Intersection of completion of subextension with algebraic closure

Question

Let $K$ be a field, $char(K)=0$, complete w.t.r. to a discrete valuation and with perfect residue field of $char =p>0$.

Now consider a subextension $K\subset L\subset \bar{K}$ and denote by $C_K$ (resp. $\hat{L}$) the completion of $\bar{K}$ (resp. $L$). I would like to prove that \begin{equation*} \hat{L}\cap \bar{K}= L. \end{equation*}

If $L=\bar{K}^H$, where $H$ is a closed subgroup, i.e. $L/K$ Galois, one can show that $\hat{L}=C_K^H$ and thus

\begin{equation*} \hat{L}\cap \bar{K}= C_K^H\cap \bar{K}=L \end{equation*} where the second equality follows from the fact that all elements in the intersection are invariant under $H$.

But how does one conclude the general case?

Answer 1

After a discussion with a friend, here is a somewhat complete (but not self-contained) proof of the statement. We want to prove the following

Theorem 1 With the above notation, $\hat{L}\cap \bar{L}=L$.

We are going to do this in several steps. First let's state a more general version of Krasner's Lemma.

General Krasner's Lemma Let $(K, |·|)$ be a non-archimedean valued field, $\mathcal{O}_K$ it's ring of of integers, $m$ its maximal ideal, $k = \mathcal{O}/m$ its residue field. Assume that $|·|$ extends to a separable closure $K^{sep}$. Then TFAE:

1. $\mathcal{O}_K$ is Henselian

2. Krasner's Lemma holds for $K$.

Proof I think one way to show this might be via a third criterion about uniqueness of valuations on finite extensions of $K$. I will leave this out for now, as this is not the main point.

Now we can state the main theorem:

Theorem 2 Let $K$ be as in Krasner's Lemma with $char(K)=0$ (i.e. I will not distinguish between algebraic and seperable closure anymore). Then TFAE:

1. $ \hat{K} \cap \bar{K}=K$

2. $\mathcal{O}_K$ is Henselian.

Proof "$1. \Rightarrow 2.$" Let $f(x)\in \mathcal{O}_K[x]$ be a polynomial, such that there exists a $b\in \mathcal{O}_K$ satisfying $f(b) \equiv 0 \mod m_K $ and $f'(b)\not \equiv 0 \mod m_K$. We have to find a root $ \tilde{b} \in\mathcal{O}_K$ of $f(x)$.

To do this, consider $f(x)$ as a polynomial inside $\mathcal{O}_{\hat{K}}[x]$. As $\hat{K}$ is complete, $\mathcal{O}_{\hat{K}}$ is Henselian and thus we find a root $\tilde{b}\in \mathcal{O}_{\hat{K}}$. Now $\tilde{b}\in \bar{K}$, as it is a root of $f$, and $\tilde{b}\in \hat{K}$ by construction. Thus $\tilde{b}\in K$. As $\tilde{b}\in \mathcal{O}_{\hat{K}}$ we conclude that $\tilde{b}\in\mathcal{O}_{K}$.

"$2. \Rightarrow 1.$"

Let $a\in \hat{K}\cap\bar{K}$ with minimal polynomial $f(x)\in K[x]$ and other roots $a_2,...,a_n$. As $K$ is dense inside $\hat{K}$, we can find $b\in K$, such that $|b-a|<|b-a_i|$. But by General Krasner's Lemma ($1. \Rightarrow 2.$) Krasner's Lemma holds and thus $K(a)\subset K(b)=K$, i.e. $a\in K$.

Lemma 3 $\mathcal{O}_L$ is Henselian.

Proof If $L/K$ is finite, this is clear, as $L$ is still complete. If $L/K$ is infinite, choosing a polynomial $f(x)\in \mathcal{O}_L[x]$, will always allow us to reduce to the finite field extension case (as there are only finitely many coefficents in $f$).

Thus Theorem 1 is proven by combining Lemma 3 with Theorem 2.