Product of ramification groups: Is $G_i= H_i H'_i$?

Asked Aug 17 '19 at 15:39

Active Aug 17 '19 at 22:54

Viewed 50 times

Let $L/K$ be a finite Galois extension of local fields with Galois group $G$. Suppose we have two linear disjoint Galois subextensions $E/K$ and $E'/K$ of $L/K$ with $EE'=L$. Let $H=G(L/E)$ and $H'=G(L/E')$. Then we get $G=H H'$.

Now consider the higher ramification groups $G_{i}= \{ \sigma \in G \mid v_L(\sigma x -x)\geq i+1\}$ where $\mathcal O_L = \mathcal O_K[x]$. $H_i$ and $H'_i$ are defined similarly.

My question: Is it true that $G_i= H_i H'_i$?

I was neither able to prove it nor to construct an example where it does not hold. Does anybody have an idea?

edited Aug 17 '19 at 22:54

asked Aug 17 '19 at 15:39

S4KUL

Let $r(\sigma,E) = \inf { v_L(\sigma(a)-a) - v(a), a \in E}$ then don't we have $r(\sigma,L) = \min(r(\sigma,E),r(\sigma,E'))$ – reuns Aug 17 '19 at 18:05
3

If this were true, then the compositum of two totally ramified extensions would be totally ramified, but this is false, e.g. $K = \mathbf{Q}_2$, $E = \mathbf{Q}_2(\sqrt{2})$ and $E' = \mathbf{Q}_2(\sqrt{-6})$. – user687721 Aug 17 '19 at 18:05
S4KUL, observe that in user687721's example neither $L/E$ nor $L/E'$ is ramified. – Jyrki Lahtonen Aug 18 '19 at 04:54
A somewhat related mistaken belief. – Jyrki Lahtonen Aug 18 '19 at 05:36
I understand the example, thanks! Is there also an example if $L/K$ is supposed to be totally ramified? – S4KUL Aug 18 '19 at 12:05

Product of ramification groups: Is $G_i= H_i H'_i$?

0 Answers0

Linked