Localization of a Dedekind Domain I have a question about this article here:
In the proof of (ii) of corollary 5.3 it says that $R/P^r\cong R_P/R_P P^r$ where $R_P$ is the localization of a Dedekind Domain R at the prime ideal P. Can someone explain this to me?
Thanks.
Edit: And I also don't really understand why the quotient of a DVR is a PID. I know that DVRs are PIDs and quotients of PIDs over prime ideals are PIDs again but $R_P P^r$ isn't a prime ideal, is it?

Let $\phi: R \rightarrow R_P /R_P P^r, \ x \mapsto x + R_P P^r$ and $\mu: R \rightarrow R_P, \ x \mapsto \frac{x}{1}$. Then we have
$$ ker(\phi) = \mu^{-1}(R_P P^r) = P^r. $$
By the first isomorphism theorem for rings you get $R/P^r \cong R_P /R_P P^r$ as rings.
Edit: 
Let $I\subseteq R_P/R_P P^r$ be an ideal and $\psi: R_P \rightarrow R_P / R_P P^r, \ x \mapsto x + R_P P^r $. As $\psi$ is surjective we have
$$ \psi \circ \psi^{-1} (I) = I.$$
Furthermore, the ideal $\psi^{-1}(I)$ is principal (as $R_P$ is a PID) and thus $I= \psi \circ \psi^{-1}(I)$ is principal as well. What does not hold is, that $R/P^r$ is a PID for $r>1$ (it is not integral, as $(p^{r-1}+ P^r)\cdot (p + P^r)= 0 + P^r$ ) and thus neither is $R_P/R_P P^r$.
