Lemma 10.106.7. Suppose $R$ is a Noetherian local ring. Let $x \in \mathfrak m$ be a nonzerodivisor such that $R/xR$ is a regular local ring. Then $R$ is a regular local ring. More generally, if $x_1, \ldots , x_ r$ is a regular sequence in $R$ such that $R/(x_1, \ldots , x_ r)$ is a regular local ring, then $R$ is a regular local ring.
Proof. This is true because $x$ together with the lifts of a system of minimal generators of the maximal ideal of $R/xR$ will give $\dim (R)$ generators of $\mathfrak m$. Use Lemma 10.60.13. The last statement follows from the first and induction. $\square$
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Your email address will not be published.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.