Bourbaki

Proposition 11

Tag 0D0Z, permanent. Cite it as [Bourbaki, Tag 0D0Z].

Commutative Algebra, chapter X, § 1. Profondeur

Let ρ:A→B\rho : A \to B be a local homomorphism of noetherian local rings, MM a finitely generated A-module and NN a finitely generated B-module; suppose that the A-module NN is flat.

a) Let (x1,…,xr)(x_1, \ldots, x_r) be a sequence of elements of mAm_A which is regular for the A-module MM, and (y1,…,ys)(y_1, \ldots, y_s) a sequence of elements of mBm_B which is regular for the B-module κA⊗AN\kappa_A \otimes_A N; then the sequence (y1,…,ys,ρ(x1),…,ρ(xr))(y_1, \ldots, y_s, \rho(x_1), \ldots, \rho(x_r)) of elements of mBm_B is regular for the B-module M⊗ANM \otimes_A N.

b) One has the equality

prof⁡B(M⊗AN)=prof⁡A(M)+prof⁡B(κA⊗AN) \operatorname{prof}_B(M \otimes_A N) = \operatorname{prof}_A(M) + \operatorname{prof}_B(\kappa_A \otimes_A N)

Let xx denote the ideal of A generated by the sequence xx and yy the ideal of B generated by yy. By Prop. 10, the sequence yy is M⊗ANM \otimes_A N-regular and N/yNN/yN is flat over A, so that the sequence ρ(x)=(ρ(x1),…,ρ(xr))\rho(x) = (\rho(x_1), \ldots, \rho(x_r)) is regular for M⊗A(N/yN)=(M⊗AN)/y(M⊗AN)M \otimes_A (N/yN) = (M \otimes_A N)/y(M \otimes_A N). This proves a).

To prove b), we may suppose that M and N are nonzero. By Nakayama's lemma, κA⊗AN\kappa_A \otimes_A N is likewise nonzero, so that prof⁡A(M)\operatorname{prof}_A(M) and prof⁡B(κA⊗AN)\operatorname{prof}_B(\kappa_A \otimes_A N) are finite (No. 4, Cor. 2 of Th. 2). Take maximal regular sequences xx and yy; then one has r=prof⁡A(M)r = \operatorname{prof}_A(M), s=prof⁡B(κA⊗AN)s = \operatorname{prof}_B(\kappa_A \otimes_A N), and there exists an injective a-linear mapping u:κA→M/xMu : \kappa_A \to M/xM and an injective b-linear mapping v:κB→κA⊗A(N/yN)v : \kappa_B \to \kappa_A \otimes_A (N/yN) (No. 4, Cor. 1 of Th. 2). Since N/yNN/yN is flat over A, the b-linear mapping (u⊗1N/yN)∘v(u \otimes 1_{N/yN}) \circ v of κB\kappa_B into (M/xM)⊗A(N/yN)=(M⊗AN)/(ρ(x)+y)(M⊗AN)(M/xM) \otimes_A (N/yN) = (M \otimes_A N)/(\rho(x) + y)(M \otimes_A N) is injective. This implies the equality prof⁡B(M⊗AN)=r+s\operatorname{prof}_B(M \otimes_A N) = r + s (loc. cit.).

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

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