Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
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

2 Answers 2

up vote 3 down vote accepted

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.

share|improve this answer
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

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.)

share|improve this answer
Dear Pete, thanks for your response! –  Waldott Feb 3 '12 at 5:10

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.