Dual of module isomorphic to module itself?

Question

If $A$ is a finitely generated $B$-module, is it true that dual of $A$ is isomorphic to $A$,i.e.,$Hom(A,B)\cong A$?

I guess given $f \in Hom(A,B)$, if $f(a)=1$, then I can identify $f$ to $a$. Am I right?

Answer 1

This is false in general. Indeed, take the $\mathbb Z$ module $\mathbb Z/n\mathbb Z$ for $n \geq 2 $. While $|\mathbb Z/n \mathbb Z| = n$, one can show that there is exactly one group homomorphism (i.e. $\mathbb Z$-linear map) $\mathbb Z/n \mathbb Z \longrightarrow \mathbb Z$, which is the zero morphism. Thus, these modules are certainly not isomorphic.

Indeed, such a map is determined entirely by its image on 1 and exists if and only if 1 is sent to an element $a \in \mathbb Z$ such that $na = 0$, and the only such $a$ is $0$. This is because of the universal property of quotients of $R$ modules, which states that an $R$ linear map $M \longrightarrow M'$ which vanishes on a submodule $N \subseteq M$ factors uniquely through $M/N$.

The result is, however, true for free modules of finite rank by the same proof as the vector space case.