algebraic closure of $\mathbb{Q}_p$ is not complete

Question

A paper I'm reading says that the algebraic closure of $\mathbb{Q}_p$ is not complete, by using for example the Baire theorem. Wikipedia says the Baire theorem says that a complete metric space is a Baire space (meaning a countable intersection of dense open sets remains dense). I don't see how this can be used to show the claim that $\overline{\mathbb{Q}_p}$ is not complete. Any other argument is also welcome.

Answer 1

The usual proof relies on Krasner's lemma. Let $ L/K $ be an infinite algebraic extension, where $ K $ is a perfect local field complete with respect to some nonarchimedean valuation. Then, there is $ a_1, a_2 , \ldots $ in $ L $ which are linearly independent over $ K $, and we may choose coefficients $ c_i $ for each $ a_i $ such that $ |c_i a_i| \to 0 $, and such that $ |c_{i+1} a_{i+1}| $ is smaller than the distance from $ s_k $, $ k \leq i $ to any of its conjugates in $ L $ for each $ i $. Then, the sums

$$ s_n = \sum_{i = 1}^n c_i a_i $$

form a Cauchy sequence in $ L $, which converges to a limit in $ L $ if $ L $ is complete. However, denoting the limit by $ s $, we have

$$ |s - s_n| \leq \max_{n+1 \leq i} |c_i a_i| \leq |s_n' - s_n| $$

where $ s_n' $ is an arbitrary $ K $-conjugate of $ s_n $. It follows from Krasner's lemma that $ s_n \in K(s) $ for all $ n $, and thus $ [K(s) : K] $ is infinite, which is a contradiction.

This proof is "constructive" in the sense that it explicitly constructs an element in $ \mathbb C_p $ that is not algebraic over $ \mathbb Q_p $. However, a nonconstructive proof using the Baire category theorem may proceed as follows: if $ L/K $ is of countably infinite dimension, it is the countable union of nowhere dense subspaces. For example, if a basis is $ \beta_1, \beta_2, \beta_3, \ldots $, one sees that this subextension is precisely

$$ \bigcup_{i = 1}^{\infty} \textrm{span}(\beta_1, \beta_2, \ldots, \beta_i) $$

This, of course, contradicts the BCT.