A question on valuation overrings of a PID

Question

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.

Answer 1

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.)

Answer 2

In this topic it's shown that every subring of $K$ containing $A$ is a ring of fractions of $A$. Now let $V$ be a valuation ring, $A\subset V\subset K$. Then there is $S\subset A$ a multiplicatively closed set such that $V=S^{-1}A$. Since $A$ is a PID we have that $S^{-1}A$ is also a PID, so $V$ is a DVR.

The prime ideals of $S^{-1}A$ are of the form $S^{-1}Q$ with $Q$ a prime ideal of $A$ such that $S\cap Q=\emptyset$. If $Q\ne 0$ we have $Q=(q)$ with $q\in A$ a prime element. Since $V=S^{-1}A$ has only one non-zero prime ideal there is only one prime element $p\in A$ such that $S\cap (p)=\emptyset$ and therefore the only prime (maximal) ideal of $S^{-1}A$ is $S^{-1}(p)$. Now use that the localization of a local ring at its maximal ideal is the ring itself and get that $V=S^{-1}A=(S^{-1}A)_{S^{-1}(p)}=A_{(p)}$.