Bourbaki

Proposition 8

Tag 0D0T, permanent. Cite it as [Bourbaki, Tag 0D0T].

Commutative Algebra, chapter X, § 1. Profondeur

Let A be a noetherian ring, F a closed subset of Spec(A)\mathrm{Spec}(A), and M a finitely generated A-module. One has

profF(M)=inf⁡p∈FprofAp(Mp)=inf⁡p∈Supp(M)∩FprofAp(Mp). \mathrm{prof}_F(M) = \inf_{p \in F} \mathrm{prof}_{A_p}(M_p) = \inf_{p \in \mathrm{Supp}(M) \cap F} \mathrm{prof}_{A_p}(M_p) .

This is clear if profF(M)=+∞\mathrm{prof}_F(M) = +\infty (Remark 2). If profF(M)=0\mathrm{prof}_F(M) = 0, there exists a prime ideal p∈Ass(M)∩Fp \in \mathrm{Ass}(M) \cap F (Remark 2); one has pAp∈Ass(Mp)pA_p \in \mathrm{Ass}(M_p) (IV, § 1, No. 2, Prop. 5), hence profAp(Mp)=0\mathrm{prof}_{A_p}(M_p) = 0 (Remark 2 of No. 1), whence the proposition in this case.

Suppose 0<profF(M)<+∞0 < \mathrm{prof}_F(M) < +\infty; let J be an ideal of A such that V(J)=FV(J) = F, and let x be an element of J such that the homothety xMx_M is injective (loc. cit.). For each prime ideal p, the homothety xMpx_{M_p} is injective. By Prop. 7 of No. 4, we therefore have

profF(M/xM)=profF(M)−1profAp((M/xM)p)=profAp(Mp)−1. \mathrm{prof}_F(M/xM) = \mathrm{prof}_F(M) - 1 \\ \mathrm{prof}_{A_p}((M/xM)_p) = \mathrm{prof}_{A_p}(M_p) - 1 .

One then concludes by induction on the integer profF(M)\mathrm{prof}_F(M).

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

Cited by

Nothing.

Cites

Nothing.

Translated with machine help, not yet reviewed by a person.