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. $\square$

Comment #3523 by Dario Weißmann on

Shouldn't we choose $g\notin \mathfrak{p}$ such that after the replacement of $R$ by $R_g$ the $f_1,\dots,f_r$ generate $\mathfrak{p}$ as well as $f_1,\dots,f_r$ is a regular sequence?

Comment #3756 by Laurent Moret-Bailly on

In the statement, "for all $\mathfrak{p}\subset R$" should be "for all $\mathfrak{p}\in\mathrm{Spec}(R)$".

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