Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $A_P/IA_P$?

share|improve this question

1 Answer 1

up vote 1 down vote accepted

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.

share|improve this answer
    
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 6:06
    
What are you having trouble with? –  Rankeya Nov 18 '12 at 6:13
    
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 6:27
    
If you write it up as another answer, I will accept it. –  Makoto Kato Nov 18 '12 at 6:30
    
I added it to the answer above. Glad to know it helped. –  Rankeya Nov 18 '12 at 6:34

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.