Bourbaki

Lemma 1

Tag 0D4X, permanent. Cite it as [Bourbaki, Tag 0D4X].

Commutative Algebra, chapter X, § 3. Profondeur et dimension homologique

Let AA be an artinian local ring. The following conditions are equivalent:

(i) AA is a Gorenstein ring; (ii) the AA-module AA is injective; (iii) the vector space HomA(κA,A)\mathrm{Hom}_A(\kappa_A, A) over κA\kappa_A is of dimension 1.

Let us recall (A, VIII, p. 3.5) that AA has a minimal nonzero ideal; such an ideal is a simple module, hence isomorphic to κA\kappa_A. The vector space HomA(κA,A)\mathrm{Hom}_A(\kappa_A, A) over κA\kappa_A is therefore nonzero; saying that it is of dimension 1 means that AA contains only one minimal nonzero ideal, which is therefore the socle of AA (A, VIII, § 4, No. 6).

The equivalence of (i) and (ii) was proved after Prop. 10. Suppose that the AA-module AA is injective. Let xx and yy be two nonzero elements of AA annihilated by mAm_A. There exists a unique AA-linear mapping φ:Ax→A\varphi : Ax \to A such that φ(x)=y\varphi(x) = y; since AA is injective, it extends to an endomorphism of AA, which implies that yy belongs to AxAx, whence (iii).

Conversely, suppose that HomA(κA,A)\mathrm{Hom}_A(\kappa_A, A) is of dimension 1. Let MM be a finitely generated AA-module; it is of finite length, hence has a composition series whose quotients are isomorphic to κA\kappa_A. One deduces by induction on the length of MM the inequality longA(HomA(M,A))⩽longA(M)\mathrm{long}_A(\mathrm{Hom}_A(M, A)) \leqslant \mathrm{long}_A(M). In the exact sequence of extension modules

0→HomA(κA,A)→HomA(A,A)→αHomA(mA,A)→ExtA1(κA,A)→0, 0 \to \mathrm{Hom}_A(\kappa_A, A) \to \mathrm{Hom}_A(A, A) \xrightarrow{\alpha} \mathrm{Hom}_A(m_A, A) \to \mathrm{Ext}_A^1(\kappa_A, A) \to 0 ,

one therefore has longA(HomA(mA,A))⩽longA(mA)=long(A)−1=longA(Im α)\mathrm{long}_A(\mathrm{Hom}_A(m_A, A)) \leqslant \mathrm{long}_A(m_A) = \mathrm{long}(A) - 1 = \mathrm{long}_A(\mathrm{Im}\, \alpha). Consequently α\alpha is surjective, ExtA1(κA,A)\mathrm{Ext}_A^1(\kappa_A, A) is zero and the AA-module AA is injective (No. 3, Prop. 4).

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

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