Lemma 10.49.3. Let $A$ be a valuation ring with maximal ideal $\mathfrak m$ and fraction field $K$. Let $x \in K$. Then either $x \in A$ or $x^{-1} \in A$ or both.

**Proof.**
Assume that $x$ is not in $A$. Let $A'$ denote the subring of $K$ generated by $A$ and $x$. Since $A$ is a valuation ring we see that there is no prime of $A'$ lying over $\mathfrak m$. Since $\mathfrak m$ is maximal we see that $V(\mathfrak m A') = \emptyset $. Then $\mathfrak m A' = A'$ by Lemma 10.16.2. Hence we can write $1 = \sum _{i = 0}^ d t_ i x^ i$ with $t_ i \in \mathfrak m$. This implies that $(1 - t_0) (x^{-1})^ d - \sum t_ i (x^{-1})^{d - i} = 0$. In particular we see that $x^{-1}$ is integral over $A$. Thus the subring $A''$ of $K$ generated by $A$ and $x^{-1}$ is finite over $A$ and we see there exists a prime ideal $\mathfrak m'' \subset A''$ lying over $\mathfrak m$ by Lemma 10.35.17. Since $A$ is a valuation ring we conclude that $A = (A'')_{\mathfrak m''}$ and hence $x^{-1} \in A$.
$\square$

## 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Your email address will not be published.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.

## Comments (4)

Comment #42 by Rankeya on

Comment #48 by Johan on

Comment #2365 by Dominic Wynter on

Comment #2428 by Johan on