Bourbaki

Exercise 10, § 4

Commutative Algebra, X, p. 164

Bourbaki marks this one as harder than the rest.

Let AA be a noetherian ring, MM a finitely generated AA-module. For every integer n⩾0n \geqslant 0, denote by XnX_n the set of elements pp of Spec(A)\mathrm{Spec}(A) such that dim⁡Ap(Mp)−profAp(Mp)⩾n\dim_{A_p}(M_p) - \mathrm{prof}_{A_p}(M_p) \geqslant n. Let kk be an integer ⩾0\geqslant 0. (a) In order that MM satisfy the property SkS_k (§1§ 1, exer. 8), it is necessary and sufficient that one have codim(Xn,Supp(M))⩾n+k\mathrm{codim}(X_n, \mathrm{Supp}(M)) \geqslant n + k for every n⩾0n \geqslant 0. (b) Suppose that XnX_n is closed in Spec(A)\mathrm{Spec}(A) for every n⩾0n \geqslant 0. Let pp be an element of Supp(M)\mathrm{Supp}(M). In order that the ApA_p-module MpM_p satisfy the property SkS_k (§1§ 1, exer. 8), it is necessary and sufficient that one have codim(Xn′,Supp(M))⩾n+k\mathrm{codim}(X'_n, \mathrm{Supp}(M)) \geqslant n + k for every n⩾0n \geqslant 0 and every component Xn′X'_n of XnX_n containing pp. (c) Suppose that the ring AA is presentable. Deduce from (b) that the set of prime ideals pp of AA such that MpM_p satisfies the property SkS_k is open in Spec(A)\mathrm{Spec}(A).

Solution

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Machine translation of the English, by gpt-5-mini. Not checked by a person. That model is a cut down version of the one the rest of this language was written by.