3
$\begingroup$

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.

  • 1
    This is a consequence of $S^{-1}A$ being flat but not faithfully flat.2012-06-07

1 Answers 1

10

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