Bourbaki

Corollary 1

Tag 0D4K, permanent. Cite it as [Bourbaki, Tag 0D4K].

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

Retain the hypotheses of theorem 1.

a) One has dpA(M)≤prof⁡(A)dp_A(M) \leq \operatorname{prof}(A). For there to be equality, it is necessary and sufficient that the maximal ideal mAm_A be associated to MM.

b) One has prof⁡A(M)≤prof⁡(A)\operatorname{prof}_A(M) \leq \operatorname{prof}(A). For there to be equality, it is necessary and sufficient that MM be free.

a) In fact, “ prof⁡A(M)=0\operatorname{prof}_A(M) = 0 ” is equivalent to “ mA∈Ass⁡(A)m_A \in \operatorname{Ass}(A) ” (§1§ 1, No. 1, remark 2).

b) In fact, “ dpA(M)=0dp_A(M) = 0 ” is equivalent to “ MM is free ”.

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

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