# $\mathbb{R}$ as a $\mathbb{Z}$ module

Is $\mathbb{R}$ free as a $\mathbb{Z}$ module? If it is, it must be of infinite rank, I suspected that some set theory will be involved, but have no clue about this fact.

• Note that $\operatorname{Hom}(\mathbb R,\mathbb Z)=0$, which is clearly the opposite of how free modules behave. – MooS Feb 15 '16 at 6:00

No. Free abelian groups don't have any nonzero divisible elements, but every element of $\mathbb{R}$ is divisible. $\mathbb{R}$ is free as a $\mathbb{Q}$-module (this requires the axiom of choice), since $\mathbb{Q}$ is a field.