How do I prove $\operatorname{Hom}_R(A, B)$ torsion-free and divisible under the given conditions, where $A$ and $B$ are abelian groups.

Question

Let $A,B$ be abelian groups. I want to show that

(a) the group $\operatorname{Hom}_R(A,B)$ is torsion-free when $A$ is divisible,

and

(b) the group $\operatorname{Hom}_R(A,B)$ is divisible when $A$ is torsion-free and divisible.

Answer 1

I assume that $R$ should be $\mathbb{Z}$?

(a) If $\text{Hom}_\mathbb{Z}(A,B)$ is not torsion-free, then for some nonzero homomorphism $\varphi:A\to B$ there is a positive integer $n$ so that $n\varphi=0$. But for any $a\in A$, there is some $a'\in A$ with $a=na'$. So $\varphi(a)=n\varphi(a')=0$. Therefore $\varphi=0$, giving a contradiction.

(b) If also $A$ is torsion-free, then for any positive integer $n$ and any $a\in A$ there is a unique $a'\in A$ with $a=na'$ and $\theta:a\mapsto a'$ is a homomorphism $A\to A$. So for any homomorphism $\varphi:A\to B$, $\varphi=n\varphi\theta$, and so $\text{Hom}_\mathbb{Z}(A,B)$ is divisible.