= Solution
Let $\alpha$ be a root of
$$
f(X)=X^3+3X+3.
$$
The polynomial is Eisenstein at $3$, so $E=\mathbb Q_3(\alpha)$ is totally ramified of degree three and $\mathcal O_E=\mathbb Z_3[\alpha]$. Its discriminant is
$$
\Delta=-4\cdot3^3-27\cdot3^2=-351=-3^3\cdot13.
$$
Its odd valuation makes $\Delta$ nonsquare in $\mathbb Q_3$, so the <Galois group of an irreducible cubic> shows that the splitting field $L$ has Galois group $S_3$. The quadratic extension obtained by adjoining $\sqrt\Delta$ is ramified, so $L/\mathbb Q_3$ is totally ramified. Therefore
$$
G_{-1}=G_0=S_3,
\qquad
G_1=A_3,
$$
because the <wild inertia group> is the unique Sylow $3$-subgroup of $S_3$.
It remains to find the wild break. Since
$$
f'(\alpha)=3(\alpha^2+1)
$$
and $\alpha^2+1$ is a unit, the <different ideal> of $E/\mathbb Q_3$ has exponent $v_E(f'(\alpha))=3$. The extension $L/E$ is a tamely ramified quadratic extension and has different exponent one. The <different in a tower> therefore gives different exponent
$$
d(L/\mathbb Q_3)=1+2\cdot3=7.
$$
On the other hand, the <different exponent from ramification groups> is
$$
7=\sum_{i\geq0}(|G_i|-1)
=5+2b,
$$
where $b$ is the last index for which $G_b=A_3$. Thus $b=1$, and
$$
G_{-1}=G_0=S_3,\qquad G_1=A_3,\qquad G_i=1\quad(i\geq2).
$$
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