Bourbaki

Lemma 2

Tag 0D2S, permanent. Cite it as [Bourbaki, Tag 0D2S].

Commutative Algebra, chapter X, § 2. Modules et anneaux macaulayens

Let G‾\overline{\mathscr{G}} be the image of G\mathscr{G} in A/Ann(M)A/\mathrm{Ann}(M). One has

ht(G‾)=codim(Supp(M/SM),Supp(M)). \mathrm{ht}(\overline{\mathscr{G}}) = \mathrm{codim}(\mathrm{Supp}(M/SM), \mathrm{Supp}(M)) .
When moreover SM≠MSM \neq M, one has
ht(G‾)≤Card(S). \mathrm{ht}(\overline{\mathscr{G}}) \leq \mathrm{Card}(S) .

Let us denote by α\alpha the annihilator of MM. By the cor. to Prop. 18 of II, § 4, No. 4, the support of the A-module M/SMM/SM is V(G+α)V(\mathscr{G} + \alpha). Its codimension in Supp(M)\mathrm{Supp}(M) is therefore equal to the codimension of V(G+α)V(\mathscr{G} + \alpha) in V(α)V(\alpha), or again to the codimension of V((G+α)/α)V((\mathscr{G} + \alpha)/\alpha) in Spec(A/α)\mathrm{Spec}(A/\alpha), which is none other than the height of G‾\overline{\mathscr{G}}.

Suppose SM≠MSM \neq M; the inequality ht(G‾)≤Card(S)\mathrm{ht}(\overline{\mathscr{G}}) \leq \mathrm{Card}(S) is obvious when SS is infinite, and follows from Prop. 4 b) of VIII, § 3, No. 3 when SS is finite.

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

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