Bourbaki

Proposition 10

Tag 0D0Y, permanent. Cite it as [Bourbaki, Tag 0D0Y].

Commutative Algebra, chapter X, § 1. Profondeur

*Let ρ:A→B\rho : A \to B be a local homomorphism of noetherian local rings, NN a finitely generated B-module and y=(y1,…,ys)y = (y_1, \ldots, y_s) a sequence of elements of mBm_B. Let y\mathfrak{y} denote the ideal of BB generated by this sequence. The following conditions are equivalent:

(i) the A-module N/yNN/\mathfrak{y}N is flat and the sequence yy is N-regular; (ii) the A-module NN is flat and the sequence yy is (κA⊗AN)(\kappa_A \otimes_A N)-regular.*

When these conditions are satisfied, for every A-module MM, the sequence yy is M⊗ANM \otimes_A N-regular.

Let us prove the equivalence of (i) and (ii) by induction on ss. The case s=0s = 0 being evident, suppose s⩾1s \geqslant 1; let y′y' denote the sequence (y1,…,ys−1)(y_1, \ldots, y_{s-1}) and y′\mathfrak{y}' the ideal of BB generated by it. By lemma 2 applied to the B-module N/y′NN/\mathfrak{y}'N and to the element ysy_s of BB, condition (i) is equivalent to

(i') the A-module N/y′NN/\mathfrak{y}'N is flat, the sequence y′y' is N-regular, and the homothety of ratio ysy_s in κA⊗A(N/y′N)=(κA⊗AN)/y′(κA⊗AN)\kappa_A \otimes_A (N/\mathfrak{y}'N) = (\kappa_A \otimes_A N)/\mathfrak{y}'(\kappa_A \otimes_A N) is injective.

This condition is equivalent to (ii) by the induction hypothesis.

The last assertion follows likewise by induction on ss from the last assertion of lemma 2.

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

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