# If the dual of a module is finitely generated and projective, can we claim that the module itself is?

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.
• 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? – Ender Wiggins Mar 25 at 16:04
• @EnderWiggins The dual is $0$. Don't confuse notations. – egreg Mar 25 at 16:17