Bourbaki

Example 2

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

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

If HA0(M)=MH_A^0(M) = M (for example if M is artinian), HAi(M)H_A^i(M) is zero for i>0i > 0. Let in fact (I, e) be an injective envelope of M. The submodule HA0(I)H_A^0(I) of I is injective (example 1) and contains e(M)e(M), hence is equal to I. Put N=Coker eN = \mathrm{Coker}\,e and consider the exact sequence 0→M→eI→pN→00 \to M \xrightarrow{e} I \xrightarrow{p} N \to 0. Since I=HA0(I)I = H_A^0(I), one has N=HA0(N)N = H_A^0(N) and the homomorphism HA0(p)H_A^0(p) is surjective. Since HAi(I)H_A^i(I) is zero for i>0i > 0 (example 1), HA1(M)H_A^1(M) is zero and HAi(M)H_A^i(M) is isomorphic to HAi−1(N)H_A^{i-1}(N) for i>1i > 1; one concludes by reasoning by induction on the integer ii.

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

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