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

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?

• 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 at 18:05
• 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 at 18:05
• S4KUL, observe that in user687721's example neither $L/E$ nor $L/E'$ is ramified. – Jyrki Lahtonen Aug 18 at 4:54
• – Jyrki Lahtonen Aug 18 at 5:36
• I understand the example, thanks! Is there also an example if $L/K$ is supposed to be totally ramified? – S4KUL Aug 18 at 12:05