1
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – Lubin Oct 7 '15 at 3:36
0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post - you can always comment on your own posts, and once you have sufficient reputation you will be able to comment on any post. $\endgroup$ – Moya Oct 5 '15 at 22:55
  • $\begingroup$ I have pointed the mistake. The author seems to have enough knowledge of the subject to realize immediately that he mistakes the equality and need no further explanations. $\endgroup$ – Heinrich Oct 5 '15 at 23:30
  • 1
    $\begingroup$ Well, maybe that's what you should've written the first time, not just an equation followed by a question mark. The answer is not just for the asker's benefit, but also for the benefit of others who might not immediately recognize the mistaken equation. $\endgroup$ – James47 Oct 6 '15 at 21:31
  • $\begingroup$ Actually, Heinrich has pointed out the very nub, the core, the kernel, the heart of OP’s misunderstanding. It might have been more appropriate for him to enter this as a comment than as an answer, but I do not agree with any downvote here. $\endgroup$ – Lubin Oct 7 '15 at 3:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.