Hom of the direct product of $\mathbb{Z}_{n}$ to the rationals is nonzero. Why is $\mathrm{Hom}_{\mathbb{Z}}\left(\prod_{n \geq 2}\mathbb{Z}_{n},\mathbb{Q}\right)$ nonzero?
Context: This is problem $2.25 (iii)$ of page $69$ Rotman's Introduction to Homological Algebra:

Prove that
  $$\mathrm{Hom}_{\mathbb{Z}}\left(\prod_{n \geq 2}\mathbb{Z}_{n},\mathbb{Q}\right) \ncong \prod_{n \geq 2}\mathrm{Hom}_{\mathbb{Z}}(\mathbb{Z}_{n},\mathbb{Q}).$$

The right hand side is $0$ because $\mathbb{Z}_{n}$ is torsion and $\mathbb{Q}$ is not.

Let $G=\prod_{n\geq2}\mathbb Z_n$ and let $t(G)$ be the torsion subgroup, which is properly contained in $G$ (the element $(1,1,1,\dots)$ is not in $t(G)$, for example) Then $G/t(G)$ is a torsion-free abelian group, which therefore embeds into its localization $(G/t(G))\otimes_{\mathbb Z}\mathbb Q$, which is a non-zero rational vector space, and in fact generates it as a vector space. There is a non-zero $\mathbb Q$-linear map $(G/t(G))\otimes_{\mathbb Z}\mathbb Q\to\mathbb Q$  (here the Choice Police will observe that we are using the axiom of choice, of course...). Composing, we get a non-zero morphism $$G\to G/t(G)\to (G/t(G))\otimes_{\mathbb Z}\mathbb Q\to\mathbb Q.$$
Remark. If $H$ is a torsion-free abelian group, its finitely generated subgroups are free, so flat. Since $H$ is the colimit of its finitely generated subgroups, it is itself flat, and tensoring the exact sequence $0\to\mathbb Z\to\mathbb Q$ with $H$ gives an exact sequence $0\to H\otimes_{\mathbb Z}\mathbb Z=H\to H\otimes_{\mathbb Z}\mathbb Q$. Doing this for $H=G/t(G)$ shows $G/t(G)$ embeds in $(G/t(G))\otimes_{\mathbb Z}\mathbb Q$, as claimed above.
