Proposition 2

Tag 0E2L, permanent. Cite it as [Bourbaki, Tag 0E2L].

Algebra, chapter X, § 9. Complexes de Koszul

Let $(L_i)_{i \in I}$ be a family of $A$ -modules, where the set $I$ is finite and totally ordered. Let $u$ be a linear form on $\bigoplus_{i \in I} L_i$ , $u_i$ its restriction to $L_i$ .

The canonical isomorphism of $A$ -algebras (III, p. 84)

g : \bigotimes_{i \in I} \Lambda(L_i) \to \Lambda(\bigoplus_{i \in I} L_i)

is an isomorphism of the complex $\bigotimes_{i \in I} K(u_i)$ (X, p. 63) onto the complex $K(u)$ .

In fact, by X, p. 64, remark 4, the differential $D$ of the complex $\bigotimes_{i \in I} K(u_i)$ is an antiderivation; the antiderivations $d_u$ and $g \circ D \circ g^{-1}$ of $\Lambda(\bigoplus L_i)$ coincide on $\bigoplus L_i$ with the mapping $x \mapsto u(x).1$ of $\bigoplus L_i$ into $\Lambda(\bigoplus L_i)$ , hence are equal (III, p. 128, cor.).

Let $C$ and $C'$ be two complexes of $A$ -modules. One has (X, p. 63 and p. 99) canonical isomorphisms of complexes

C \otimes_A (C' \otimes_A K(u)) \to (C \otimes_A C') \otimes_A K(u)

\mathrm{Homgr}_A(C', \mathrm{Homgr}_A(K(u), C)) \to \mathrm{Homgr}_A(C' \otimes_A K(u), C),

that is, isomorphisms

(4)

C \otimes_A K^A(u, C') \to K^A(u, C \otimes_A C')

(5)

\operatorname{Homgr}_A(C', K^i_A(u, C)) \to \operatorname{Homgr}_A(K^A(u, C'), C)

Dans (4) and (5), take $C' = K(u')$ , where $u' : L' \to A$ is a linear form on an $A$ -module $L'$ , and note that $K^A(u, K(u'))$ which is equal by definition to $K(u') \otimes_A K(u)$ is identified, according to Prop. 2, with $K(u' \oplus u)$ where $u' \oplus u : L' \oplus L \to A$ is the linear form $(x', x) \mapsto u'(x') + u(x)$ . We then obtain isomorphisms of complexes

(6)

K^A(u' \oplus u, C) \to K^A(u, K^A(u', C))

(7)

K^i_A(u', K^i_A(u, C)) \to K^i_A(u' \oplus u, C)

By passing to homology, we deduce isomorphisms of $A$ -modules

H^A_r(u' \oplus u, C) \to H^A_r(u, K^A(u', C)) , \qquad r \in \mathbf{Z} ,

H^r_A(u', K^i_A(u, C)) \to H^r_A(u' \oplus u, C) , \qquad r \in \mathbf{Z} .

Finally note that the homomorphism deduced from the product in the algebra $\Lambda(L)$

m : K^A(u) \otimes_A K^A(u) \to K^A(u)

is a morphism of complexes (since $d_u$ is an antiderivation). Supposing that $L$ is free of rank n and composing with the morphism of complexes $K^A(u) \to \Lambda^n L(-n)$ which is the identity in degree $n$ , we deduce a morphism of complexes

\chi : K^A(u) \otimes_A K^A(u) \to \Lambda^n L(-n) ;

to this morphism corresponds canonically, according to X, p. 99, Prop. 12, a morphism of complexes

\varphi : K^A(u) \to \operatorname{Homgr}_A(K^A(u), \Lambda^n L(-n))

which is bijective (III, p. 87, formula (20)). For every complex $C$ , we deduce a composed isomorphism

K^A(u, C) = C \otimes_A K^A(u) \xrightarrow{1 \otimes \varphi} C \otimes \operatorname{Homgr}_A(K^A(u), \Lambda^n L(-n)) \to \\ \to \operatorname{Homgr}_A(K^A(u), C \otimes_A \Lambda^n L(-n)) = K^i_A(u, C \otimes_A \Lambda^n L(-n)) .

By passing to homology, we therefore have canonical isomorphisms

(8)

H^A_r(u, C) \to H^{n-r}_A(u, C \otimes_A \Lambda^n L) , \qquad r \in \mathbf{Z} .

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

Cited by

Nothing.

Cites

Nothing.