Remark 2

Tag 0D18, permanent. Cite it as [Bourbaki, Tag 0D18].

Commutative Algebra, chapter X, § 1. Profondeur

One has $\sup_{p \in \operatorname{Ass}(M)} \dim(A/p) = \dim(M)$ (VIII, § 1, No. 4, remark 2), and one recovers the inequality $prof(M) \leq \dim(M)$ for $M \neq 0$ (No. 4, cor. 2 of th. 2).

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

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