# Compute Hom(q,q/z)

• July 1st 2011, 02:49 AM
seventhson
Compute Hom(q,q/z)
$Hom (\mathbb{Z} , \mathbb{Q} / \mathbb{Z} )$ is trivially isomorphic to $\mathbb{Q} / \mathbb{Z}$ since any morphism is determined by where it sends $1$. To prove that $\mathbb{Q}$ is injective take any morphism $f: n\mathbb{Z} \rightarrow \mathbb{Q}$ with $f(n)=a$ then, since $\mathbb{Q}$ is divisible there exists $b\in \mathbb{Q}$ with $nb=a$ take $g: \mathbb{Z} \rightarrow \mathbb{Q}$ with $g(1)=b$ then $gi=f$ where $i :n\mathbb{Z} \rightarrow \mathbb{Z}$ is the inclusion, conclude by Baer's criterion. I don't have the answer to the second but maybe the fact that $\mathbb{Q} / \mathbb{Z} \cong \oplus_{p \ prime } \mathbb{Z}_{p^\infty}$ is of use.