Let $A$ be a PID and let $K$ be its quotient field. Let $V$ be a valuation ring of $K$ containing $A$ and assume $V\neq K$. Show that $V$ is a local ring $A_{(p)}$ for some prime element $p$.

I know that a valuation ring is always a local ring. So $V$ has a unique maximal ideal, say $\mathcal M $. Again $A$ is a PID, so a maximal ideal is of the form $\langle p\rangle$ for some prime element $p$.

After this how should I proceed?

Thanks in advance.


If $R$ is an integral domain with fraction field $K$, then an overring of $R$ is a ring $T$ with $R \subset T \subset K$.

Easy examples of overrings are given by localization at a multiplicatively closed subset $S$ of $R \setminus \{0\}$. For a general integral domain there are overrings which are not obtained by localization: this comes down to the fact that in general adjoining sets of elements $\frac{x}{y}$ cannot be reduced to adjoining sets of elements $\frac{1}{y}$.

However, for a PID it is easy to show that every overring is actually a localization. Hint: given $\frac{x}{y} \neq 0$, we may find $x'$ and $y'$ such that $\frac{x}{y} = \frac{x'}{y'}$ and $x'$ and $y'$ generate the unit ideal.

Then you have to show that every local localization of a PID which is not a field is obtained by localizing via $S = R \setminus (p)$ for some prime element $p$.

If you have difficulty establishing either of these facts, let us know.

Added: I apologize for not yet addressing the work you have done. With respect to that: I think you mean to claim that $\alpha$ is a PID. How do you know that? (It's true -- that's part of what you're trying to show -- but it seems to require proof.)

  • $\begingroup$ From this link in the 2nd answer we are able to see that T is a PID, math.stackexchange.com/questions/137876/… $\endgroup$
    – Germain
    Feb 18 '14 at 20:49
  • $\begingroup$ @analysis89: agreed that that establishes it, just as I was suggesting above. This brings you to the second half of what I asked you to show: a local localization of a PID is obtained by localizing at the complement of a prime element. How are you doing with that? E.g. do you know how to determine the prime ideals in a localization $S^{-1} R$ given the prime ideals of $R$ and $S$? Or just think about what happens to a factorization into prime elements upon passage to the localization. $\endgroup$ Feb 19 '14 at 5:36
  • $\begingroup$ Ok now I got that every overring is actually a localisation. But not getting an enough clear idea about the second one. Please help. $\endgroup$
    – Germain
    Feb 21 '14 at 19:40
  • $\begingroup$ Try classifying all of the saturated multiplicatively closed subsets of a PID. The answer will come out in terms of prime elements. The prime elements that generate your subset $S$ are exactly the ones that become units in the localization. So if you want only one prime element (up to units in your localization), what does that say about $S$? $\endgroup$ Feb 22 '14 at 23:52

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