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 $S^{-1}M \cong S^{-1}N \Rightarrow M \cong N$.


  • 1
    $\begingroup$ This is a consequence of $S^{-1}A$ being flat but not faithfully flat. $\endgroup$ – ashpool Jun 7 '12 at 15:46

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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