Assume that $R$ is a commutative ring and that $M$ is a (left) $R$-module. Assume also that we know for some reason that $M^*:=\mathsf{Hom}_R(M,R)$ is finitely generated and projective as (right) $R$-module. Can we claim that $M$ itself is finitely generated and projective?

It is well-known that the converse is true, but I am able neither to prove nor to disprove the foregoing implication.

Of course, $M^{**}$ is finitely generated and projective, but in general the canonical morphism $j:M\to M^{**}$ is not injective, whence I don't know how to use this fact. Can somebody give me a hint, either in proving the statement or in finding a counterexample?


If $\operatorname{Hom}_R(M,R)=M^*$ is finitely generated and projective, then $R^n\cong M^*\oplus N$, so we have $$ M^{**}\oplus N^*\cong R^n $$ so $M^{**}$ is finitely generated and projective. Unfortunately, the canonical homomorphism $M\to M^{**}$ is neither injective nor surjective, in general.

A trivial example is $M=\mathbb{Q}$, with $R=\mathbb{Z}$. You can complicate the situation at will.

Just to give the flavor, suppose $R$ is a PID and that $M$ is finitely generated. Then $M^*$ is finitely generated and free: you lose all information about the torsion part, when doing the dual.

  • $\begingroup$ Thanks egreg but I am afraid I didn't get your point. Why $\mathbb{Q}^*$ should be finitely generated and projective as $\mathbb{Z}$-module? If I am not mistaken, an element in $\mathbb{Q}^*$ is uniquely determined by its images over the rationals of the form $\frac{1}{p}$ for $p$ prime, isn't it? $\endgroup$ – Ender Wiggins Mar 25 at 16:04
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    $\begingroup$ @EnderWiggins The dual is $0$. Don't confuse notations. $\endgroup$ – egreg Mar 25 at 16:17
  • $\begingroup$ Oh, you're right, my bad. I realized it just know. Thanks. $\endgroup$ – Ender Wiggins Mar 25 at 16:23

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