# A trace formula for an extension of complete valued fields

I'm trying to understand the following fact.

Proposition. Let $L/K$ be a finite separable extension of complete valued fields, with respective valuations $w$ and $v$, valuation rings $\mathcal{O}_L$ and $\mathcal{O}_K$, residue fields $\ell$ and $k$, and with ramification index $e$. Then for $x \in \mathcal{O}_L$, $$\overline{\mathrm{Tr}_{L/K}(x)} = e\mathrm{Tr}_{\ell/k}(\overline{x})$$ in $k$. (The bar denotes reduction to whichever residue field.)

There's a rather finicky, linear-algebra sort of proof of this in Chevalley's Introduction to the Theory of Algebraic Functions of One Variable. It seems to me that there should be a more intuitive explanation. Specifically, if $p$ is the minimal polynomial of $x$ over $K$, then $p$ has coefficients in $\mathcal{O}_K$, and so $\overline{p}$ is a polynomial over $k$ with $\overline{p}(\overline{x}) = 0$. If $\overline{q}$ is the minimal polynomial for $\overline{x}$ over $k$, then I'd guess that $\overline{q}^e = \overline{p}$. This gives the above trace formula just by looking at coefficients.

Let's assume $L/K$ is Galois first, the general case can be done by descending back. Then trace should be sum of Galois conjugates, so you want to connect $Gal(L/K)$ and $Gal(\ell/k)$. But there is an exact sequence $0 \to I(L/K) \to Gal(L/K) \to Gal(\ell/k) \to 0$, which implies what you say since the order of the inertia group is $e$. –  user27126 Mar 14 '13 at 8:15
@Sanchez: It can happen that $\ell/k$ is inseparable. –  user18119 Mar 15 '13 at 0:29