Bourbaki

Remark 1

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

Commutative Algebra, chapter X, § 10. Cohomologie locale, dualité de Grothendieck

Let XX be the topological space Spec(A)\mathrm{Spec}(A), OX\mathscr{O}_X the structural sheaf of rings and M~\widetilde{M} the OX\mathscr{O}_X-module associated with MM. The graded AA-module HA(M)H_A(M) is identified with the module H{mA}(X,M~)H_{\{m_A\}}(X, \widetilde{M}) of cohomology with support in the closed point mAm_A of XX.

For every homomorphism f:M→Nf : M \to N of AA-modules, the mappings ExtA(1A/a,f):ExtA(A/a,M)⟶ExtA(A/a,N)\mathrm{Ext}_A(1_{A/a}, f) : \mathrm{Ext}_A(A/a, M) \longrightarrow \mathrm{Ext}_A(A/a, N) form an inductive system of graded linear mappings. By passing to the inductive limit, one obtains a graded homomorphism HA(f):HA(M)→HA(N)H_A(f) : H_A(M) \to H_A(N). For every sequence M→fN→gPM \xrightarrow{f} N \xrightarrow{g} P of AA-modules and homomorphisms, one has HA(g∘f)=HA(g)∘HA(f)H_A(g \circ f) = H_A(g) \circ H_A(f). Let

0→M→fN→gP→0 \begin{array}{ccccccccc} 0 & \to & M & \xrightarrow{f} & N & \xrightarrow{g} & P & \to & 0 \end{array}

be an exact sequence of AA-modules. By AA, X, p. 90, Prop. 8, the connecting homomorphisms of the extension modules ExtA(A/a,P)⟶ExtA(A/a,M)\mathrm{Ext}_A(A/a, P) \longrightarrow \mathrm{Ext}_A(A/a, M) form an inductive system of AA-linear mappings, graded of (ascending) degree +1. By passing to the inductive limit, one deduces an AA-homomorphism ∂(E):HA(P)→HA(M)\partial(\mathscr{E}) : H_A(P) \to H_A(M), graded of degree +1, which makes exact the sequence of homomorphisms

…⟶HAn−1(P)→∂n−1(E)HAn(M)→HAn(f)HAn(N)→HAn(g)HAn(P)→∂n(E)HAn+1(M)⟶… \ldots \longrightarrow H_A^{n-1}(P) \xrightarrow{\partial^{n-1}(\mathscr{E})} H_A^n(M) \xrightarrow{H_A^n(f)} H_A^n(N) \xrightarrow{H_A^n(g)} H_A^n(P) \xrightarrow{\partial^n(\mathscr{E})} H_A^{n+1}(M) \longrightarrow \ldots

Let M be an a-module. For every ideal α\alpha of A, the A-module HomA(A/α,M)\mathrm{Hom}_A(A/\alpha, M) is canonically identified with the submodule of M consisting of the elements annihilated by α\alpha. Thus HA0(M)H_A^0(M) is identified with the submodule of M consisting of the elements mm which are annihilated by a power of mA\mathfrak{m}_A, that is to say such that longA(Am)<+∞\mathrm{long}_A(Am) < +\infty. In particular one has HA0(M)=MH_A^0(M) = M when M is artinian.

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

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