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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 $\mathbb Q_p$.

Guest
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    $Q_p$ is not finite dimensional over $Q$ (because it is uncountable) so it i hard to explain why it is... – Mariano Suárez-Álvarez Aug 19 '17 at 18:35
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    I don't believe that $\Bbb Z_p$ is free over $\Bbb Z$; certainly $\Bbb Q_p$ is not finite-dimensional over $\Bbb Q_p$. – Angina Seng Aug 19 '17 at 18:35
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    And $Q_p$ is not an algebraic number field. – Mariano Suárez-Álvarez Aug 19 '17 at 18:36
  • @ LordSharktheUnknown Another non-free $\mathbb{Z}$-module is $\mathbb{Z}^{\mathbb{N}}$, but the proof is much less intuitive than yours for $\mathbb{Z}_p$ – reuns Aug 19 '17 at 19:25
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    $\mathbb{Z}_p$ is the simplest nontrivial example of a free pro-p group -- that on a single generator . Thus, for instance, any surjective morphism of a pro-p group onto $\mathbb{Z}_p$ splits, and also, given an element $g$ of an arbitrary pro-p group $G$, we can find a continuous morphism from $\mathbb{Z}_p$ to $G$ that maps $1$ to $g$. Could that be the context you were in on that Wikipedia page? – Barry Smith Aug 20 '17 at 17:10

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$\Bbb Z_p$ 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 $G=\{0\}$.

Angina Seng
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  • Hmmm, well that explains why I couldn't figure it out, lol. Now I can't seem to find where on wikipedia I thought I saw that it was free... – Guest Aug 19 '17 at 19:22