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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.

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

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