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For the standard representation of the symmetric group $S_3$, the unramified local factor of an Artin L-function, with $t=p^{-s}$, is respectively the displayed factor for identity, transposition, and three-cycle Frobenius cycle types. These follow by removing the trivial summand from the three-point permutation representation. If the inertia group contains a three-cycle, its invariant subspace in the standard representation is zero, so the ramified local factor is one.

Modulo two the polynomial becomes $h=X^5+X^2+1$. It has neither zero nor one as a root. The only irreducible quadratic over ${\mathbb{F}}_2$ is $X^2+X+1$; modulo it, $X^3=1$ and $h\equiv 1$. A reducible quintic has a factor of degree at most two, so $h$ is irreducible. It is also separable: $h^{\prime }=X^4$, which is coprime to $h$. Thus the rational polynomial is irreducible, and its group $G\le S_5$ is transitive.

Modulo three there is the exact factorizationThe quadratic has no root. The cubic takes value one at each of $0,1,2$, so it too has no root and is irreducible. The two factors are distinct and coprime. The Frobenius cycle type therefore supplies an element with disjoint cycles of lengths two and three. Its cube is a single transposition.

A transitive action of prime degree is primitive: a block size divides five, so the only sizes are one and five. To finish without classifying all degree-five groups, apply the primitive group with a transposition argument. Form the graph of all conjugate transpositions in $G$. Its connected components are blocks, so it is connected. Transpositions along the edges of a connected graph generate every transposition: along a path, conjugating adjacent swaps yields the swap of its endpoints. Hence they generate $S_5$, and

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

Modulo three, the quintic has no linear factor: its value is $-1$ at every element of ${\mathbb{F}}_3$. The monic irreducible quadratics over this field are $t^2+1$, $t^2+t+2$ and $t^2+2t+2$. Division leaves respectively $-1$, $t-1$ and $t-1$, all nonzero. A reducible quintic has a factor of degree at most two, so it is irreducible modulo three and therefore over $\mathbb{Q}$. Its Galois group is transitive on five roots.

Modulo two its squarefree factorization isBoth factors are irreducible, so the Frobenius cycle type theorem gives an element that is a disjoint transposition times a three-cycle. Cubing it gives a transposition.

The prime-degree transposition criterion can be proved here: a transitive subgroup of prime degree containing a transposition is the whole symmetric group. Construct a graph on the five roots whose edges are the pairs swapped by all conjugates of this transposition. The group acts transitively on vertices and permutes connected components, so all component sizes are equal and divide five. An edge rules out size one; the graph is connected. Edge transpositions of any connected graph generate the full symmetric group, for instance by transporting swaps along a spanning tree. All belong to the Galois group, provingThe prime-degree condition is essential to the short transposition argument.

The quartic is irreducible by Eisenstein criterion at two, so its Galois group is transitive on its four roots. Its discriminant is $1616=16\cdot 101$, so five and seven are unramified primes for the polynomial. The squarefree factorizations areThe cubic modulo five has no root in ${\mathbb{F}}_5$, hence is irreducible. The quadratic modulo seven has discriminant five, a nonsquare there. The Frobenius cycle type theorem therefore supplies a three-cycle and a transposition in the group.

Transitivity implies that its order is divisible by four, and the three-cycle implies divisibility by three. Its order divides $24$, so it is $12$ or $24$. The only subgroup of index two in $S_4$ is $A_4$: an index-two subgroup is the kernel of a group homomorphism of a nontrivial homomorphism to $C_2$, and all transpositions are conjugate and generate $S_4$, forcing this homomorphism to be the sign. The exhibited transposition excludes $A_4$. Hence

Orbits of the root action are precisely roots belonging to the same irreducible factor, proving transitivity iff irreducibility.

For a same-degree counterexample, let $L=\mathbb{Q}(\sqrt{2},\sqrt{3})$,Both split in $L$, but the first action has two orbits of size two while the second is transitive, so the subgroups are not conjugate in $S_4$.

Over a finite field, Frobenius cycle type proves the assertion. The given quartic reduces over ${\mathbb{F}}_2$ to $X^4+X^3+1$, which is irreducible, so its group there is cyclic of order four generated by a $4$-cycle.

Over $\mathbb{Q}$ the polynomial is irreducible by the same reduction. Modulo $3$ it factors as linear times irreducible cubic, so Dedekind factorization cycle test gives a $3$-cycle; modulo $2$ it gives a $4$-cycle. Its discriminant is $229$, and modulo $23$ its factor degrees are $1,1,2$, giving a transposition. These cycle types force the transitive group to be $S_4$.

Let $R$ be an integrally closed domain and $f\in R[X]$ monic, with discriminant outside a prime ideal $P$. The roots stay distinct after reduction. In the splitting algebra choose a prime above $P$. Its decomposition group maps onto the Galois group of the residue splitting field; its inertia group fixes every reduced root and is trivial because reduction bijects the distinct root sets. Thus the residue Galois group embeds in the generic group as a permutation group. For finite residue fields, irreducible factor degrees give the Frobenius cycle type.