# Separable extensions of complete valued fields are automatically totally ramified?

I come up with the following argument which seems to be too good to be true:

Suppose that $L|K$ is finite separable extension of complete valued fields. Let $\nu, \nu'$ be the valuation on $L$ and $K$ respectively where $\nu$ extends $\nu'$ and let $\pi$ be prime element of $K$ and we know that $\nu(\pi)$ is the ramification index of this extension. Let us further assume that $L|K$ is Galois for a moment. We have $$\nu(\pi) = \nu'(N_{L|K}(\pi))$$ where $N_{L|K}$ denote the norm which can be computed by $$N_{L|K}(\pi) = \prod_{\sigma \in G} \sigma(\pi) = \prod_{\sigma \in G} \pi = \pi^{|G|}$$ where $G = \text{Gal}(L|K)$. The second equality is because $\pi \in K$ and so it is fixed by every element of the Galois group $G$. This shows that $$\nu(\pi) = \nu'(\pi^{|G|}) = |G|$$ which means that the ramification index of $L|K$ is the same as the degree of the field extension $[L : K]$ because $[L : K] = |G|$ from basic Galois theory. In other words, the extension $L|K$ is totally ramified and we further have $\kappa_L = \kappa_K$ where $\kappa_F$ is residue class field of $F$.

Now if $L|K$ is only assumed to be finite separable, we can get a Galois closure $N$ for it. By above reasoning, $N|K$ is totally ramified, $\kappa_N = \kappa_K$. But since $\kappa_L$ must be an intermediate field between $\kappa_N|\kappa_K$, we must have $\kappa_L = \kappa_K = \kappa_N$. So $L|K$ is also totally ramified.

To sum up, we have shown that any finite separable extension of complete value fields is totally ramified. This is sort of bizarre to me! So the question is of course what is wrong in the argument?

• Have you looked at any examples? I suggest the example of $K=\Bbb Q_2$ and $L=K(\omega)$, where $\omega^2+\omega+1=0$, in other words $\omega$ is a primitive cube root of unity. Since your claim is emphatically false, you should check each step of your argument against this example. For me at least, mathematical understanding proceeds out of examples. It may be different for you, but I recommend this attitude to you most strongly. – Lubin Oct 7 '15 at 3:36

$\nu(\pi) = \nu'(N_{L|K}(\pi))$ ?