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 $A_P/IA_P$?
Asked
Active
Viewed 1,766 times
2 Answers
8
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.
Rankeya
- 8,826
-
I would like to know rather a direct proof. How do you prove that the canonical morphism $\bar A_{\bar P} \rightarrow A_P/IA_P$ is an isomorphism? – Makoto Kato Nov 18 '12 at 06:06
-
What are you having trouble with? – Rankeya Nov 18 '12 at 06:13
-
The canonical map 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, $ta + I = 0$ which implies that $a+I/1+I = 0$, and from this you show that $a+I/s+I = 0$. – Rankeya Nov 18 '12 at 06:27
-
1I added it to the answer above. Glad to know it helped. – Rankeya Nov 18 '12 at 06:34
0
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.
Ken
- 2,544