integral closure of finite totally ramified extension of a DVR

Question

I want to prove the following:

Let $K$ be a local field. If $E \subset K^{sep}$ be a finitely ramified extension of $K$, then $\mathcal{O}_E$ is a DVR.

Here $E/K$ is finitely ramified extension if $E$ is a finite extension of some (not necessarily finite) unramified extension $F$ of $K$.

It is not so hard to see that $\mathcal{O}_F$ is a DVR. We first see that $$ \mathcal{O}_F = \bigcup_{F'/K} \mathcal{O}_{F'}$$ where $F'/K$ is a finite intermediate extension of $F/K$. They are all unramified, so the maximal ideals of $\mathcal{O}_{F'}$ is all of the form $(\pi)\mathcal{O}_{F'}$. By computing the units, we see that $\mathcal{O}_F$ is a local ring. Also, we can see that the ideals are all of the form $(\pi^m) \mathcal{O}_F$ via similar reasoning. A Local PID is DVR.

Then the argument that I saw in a paper says that $\mathcal{O}_E$ is a DVR because $\mathcal{O}_E$ is an integral closure of $\mathcal{O}_F$ in $E$. I know that this is true if $\mathcal{O}_F$ is a complete DVR.

What is special about $\mathcal{O}_F$ so that $\mathcal{O}_E$ becomes a DVR? (i.e. how should we prove this?)

Answer 1

Short answer: $\mathcal{O}_F$ is Henselian DVR, so the valuation of $F$ extends uniquely to a discrete valuation of $E$.

Long answer: There are two possible approaches (I) and (II)

(I)

Consider a monic polynomial $f(X)$ in $\mathcal{O}_F$. Let $\kappa_F$ be the residue field of $F$ and suppose that $$ \overline{f}(X) = \overline{g}(X) \overline{h}(X) $$ Because there is only finitely many coefficients for $f(X)$, $\overline{g}(X)$ and $\overline{h}(X)$, there exists an intermediate field $F'$ satisfying $F/F'/K$ with $F'/K$ finite such that all the coefficients of $f(X)$ and the lifts of the coefficients of $\overline{g}(X)$ and $\overline{h}(X)$ is contained inside $F'$.

It is not hard to find a finite extension of $F$ containing one coefficient. Therefore, we take the finite compositum of such to obtain $F'$.

Therefore, $f(X)$ can be considered as a polynomial in $\mathcal{O}_{F'}$ and may assume that $\overline{f}(X)$ factors inside $\kappa_{F'}$. As $K$ is a complete discrete valuation field (CDVF), $F'$ is also a CDVF. Therefore, Hensel's lemma holds. $\mathcal{O}_F$ is a Henselian DVR.

Or, simply, use the

(II)

There is a characterization of Henselian field. $K$ is henselian if and only if for every algebraic extension $L/K$, the valuation of $v$ of $K$ extends uniquely to a valuation of $L$. It follows that any algebraic extension of Henselian field is a Henselian field.

Let $E$ be an extension of $L/K$ with $w, w'$ extension of $w|_{E} = w'|_{E}$. In turn, extensions of $w|_K = w'|_{K}$, hence $w=w'$.

Therefore $\mathcal{O}_{F}$ is a Henselian DVR.

Then

We know that there exists an explicit formula for finite extension $E/F$ of Henselian field, namely

$$ w(x) = \frac{1}{n} v(N_{E/F}(x)) $$

Therefore, $w$ must be a discrete valuation. This shows that $\mathcal{O}_E$ is a DVR.