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The Stacks project

Lemma 15.47.3. Let R be a Noetherian ring. Let X = \mathop{\mathrm{Spec}}(R). Assume that for all primes \mathfrak p \subset R the ring R/\mathfrak p is J-0. Then R is J-1.

Proof. We will show that the criterion of Lemma 15.47.2 applies. Let \mathfrak p \in \text{Reg}(X) be a prime of height r. Pick f_1, \ldots , f_ r \in \mathfrak p which map to generators of \mathfrak pR_\mathfrak p. Since \mathfrak p \in \text{Reg}(X) we see that f_1, \ldots , f_ r maps to a regular sequence in R_\mathfrak p, see Algebra, Lemma 10.106.3. Thus by Algebra, Lemma 10.68.6 we see that after replacing R by R_ g for some g \in R, g \not\in \mathfrak p the sequence f_1, \ldots , f_ r is a regular sequence in R. After another replacement we may also assume f_1, \ldots , f_ r generate \mathfrak p. Next, let \mathfrak p \subset \mathfrak q be a prime ideal such that (R/\mathfrak p)_\mathfrak q is a regular local ring. By the assumption of the lemma there exists a non-empty open subset of V(\mathfrak p) consisting of such primes, hence it suffices to prove R_\mathfrak q is regular. Note that f_1, \ldots , f_ r is a regular sequence in R_\mathfrak q such that R_\mathfrak q/(f_1, \ldots , f_ r)R_\mathfrak q is regular. Hence R_\mathfrak q is regular by Algebra, Lemma 10.106.7. \square


Comments (4)

Comment #3523 by Dario Weißmann on

Shouldn't we choose such that after the replacement of by the generate as well as is a regular sequence?

Comment #3756 by Laurent Moret-Bailly on

In the statement, "for all " should be "for all ".

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  • 5 comment(s) on Section 15.47: The singular locus

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