Remark

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

Algebra, chapter X, § 7. Produit de composition

Let $\theta \in \mathrm{Ext}^n_A(M, N)$ , resp. $\theta' \in \mathrm{Ext}^n_A(M', N')$ , be the class of an exact sequence

0 \to N \xrightarrow{f_{n+1}} R_n \to \ldots \to R_1 \xrightarrow{f_1} M \to 0,

resp.

0 \to N' \xrightarrow{f'_{n+1}} R'_n \to \ldots \to R'_1 \xrightarrow{f'_1} M' \to 0.

Let $i_N,\ i_{N'}$ , be the canonical injections of $N$ and $N'$ into $N \oplus N'$ , $q_M,\ q_{M'}$ , the projections of $M \oplus M'$ onto $M$ and $M'$ . Consider the homomorphism

m = \mathrm{Ext}\,(q_M,\ i_N) \oplus \mathrm{Ext}\,(q_{M'},\ i_{N'})

from $\mathrm{Ext}_A(M, N) \oplus \mathrm{Ext}_A(M', N')$ into $\mathrm{Ext}_A(M \oplus M', N \oplus N')$ . The element

m(\theta, \theta') = i_N \circ \theta \circ q_M + i_{N'} \circ \theta' \circ q_{M'}

is the class of the exact sequence

0 \to N \oplus N' \xrightarrow{f_{n-1} \oplus f'_{n-1}} R_n \oplus R'_n \to \ldots \to R_1 \oplus R'_1 \xrightarrow{f_1 \oplus f'_1} M \oplus M' \to 0 .

Indeed, if one denotes this class by $\theta''$ , it follows from prop. 4 that one has

\theta'' \circ i_M = i_N \circ \theta = m(\theta, \theta') \circ i_M \quad \text{and} \quad \theta'' \circ i_{M'} = i_{N'} \circ \theta = m(\theta, \theta') \circ i_{M'} ;

by X, p. 89, prop. 7, this entails $\theta'' = m(\theta, \theta')$ .

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

Cited by

Nothing.

Cites

Nothing.