Quotient ring of a localization of a ring Let $A$ be a commutative ring.
Let $P$ be a prime ideal of $A$.
Let $I$ be an ideal of $A$ such that $I \subset P$.
Let $\bar A = A/I$.
Let $\bar P = P/I$.
Is $\bar A_{\bar P}$ isomorphic to $A_P/IA_P$?
 A: Yes. This is because $A_p/IA_p \cong (A-p)^{-1}(A/I)$, and the latter ring is isomorphic to $(A/I)_{p/I}$ because the image of $A-p$ under the projection $A \rightarrow A/I$ is $A/I - p/I$.
One way to observe the first isomorphism is to note that $A_p/IA_p \cong A/I \otimes_A A_p \cong (A-p)^{-1}(A/I)$ (The latter isomorphism follows from the fact that given an $A$-module $M$, $S^{-1}A \otimes_A M \cong S^{-1}M$). 
Later edit at OP's request: (To show the canonical map is an isomorphism)
The canonical map $\overline{A}_{\overline{p}} \rightarrow A_p/IA_p$ is clearly surjective. Let $\frac{a+I}{s+I}$ be mapped to $0$. Then $\frac{a}{s} \in IA_p$. Then $ta \in I$ for some $t \in A-p$. So, $ta + I = (t+I)(a+I) = 0$, and from this you show that $\frac{a+I}{s+I} = 0$, proving injectivity.
A: Yes. Check that the canonical maps $A\to A_P/IA_P$ and $A\to(A/I)_{P/I}$ has the following property: any map $A\to B$ of commutative rings such that

*

*$I$ lies in the kernel, and

*each $x\in A\setminus P$ is mapped to a unit

uniquely factors through both of these maps.
