A question on valuation overrings of a PID 
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.
 A: 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.)
