# Galois action on a uniformizer of a non-archimedean local field

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 $$2$$?

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.