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Could you give me an example of a finitely generated module that is flat over a local ring but not projective?

For a non finitely generated I took $\mathbb{Q}$ over $\mathbb{Z}_p$, but I cannot find an example of a finitely generated one. Of course it should be a module over a local non-noetherian ring. I don't know a lot of local non-noetherian rings, the first that came to my mind was $k[x_1,x_2,\ldots]/(x_1,x_2^2,x_3^3,\ldots)$, since localization is a good way to find flat modules I wanted to localize this ring, but it has only one prime ideal (the maximal ideal); so I really don't know what to do. Any help?

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  • $\begingroup$ Power series rings in infinitely many indeterminates might be a better place to look for primes. $\endgroup$ Oct 26, 2012 at 4:29
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    $\begingroup$ mathoverflow.net/questions/32847/… $\endgroup$
    – user26857
    Oct 26, 2012 at 12:37
  • $\begingroup$ See Manny Reyes' comment to the un-accepted answer in navigetor's link! That's the best way to think of it... $\endgroup$
    – rschwieb
    Oct 26, 2012 at 19:22

1 Answer 1

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Over a (commutative) local ring (non necessarily noetherian), any finitely generated flat module is free (Matsumura, Commutative Algebra, Prop. 3.G, p. 21), hence projective.

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  • $\begingroup$ I cannot find this theorem, you said page 21 proposition 3, right? $\endgroup$
    – Chris
    Oct 26, 2012 at 16:28
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    $\begingroup$ in that proposition $M$ is supposed to be finitely presented $\endgroup$
    – Chris
    Oct 26, 2012 at 18:31
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    $\begingroup$ @Chris: yes it is page 21, Proposition 3.G (not 3). You are probably reading "Commutative ring theory". The same result is there, page 51, Theorem 7.10. The module is just of finite type. $\endgroup$
    – user18119
    Oct 26, 2012 at 20:24

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