This is from Liu 1.2.9.

Let $A$ be an integral domain, and $K$ its field of fractions. Let $M$ be a finitely generated sub-$A$-module of $K$. Why do $M$ is flat if and only if $M_{\mathfrak p}$ is free of rank $1$ over $A_\mathfrak p$ for every prime ideal $\mathfrak p$ of $A$?


For a finitely generated module over a local ring being flat is the same as being free. Can the fraction field of a domain have two elements linearly independent over the domain? Try this for $\mathbf{Z}$, for example.

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    $\begingroup$ "The finitely generated flat modules over a local ring are exactly the free modules of finite rank". $\endgroup$ – user26857 Jan 22 '15 at 20:46
  • $\begingroup$ @user26857 I thought I had written that, whoops. Too much editing. Thanks for the correction. $\endgroup$ – Hoot Jan 22 '15 at 21:44
  • $\begingroup$ So there is a mistake in Liu? I understand the Theorem 2.16 on page 11 that a finitely generated flat module over a local ring is free over that ring and $\operatorname{Frac}(\mathbb Z_{p\mathbb Z})$ is free over $\mathbb Z_{p\mathbb Z}$ but not finitely generated. $\endgroup$ – Jaakko Seppälä Jan 22 '15 at 22:13
  • $\begingroup$ @user2219869 I made a small correction about an hour ago thanks to user26857 but even given that I don't understand your question. I don't think $\mathbf Q$ is free over $\mathbf Z_{(p)}$. Why would that be? $\endgroup$ – Hoot Jan 22 '15 at 23:03
  • $\begingroup$ @Hoot It is on page 11. I understood the text before 2.16 that the converse of 2.16 is not true in general. By taking those structures one can find a module that is free over a local ring but not finitely generated and flat. I haven't seen a proof that if M is free over a local ring A, then M is flat over A. Maybe the converse means that M is flat but not finitely generated. $\endgroup$ – Jaakko Seppälä Jan 22 '15 at 23:39

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