Free modules have no infinitely divisible elements Let $F$ be a free $\mathbb Z$-module. How can we show that $F$ has no non-zero infinitely divisible element? (An element $v$ in $F$ is called infinitely divisible if the equation $nx = v$ has solutions $x$ in $F$ for infinitely many integers $n$.)

By definition, $F$ has a basis $(b_i)_{i \in I}$. Suppose
$$
v = a_1 b_1 + \dots + a_k b_k,
$$
for $a_i \in \Bbb{Z}$, is divisible by infinitely many $n$. Choose $n$ positive, larger than all the $\lvert a_i \rvert$, so that $v$ is divisible by $n$. If $v = n x$ for
$$
x = x_1 b_1 + \dots + x_k b_k
$$
then $n x_i = a_i$ for all $i$, as the $b_i$ are a basis, which implies all $a_i = 0$, as $n > \lvert a_i \rvert$. Thus $v = 0$.
