Isomorphism between quotient ring and its localization Let $R$ be a domain, $P$ a prime ideal of $R$, and $k$ an positive integer. I am wondering if we have the isomorphism:
$$
R/P^k\cong R_P/(PR_P)^k
$$
where $R_P$ is the localization of $R$ at $P$.
If it does not hold in general, at least I think it hold if $R$ is a Dedekind domain.
I can show the map:
$$
f:R/P^k\rightarrow R_P/(PR_P)^k
$$
given by
$$
r+P^k\mapsto r+(PR_P)^k
$$
is injective. But fail to show it is also surjective.
 A: 
Let $A$ be a commutative ring, $\newcommand\ideal[1]{\mathfrak{#1}}$$\ideal p$ be a prime ideal in $A$, $\ideal a$ be an ideal in $A$.
  If $A_{\ideal p}$ is the localization of $A$ at $\ideal p$, then $A_{\ideal p}/\ideal aA_{\ideal p}$ is the localization of the ring $A/\ideal a$ at $\ideal p$.
  If, moreover, $\ideal p=\sqrt{\ideal a}$ is a maximal ideal of $A$, then $A/\ideal a\cong A_{\ideal p}/\ideal aA_{\ideal p}$.

First note that there exists one and only one ring homomorphism $\xi$ making commutative the following diagram:

Consider the following commutative diagram of $A$-module homomorphisms with exact rows.

Then there exists an ring isomorphism which makes commutative the bottom triangle.
Since $\pi$ is an ring epimorphism, the right-handed triangle is commutative as well and this prove that $A_{\ideal p}/\ideal aA_{\ideal p}$ is the localization of the ring $A/\ideal a$ at $\ideal p$.
If $\ideal p=\sqrt{\ideal a}$ is a maximal ideal, then $\pi^{-1}(A/\ideal a)^\times=A-\ideal p$, hence $\pi$ factors through the localization $A\to A_{\ideal p}$ giving rise to a ring homomorphism $A_{\ideal p}\to A/\ideal a$ making the commutative upper triangle in the following diagram.

Since $A\to A_{\ideal p}$ is a ring epimorphism, the triangle on the bottom commutes as well thus making $\xi$ an $A_{\ideal p}$-algebra homomorphism.
Since $\xi$ is a localization at $\ideal p$ it must to be a ring isomorphism.
