 Solution

Write $\alpha =\sqrt[3]{3}$ and $\zeta ={\zeta }_3$. The cubic is Eisenstein and adjoining $\zeta$ adds at most a quadratic extension, so $[L:{\mathbb{Q}}_3]\le 6$. SetUsing ${\zeta }^2+\zeta +1=0$ and ${\alpha }^3=3$, compute ${\pi }^3=1+2\zeta$ and ${\pi }^6=-3$. Thus $\pi$ satisfies the Eisenstein polynomial $X^6+3$. This proves $L={\mathbb{Q}}_3(\pi )$, degree six, with $\pi$ a uniformiser and the extension totally ramified. It is the splitting field of $X^3-3$, so its Galois group is $S_3$. This is the Eisenstein sextic presentation of the splitting field of X3 minus 3 over Q3.

The six automorphisms have $\sigma (\alpha )={\zeta }^a\alpha$, $a\in \{0,1,2\}$, and $\sigma (\zeta )={\zeta }^b$, $b\in \{1,2\}$. Since $v_L(3)=6$, we have $v_L(\alpha )=2$ and $v_L(\zeta -1)=3$.

If $b=1$ and $a\ne 0$, then $\sigma (\pi )={\zeta }^{-a}\pi$, whence $v_L(\sigma (\pi )-\pi )=1+3=4$. These are the two nonidentity elements of the cyclic subgroup $C_3$.

If $b=2$, use ${\zeta }^2-1=(\zeta -1)(\zeta +1)=-(\zeta -1){\zeta }^2$. Then $\sigma (\pi )=-{\zeta }^{2-a}\pi$. The coefficient $-{\zeta }^{2-a}-1$ reduces to $-2\ne 0$ in the residue field ${\mathbb{F}}_3$, so the difference has valuation one. These three automorphisms are the transpositions.

The uniformizer criterion for lower ramification groups consequently givesAs a consistency check, the different exponent is $(6-1)+3(3-1)=11$. The derivative of the monogenic Eisenstein polynomial gives the same value, $v_L(6{\pi }^5)=6+5=11$. In particular, stopping the wild filtration at $G_1$ would give the wrong different.