Prove that $\operatorname{Hom}_{\Bbb{Z}}(\Bbb{Q},\Bbb{Z}) = 0$ and show that $\Bbb{Q}$ is not a projective $\Bbb{Z}$-module. 
1) Prove that $\operatorname{Hom}_{\Bbb{Z}}(\Bbb{Q},\Bbb{Z}) = 0$. 
2) Show that $\Bbb{Q}$ is not a projective $\Bbb{Z}$-module. 

1) We know that $\Bbb{Q}$ is an injective $\Bbb{Z}$-module.
This implies that every short exact sequence $$0 \rightarrow \Bbb{Q} \xrightarrow{f} A \xrightarrow{g} B \rightarrow 0$$ is split exact. 
In particular, 
$$0 \rightarrow \Bbb{Q} \xrightarrow{f}  \Bbb{Z} \xrightarrow{g} A \rightarrow 0$$
is split exact. 
But this implies that we have a homomorphism $k: \Bbb{Z} \rightarrow \Bbb{Q}$  such that $kf = 1_{\Bbb{Q}}$.
So we need to look at what possible homomorphisms we could have from $\Bbb{Z}$ to $\Bbb{Q}$. Since $\Bbb{Z}$ is cyclic, the homomorphism $k$ is determined by where it sends $1$. Suppose that we send $1$ to $q \in \Bbb{Q}$ such that $q \not= 1$ and $q \not= 0$. Let $m \in \Bbb{Z}$. Then
$$k(m) = mk(1) = mq$$ 
But then 
$$qmq = k(1)k(m) \not= k(m) = mq$$ 
So either $1 \in \Bbb{Z}$ must be sent to $1 \in \Bbb{Q}$ or $k$ can be the trivial homomorphism.
But if $1_{\Bbb{Z}}$ is sent to $1_{\Bbb{Q}}$, then $k$ must be the inclusion map. However in order to for $kf = 1_{\Bbb{Q}}$ to hold, $f$ must also sent $z$ to $z$ for all $z \in \Bbb{Q}$. But if that's the case, then for any $a/b \in \Bbb{Q}$ where $b \not= 0, 1$, we have (for $n \in \Bbb{Z}$) $$f(a/b) = n = f(n)$$
and this contradicts the fact that $f$ must be injective (since the sequence is exact).    
So the only possible $\Bbb{Z}$-module map from $\Bbb{Z}$ to $\Bbb{Q}$ is $0$. 
Do you think my answer is correct? 
2) I was wondering if anybody could give a hint on this one, because I couldn't really get started.
Thanks in advance. 
 A: 1) I think you made this much harder than it has to be. $\mathbb{Q}$ is a divisible $\mathbb{Z}$-module. This implies that if $f:\mathbb{Q}\to\mathbb{Z}$, then $f(\mathbb{Q})$ is a divisible $\mathbb{Z}$-submodule of $\mathbb{Z}$. How many of those do you know?
2) Hint: PIDs have the nice property that a submodule of a free module is free. With this, what can you say about projective $\mathbb{Z}$-modules? Namely note that they are all (BLANK). After you've done this, explain why $\mathbb{Q}$-can't be (BLANK) by 1) and thus can't be projective. 
A: 2) If $\mathbb{Q}$ is a projective $\mathbb{Z}$-module, then it is direct summand of a free module, which is isomorphic to $\mathbb{Z}^{(I)}:=\oplus_{i\in I}\mathbb{Z}$; for some set $I$.
In this case there is a injective homomorphism $\iota:\mathbb{Q}\hookrightarrow \mathbb{Z}^{(I)}$. But note that, for each $i\in I$, if $\pi_i:\mathbb{Z}^{(I)}\to\mathbb{Z}$ is the $i$-ith projection, $\pi_i\iota:\mathbb{Q}\to\mathbb{Z}$ is a homomorphism and, therefore it is the null homomorphism. 
Thus, for any $x\in \mathbb{Q}$, $\pi_{i}\iota(x)=0$ for every $i\in I$. Then $\iota(x)=0$. Therefore $\iota$ is also the null map - which contradicts the hypothesis that it is a injective homomorphism.
