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Suppose $A$ is a PID, with quotient field $k$. Note here that I'm not assuming that $A$ itself is a valuation ring.

Is it true that $A_\mathfrak{p}$ is a valuation ring in $k$ for any prime $\mathfrak{p}\subset A$, and any valuation subring $B\supseteq A$ of $k$ is equal to $A_\mathfrak{p}$ for some prime ideal $\mathfrak{p}\subseteq A$?

My thoughts were something like this: If $B$ is a valuation ring containing $A$, the let $\mathfrak{m}$ be maximal in $B$. Put $\mathfrak{p}=\mathfrak{m}\cap A$. If $x\in A-\mathfrak{p}$, then $x\in B-\mathfrak{m}$, hence $x$ is a unit in $B$. Now $A_\mathfrak{p}\subseteq B$. As a valuation ring, $B$ is a local ring, and so any arbitrary element has form $x/y\in B$, for $x,y\in B$, and $y\notin\mathfrak{m}$. Since $k$ is the field of fractions, taking $x,y\in A$ with $x\notin\mathfrak{p}$ shows $x/y\in A_\mathfrak{p}$, so $A_\mathfrak{p}=B$. Is this argument correct?

On the other hand, if $\mathfrak{p}$ is a prime ideal of $A$, then I have a tower $A\subset A_\mathfrak{p}\subset k$. Is there a good way to conclude $A_\mathfrak{p}$ is a valuation ring in $k$?

I'm motivated to ask this question since I've seen similar ideas expressed when $A$ is a valution ring, but does this still work when $A$ is assumed to be a PID? If possible, I would appreciate seeing a correct formulation of the proof. Many thanks!

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I think your proof of $A_\mathfrak{p} = B$ is correct. Here's how you show that $A_{\mathfrak{p}}$ is a discrete valuation ring. Note that $A_{\mathfrak{p}}$ is $1$-dimensional. (Here I assume that $\mathfrak{p}$ is not the zero ideal.) Furthermore, it is clearly noetherian. (Take an ideal. Then it is clearly finitely generated.) Now, you use that a PID is integrally closed in its fraction field. Since the localization of a PID is again a PID you're done because you just showed that $A_{\mathfrak{p}}$ is a local PID. (You can replace PID by Dedekind domain.) –  Ali Jan 27 '12 at 22:02
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Let $A$ be a PID and let $B$ be any overring of $A$ (ie $A\subseteq B \subseteq k$ where $k$ is the quotient field of $A$).

If we have $\frac{x}{y}\in B$ with $gcd(x,y)=1$ then since $A$ is a PID, there exist $\alpha,\beta\in A$ such that $x\alpha+y\beta=1$. Thus $\frac{1}{y}=\beta+\alpha \frac{x}{y}\in B$ and thus $y\in U(B)$. From there, it's relatively easy to show that $B$ is, in fact, a localization of $A$ (just localize at the multiplicative subset of $A$ generated by a carefully selected set of prime elements).

So, since every overring of a PID is a localization and since the prime ideals of a valuation ring are linearly ordered by inclusion, the only (non-trivial) valuation overrings of $A$ are of the form $A_P$ for a prime ideal $P$ of $A$.

As for showing that $A_P$ is a valuation domain, well, letting $P=pA$ for some prime element $p$ of $A$ and choosing $\frac{x}{y}\in A_P$, we clearly have $p\nmid y$ (since $y\notin P$). Since $A$ is a UFD, we may write $x=p^n t$ for $n\geq 0$ and $t\in A$ with $p\nmid t$ (ie we yank all of the factors of $p$ out of $x$ that we can). Then $\frac{x}{y}=p^n \frac{t}{y}$ and since $t\notin P$, we have $u:=\frac{t}{y}\in U(A_P)$.

So, given any two nonzero nonunits $\alpha=p^n u$ and $\beta=p^m v$ of $A_P$ (for $n,m\geq 0$ and $u,v\in U(A_P)$) then (in $A_P$) either $\alpha \vert \beta$ or $\beta \vert \alpha$ depending on whether or not $n\leq m$ or $m\leq n$ (of course if $m=n$ then $\alpha$ and $\beta$ are associates in $A_P$). Therefore $A_P$ is a valuation domain.

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Thank you Jack. The one thing that I'm not following is the last paragraph on how $A_\mathfrak{p}$ is a valuation ring. Do you mind adding a bit more detail about why everything is so easy to see? –  Waldott Jan 28 '12 at 1:08
    
@Waldott - You're welcome. See the edited post above. In fact, the latter part of my answer can be generalized a bit: if $A$ is a domain with a prime element $p$ and if $P=pA$, then $A_P$ is a valuation domain (ie you always get a valuation domain if you localize at a principal prime). –  user5137 Jan 28 '12 at 1:13
    
I see now, thanks kindly for the edit! (By the way, I notice our proofs of the other claim are a little different, at least on the surface. Is mine correct as well? I'd like to know if I was at least on right on that point, since I haven't gotten any confirmation on it one way or another.) –  Waldott Jan 28 '12 at 1:18
    
@Waldott - Sorry, I didn't see your comment in my inbox... It looks like you're going down the right path. However, how do you know that $\mathcal{m}\cap A\not=0$? –  user5137 Feb 2 '12 at 1:31
    
No worries Jack. Since $B$ contains $A$, any nonunit of $A$ is contained in some maximal ideal of $B$, so I can take $\mathfrak{m}$ to be such an ideal? –  Waldott Feb 3 '12 at 5:09
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The result holds for any Dedekind domain $A$. Indeed, by Krull-Akizuki, any ring intermediate between a Dedekind domain and its fraction field is again a Dedekind domain, so $B$ is a local Dedekind domain and thus the valuation ring of a discrete valuation $v$ on $K$ with $A \subset B = R_v$. Now apply Theorem 13b) of these notes to conclude that $v = v_{\mathfrak{p}}$ for a unique nonzero prime ideal $\mathfrak{p}$ of $A$.

(This is not the first question on this site I've answered by referencing "Theorem 13". Although it is a simple result that most or all experts in the field must know, it seems to be missing from most expositions of algebraic number theory / valuation theory.)

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Dear Pete, thanks for your response! –  Waldott Feb 3 '12 at 5:10
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