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. $\square$
Post a comment
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$
). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (4)
Comment #3523 by Dario Weißmann on
Comment #3663 by Johan on
Comment #3756 by Laurent Moret-Bailly on
Comment #3890 by Johan on
There are also: