Bourbaki

Proposition 9

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

Commutative Algebra, chapter VIII, § 7. Multiplicités

Let us denote by p1,...,prp_1, ..., p_r those prime ideals of the graded ring gr(A)=⨁n(qn/qn+1)\mathrm{gr}(A) = \bigoplus_n (q^n/q^{n+1}) which are associated with the graded module gr(M)=⨁n(qnM/qn+1M)\mathrm{gr}(M) = \bigoplus_n (q^n M/q^{n+1} M) and do not contain gr1(A)=q/q2\mathrm{gr}_1(A) = q/q^2. Let δ\delta be an integer >0> 0, ξ\xi an element of grδ(A)\mathrm{gr}_\delta(A), and φ:gr(M)→gr(M)\varphi : \mathrm{gr}(M) \to \mathrm{gr}(M) the homothety of ratio ξ\xi in gr(M)\mathrm{gr}(M). For φn\varphi_n to be injective for every nn sufficiently large, it is necessary and sufficient that ξ\xi belong to none of the pip_i. In fact, the prime ideals associated with the gr(A)\mathrm{gr}(A)-module Ker φ\mathrm{Ker}\, \varphi are those of the prime ideals associated with gr(M)\mathrm{gr}(M) which contain ξ\xi (IV, § 1, No. 1, Def. 1). By Lemma 3, (Ker φ)n(\mathrm{Ker}\, \varphi)_n is zero for nn sufficiently large, if and only if all these ideals contain gr+(A)\mathrm{gr}_+(A) (or, what amounts to the same, gr1(A)\mathrm{gr}_1(A)), whence the proposition.

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

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