Prove that $\operatorname{Hom}_\mathbb Z(K, T') = 0$ for $K = \mathbb Q / \mathbb Z$ and $T'$ reduced torsion.

Question

How does one prove that $\operatorname{Hom}_\mathbb Z(K, T') = 0,$ where $K = \mathbb Q / \mathbb Z$ and $T'$ is a torsion whose maximal divisible subgroup is trivial ? Thank you for your help in advance!

Answer 1

One can say more generally that $\operatorname{Hom}_\mathbb Z(D, T) = 0$ whenever $D$ is a divisible $\mathbb Z$-module and $T$ is a reduced torsion $\mathbb Z$-module. Like @AnginaSeng points out, the homomorphic image of a divisible $\mathbb Z$-module is divisible*. Considering that $\varphi$ is in $\operatorname{Hom}_\mathbb Z(D, T)$ only if $\varphi(D) \subseteq \tau(T)$**, where $\tau(T) = T$ is the torsion $\mathbb Z$-submodule of $T,$ we conclude that $\varphi(D) = 0$ by hypothesis that $T$ is reduced.

Proof. (*) We must show that $n \varphi(D) = \varphi(D)$ for all nonzero integers $n.$ Given any element $\varphi(d)$ in $\varphi(D),$ by hypothesis that $D$ is divisible, we have that $d = nd'$ for some element $d'$ of $D.$ Consequently, we have that $\varphi(d) = \varphi(nd') = n \varphi(d').$ But this holds for all integers $n$ and all elements $\varphi(d)$ of $\varphi(D).$ QED.

Proof. (**) We claim more generally that for any $\mathbb Z$-modules $M$ and $M'$ such that $M$ is torsion, $\varphi$ is in $\operatorname{Hom}_\mathbb Z(M, M')$ only if $\varphi(M) \subseteq \tau(M').$ Observe that if $\varphi$ is in $\operatorname{Hom}_\mathbb Z(M, M'),$ then $\varphi(nm) = n \varphi(m)$ for all integers $n$ and all elements $m$ of $M.$ Consequently, if $\varphi$ is in $\operatorname{Hom}_\mathbb Z(M, M'),$ then we have that $n \varphi(m) = \varphi(nm) = \varphi(0) = 0$ whenever $nm = 0,$ hence we have that $\varphi(M) \subseteq \tau(M').$ QED.