## Tag `00IB`

Chapter 10: Commutative Algebra > Section 10.49: Valuation rings

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 $x^{-1} \in A$. $\square$

The code snippet corresponding to this tag is a part of the file `algebra.tex` and is located in lines 11194–11199 (see updates for more information).

```
\begin{lemma}
\label{lemma-valuation-ring-x-or-x-inverse}
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.
\end{lemma}
\begin{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 \ref{lemma-Zariski-topology}.
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 \ref{lemma-integral-overring-surjective}. Since $A$
is a valuation ring we conclude that $A = (A'')_{\mathfrak m''}$
and hence $x^{-1} \in A$.
\end{proof}
```

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