Bourbaki

Remark

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

Commutative Algebra, chapter X, § 8. Dualité des modules de longueur finie

Suppose the ring AA artinian. One has long⁡A(I)=long⁡A(DA(A))=long⁡(A)\operatorname{long}_A(I) = \operatorname{long}_A(D_A(A)) = \operatorname{long}(A) (Theorem 2, c)). Let MM be a finitely generated AA-module; it admits an injective envelope isomorphic to IsI^s, where ss is the dimension of the socle of MM (Lemma 4). Hence one has long⁡A(M)≤slong⁡(A)\operatorname{long}_A(M) \leq s \operatorname{long}(A); for equality to hold, it is necessary and sufficient that MM be injective. In particular, for the AA-module AA to be injective, it is necessary and sufficient that its socle be of dimension 1; thus one recovers Lemma 1 of § 3, No. 7.

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

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