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I was wondering if the following holds:

If $A$ is a valuation ring with maximal ideal $m_A$ and $B$ is a ring extension of $A$ contained in $\mathrm{Frac}(A)$ then the only maximal ideal of $B$ is still $m_A$.

I am convinced that this result is true. But then I can't understand a result stated in Atiyah and Macdonald that says that being $S$ the set of all local subrings of $Frac(A)$ ordered by the relation of domination ($A$, $B$ local rings, $B$ dominates $A$ if $B \supseteq A$ and $m_B=m_A\cap A$) then $A\in S$ is maximal if and only if $A$ is a valuation ring of $Frac(A)$.

If the previous result is true I can't see why a valuation ring should be maximal in $S$ when it would be enough to take any ring extension of $A$ contained in $Frac(A)$.

Thanks for the attention.

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2 Answers 2

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Let $A \subset B \subset Frac(A)$ be an extension of rings with $A$ a valuation ring of $Frac(A)$. Suppose that the inclusion $A \subsetneq B$ is strict. Let $x \in B \setminus A$ then $ x^{-1} \in A$. Furthermore since $x \notin A$ we have that $x^{-1} \in \mathfrak m_A$. In particular the extension of $\mathfrak m_A$ to $B$ is $B$ because $1=xx^{-1} \in \mathfrak m_A^e$.

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  • $\begingroup$ So are you saying that there exists no proper ring extension of a valuation ring $A$ properly contained in $Frac(A)$? What about the localisations of $A$? $\endgroup$
    – Corra
    Jan 19, 2013 at 22:04
  • $\begingroup$ @Corra No, certainly any proper subring of $Frac(A)$ containing $A$ is also a valuation ring. But its maximal ideal won't lie over $\mathfrak m_A$. Which shouldn't be surprising because any valuation ring is integrally closed, so any such extension can't be intgral. $\endgroup$
    – JSchlather
    Jan 19, 2013 at 23:16
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For a counterexample to your claim take $B = \mathrm{Frac} \ A$. The only maximal ideal of $B$ is then the zero ideal.

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