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Is there a non-projective flat module over a local ring? Here I assume the ring is commutative with unit.

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up vote 14 down vote accepted

$\mathbb{Q}$ is flat over $\mathbb{Z}_p$, but not projective.

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Well done! So is $k((x))$ over $k[[x]]$... –  Bugs Bunny Jul 21 '10 at 21:39
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So is the $\mathfrak{m}$-adic completion of any non-Artinian local ring. –  Graham Leuschke Jul 21 '10 at 22:25
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It is related to Bass' theorem. Flat modules are projective iff the ring is perfect. $p$-adic integers or formal power series are examples of local rings which are not perfect and have nonprojective flat modules.

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But we can also make a slightly positive statement. Every local ring is semiperfect, and every finitely generated flat module over a semiperfect ring is projective. Moreover, since projective modules over a local ring are free, we see that every finitely generated flat module over a local ring is free. –  Manny Reyes Jul 21 '10 at 21:54
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For those, who -- like me -- didn't know what a perfect ring was, see: en.wikipedia.org/wiki/Perfect_ring. –  Pete L. Clark Jul 22 '10 at 3:26
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