I found out that $\mathbb{Q}$ is not a projective $\mathbb{Z}$-module, because it can not be a direct summand of a free $\mathbb{Z}$-module. What about $\mathbb{R}$, is $\mathbb{R}$ a projective $\mathbb{Z}$-module?
I would say that $\mathbb{R}$ is not a projective $\mathbb{Z}$-module with the same argument as for $\mathbb{Q}$, but I'm not sure, or is it wrong?
On
Recall that $\Bbb{Q}$ is a direct summand of $\Bbb{R}$: hence $\Bbb{R}$ is not projective.
To show that $\Bbb{Q}$ is a direct summand of $\Bbb{R}$, you need the Axiom of Choice: since $\Bbb{R}$ is a $\Bbb{Q}$-vector space, pick a basis $B$ such that $1 \in B$. Then $$\Bbb{R} = \Bbb{Q} \oplus \Bbb{Q}(B \setminus \{ 1\})$$ as $\Bbb{Q}$-vector spaces. This implies that this direct sum is valid as $\Bbb{Z}$-modules as well.
Note that $\mathbb{R}$ doesn't even embed in a free group, so it cannot be a summand thereof.
Indeed, let's prove that the only divisible subgroup of a free group is trivial. Let $F=\mathbb{Z}^{(X)}$ be free. Take a nonzero element $t=(n_x)_{x\in X}$ and consider the positive lowest common multiple $m$ of all the coefficients $n_x$. Then there is no element $u\in F$ such that $2mu=t$. Thus $t$ belongs to no divisible subgroup of $F$.