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Suppose $A$ noetherian and define

$G(A):=\{M: M$ is an $A$-module reflexive and Ext$^i_A(M,A)=$Ext$^i_A(M^*,A)=0$ for $i\geq1\}$

Why are projective modules contained in this class? Of course if $M$ is projective, then $\mathrm{Ext}^i_A(M,A)=0$. I don't understand why the other conditions hold. Could you tell me why?

(It's not clear to me that we need to suppose $M$ finitely generated.)

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1 Answer 1

up vote 4 down vote accepted

Not all projectives are in that class, for not all projectives are reflexive; for an example, take $A$ to be a field and let $M$ be an infinite dimensional vector space.

Finitely generated ones are, though, and then one can check easily that their duals are also projective.

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Could you show me a proof of why finitely generated projective modules are reflexive? and why their dual is projective? –  Chris Sep 30 '12 at 4:13
    
I suggest you try to prove that yourself :-) You could start by proving that a finite direct sum of reflexives is reflexive, and that a direct summand of a reflexive module is reflexive. –  Mariano Suárez-Alvarez Sep 30 '12 at 4:13
    
oh right thank you –  Chris Sep 30 '12 at 4:51

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