I have a commutative ring with identity $R$, and an $R$-module $M$. Next I have an $R$-submodule $N$ of $M$. Finally, I have a multiplicatively closed subset $S$ of $R$.
I am asked to show that $S^{-1}(M/N)$ is isomorphic to $(S^{-1}M)/(S^{-1}N)$.
I guessed at the following mapping:
Define $\phi:S^{-1}(M/N)\to (S^{-1}M)/(S^{-1}N)$ by $\phi(\frac{m+N}{s}) = \frac{m}{s} + S^{-1}N$.
I'm stuck trying to show that my map is well defined. Let's assume that $\frac{m_1+N}{s_1} = \frac{m_2+N}{s_2}$. Then I want to show that $\frac{m_2}{s_2} + S^{-1}N = \frac{m_2}{s_2} + S^{-1}N$
By the definition of the equivalence of elements of $S^{-1}(M/N)$, there is an $s\in S$ such that
$s[s_2(m_1 + N) - s_1(m_2 + N)] = N$ (since $N$ is the $0$ element of $M/N$).
Now I get from this that
$s[\{(s_2m_1) + N\} - \{(s_1m_2) + N\}] = N$
and thus
$s[(s_2m_1 - s_1m_2) + N] = N$
Then
$s(s_2m_1 - s_1m_2) + N = N$
which implies
$s(s_2m_1 - s_1m_2) \in N$
or
$ss_2m_1 - ss_1m_2 \in N$
which is the definition of
$ss_2m_1 + N = ss_1m_2 + N$.
Now somehow from here I need to get to the conclusion that $\frac{m_1}{s_1} + S^{-1}N = \frac{m_2}{s_2} + S^{-1}N$. That is, $\frac{m_1}{s_1} - \frac{m_2}{s_2} \in S^{-1}N$.
I've been stuck for a while now, any advice? or have I gone wrong somewhere?