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Thread: Quasi-local domain is a PID if its localizations at maximal ideals are PIDS

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    MHF Contributor Bruno J.'s Avatar
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    Quasi-local domain is a PID if its localizations at maximal ideals are PIDS

    Let $\displaystyle A$ be a quasi-local domain (i.e. a commutative domain with finitely many maximal ideals). Suppose that for every maximal ideal $\displaystyle M$ of $\displaystyle A$, the localization $\displaystyle A_M$ is a P.I.D. Show that $\displaystyle A$ is a P.I.D.
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    I think most texts define quasi-local to mean that there's only one maximal ideal. Do you mean semi-quasilocal? What's the reference for this problem?
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    MHF Contributor Bruno J.'s Avatar
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    Thank you for your interest. A local ring is a ring having a single maximal ideal. A quasi-local ring is a ring having a finite number of maximal ideals.

    The reference is Lorenzini's "An introduction to arithmetic geometry", AMS.
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    Specifically, where in Lorenzini's book is this problem?
    Last edited by ojones; Jun 2nd 2011 at 06:56 PM.
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    MHF Contributor Bruno J.'s Avatar
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    Chapter III, problem 3 on page 126.
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    I'm still thinking about this but here's my approach. First consider only local domains for which there's only one maximal ideal. Let $\displaystyle I\subset A$ be an ideal. Then the expansion $\displaystyle I^E$ of $\displaystyle I$ to $\displaystyle A_M$ is principal, by assumption. Let $\displaystyle a/s$ be its generator, where $\displaystyle a\in I$ and $\displaystyle s\in S=A\setminus M$. Now $\displaystyle a$ is a candidate for the generator of $\displaystyle I$ but now to prove it. Obviously the maximality of $\displaystyle M$ is critical.
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    OK, I think I have a proof. It may not be the most direct, however. We know that any local domain who's maximal ideal is principal must be a discrete valuation ring. Now, let $X$ be the maximal spectrum of $A$. Then,

    $\displaystyle A=\cap_{M\in X}A_M.$

    Since each $\displaystyle A_M$ is a DVR and the intersection is finite, we have that $\displaystyle A$ is a PID.
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    Quote Originally Posted by ojones View Post
    OK, I think I have a proof. It may not be the most direct, however. We know that any local domain who's maximal ideal is principal must be a discrete valuation ring. Now, let $X$ be the maximal spectrum of $A$. Then,

    $\displaystyle A=\cap_{M\in X}A_M.$

    Since each $\displaystyle A_M$ is a DVR and the intersection is finite, we have that $\displaystyle A$ is a PID.
    Why is the statement $\displaystyle A=\cap_{M\in X}A_M.$ true?
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    This is proposition 9.7 (page 74) in Lorenzini's book.
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