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With reference to my older question Product of ramification groups: Is $G_i= H_i H'_i$?, I want to specify it in the totally ramified case:

Let $L/K$ be a finite Galois totally ramified 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.

Is it now true that $G_i= H_i H'_i$, if $L/K$ is totally ramified?

The counterexample in the referred older question uses the fact that the composite of two totally ramified extensions is not totally ramified in general.
Now this method can not be used to construct a counterexample. Has anybody an idea for a counterexample or a proof of the assertion?

S4KUL
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    Have you even tried to compute a single example? Last time I gave a counterexample, and it's clear you just moved on rather than trying to reflect on it. There are seven quadratic extensions of $\mathbf{Q}_2$ (six are ramified). I suggest computing what happens in all of the $15 = \binom{6}{2}$ cases and actually thinking about it yourself. – user687721 Aug 19 '19 at 11:38

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