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Realize the two-dimensional irreducible representation $\rho$ as the standard representation of the symmetric group onwith $S_3$ permuting coordinates. It is the complement of the trivial summand in the three-point permutation representation. For each prime the local factor of an Artin L-function iswhere $I_p$ is the inertia group and Frobenius means any lift of the residue-field Frobenius automorphism. On $V^{I_p}$ the lift is independent of that choice. At an unramified prime, put $t=p^{-s}$. The Local Artin factors of the standard representation of S3 areThe polynomial discriminant of $X^3-3$ is $-243=-3^5$, so only $3$ can ramify in its splitting field. For $p\ne 3$, the squarefree factorization of $X^3-3$ modulo $p$ determines the Frobenius cycle type on its roots.

At $p=2$,The quadratic is irreducible, so Frobenius is a transposition andThus $a_2=0$, $a_4=1$ and $a_8=0$. At $p=5$,The quadratic discriminant is $3(\mathrm{mod}5)$, a nonsquare, so this is again a transposition. Consequently $L_5(\rho ,s)=(1-5^{-2s}{)}^{-1}$ and $a_5=0$.

At $p=7$, the possible cubes are $0,1,-1$, so $X^3-3$ has no root modulo $7$. Being cubic, it is irreducible; Frobenius is a three-cycle. Thereforeand $a_7=-1$.

The ramified factor at $3$ must be computed on inertia group invariants. Write $\alpha =3^{1/3}$ and set $\pi =({\zeta }_3-1)/\alpha$. Using $({\zeta }_3-1{)}^2=-3{\zeta }_3$ givesThe Eisenstein polynomial $X^6+3$ over ${\mathbb{Q}}_3$ has degree six, so ${\mathbb{Q}}_3(\pi )$ is a totally ramified degree-six subfield of ${\mathbb{Q}}_3({\zeta }_3,\alpha )$. The latter has degree at most $[F:\mathbb{Q}]=6$, so the fields agree. This Eisenstein sextic presentation of the splitting field of X3 minus 3 over Q3 shows $D_3=I_3=S_3$. Since the only permutation-invariant vectors in ${\mathbb{C}}^3$ have all coordinates equal, their intersection with $V$ is zero. ThusIn particular $a_3=a_9=0$; an unramified degree-two factor at $3$ would give the wrong coefficients.

Finally the Euler product makes its Dirichlet series coefficients multiplicative for coprime indices. Its constant coefficient is $a_1=1$, and $a_6=a_2{a}_3=0$, $a_{10}=a_2{a}_5=0$. Every integer up to ten is now covered, giving

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.

Let $a=\sqrt[3]{3}$ and let $\zeta$ be a primitive cube root of unity. The splitting field is $L={\mathbb{Q}}_3(a,\zeta )$. The polynomial $X^3-3$ is an Eisenstein polynomial, so $[{\mathbb{Q}}_3(a):{\mathbb{Q}}_3]=3$. Meanwhile $(1+2\zeta {)}^2=-3$ shows that ${\mathbb{Q}}_3(\zeta )$ is quadratic: $-3$ is not a square because its valuation is odd. A quadratic field cannot lie in the degree-three field. Consequently $[L:{\mathbb{Q}}_3]=6$, and the action on the three roots identifies its Galois group with the symmetric group $S_3$.

A useful uniformizer isSince $(\zeta -1{)}^3=-3\zeta (\zeta -1)$, we obtainThus $\pi$ is a root of the Eisenstein polynomial $X^6+3$. Its degree is six, so $L={\mathbb{Q}}_3(\pi )$, the extension is totally ramified, and $\pi$ is a uniformizer. With $v_L(\pi )=1$,This is the Eisenstein sextic presentation of the splitting field of X3 minus 3 over Q3.

Let $\tau (a)=\zeta a$, $\tau (\zeta )=\zeta$, and let $s(a)=a$, $s(\zeta )={\zeta }^{-1}$. Then $\tau$ has order three, $s$ has order two, and $s\tau s={\tau }^{-1}$. Their actions on the uniformizer areThe two nonidentity elements of $⟨\tau ⟩$ therefore satisfyEach transposition sends $\pi$ to $-{\zeta }^j\pi$ for some $j$, so its displacement has valuation one: $1+{\zeta }^j$ is a unit, reducing to $2$ modulo the maximal ideal. The uniformizer criterion for lower ramification groups now givesFor real indices, this is $S_3$ on $[-1,0]$, $C_3$ on $(0,3]$, and $1$ for $t>3$. In particular the lower breaks are $0$ and $3$.

On $(0,3]$ the index $[G_0:G_t]$ is two; beyond three it is six. Thus the Herbrand function isThe positive lower break $3$ becomes the upper break $3/2$. ThereforeThe fractional upper break is allowed because the full extension is nonabelian. As a check, the different exponent from ramification groups is $5+3\cdot 2=11$. The derivative of the minimal polynomial $X^6+3$ gives the same result: $v_L(6{\pi }^5)=6+5=11$. This is the $p=3$ case of ramification groups of the splitting field of Xp minus p over the p-adic numbers.