Bourbaki

Proposition 8

Tag 0E3D, permanent. Cite it as [Bourbaki, Tag 0E3D].

Algebra, chapter X, § 9. Complexes de Koszul

Suppose the sequence x is regular for M. The element of ExtAn(M/(x)M,M)\mathrm{Ext}_A^n(M/(x)M, M) associated with the exact sequence (23) is (−1)n(n+1)/2θx(-1)^{n(n+1)/2} \theta_x. For i=0,1,…,ni = 0, 1, \ldots, n, define an A-linear mapping

ai:Ki(x,M)→Mi=M/(x1M+⋯+xi−1M) a^i : \mathbf{K}^i(x, M) \to M_i = M/(x_1 M + \cdots + x_{i-1} M)
as follows: if m∈Ki(x,M)m \in \mathbf{K}^i(x, M), ai(m)a^i(m) is the class in MiM_i of the element m(1,2,…,i)m(1, 2, \ldots, i) of MM. It is clear that a0a^0 is the identity mapping of MM and that p∘an=qp \circ a^n = q. Moreover ai+1∘∂i(m)a^{i+1} \circ \partial^i(m) is the image in Mi+1M_{i+1} of the element
∑k=1i+1(−1)k+1xkm(1,2,…,k−1,k+1,…,i+1). \sum_{k=1}^{i+1} (-1)^{k+1} x_k m(1, 2, \ldots, k-1, k+1, \ldots, i+1) .
Since xkx_k annihilates Mi+1M_{i+1} for k=1,…,ik = 1, \ldots, i, ai+1∘∂i(m)a^{i+1} \circ \partial^i(m) is also the image of
(−1)ixi+1m(1,2,…,i), (-1)^i x_{i+1} m(1, 2, \ldots, i) ,
hence
ai+1∘∂i=(−1)ix‾i+1∘ai. a^{i+1} \circ \partial^i = (-1)^i \overline{x}_{i+1} \circ a^i .
By X, p. 120, cor. 1 and 2, the element of ExtAn(M/(x)M,M)\mathrm{Ext}_A^n(M/(x)M, M) associated with (23) is equal to ∏i=1n(−1)i⋅θx\prod_{i=1}^n (-1)^i \cdot \theta_x, whence the assertion.

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

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Translated with machine help, not yet reviewed by a person.