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.


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

  • $\begingroup$ Okay, I think I understand- using the section you have linked to as a guide, letting $E$ be the generating set for $\mathbb{Q}$ and $R^{(E)}=\mathbb{Q}$, in particular we can use the fact that $\mathbb{Z}$ is a $\mathbb{Z}$-module over itself, so let $M=\mathbb{Z}$ and then given an arbitrary map $\varphi:E\to\mathbb{Z}$, there exists a unique homomorphism $\psi:\mathbb{Q}\to\mathbb{Z}$ such that $\varphi=\psi\circ i$, where $i:E\to\mathbb{Q}$ is the inclusion map. However, we have just shown that the only $\psi$ that exists is $0$, and $0\circ i=0\neq\varphi$. $\endgroup$ – MathStudent1324 May 3 '18 at 23:47
  • $\begingroup$ @MathStudent1324 Right. (Although to be precise we have to choose a map $\phi : E \to \mathbb{Z}$ which has some nonzero elements in its image.) $\endgroup$ – Lorenzo Najt May 3 '18 at 23:50
  • $\begingroup$ that makes sense, thank you! I do have one more general question- once we have arrived at the fact that $\mathbb{Q}$ is a free $\mathbb{Z}$-module, is it possible to directly contradict this instead of contradicting the assumption that $\mathbb{Q}$ is projective over $\mathbb{Z}$? Or would the proof of this be essentially the same? $\endgroup$ – MathStudent1324 May 3 '18 at 23:53
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    $\begingroup$ @MathStudent1324 Free modules are projective, so you've shown that $\mathbb{Q}$ is not projective, and in particular it's not free. If you are asking about the logic of the proof by contradiction - then once you arrive at the deduction that $\mathbb{Q}$ is free, if you know this is false for some reason (say the proof you just outlined), then you can contradict the assumption that $\mathbb{Q}$ was projective. But the proof is essentially the same; just pick whichever packaging makes the most sense to you. $\endgroup$ – Lorenzo Najt May 3 '18 at 23:58

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