Does localisation (of modules) cancel.

Question

If I have two modules $R$ and $M$ and I take the localisation of both then do we have,

$$\frac{S^{-1}R}{S^{-1}M}\cong\frac{R}{M}$$

Answer 1

No in general since localization is an exact functor one has it commutes with quotients, so \begin{equation} \frac{S^{-1}(M)}{S^{-1}(N)}\cong S^{-1}\left(\frac{M}{N}\right) \end{equation} So an easy counter example is taking $N=0$ and $M=\mathbb{Z}/(2)$ over the ring $\mathbb{Z}$ them one has: \begin{equation} \frac{S^{-1}(M)}{S^{-1}(N)}\cong0, \frac{M}{N}\cong\mathbb{Z}/(2) \end{equation}