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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}

    Comments (4)

    Comment #42 by Rankeya on August 30, 2012 a 12:09 am UTC

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

    Comment #48 by Johan (site) on August 30, 2012 a 11:36 am UTC

    Fixed. Thanks!

    Comment #2365 by Dominic Wynter on February 6, 2017 a 2:29 am UTC

    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 Johan (site) on February 17, 2017 a 2:32 pm UTC

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

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