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 $M, N$ be $A$-modules, where $A$ is a commutative ring with identity. Let $S$ be a multiplicative subset of $A$ that contains no zero divisors and contains the identity of $A$. I am looking for a counterexample to the statement $S^{-1}M \cong S^{-1}N \Rightarrow M \cong N$.

Thanks.

share|improve this question
1  
This is a consequence of $S^{-1}A$ being flat but not faithfully flat. –  ashpool Jun 7 '12 at 15:46

1 Answer 1

up vote 9 down vote accepted

$A=M=\mathbb Z$, $N=\mathbb Q$ and $S=\mathbb Z-\{0\}$. Then $$S^{-1}M=S^{-1}N=\mathbb Q$$

share|improve this answer

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.