Is there $f$ in $\operatorname{Hom}(\mathbb{Q},\mathbb{Q})$ with kernel $\mathbb{Z}$? Is there a group homomorphism from $\mathbb{Q}$ (the group of rationals) to $\mathbb{Q}$ whose kernel is $\mathbb{Z}$?
 A: The hint in in the same vein as that given by Tobias in the comments above, but a bit more explicit. Consider where $1/2$ would be sent to, given that $1/2 + 1/2 = 1$.
A: Homological algebra can also solve this problem.
Suppose $f:\Bbb Q\to \Bbb Q$ has kernel $\Bbb Z$. Then there is an exact sequence of abelian groups
$$
0\to \Bbb Z\to \Bbb Q\to G\to 0\tag{1}
$$
where $G=\operatorname{im}f$. Since $\Bbb Z\not\simeq\Bbb Q$, (1) implies $G\neq 0$.
Taking ranks in (1) gives
$$
\DeclareMathOperator{rk}{rk}\rk \Bbb Q=\rk \Bbb Z+\rk G\tag{2}
$$
But $\rk\Bbb Q=\rk\Bbb Z=1$ so (2) is equivalent to $\rk G=0$. That is, $G$ must be torsion, a contradiction since $\Bbb Q$ has no nontrivial torsion subgroups. 
Here, $\rk A=\dim_{\Bbb Q}\left( A\otimes_{\Bbb Z}\Bbb Q\right)$. Since $\Bbb Q$ is a flat $\Bbb Z$-module, applying $-\otimes_{\Bbb Z}\Bbb Q$ to an exact sequence of abelian groups
$$
0\to A\to B\to C\to 0
$$
gives an exact sequence of $\Bbb Q$-vector spaces
$$
0\to A\otimes_{\Bbb Z}\Bbb Q\to B\otimes_{\Bbb Z}\Bbb Q\to C\otimes_{\Bbb Z}\Bbb Q\to 0
$$
Taking dimensions justifies the formula in (2). In particular we use that $\Bbb Z\otimes_{\Bbb Z}\Bbb Q\simeq \Bbb Q\otimes_{\Bbb Z}\Bbb Q\simeq\Bbb Q$.
