2

1

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

flag

2 Answers

14

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

link|flag
2 
Well done! So is $k((x))$ over $k[[x]]$... – Bugs Bunny Jul 21 2010 at 21:39
1 
So is the $\mathfrak{m}$-adic completion of any non-Artinian local ring. – Graham Leuschke Jul 21 2010 at 22:25
3

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.

link|flag
2 
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 2010 at 21:54
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 2010 at 3:26

Your Answer

Get an OpenID
or

Not the answer you're looking for? Browse other questions tagged or ask your own question.