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Let $R$ be a commutative ring, $M$ an $R$-module, $M^*=Hom_R(M,R)$ its dual. What are sufficient (and possibly necessary) conditions on $M$ that ensure that $M^*$ is flat? Is there a name for such modules?

PS I would call such a module coflat if this term were not already used for something else.

PPS As $M^\ast$ is clearly torsion-free, I already know, thanks to this beautiful website, some conditions on $R$ that make all $M^*$ flat. I also know about reflexive modules.

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1 Answer 1

Bugs, I'm going to restrict myself to the case when $M$ is finitely generated and $R$ is Noetherian (I wonder if you need this though, I don't see any non-finitely generated examples which clearly fail my test off the top of my head).
Based on what you said, you probably already know this though.

Proposition: $M$ has flat dual if and only if the reflexive hull, $M^{\ast \ast}$, is flat.

Proof: Since $M$ and thus $M^\ast$ is finitely generated, $M^{\ast}$ is flat if and only if it is locally free. But $M^\ast$ is reflexive, so it is locally free if and only if $M^{\ast \ast}$ is.

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Thanks, but $M^*$ is not finitely generated, in general. You need $R$ to be Noetherian to conclude this... Besides, finite generation of $M$ is a bad assumption for me. –  Bugs Bunny May 4 '11 at 14:21
    
Ah, yes, I was implicitly assuming that $R$ was Noetherian, sorry about that. I'll fix my answer in that regard. –  Karl Schwede May 4 '11 at 14:23
    
Bugs, do you care primarily about the Noetherian case or non-Noetherian case. –  Karl Schwede May 4 '11 at 14:23
    
A criterion for a Noetherian ring would be useful, indeed. Finite generation, on the other hand, is too strong. –  Bugs Bunny May 4 '11 at 14:26

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