Bourbaki

Proposition 2

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

Algebra, chapter X, § 7. Produit de composition

We have a,b∈Zn(HomgrA(L(M),I(N)))a, b \in Z^n(\mathrm{Homgr}_A(L(M), I(N))). Moreover, the classes a‾\overline{a} and b‾\overline{b} of aa and bb in Hn(HomgrA(L(M),I(N)))=ExtAn(M,N)H^n(\mathrm{Homgr}_A(L(M), I(N))) = \mathrm{Ext}_A^n(M, N) depend only on the exact sequence (4) and are equal.

By prop. 1, applied to the two extreme rows of (5) and to the composed vertical arrows, with p=n,q=0p = n, q = 0, we have a,b∈Zn(HomgrA(I(M),L(N)))a, b \in Z^n(\mathrm{Homgr}_A(I(M), L(N))) and

a−b=a−(−1)(n+1)nb∈Bn(HomgrA(L(M),I(N))). a - b = a - (-1)^{(n+1)n} b \in B^n(\mathrm{Homgr}_A(L(M), I(N))) .

Since aa (resp. bb) is independent of the choice of uu (resp. vv), the element a‾=b‾\overline{a} = \overline{b} of ExtAn(M,N)\mathrm{Ext}_A^n(M, N) is independent of the choice of the morphisms uu and vv, whence the proposition.

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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