Bourbaki

Proposition 1

Tag 0D6U, permanent. Cite it as [Bourbaki, Tag 0D6U].

Commutative Algebra, chapter X, § 5. Intersections complètes

Let AA be a ring, JJ an ideal of AA, generated by a finite sequence x=(x1,…,xr)x = (x_1, \ldots, x_r) of elements of AA. The following conditions are equivalent:

(i) the sequence xx is completely secant for AA ; (ii) (resp. (ii')) for every prime ideal (resp. maximal ideal) p∈V(J)p \in V(J), the image of xx in ApA_p is completely secant for ApA_p ; (iii) (resp. (iii')) for every prime ideal (resp. maximal ideal) p∈V(J)p \in V(J), the ideal JJ is completely secant at pp and the image of xx in the κ(p)\kappa(p)-vector space κ(p)⊗AJ\kappa(p) \otimes_A J forms a basis of it.

When AA is noetherian, these conditions are also equivalent to:

(iv) for every integer n≥0n \geq 0, the A/JA/J-module Jn/Jn+1J^n / J^{n+1} is free, and the images of the monomials x1α1…xrαrx_1^{\alpha_1} \ldots x_r^{\alpha_r} of total degree nn form a basis of it.

Let pp be a prime ideal of AA; let us denote by xpx_p the image of the sequence xx in ApA_p. For every integer i≥0i \geq 0, the ApA_p-module Hi(xp,Ap)H_i(x_p, A_p) is isomorphic to (Hi(x,A))p(H_i(x, A))_p (A, X, p. 151, 2)) ; it is zero if pp does not contain JJ, since then one has Jp=ApJ_p = A_p (A, X, p. 148, cor. 2). The equivalence of (i), (ii) and (ii') follows from this and from II, § 3, No. 3, cor. 2 of th. 1. On the other hand, the equivalence of (ii) and (iii), and that of (ii') and (iii'), follow from § 1, No. 3, cor. 2 of th. 1.

The implication (i) ⇒\Rightarrow (iv) is a consequence of th. 1 of A, X, p. 160. Finally, condition (iv) implies by localization the analogous condition for the (Ap/Jp)(A_p/J_p)-modules Jpn/Jpn+1J_p^n / J_p^{n+1} for every p∈V(J)p \in V(J), and the latter entails (ii) by cor. 1 of A, X, p. 160.

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

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