= Solution
Write $\alpha=\sqrt[3]{3}$ and $\zeta=\zeta_3$. The cubic is <Eisenstein> and adjoining $\zeta$ adds at most a quadratic extension, so $[L:\mathbb Q_3]\leq6$. Set
$$
\pi=\frac{\zeta-1}{\alpha}.
$$
Using $\zeta^2+\zeta+1=0$ 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 $a\ne0$, 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 $-2\ne0$ 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 gives
$$
\boxed{G_{-1}=G_0=S_3,\qquad G_1=G_2=G_3=C_3,\qquad G_i=\{1\}\ (i\geq4).}
$$
As 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.
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