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Hi I have the following problem:

Let $R$ be a ring and $S\subset R$ multiplicatively closed and $M$ be an $R$-module.

Show: $M$ is flat $\Leftrightarrow S^{-1}M$ is a flat $S^{-1}R$-module.

I think I can show $"\Rightarrow$" but for $"\Leftarrow"$ I don't know how to start. Can someone help me please?

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Tobi92sr
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  • Don't you need more than this? Should it be $S^{-1}M$ is flat over $S^{-1}R$ for all $S$ or something like that? – TomGrubb Jun 06 '18 at 16:22
  • I don't think that is a dublicate, since in the other question we look at $M$ as an $R$-modul and then at $M$ as an $S^{-1}R$-modul. But in my question I want to show that $S^{-1}M$ is flat not $M$. – Tobi92sr Jun 06 '18 at 19:02

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The statement as you wrote it is not correct.
Consider $R=\Bbb Z$ and $S= \Bbb Z \setminus \{0\}$, then $S^{-1}R= \Bbb Q$, and clearly any $S^{-1}R$ module is flat in this case, as $\Bbb Q$ is a field. But not every $\Bbb Z$-module is flat.

Lukas Heger
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