Lemma 10.119.13. Let $R$ be a Noetherian local domain with fraction field $K$. Assume that $R$ is not a field. Let $K \subset L$ be a finitely generated field extension. Then there exists discrete valuation ring $A$ with fraction field $L$ which dominates $R$.

Proof. If $L$ is not finite over $K$ choose a transcendence basis $x_1, \ldots , x_ r$ of $L$ over $K$ and replace $R$ by $R[x_1, \ldots , x_ r]$ localized at the maximal ideal generated by $\mathfrak m_ R$ and $x_1, \ldots , x_ r$. Thus we may assume $K \subset L$ finite.

By Lemma 10.119.1 we may assume $\dim (R) = 1$.

Let $A \subset L$ be the integral closure of $R$ in $L$. By Lemma 10.119.12 this is Noetherian. By Lemma 10.36.17 there is a prime ideal $\mathfrak q \subset A$ lying over the maximal ideal of $R$. By Lemma 10.119.7 the ring $A_{\mathfrak q}$ is a discrete valuation ring dominating $R$ as desired. $\square$

## 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Your email address will not be published.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).

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 00PH. Beware of the difference between the letter 'O' and the digit '0'.