Lemma 10.9.15. Any submodule $N'$ of $S^{-1}M$ is of the form $S^{-1}N$ for some $N\subset M$. Indeed one can take $N$ to be the inverse image of $N'$ in $M$.
Proof. Let $N$ be the inverse image of $N'$ in $M$. Then one can see that $S^{-1}N\supset N'$. To show they are equal, take $x/s$ in $S^{-1}N$, where $s\in S$ and $x\in N$. This yields that $x/1\in N'$. Since $N'$ is an $S^{-1}R$-submodule we have $x/s = x/1\cdot 1/s\in N'$. This finishes the proof. $\square$
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