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?
Best Answer
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.