Subfield of a complete valuation field also complete

Question

If we have a complete field $K$ with respect to a discrete valuation $v$ and $L\subset K$ is a finite separable subfield, is $L$ complete with respect to the restriction of the valuation $v$?

Answer 1

Yes, $L$ is complete. The solution below was inspired by answers to similar questions asked before on MO and MSE here, here, and here.

I prefer to think about this using absolute values rather than valuations, so we are told that $K$ is complete with respect to a nontrivial (!) nonarchimedean absolute value $|\cdot|$ with discrete value group and $L$ is a subfield of finite codimension in $K$ such that $K/L$ is separable. Our goal is to show $L$ is complete with respect to $|\cdot|$.

Since $|\cdot|$ on $K$ has a nonzero discrete value group, $K$ is not algebraically closed: a nontrivial nonarchimedean absolute value on an algebraically closed field doesn't have a discrete value group since we can take $n$th roots in $K$ for all $n \geq 1$, so the value group on $K^\times$ is divisible.

Since $K/L$ is finite separable, we can enlarge $K$ to a finite Galois extension $M$ of $L$. The absolute value $|\cdot|$ on $K$ extends uniquely to an absolute value on $M$ by completeness of $|\cdot|$ on $K$, and this extension is a discrete and complete absolute value on $M$. Write this absolute value on $M$ as $|\cdot|$. Its discreteness on $M$ implies $M$ is not algebraically closed too.

Lemma. The absolute value $|\cdot|$ on $L$ is nontrivial.

Proof. If $|\cdot|$ is trivial on $L$ then $L$ is complete with respect to $|\cdot|$. By the unique extension of a complete absolute value on a field to a finite extension, the trivial absolute value on $L$ has only one extension to an absolute value on $K$. Therefore the original absolute value $|\cdot|$ on $K$ is trivial, which contradicts $|\cdot|$ being nontrivial on $K$. We have proved the lemma.

Theorem. For each $\sigma \in {\rm Gal}(M/L)$, $|\sigma(\alpha)| = |\alpha|$ for all $\alpha \in M$.

Proof. On $M$ we can define two absolute values $|\cdot|_1$ and $|\cdot|_2$ by $|\alpha|_1 = |\alpha|$ and $|\alpha|_2 = |\sigma(\alpha)|$ for $\alpha \in M$. The field $M$ is complete with respect to $|\cdot|_1$ by the way $|\cdot|$ is defined on $M$. Let's show $M$ is also complete with respect to $|\cdot|_2$: if $\{\alpha_n\} \subset M$ is a Cauchy sequence with respect to $|\cdot|_2$ then $\{\sigma(\alpha_n)\} \subset M$ is a Cauchy sequence with respect to $|\cdot|_1 = |\cdot|$. Then completeness of $M$ with respect to $|\cdot|$ tells us there is $\beta \in M$ such that $|\sigma(\alpha_n) - \beta|_1 \to 0$ as $n \to \infty$, so $|\alpha_n - \sigma^{-1}(\beta)|_2 \to 0$ as $n \to \infty$.

Since $M$ is complete with respect to two nonarchimedean absolute values, we can now appeal to a theorem of F. K. Schmidt, which says when a field satisfies Hensel's lemma with respect to two nonarchimedean absolute values, then either the field is separably closed or the two absolute values on the field are equivalent (give the field the same topology): see Schmidt's theorem in Arno Fehm's answer here. Complete nonarchimedean absolute values on a field satisfy Hensel's lemma, so using Schmidt's theorem on $M$ tells us that either $M$ is separably closed or $|\cdot|_1$ and $|\cdot|_2$ on $M$ are equivalent.

A complete valued field that is separably closed is algebraically closed (see my first two comments to Arno Fehm's answer here), and we showed earlier that $M$ is not algebraically closed, so $|\cdot|_1$ and $|\cdot|_2$ are equivalent on $M$. Equivalent absolute values on a field are powers of each other: there is some $t > 0$ such that $|\alpha|_2 = |\alpha|_1^t$ for all $\alpha \in M$ (see Definition 1.2 and Theorem 2.1 here). Thus $$ |\sigma(\alpha)| = |\alpha|^t $$ for all $\alpha \in M$, so $|\alpha| = |\alpha|^t$ for all $\alpha \in L$. By the lemma, $|\cdot|$ is nontrivial on $L$, so picking $\alpha \in L$ such that $|\alpha| \not= 0$ or $1$ shows $t = 1$. Thus $$ |\sigma(\alpha)| = |\alpha| $$ for all $\alpha \in M$. That finishes the proof of the theorem.

Corollary. The field $L$ is complete with respect to $|\cdot|$.

Proof. Let $\{a_n\} \subset L$ be Cauchy with respect to $|\cdot|$. By completeness of the larger field $K$, there is a limit $a$ in $K$, so $|a_n - a| \to 0$ as $n \to \infty$. To show $a \in L$, we'll show $\sigma(a) = a$ for each $\sigma \in {\rm Gal}(M/L)$: $$ |a_n - \sigma(a)| = |\sigma(a_n) - \sigma(a)| = |\sigma(a_n - a)| = |a_n - a|, $$ where the last equation uses the theorem above, so $|a_n - \sigma(a)| \to 0$ as $n \to \infty$. By uniqueness of limits, $\sigma(a) = a$. This holds for all $\sigma \in {\rm Gal}(M/L)$, so $a \in L$. That proves the corollary, which answers the OP's question.

Remark. The only way we used the condition that the absolute value on $K$ is discrete is to show $K$, and then $M$, is not algebraically closed as preparation for our use of Schmidt's theorem on $M$. If we drop the assumption that the absolute value on $K$ is discrete and replace it with the weaker assumption that $K$ is not algebraically closed, then we can still deduce that $M$ is not algebraically closed by the following method. First of all, to prove $L$ is complete we of course may assume $L \not= K$. Then $1 < [K:L] < \infty$, so $[M:L]= [M:K][K:L] > 1$. If $M$ is algebraically closed, then the Artin-Schreier theorem implies $[M:L] = 2$, so $[M:K] = 1$ and thus $M = K$, which contradicts the assumption that $K$ is not algebraically closed. So $M$ is not algebraically closed and the rest of the proof goes through as before. This means that in the OP's question, instead of $K$ being a complete discretely valued field, it can be a non-algebraically closed field complete with respect to some nontrivial nonarchimedean absolute value and it's still true that $L$ is complete. (There are counterexamples using Zorn's lemma if we allow $K$ to be algebraically closed, such as $K = \mathbf C$ and $L = \mathbf R$. For a prime $p$, $\mathbf C \cong \mathbf C_p$ as abstract fields by Zorn's lemma, so in a nonconstructive way $\mathbf C$ is complete with respect to an extension of the $p$-adic absolute value. The subfield $\mathbf R$ of finite codimension in $\mathbf C$ is not complete for that extension of the $p$-adic absolute value because otherwise $\mathbf R$ would contain $\mathbf Q_p$, and many negative integers are squares in $\mathbf Q_p$ while they are not squares in $\mathbf R$.)