Lemma 10.50.3. 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 10.17.2. 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 10.36.17. Since $A$ is a valuation ring we conclude that $A = (A'')_{\mathfrak m''}$ and hence $x^{-1} \in A$. $\square$

Comment #42 by Rankeya on

Let $A'$ be a subring of $A$ should be let $A'$ be a subring of $K$.

Comment #2365 by Dominic Wynter on

Is it accurate to assume that in the expression $\sum_{i=0}^d t_i x^i$, $t_i\in\mathfrak{m}$ for all $i$, or do we just know that $t_0\in\mathfrak{m}$? I have only succeeded in proving that second statement (and in fact that second statement is all that is necessary).

Comment #2428 by on

Thanks for your question. I have clarified the argument. See here.

Comment #7895 by Mingchen on

This is not a big deal, but strictly speaking, one should say for any non-zero $x\in K$, blabla as you talk about $x^{-1}$. The same issue persists in the next lemma.

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