# Needing a nudge with a Commutative Algebra Question

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?

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Do you know that $S^{-1}A \otimes_A M \cong S^{-1}M$? The proof using that is nice too. –  Dylan Moreland Jan 20 '12 at 4:13

Remember that $$\frac{m_1}{s_1}-\frac{m_2}{s_2} = \frac{s_2m_1-s_1m_2}{s_1s_2} = \frac{s(s_2m_1-s_1m_2)}{ss_1s_2}.$$ And you know something about $s(s_2m_1-s_1m_2)$, right?
You've clearly given me the hint I asked for. I'm embarassed to say that I don't quite see the point, though. But yes I know that the numerator is in $N$ at least. I'm a bit raw with this I will give it some more thought. Thanks very much. :) –  roo Jan 20 '12 at 4:19
Oh... since $S$ is multiplicatively closed, I have $ss_1s_2\in S$ and therefore the RHS is in $S^{-1}N$, right? –  roo Jan 20 '12 at 4:21
@borninthe80s: Since $\frac{m_1}{s_1}-\frac{m_2}{s_2}$ can be written in the form $\frac{n}{t}$ with $n\in N$ and $t\in S$, then by definition it lies in $S^{-1}N$, so yes, this shows that $\frac{m_1}{s_1}-\frac{m_2}{s_2}\in S^{-1}N$, as needed. –  Arturo Magidin Jan 20 '12 at 4:26