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Suppose 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 $R$.

An element $s\in S$ can be multiplied onto an element $m\in M$ by the way $M$ was defined.

What about a coset? Is the natural way to multiply $s$ by an element $m + N$ of $M/N$ just the obvious one?

That is, $s(m + N) = (sm) + N$?

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Forgetting multiplicative subsets, this is how you get an $R$-module structure on $M/N$. –  Dylan Moreland Jan 20 '12 at 2:21
Sorry, I was using all the hypothesis of a homework problem, $S$ was not necessary for my question. Thanks for clarifying. –  Kyle Schlitt Jan 20 '12 at 2:22
[To prove this: if $x, y \in M$ and $x - y \in N$, so they represent the same coset, then for $r \in R$ we have $rx - ry = r(x - y) \in N$, because $N$ is an $R$-module.] –  Dylan Moreland Jan 20 '12 at 2:26

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