The Stacks project

Lemma 10.106.8. Let $(R_ i, \varphi _{ii'})$ be a directed system of local rings whose transition maps are local ring maps. If each $R_ i$ is a regular local ring and $R = \mathop{\mathrm{colim}}\nolimits R_ i$ is Noetherian, then $R$ is a regular local ring.

Proof. Let $\mathfrak m \subset R$ be the maximal ideal; it is the colimit of the maximal ideal $\mathfrak m_ i \subset R_ i$. We prove the lemma by induction on $d = \dim \mathfrak m/\mathfrak m^2$. If $d = 0$, then $R = R/\mathfrak m$ is a field and $R$ is a regular local ring. If $d > 0$ pick an $x \in \mathfrak m$, $x \not\in \mathfrak m^2$. For some $i$ we can find an $x_ i \in \mathfrak m_ i$ mapping to $x$. Note that $R/xR = \mathop{\mathrm{colim}}\nolimits _{i' \geq i} R_{i'}/x_ iR_{i'}$ is a Noetherian local ring. By Lemma 10.106.3 we see that $R_{i'}/x_ iR_{i'}$ is a regular local ring. Hence by induction we see that $R/xR$ is a regular local ring. Since each $R_ i$ is a domain (Lemma 10.106.1) we see that $R$ is a domain. Hence $x$ is a nonzerodivisor and we conclude that $R$ is a regular local ring by Lemma 10.106.7. $\square$


Comments (0)

There are also:

  • 5 comment(s) on Section 10.106: Regular local rings

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Your email address will not be published.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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 07DX. Beware of the difference between the letter 'O' and the digit '0'.