Is every $S^{-1}R$-module flat?

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Here is this week's POTW:

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Let $R$ be a commutative ring with unity. Show that if $S$ is multiplicatively closed in $R$ and if every $R$-module is flat, then every $S^{-1}R$-module is flat.

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No one answered this week's problem. You can read my solution below.

Spoiler

Let $0 \to M' \to M \to M'' \to 0$ be a short exact sequence of $S^{-1}R$-modules. Given an $S^{-1}R$-module $X$, let $_RX$ be $X$ viewed as an $R$-module by restriction of scalars via the map $R \to S^{-1}R$, $r \to r/1$. There is an $S^{-1}R$-isomorphism $S^{-1}{}_RX \approx X$. Now, given an $S^{-1}R$-module $N$, $_RN$ is by hypothesis a flat $R$-module. Tensoring the exact sequence $0 \to {}_RM' \to {}_RM \to _RM'' \to 0$ with $_RN$ yields the exact sequence $0 \to {}_RM' \otimes_R {}_RN \to _RM \to {}_RM'' \otimes_R {}_RN \to 0$. Exactness of $S^{-1}$ produces the exact sequence $0 \to S^{-1}(_RM' \otimes_R {}_RN) \to S^{-1}({}_RM \otimes_R {}_RN) \to S^{-1}({}_RM'' \otimes_R {}_RN) \to 0$, which in turn yields the sequence $$0 \to S^{-1}{}_RM' \otimes_{S^{-1}R} S^{-1}{}_RN \to S^{-1}{}_RM \otimes_{S^{-1}R} S^{-1}{}_RN \to S^{-1}{}_RM'' \otimes_{S^{-1}R} S^{-1}{}_RN \to 0.$$ Thus, we obtain the exact sequence $$0 \to M' \otimes_{S^{-1}R} N \to M \otimes_{S^{-1}R} N \to M'' \otimes_{S^{-1}R} N \to 0.$$ Consequently, $N$ is a flat $S^{-1}R$-module.