3; on the… 0DY4
Remark 3 With the notation of b), suppose either that E is injective, or that R, R', R'' are projective; then on the one hand sequence (4) is exact (X, p. 83, prop. 2) and one may apply pro… 0DY5
Proposition 4 Let M be a left A-module, a : R → M a left resolution of M such that the sequence 0DY6
Remark 4 If R is projective or if E, E', E'' are injective, the sequence (7) is exact (λ, p. 83, Prop. 2); moreover, φ(R, E'') is bijective (Th. 1); hence 0DY7
Remark 5 We leave it to the reader to state and prove the propositions analogous to Props. 3 and 4 and concerning the homomorphisms ψ₁ and φ₁. 0DY8
Proposition 5 If A is left noetherian and if M is a finitely generated A-module, the canonical homomorphism (8) is bijective. 0DY9
Proposition 6 Let B be a ring and N an (A, B)-bimodule, which is a noetherian B-module (resp. of finite length). 0DYA
Corollary Let ρ : A → B be a homomorphism of noetherian commutative rings, M a finitely generated A-module, N a B-module. If N is a finitely generated B-module (resp. of finite length), the… 0DYB
Proposition 7 a) If the A-module Q is flat, homomorphism (9) is bijective. b) If the A-module Q is finitely generated projective, homomorphism (10) is bijective. 0DYC
Proposition 8 a) If N is flat over A and over B, homomorphism (11) is bijective. 0DYD
Proposition 9 If B is flat over A, homomorphism (13) is bijective. Indeed ψ'(L(E) ⊗_A B, B ⊗_A L(F)) is bijective (X, p. 102, prop. 1) and γ is bijective (X, p. 66, cor. 2). 0DYE
Proposition 10 Homomorphism (14) is bijective in the following cases: a) B is a finitely generated projective A-module; b) B is a flat A-module, A is noetherian, and E is a finitely generated A-m… 0DYF
Definition 1 Let M be a left Z⁽ᴳ⁾-module (resp. right), n an integer geqslant 0. The group Ext_Z⁽ᴳ⁾ⁿ(Z, M) (resp. Torₙ^Z⁽ᴳ⁾(M, Z)) is denoted by Hⁿ(G, M) (resp. Hₙ(G, M)) and is called the nth… 0DYG
Example 1 It follows directly from the definition that H⁰(G, M) is isomorphic to the submodule of elements of M invariant under the action of G, and H₀(G, M) to the quotient module of M by t… 0DYH
Example 2 It follows from the preceding that H¹(G, M) is isomorphic to the Z-module Z¹(G, M)/B¹(G, M), where Z¹(G, M) is the Z-module of mappings f of G into M satisfying: 0DYI
Example 3 Analogously H²(G,M) is isomorphic to the Z-module Z²(G,M)/B²(G,M), where Z²(G,M) is the Z-module of mappings f of G× G into M, satisfying: 0DYJ
Example 4 Let M be a Z-module, considered as a right Z⁽ᴳ⁾-module by making G operate trivially. The group H₁(G,M) is isomorphic to the quotient of M⊗_ZZ⁽ᴳ⁾ by the sub-Z-module generated by t… 0DYK
Lemma 1 Let M be a Z-module; let Mᴳ denote the group Hom_Z(Z⁽ᴳ⁾, M) endowed with its natural structure of left Z⁽ᴳ⁾-module. Then: 0DYL
Proposition 11 Let L be a finite-degree galois extension of a commutative field K, with galois group G. a) One has Hⁱ(G, L) = 0 for i geqslant 1. b) One has H¹(G, L^*) = 0. c) The group H²(G, L^*… 0DYM

§ 7. Composition product

Example 1 If u ∈ Hom_A(N, P), v ∈ Hom_A(M, N), then u ∘ v is the composite of u and v. 0DZG
Example 2 If u ∈ Hom_A(N, P), v ∈ Ext_A(M, N), then 0DZH
Example 3 Let M, N, M', N' be four left A-modules. If u ∈ Ext_A(M, N), f ∈ Hom_A(M', M), g ∈ Hom_A(N, N'), then 0DZI
Proposition 1 Let (C,d) and (C',d') be two complexes of left A-modules and n,p,q three integers such that p ≥ q. For p ≥ i ≥ q-1, let fᵢ:Cᵢlongrightarrow C'_i+n+1 be a homomorphism of A-modules… 0DZJ
Proposition 2 We have a, b ∈ Zⁿ(Homgr_A(L(M), I(N))). Moreover, the classes overlinea and overlineb of a and b in Hⁿ(Homgr_A(L(M), I(N))) = Ext_Aⁿ(M, N) depend only on the exact sequence (4) and… 0DZK
Definition 1 We call the class associated with the exact sequence (4) the element θ of Ext_Aⁿ(M, N) defined by θ = (-1)^n(n+1)/2 overlinea = (-1)^n(n+1)/2 overlineb. 0DZL
Remark 1 Let (P, p) be a projective resolution of M. By X, p. 49. prop. 3, there exists a commutative diagram 0DZM
Remark 2 When n = 0, the exact sequence (4) reads 0 → N xrightarrowf M → 0, and the associated class is f⁻¹ ∈ Hom_A(M, N) = Ext⁰_A(M, N). 0DZN
Proposition 3 Let 0DZO
Corollary 1 Consider a commutative diagram with exact rows 0DZP
Corollary 2 Let 0DZQ
Corollary 3 Let 0 → N xrightarrowfₙ₋₁ Rₙ xrightarrowfₙ ⋯ → R₁ xrightarrowf₁ M → 0 be an exact sequence, θ its class in Ext_Aⁿ(M, N), u colon M' → M and v colon N → N' homomorphisms of A-module… 0DZR
Remark Let θ ∈ Extⁿ_A(M, N), resp. θ' ∈ Extⁿ_A(M', N'), be the class of an exact sequence 0DZS
Theorem 1 Let n be an integer geqslant 1, M and N two A-modules. 0DZT
Remark 1 If the ring A is Noetherian, and if the A-modules M and N are of finite type, it follows from the proof of a) that every element of Ext_Aⁿ(M, N) is the class associated with an exa… 0DZU
Corollary Let 0 → N xrightarrowf R xrightarrowg M → 0 and 0 → N xrightarrowf' R' xrightarrowg' M → 0 be two exact sequences, θ and θ' their associated classes in Ext¹(M, N). For θ = θ' to ho… 0DZV
Remark 2 Theorem 1 provides a description of Ext_Aⁿ(M, N) as a set of equivalence classes of exact sequences; it is easy to describe the group law obtained on this set by transport of struc… 0DZW
Proposition 5 Let 0DZX
Corollaire 1 a) The connecting homomorphism Hom_A(M'', M'') → Ext_A¹(M'', M') sends 1_M'' to θ. 0DZY
Corollaire 2 Consider two short exact sequences of left A-modules 0DZZ
Corollary 3 If θ∈Ext_Aⁿ(M,N) is the class of the exact sequence (S), we have 0E00
Corollary 4 If each module Rᵢ,\ i = 1, ..., n, is injective (resp. projective), the mapping α ↦ θ ∘ α (resp. α ↦ α ∘ θ) from Ext_Aᵐ(P, M) to Ext_Aᵐ⁺ⁿ(P, N) (resp. from Ext_Aᵐ(N, P) to Ext_Aᵐ⁺ⁿ… 0E01
Remark The definitions and propositions of Nos. 3 to 6 apply to A right modules, considered as left modules over the ring A^∘ opposite to A. 0E02
Remark 1 If one uses the isomorphisms ψ of X, p. 69, one can also define the composition product by the commutative diagram 0E03
Remark 2 One can also use the resolutions l(P) and l(Q). 0E04
Proposition 6 Let K, M, N be left A-modules, P, Q, R right A-modules, α ∈ Ext_Aⁿ(P, Q), β ∈ Ext_Aᵖ(Q, R), λ ∈ Ext_Aʳ(K, M), μ ∈ Ext_Aˢ(M, N), γ ∈ Torₘᴬ(P, K). Then (13) (β ∘ α) ∘ γ = β ∘ (α ∘ γ)… 0E05
Proposition 7 a) Let 0E06
Corollary 1 Let 0 → P' → P → P'' → 0 be an exact sequence of right A-modules, 0 → M' → M → M'' → 0 be an exact sequence of left A-modules. Then the composed homomorphisms of connecting homomor… 0E07
Corollary 2 If θ ∈ Ext_Aⁿ(M, N) is the class associated with the exact sequence (S), we have hat∂ₘ(P, S)(α) = θ ∘ α for all α ∈ Torₘᴬ(P, M). 0E08
Corollary 3 If all the modules Rᵢ, i = 1, ..., n, are flat, the mapping α ↦ θ ∘ α from Torₘ₊ₙᴬ(P, M) to Torₘᴬ(P, N) is bijective for every right A-module P and every integer m > 0. 0E09
Corollary 4 If θ₁ ∈ Ext_Aⁿ(P, Q) is the class associated with the exact sequence (S₁), we have ∂ᵐ(S₁, M)(α) = θ₁ ∘ α for all α ∈ Torₘᴬ(P, M). 0E0A
Proposition 8 a) The following diagram, where h_θ(α) = θ ∘ α, is commutative 0E0B
Remark Via the commutation isomorphisms, we deduce from a) an analogous statement in the case of an exact sequence (16) of right A-modules. 0E0C
Proposition 9 The following diagram, where γ_θ(α) = θ ∘ α, is commutative: 0E0D

§ 8. Homological dimension

Definition 1 Let M be an A-module. The projective dimension of M, denoted by dp_A(M), is the greatest lower bound in overlineZ of the lengths of projective resolutions of M (X, p. 48). 0E0P
Lemma 1 If dp_A(M) < n < +∞, then Ext_Aⁿ(M, N) = 0 for every A-module N and Torₙᴬ(P, M) = 0 for every righthand A-module P. This follows immediately from the fact that M has a projective r… 0E0Q
Proposition 1 Let M be an A-module and n an integer ≥ 0. The following conditions are equivalent : (i) dp_A(M) ≤ n (i.e. (def. 1), M has a projective resolution of length ≤ n); (ii) Ext_Aʳ(M, N)… 0E0R
Corollary 1 Let (Mᵢ)_i ∈ E be a family of A-modules. We have dp_A ( ⨁_i ∈ E Mᵢ ) = sup_i ∈ E dp_A(Mᵢ) . This follows from the equivalence of conditions (i) and (iii) of Proposition 1 and Propo… 0E0S
Corollary 2 Let 0 → M' → M → M'' → 0 be an exact sequence of A-modules. 0E0T
Example Let a be an element of A which is neither invertible nor a right zero divisor. Then dp_A(A/Aa)=1. 0E0U
Proposition 2 Assume A left noetherian. Let M be a finitely generated a-module and n an integer ≥0. The following conditions are equivalent: 0E0V
Corollary Suppose A left noetherian and let C₀ (resp. C) be the set of classes of A-modules projective of finite type (resp. of A-modules of finite projective dimension and of finite type)… 0E0W
Proposition 3 a) If dp_A(M) leqslant n, the canonical homomorphism (3) is injective. b) If dp_A(M) leqslant n, if A is left noetherian and if M is of finite type, the canonical homomorphism (3)… 0E0X
Definition 2 We call the homological dimension of A and denote by dh(A) the supremum in Z of the set of integers n for which there exist two A-modules M and N such that Ext_Aⁿ(M,N)≠ 0. 0E0Y
Proposition 4 Let n be an integer ≥ 0. The following conditions are equivalent: 0E0Z
Remarque 1 If dh(A) ≤ n < +∞, we have Torₙ₊₁ᴬ(P, M) = 0 for every A-module M and every right A-module P, since dp_A(M) ≤ n (cf. lemma 1). 0E10
Remarque 2 For dh(A) to be finite, it is necessary and sufficient that dp_A(M) be finite for every nonzero A-module M. This indeed follows from the preceding and from X, p. 135, cor. 1. 0E11
Corollaire Suppose A is left noetherian and let n be an integer > 0. The following conditions are equivalent: 0E12
Remark By the equivalence of (i) and (iii), we have dh(A) = dh(A^∘) if A is left and right noetherian. This equality is not valid in general (X, p. 204, Exercise 20). 0E13
Proposition 5 Suppose A is left noetherian and of finite homological dimension and let C₀ (resp. C) be the set of equivalence classes of finitely generated projective a-modules (resp. of a-modul… 0E14
Proposition 6 The following conditions are equivalent : (i) every a-module is projective, (ii) every a-module is injective, (iii) every ideal of A is an injective module, (iv) dh(A) ≤ 0, (v) A i… 0E15
Proposition 7 The following conditions are equivalent : (i) dh(A) ≤ 1, (ii) every submodule of a projective module is projective, (ii') every ideal of A is projective, (iii) every quotient of an… 0E16
Example If A is principal, then dh(A) ≤ 1. 0E17
Remark If A is (commutative) a domain, the preceding conditions are also equivalent to the following : 0E18
Corollary Let A be a ring of homological dimension ≤ 1, C a complex of projective a-modules, widetildeC a complex of projective right a-modules, C' a complex of injective a-modules, P a righ… 0E19
Lemma 2 Let ρ : A→ A' be a ring homomorphism, M an a-module, M′ an a′-module. If the a-module A'_d is flat, one has dp_A(A'⊗_A M)≤dp_A(M). If the a-module A'ₛ is projective, one has dp_A(M… 0E1A
Lemma 3 Assume A commutative and let M be an a[X]-module. a) One has dp_A(M) ≤ dp_A[X](M) ≤ dp_A(M) + 1. b) If the homothety X_M is injective, one has dp_A(M/XM) ≤ dp_A[X](M). c) If XM = 0… 0E1B
Theorem 1 Suppose A commutative. Then 0E1C
Corollary 1 Suppose A commutative. We have 0E1D
Corollary 2 Let K be a commutative field (resp. a principal ideal domain or a Dedekind ring which is not a field). Then dh(K[X₁, ..., Xₙ]) is equal to n (resp. n + 1). This follows from the fa… 0E1E
Lemma 4 Let M be a graded A-module bounded below (X, p. 56). If A₀ ⊗_A M = 0, then M = 0. 0E1F
Lemma 5 Let M be a graded A-module bounded below, and let s : A ⊗_A₀ M/J₀ M → M be a graded A-homomorphism such that 1 ⊗_A s : A₀ ⊗_A (A ⊗_A₀ M/J₀ M) → A₀ ⊗_A M is the canonical isomorphis… 0E1G
Proposition 8 Let M be a graded A-module bounded below. 0E1H
Corollary 1 Let M be a graded A-module of finite type. If the A₀-module M/J₀M is projective (resp. graded free, resp. flat) and if Tor₁ᴬ(A₀,M)=0, then the A-module M is projective (resp. grade… 0E1I
Corollary 2 0E1J
Corollary 3 (Hilbert’s Syzygy Theorem) Assume that A₀ is a commutative field and that A is generated as an A₀-algebra by n homogeneous elements of degrees 0 and algebraically independent. For every graded a-module M bou… 0E1K
Remark Corollary 2 also applies in the following cases: a) A₀ is principal and A = A₀[X₁, ..., Xₙ₋₁]; b) A₀ is a regular local noetherian ring of dimension r and A = A₀[X₁, ..., Xₙ, r]. 0E1L
Corollary 4 Assume that A₀ is semisimple. Let M be a graded a-module bounded below and let n be an integer ≥ 0. In order that dp_A(M) ≤ n, it is necessary and sufficient that Torₙ₊₁ᴬ(A₀, M) =… 0E1M
Corollary 5 Suppose the ring A₀ semisimple. If Torₙ₊₁ᴬ(A₀, A₀) = 0, one has dp_A(M) ≤ n for every graded A-module bounded below. 0E1N

§ 9. Koszul complexes

Definition 1 The complex (Λ(L), dᵤ) is denoted by Kᴬ(u) or K(u). 0E2G
Lemma 1 If the complex C is zero on the right (resp. on the left), then Kᴬ(u, C) (resp. K^*_A(u, c)) is zero on the right (resp. on the left), and γ₀ (resp. λ⁰) is bijective. 0E2H
Proposition 1 Let x ∈ L; denote by Rₓ : y ↦ x ∧ y the left multiplication by x in the algebra Λ(L). Then dᵤ ∘ Rₓ + Rₓ ∘ dᵤ = u(x) · 1_Λ(L) = u(x)_Λ(L). 0E2I
Corollary 1 If u is surjective, K(u) is homotopic to zero (X, p. 34) as well as Kᴬ(u, C) and K_A^*(u, C) for every complex C. 0E2J
Corollary 2 Let C be a complex, Ann (C) its annihilator. Then q + Ann(C) annihilates Hᴬ(u, C) and H_A^*(u, C). 0E2K
Proposition 2 Let (Lᵢ)_i ∈ I be a family of A-modules, where the set I is finite and totally ordered. Let u be a linear form on ⨁_i ∈ I Lᵢ, uᵢ its restriction to Lᵢ. 0E2L
Remarks 1) \ The above remains valid when L is projective of rank n. \ 0EEQ
Proposition 3 If the A-module L is flat, the sequences (Eₙ) above are exact for n > 0. 0E2M
Remark 1 We shall see below (X, p. 158, example) another proof of part c) above. 0E2N
Remark 2 If A is a Q-algebra, the conclusion of prop. 3 remains true without assumption on L (cf. X, p. 206, exercise 1). 0E2O
Remark 3 Let G be a group and ρ : G → GL(L) a linear representation of G in a flat A-module L. Then the (Eₙ) are exact sequences of linear representations. Suppose L finitely generated proj… 0E2P
Lemma 2 If the k-module L₀ is free with basis (eᵢ)_i ∈ I, the k-linear mappings g and h are bijective. This follows from III, p. 79, th. 1. 0E2Q
Remark Conversely, let B be a k-algebra, L a free B-module with basis (eᵢ)_i ∈ I, and M a B-module. The datum of a linear form u : L → B is equivalent to that of a family x = (xᵢ)_i ∈ I o… 0E2R
Example Take M = k[T₁, ..., Tₙ]; the Koszul complex K (∂/∂ T, M) associated with the sequence of endomorphisms (∂/∂ T₁, ..., ∂/∂ Tₙ) is identified with the de Rham complex of k[x₁, ..., xₙ… 0E2S
Lemma 3 Let K be a complex such that Kᵢ = 0 for i ≠ 0, 1, and let C be a complex and p an integer. 0E2T
Proposition 4 For every complex C, there are exact sequences . 0E2U
Corollary 1 For Hₚ(x,C)=0 to hold, it is necessary and sufficient that the homothety with ratio x in Hₚ(C) be surjective and that the homothety with ratio x in Hₚ₋₁(C) be injective. 0E2V
Corollary 2 Let x=(x₁,…,xₙ) be a sequence of elements of A, M an A-module, x' the sequence (x₁,…,xₙ₋₁). We have exact sequences 0E2W
Corollary 3 For Hᵢ(x,M) to be zero for all i>0, it is necessary and sufficient that the homothety with ratio xₙ in Hᵢ(x',M) be bijective for i>0, and that the homothety with ratio xₙ in M/(x')… 0E2X
Definition 2 The family x is said to be completely transverse for M if one has Hᵢ(x,M)=0 for i>0. 0E2Y
Proposition 5 Let x=(x₁,…,xₙ) be a sequence of elements of A. If for i=1,…,n, the homothety of ratio xᵢ in the A-module M/(x₁M+⋯+xᵢ₋₁M) is injective, the sequence x is completely secant for M. 0E2Z
Example Let k be a ring; take A = k[X₁, ..., Xₙ] and x = (X₁, ..., Xₙ). The sequence x is A-regular and the proposition gives again the acyclicity in degrees > 0 of the complex Sₖ(kⁿ) ⊗ₖ Λ… 0E30
Proposition 6 a) If ∑_i ∈ I xᵢ A = A, the family (xᵢ)_i ∈ I is completely secant for M. b) Let x = (x₁, ..., xₙ) be a sequence of elements of A. Let (aᵢⱼ) ∈ GLₙ(A); put yᵢ = ∑ⱼ aᵢⱼ xⱼ. If the se… 0E31
Remark 1 The assertion analogous to c) for regular sequences is true (X, p. 207, exercise 5). 0E32
Proposition 7 a) Let N be a flat A-module. If the family x is completely secant for M, it is so for M ⊗_A N. 0E33
Remark 2 The analogous assertions to a) and b) for regular sequences are immediate. 0E34
Remark 3 If the family x is completely transverse for A, the complex K.(x, A) defines a free resolution of the A-module A/x, with x = ∑_i ∈ I xᵢ A; therefore for every integer i geqslant 0… 0E35
Remark 4 One says that the sequence x = (x₁, ..., xₙ) is M-coregular if (denoting by (xᵢ)_M the homothety of ratio xᵢ in M) the homothety of ratio xᵢ in the module 0E36
Theorem 1 Consider the following conditions: (i) The sequence x is M-regular (X, p. 158). (ii) The sequence x is completely secant for M (X, p. 157, def. 2). (iii) One has H₁(x, M) = 0. (iv)… 0E37
Corollary 1 If A is noetherian, if the A-module M is of finite type and if the xᵢ belong to the radical of A, the conditions (i) to (v) of the theorem are equivalent. Indeed, on each of the mo… 0E38
Corollary 2 Suppose that A is a graded ring with positive degrees, M a bounded below graded A-module, and the xᵢ homogeneous elements of degree 0 of A. Then the conditions (i) to (v) of the th… 0E39
Corollary 3 Suppose that the modules M/(x₁M+⋯+xᵢ₋₁M) are separated for the 𝔵-adic topology (1≤ i≤ n) ; let p be an integer between 1 and n. In order that the sequence 𝔵 be completely secant fo… 0E3A
Remark Let 𝔵=(x₁,…,xₙ) be a sequence of elements of A such that H₁(𝔵,A)=0; then the kernel of the surjective homomorphism u:Aⁿ→𝔵 such that u(∑ aᵢ eᵢ)=∑ aᵢ xᵢ is generated by the elements… 0E3B
Lemma 4 Let A₀ be the ring Z[T₁,…,Tₙ], and let u:A₀[X₁,…,Xₙ]→ A₀[U] be the homomorphism of A₀-algebras such that u(Xᵢ)=TᵢU. The kernel of u is the ideal 𝔡₀ of A₀[X₁,…,Xₙ] generated by the… 0E3C
Proposition 8 Suppose the sequence x is regular for M. The element of Ext_Aⁿ(M/(x)M, M) associated with the exact sequence (23) is (-1)^n(n+1)/2 θₓ. For i = 0, 1, …, n, define an A-linear mappin… 0E3D
Corollary Suppose moreover that the modules M/(x₁ M + ⋯ + xᵢ₋₁ M) are separated for the (x)-adic topology, and let (aᵢⱼ) ∈ GLₙ(A). Put yᵢ = ∑ⱼ aᵢⱼ xⱼ and y = (y₁, …, yₙ) . Then the sequence… 0E3E
Proposition 9 Suppose the sequence x regular for M. If N is an A-module such that (x)N = 0, one has Ext_Aⁱ(N, M) = 0 for i < n, and the mapping α ↦ θₓ ∘ α from HomA (N, M/(x) M) into Ext^nA (N… 0E3F