# $S^{-1}M \cong S^{-1}N$ does not imply $M \cong N$

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.

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

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