My professor said in his lecture, "For Abelian groups $A$, $B$, $C$ and field $F$ $\operatorname{Hom}(A,B)=\operatorname{Hom}_\mathbb{Z} (A, B)$ but $\operatorname{Hom}(C, F) \neq \operatorname{Hom}_F (C\otimes_\mathbb{Z} F, F)$ in general, the equality holds when $C$ is free." Where $\operatorname{Hom}(,)$ contains all of the group homomorphism and $\operatorname{Hom}_F(,)$ contains all F-linear maps.
I can easily see that the former is right since Abelian groups are naturally $\mathbb{Z}$-modules. I'm new to the tensor product, so for the latter, I can't understand what's going on. Why is $C\otimes F$ an $F$-module and why does the equality hold when $C$ is free? Can you also give me an example when $C$ is not free and the equality doesn't hold?
$\endgroup$ 11 Answer
$\begingroup$- The $F$-module multiplication on $C\otimes_{\mathbb{Z}} F$ is given by $a (e\otimes b)=e\otimes (ab)$ for $a,b \in F$ and $e\in C$.
- When $C$ is free, if $\{e_i\}_{i\in I}$ is a basis for $C$ as a $\mathbb{Z}$-module, then $\{e_i \otimes 1_F\}_{i\in I}$ is a basis for $C\otimes_{\mathbb{Z}} F$ as an $F$-module. Note that $F$ is a field, so an $F$-module is always free, but when $C$ is free, there is a basis of this special form.
- Tensoring $C$ with a field $F$ of characteristic zero kills off all torsion elements in $C$.
Edit:
According to wikipedia $\operatorname{Hom}_{\mathbb{Z}}(C, F)$ is naturally isomorphic to $\operatorname{Hom}_F (C\otimes_\mathbb{Z} F, F)$ as $F$-modules for any $\mathbb{Z}$-module $C$ (read the last section "Relation between the extension of scalars and the restriction of scalars".)
