I have seen a few similar questions to this one, namely this one: Prove that $\operatorname{Hom}_{\Bbb{Z}}(\Bbb{Q},\Bbb{Z}) = 0$ and show that $\Bbb{Q}$ is not a projective $\Bbb{Z}$-module., but I have not seen a direct answer to part of the question that I would like resolved.
I want to show using the fact that $\text{Hom}_\mathbb{Z}(\mathbb{Q},\mathbb{Z})=0$ that $\mathbb{Q}$ is not a projective $\mathbb{Z}$-module. I have the following idea:
Assume on the contrary that $\mathbb{Q}$ is a projective $\mathbb{Z}$-module. Then there exists a free module $F$ such that $F\cong\mathbb{Q}\oplus M$ for some $\mathbb{Z}$-module $M$. In particular, $\mathbb{Q}$ must be a submodule of $F$. Then, because $\mathbb{Z}$ is a principal ideal domain and $F$ is a free module, $\mathbb{Q}$ must be a free module as well.
I think I am headed in the right direction, but I do not see at this point how to incorporate the information about $\text{Hom}(\mathbb{Q},\mathbb{Z})$ to come to the correct conclusion.
Best Answer
Once you've concluded that $\mathbb{Q}$ is a free module, then you can pick some basis and produce nonzero maps to $\mathbb{Z}$, which gives your contradiction.
See also: https://en.wikipedia.org/wiki/Free_module#Universal_property