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Tag 052K

Chapter 10: Commutative Algebra > Section 10.49: Valuation rings

Lemma 10.49.4. Let $A \subset K$ be a subring of a field $K$ such that for all $x \in K$ either $x \in A$ or $x^{-1} \in A$ or both. Then $A$ is a valuation ring with fraction field $K$.

Proof. If $A$ is not $K$, then $A$ is not a field and there is a nonzero maximal ideal $\mathfrak m$. If $\mathfrak m'$ is a second maximal ideal, then choose $x, y \in A$ with $x \in \mathfrak m$, $y \not \in \mathfrak m$, $x \not \in \mathfrak m'$, and $y \in \mathfrak m'$ (see Lemma 10.14.2). Then neither $x/y \in A$ nor $y/x \in A$ contradicting the assumption of the lemma. Thus we see that $A$ is a local ring. Suppose that $A'$ is a local ring contained in $K$ which dominates $A$. Let $x \in A'$. We have to show that $x \in A$. If not, then $x^{-1} \in A$, and of course $x^{-1} \in \mathfrak m_A$. But then $x^{-1} \in \mathfrak m_{A'}$ which contradicts $x \in A'$. $\square$

    The code snippet corresponding to this tag is a part of the file algebra.tex and is located in lines 11191–11196 (see updates for more information).

    \begin{lemma}
    \label{lemma-x-or-x-inverse-valuation-ring}
    Let $A \subset K$ be a subring of a field $K$ such that for all
    $x \in K$ either $x \in A$ or $x^{-1} \in A$ or both.
    Then $A$ is a valuation ring with fraction field $K$.
    \end{lemma}
    
    \begin{proof}
    If $A$ is not $K$, then $A$ is not a field and there is a nonzero
    maximal ideal $\mathfrak m$.
    If $\mathfrak m'$ is a second maximal ideal, then choose $x, y \in A$
    with $x \in \mathfrak m$, $y \not \in \mathfrak m$,
    $x \not \in \mathfrak m'$, and $y \in \mathfrak m'$ (see
    Lemma \ref{lemma-silly}). Then neither $x/y \in A$ nor $y/x \in A$
    contradicting the assumption of the lemma. Thus we see that $A$ is
    a local ring. Suppose that $A'$ is a local ring contained in $K$ which
    dominates $A$. Let $x \in A'$. We have to show that $x \in A$. If not, then
    $x^{-1} \in A$, and of course $x^{-1} \in \mathfrak m_A$. But then
    $x^{-1} \in \mathfrak m_{A'}$ which contradicts $x \in A'$.
    \end{proof}

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