The Stacks project

Lemma 10.162.11. Let $R$ be a Noetherian local ring. Let $\mathfrak p \subset R$ be a prime. Assume

  1. $R_{\mathfrak p}$ is a discrete valuation ring, and

  2. $\mathfrak p$ is analytically unramified.

Then for any associated prime $\mathfrak q$ of $R^\wedge /\mathfrak pR^\wedge $ the local ring $(R^\wedge )_{\mathfrak q}$ is a discrete valuation ring.

Proof. Assumption (2) says that $R^\wedge /\mathfrak pR^\wedge $ is a reduced ring. Hence an associated prime $\mathfrak q \subset R^\wedge $ of $R^\wedge /\mathfrak pR^\wedge $ is the same thing as a minimal prime over $\mathfrak pR^\wedge $. In particular we see that the maximal ideal of $(R^\wedge )_{\mathfrak q}$ is $\mathfrak p(R^\wedge )_{\mathfrak q}$. Choose $x \in R$ such that $xR_{\mathfrak p} = \mathfrak pR_{\mathfrak p}$. By the above we see that $x \in (R^\wedge )_{\mathfrak q}$ generates the maximal ideal. As $R \to R^\wedge $ is faithfully flat we see that $x$ is a nonzerodivisor in $(R^\wedge )_{\mathfrak q}$. Hence we win. $\square$


Comments (0)

There are also:

  • 2 comment(s) on Section 10.162: Nagata 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Your email address will not be published.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 032Z. Beware of the difference between the letter 'O' and the digit '0'.