I'm looking to understand why $\mathbb Z_p$ is a free $\mathbb Z$-module. According to wikipedia this is because $\mathbb Z_p$ is the ring of integers for the algebraic number field $\mathbb Q_p$. The proof that such a ring of integers is a free module is easy enough, but it requires that $\mathbb Q_p$ be a finite dimensional $\mathbb Q$-vector space.
Can someone explain to me why $\mathbb Q_p$ is a finite dimensional $\mathbb Q$-vector space? Or alternatively, explain why $\mathbb Z_p$ is a free $\mathbb Z$-module in a way that does not assume that finite dimensionality?
For reference, I take the algebraic viewpoint that $\mathbb Z_p$ is an inverse limit of $\mathbb Z/p^i$ whose elements can be written as power series in $p$ and whose fraction field is $\mathbb Q_p$.
$\Bbb Z_p$ is not free over $\Bbb Z$. Observe that $(p+1)\Bbb Z_p=\Bbb Z_p$. For a free Abelian group $G$, $mG=G$ for some integer $m>1$ implies that $G=\{0\}$.