Suppose that $k$ is a non-archimedean local field. Let $K/k$ be a tamely, totally ramified degree-$3$ extension of $k$. If $\varpi$ is a uniformizer of $K$, I want to understand the relationship between $\varpi$ and $\gamma(\varpi)$ is mod $P_K^2$ for all $\gamma\in\Gamma_{K/k}$.

Based off my understanding of ramification groups, I know that every element in $\Gamma_{K/k}$ must act trivially on $O_K/P_K$, but no element acts trivially on $O_K/P_K^2$. Am I able to say more about how $\Gamma_{K/k}$ acts on the uniformizer $\varpi$ in general, or does this require a case-by-case investigation? What if I specifically reduce to the case that the residue field is of order $2$?

## Best Answer

There seems to be an implicit assumption this extension is also Galois, since otherwise there will be no non-trivial automorphisms. Let me change your notation, so that the extension is $L/K$, and the residue fields of $L$ and $K$ are $l$ and $k$ respectively. The assumption that the extension is totally ramified implies that $l=k$. Choose a uniformizer $\varpi$ of $L$. Under your assumptions, there is an injection

$$\mathrm{Gal}(L/K) \rightarrow k^{\times}$$

Given by

$$\sigma \mapsto \frac{\sigma \varpi}{\varpi}.$$

This reflects that knowing $\sigma \varpi \bmod \varpi^2$ determines $\sigma$ because there is no wild ramification. We see that the only possibility is that the generator $\sigma$ satisfies

$$ \sigma \varpi = \zeta \varpi \bmod \varpi^2,$$

where $\zeta$ is a non-trivial $3$-rd root of unity. In particular, this is true regardless of the choice of uniformizer.

Note that a necessary condition for this to be possible is that $|k^{\times}|$ is divisible by $3$. In particular, if $k$ has order $2$, then there are no such extensions $L/K$. For example, if $K = \mathbf{Q}_2$, there are no Galois tamely ramified extensions $L/K$ of degree $3$. There are non-Galois extensions like $\mathbf{Q}_2(2^{1/3})$ but they are not Galois and have no non-trivial automorphisms.