Remark 2

Tag 0DV3, permanent. Cite it as [Bourbaki, Tag 0DV3].

Algebra, chapter X, § 4. Produits de torsion

For every integer $r$ , we have $(C\otimes_A C')(r)=C(r)\otimes_A C'$ , but $(C\otimes_A C')(r)$ and $C\otimes_A C'(r)$ do not in general have the same differential.

Let $p, q$ be two integers, $x \in Z_p(C), x' \in Z_q(C')$ ; then the element $x \otimes x'$ of $C_p \otimes C'_q$ belongs to $Z_{p+q}(C \otimes C')$ by (1); moreover, if $y \in C_{p+1}, y' \in C'_{q+1}$ , we have

(x + dy) \otimes (x' + d'y') = x \otimes x' + D(y \otimes x' + (-1)^p (x + dy) \otimes y');

by passing to the quotients, we deduce a

k

-linear mapping, called canonical

\gamma_{p,q}(C, C') : H_p(C) \otimes_A H_q(C') \to H_{p+q}(C \otimes_A C');

if we endow

H(C) \otimes_A H(C')

with the grading such that

(H(C) \otimes H(C'))_n = \sum_{p+q=n} H_p(C) \otimes_A H_q(C'),

the

\gamma_{p,q}

define a graded

k

-linear mapping of degree 0

\gamma(C, C') : H(C) \otimes_A H(C') \to H(C \otimes_A C').

Prop. 6 of II, p. 59, can be reformulated as follows:

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

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Cites

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