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I know that the algebraic closure of $\mathbb Q_p$, which I'll denote $\overline{\mathbb Q_p}$, is not metrically complete: there are $p$-adic Cauchy sequences that do not converge. (The example I know is $\sum_{n > 0, (n,p) = 1} \zeta_n p^n$ where $\zeta_n$ is a primitive $n^{th}$ root of unity; there are plenty of others).

However, there is a proof via Krasner's Lemma - mentioned on Wikipedia here, with a reference to Proposition 8.1.5 of Neukirch, Schmidt, and Wingberg's Cohomology of Number Fields - that the $p$-adic completion of the separable algebraic closure of a global field $k$ is equal to the separable algebraic closure of its completion. That is, for $k = \mathbb Q$, $\overline{(\mathbb Q_p)} = (\overline{\mathbb Q})_p$.

Why doesn't this mean that the algebraic closure of $\mathbb{Q}_p$ is complete? The example I gave above, $\sum_{n > 0, (n,p) = 1} \zeta_n p^n$ is a $p$-adic Cauchy sequence in $\overline{\mathbb{Q}}$, so it should converge to an element of $(\overline{\mathbb{Q}})_p$.

EDIT This is answered here. For reference, $(\overline{\mathbb{Q}})_p$ denotes the union of all completions of number fields with respect to valuations extending the $p$-adic valuation on $\mathbb{Q}$.

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Dorebell
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  • I don't think that $p$ remains prime in $\overline{\mathbb{Q}}$. Without that, you can't do $p$-adic completion. On the other hands, I have seen somewhere that $\overline{\mathbb{Q}_p} = \mathbb{C}$ and it is unlikely that $p$-adic completion of something will gives us $\mathbb{C}$. – An Hoa Dec 10 '15 at 00:15
  • Hmm - good point. I know that the $p$-adic valuation can be extended non-uniquely from $\mathbb{Q}$ to a number field by choosing some prime lying above $\mathbb{Q}$. I'm not sure how to define a valuation on $\overline{\mathbb{Q}}$. – Dorebell Dec 10 '15 at 00:56

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