How to prove that any infinite algebraic extension of a complete field is never complete?

Question

My first idea is using Baire category theorem since I thought an infinite algebraic extension should be of countable degree. However, this is wrong, according to this post.

This approach may still work if it is true that infinite algebraic extensions of complete fields have countable degree. For instance, infinite algebraic extensions of local fields are of countable degree. I wonder if this is true for general case.

Another possible approach is to generalize the proof in Bosch, Güntzer, Remmert: Non-Archimedean Analysis, Lemma 1, Section 3.4.3. Where they use Krasner's Lemma to prove that if the algebraic closure of a complete field is of infinite degree then it is not complete. However, Krasner's Lemma only works for separable elements, so I wonder if this proof can be used for infinite purely inseparable extensions of complete fields.

Answer 1

I don't see how to generalize the proof in Bosch-Güntzer-Remmert's book. The same goes for the Baire category approach, which indeed seem very natural. I think however that this should work. The argument is not mine. I will basically write it out, in case anyone deletes the .pdf in the future. Any errors that occurs in this post are due to me and I would recommend reading the .pdf above since it is written in a more coherent way than this post.

Note that there is no restriction to consider infinite fields. I will assume throughout that the valuation is non-archimedean (the archimedean case is of course trivial).

Let $K$ be any infinite field and consider an infinite algebraic extension $L$ of $K.$ Then you have that $|L| = |K|.$This is discussed in this question.

From this, I claim that if $K \subset L$ is an infinite algebraic extension and $K$ and $L$ are complete, then $|K| < |L|,$ which gives a contradiction.

Indeed, let us take a (countably) infinite linearly independent subset $\{x_i\}$ of $L$ over $K.$ Now, we can assume that we choose $x_i$'s such that $|x_i|/2 > |x_{i+1}|.$ Then we have that the sums $$\omega_{\epsilon} = \sum \epsilon_i x_i $$ where $\epsilon_i \in \{0,1\}$ are in $L$ and further are distinct.

That all sums $\omega_{\epsilon}$ are in $L$ is clear - the terms form a null-sequence and for non-archimedean fields this enough. Further, let us take two sums $\omega_{\epsilon} = \sum_i \epsilon_i x_i$ and $\omega_{\epsilon'} = \sum_i \epsilon'_i x_i$ where for some $i$ $\epsilon_i \neq \epsilon'_i.$ Then if we consider the difference $\omega_{\epsilon}-\omega_{\epsilon'} = \sum_i (\epsilon_i-\epsilon'_i)x_i$ there is a smallest $i_0$ such that $\epsilon_{i_0} \neq \epsilon'_{i_0}.$ Then we have that $$|(\epsilon_{i_0} - \epsilon'_{i_0})x_{i_0}| = |\sum_{k=i_0}^\infty (\epsilon_k-\epsilon'_k) x_k| < |\sum_{k=1}^\infty (\epsilon_{i_0}-\epsilon'_{i_0})x_{i_0}/2^k = |(\epsilon_{i_0} - \epsilon'_{i_0})x_{i_0}| .$$ This yields a contradiction and thus all the $\omega_{\epsilon}$ are distinct. So using these $\omega_\epsilon$ we see that $|K| < 2^{|K|} \leq |L|$ which is a contradiction.

Answer 2

The separable case does indeed follow from Krasner's lemma: see here.

I think I have a counterexample to the purely inseparable case. The idea is that we adjoin all possible $p$-th roots of elements in the ground field $K$, where $p$ is the characteristic, resulting in a new field $L$ to which the original valuation extends uniquely (because $L/K$ is algebraic). Then all $p$-th powers of $\ell\in L$ lie in $K$, essentially by the freshman's dream, and in fact the map $L\rightarrow K$ given by $\ell\rightarrow\ell^p$ is a uniformly continuous homeomorphism with uniformly continuous inverse. Therefore, $L$ is complete as long as $K$ is.

We just need to make sure that we can do this with $K$ a complete field and have enough $p$-th roots to make $[L:K]=\infty$. To ensure this, we'll just adjoin a bunch of dummy variables in forming $K$, and the $p$-th roots of these dummy variables will be linearly independent. We begin with the field of rational functions in infinitely many variables over $\mathbb{F}_p, $ namely $\mathbb{F}_p(x_1,x_2,\dots).$ As norm we take the one induced by $|x_1|=\frac{1}{2}$ and $|R(x_2,x_3,\dots)|=1$ for all rational functions $R$, essentially the $x_1$-adic valuation. We still need to complete this, and its easy to see that the completion is the field of infinite Laurent series in $x_1$, expressions of the form $\sum\limits_{k=-n}^{\infty} (x_1^k)R_k(x_2,x_3,\dots)$ where $R_k$ are arbitrary rational functions. This completion is our ground field $K$.

After adjoining all $p$-th roots to form $L$ as described above, we just need to check that $(x_i)^{\frac{1}{p}}$ are linearly independent over $K$; this will show that $[L:K]=\infty$. Well, this is pretty easy: assume not! Then taking $p$-th powers and using the freshman's dream again, we get a non-trivial identity of the form $\sum\limits_{i=1}^m x_i P_i(x_1^p,x_2^p,\dots)=0$ for $P_i \in K$. We can consider the smallest $j$ such that one of the $x_iP_i(x_1^p,x_2^p,\dots)$ series has a non-zero $x_1^j$-term. Taking the $x_1^j$-coefficient of our identity we have $0=\sum\limits_{i=2}^m x_i S_i(x_2^p,x_3^p,\dots)$ where $S_i$ is the $x_1^j$-coefficient of $P_i$ (it's clear that the $x_1P_1(x_1^p,x_2^P,\dots$) doesn't contribute an $x_1^j$ term; if it did, it would be the only one to do so!). The $S_i$ are rational functions, so by clearing denominators they are wlog polynomials, since we have finitely many $S_i$ each using only finitely many variables. But now the monomial terms clearly don't match up at all, so we have a contradiction, and we've established linear independence.