If $M$ is a finitely generated right module over $R$, then can we conclude that the canonical map $f: M^{*} \to M^{***}$ is bijective?

Question

Let $R$ be a ring with unity. If $M$ is a right module over $R$, then we can conclude that $M^{*}=\text{Hom}_R(M, R)$ is a left torsionless module over $R$, i.e., the canonical map $f: M^* \to M^{***}$ is injective, where $*$ is the functor $\text{Hom}_R(-, R)$.

Now the question arises: If $M$ is a finitely generated right module over $R$, then can we conclude that the canonical map $f: M^{*} \to M^{***}$ is bijective, where $*$ is the functor $\text{Hom}_R(-, R)$?

Yes, it has answered my question. Thank you very much. – Liang Chen Jun 07 '24 at 10:53 — Liang Chen, Jun 07 '24 at 10:53

score 1 · Answer 1 · answered Jun 06 '24 at 17:09

We can if $R$ is integral domain. See for a proof Proposition 5.4.11 of Differential geometry of complex vector bundles by Kobayashi.

I don't know about the general case (and in the integral case, I guess you can find proofs in books more centered on commutative algebra, I don't know, I am a geometer).

If $M$ is a finitely generated right module over $R$, then can we conclude that the canonical map $f: M^{*} \to M^{***}$ is bijective?

1 Answers1