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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.

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

The three quadratic subfields are $\mathbb{Q}(\sqrt{-2})$, $\mathbb{Q}(\sqrt{-3})$ and $\mathbb{Q}(\sqrt{6})$, with respective fundamental discriminants $-8,-3,24$. Their associated quadratic Dirichlet characters are the Kronecker symbolsThe regular representation of $C_2\times C_2$ is the direct sum of the trivial character and these three nontrivial characters. At an unramified prime the character is the sign by which Frobenius automorphism acts on the square root defining its quadratic subfield; at a ramified prime the linear character has no inertia group invariants. The local factor of an Artin L-function consequently givesThe zero values of a quadratic Dirichlet character at conductor primes mean that its local factor is one, not that the whole Artin L-function vanishes there.

Only the rational primes $2,3,5,7$ can contribute to coefficients with index at most ten. Their character values areFor the odd unramified primes these entries follow by checking squares: modulo five, both $-8$ and $-3$ are the nonsquare $2$, while $24$ is the square $4$; modulo seven they are respectively $6,4,3$. At $3$, $-8\equiv 1$ is a square. At $2$, the $-3$ character is $-1$ because the integral quadratic generator has polynomial $X^2+X+1$, irreducible modulo two. The other entries at $2$ and $3$ are zero exactly where the corresponding fundamental discriminant is divisible by that prime. This verifies every local factor, including the ramified ones.

The Euler product makes $a_n$ a multiplicative arithmetic function. Expanding the first two factors givesTherefore $a_2=0$, $a_4=1$, $a_8=0$, $a_3=2$ and $a_9=3$. Both factors for $5$ and $7$ have zero coefficient of $T$, so $a_5=a_7=0$. Multiplicativity gives $a_6=a_2{a}_3=0$ and $a_{10}=a_2{a}_5=0$, while $a_1=1$. The answer isAs an arithmetic check, the regular representation Artin L-function is the Dedekind zeta function of $F$. The local factors say there is one prime of norm four above two and two primes of norm three above three. Their powers give respectively one ideal of norm four and three ideals of norm nine, matching the nonzero coefficients above.

We use the Dirichlet density form of the Chebotarev density theorem, the version directly proved by the permitted information at $s=1$. If $L/K$ is a finite Galois extension of number fields with group $G$ and $C$ is a conjugacy class, let $S_C$ be the unramified prime ideals of $K$ whose Frobenius conjugacy class is $C$. ThenThe same conclusion holds for a union of conjugacy classes, replacing $|C|$ by the size of that union. The denominator may equivalently be the sum over all primes of $K$. Here and below $s>1$ is real. This specifies the density being proved; an unweighted prime-counting asymptotic is a stronger formulation and is not being inferred merely from a limit at one.

First establish the necessary pole for a general base field. The Dedekind zeta function as a permutation Artin L-function gives it using exactly the allowed assumptions. Take a finite normal closure $E$ of $K/\mathbb{Q}$, and let $W$ be the permutation representation on ${\mathrm{Hom}}_{\mathbb{Q}}(K,E)$. Its trivial constituent has multiplicity one: an invariant function on the transitive set of embeddings is constant. At a rational prime, a basis of $W^I$ consists of the sums over inertia orbits of embeddings. Frobenius automorphism permutes these orbit sums. Within the decomposition-group orbit belonging to a prime $\mathfrak{p}$ of $K$, the resulting cycle has length $f_{\mathfrak{p}}$, because the residue embeddings are permuted cyclically by finite-field Frobenius. A cycle of length $f$ has determinant $\det (1-T\rho (\mathrm{Frob}))=1-T^f$. Consequently the local factor of an Artin L-function givesincluding at ramified primes. Decomposing $W=1\oplus {⨁}_{\psi \ne 1}{m}_{\psi }\psi$ givesAll factors after $\zeta$ are holomorphic and nonzero at one by the allowed hypothesis. Thus ${\zeta }_K$ has a simple pole there andThis argument also explains why the trivial relative Artin L-function is ${\zeta }_K$, rather than always the Riemann zeta function.

Now let $\chi$ be an irreducible character of $G$, of degree $d_{\chi }$. At an unramified prime the Euler product has local factor $\det (1-{\rho }_{\chi }(g_{\mathfrak{p}})(N\mathfrak{p}{)}^{-s}{)}^{-1}$, where $g_{\mathfrak{p}}$ is a Frobenius representative. The eigenvalues of ${\rho }_{\chi }(g)$ are roots of unity, since $g$ has finite order. Expanding the logarithm for $s>1$ therefore giveswhere the finitely many ramified local factors contribute a function bounded as $s\downarrow 1$. They are nonzero and finite near one because their eigenvalues have modulus one while $(N\mathfrak{p}{)}^{-1}<1$.

The terms $m\ge 2$ are uniformly bounded. Indeed $|\chi (g)|\le d_{\chi }$, and for $N\mathfrak{p}\ge 2$ and $s\ge 1$,There are at most $[K:\mathbb{Q}]$ primes of $K$ above each rational prime and each has norm at least that prime. Hence ${\sum }_{\mathfrak{p}}(N\mathfrak{p}{)}^{-2}<\infty$. Isolating $m=1$ proves the character prime sums in the Chebotarev proof identityFor nontrivial $\chi$, the allowed holomorphy and nonvanishing at one give a local holomorphic logarithm and a bounded right side. The logarithm specified by the Euler series differs from that local branch by a constant multiple of $2\pi i$ on the real interval near one, so it too is bounded. For the trivial character, the right side is $\log (1/(s-1))+O(1)$ by the pole of ${\zeta }_K$. Thus

Expand the conjugacy-class indicator by character orthogonality. For $c\in C$,To see the coefficient, take its inner product with $\chi$: $|G{|}^{-1}{\sum }_{g\in C}\overline{\chi (g)}=|C|\overline{\chi (c)}/|G|$. The irreducible characters form an orthonormal basis of class functions, so this is the full expansion. Substituting the prime-sum estimates leaves only the trivial character's logarithmic term:Division by the diverging logarithm proves the theorem. The same trivial-character calculation shows that all primes of $K$ have density one; adding or removing finitely many primes does not change a Dirichlet density.

There is a necessary qualification for the final polynomial application: the printed claim is false for degree one. For example, $X$ is monic and irreducible in $\mathbb{Z}[X]$, yet has the root zero modulo every prime. We prove the intended assertion for degree $n\ge 2$.

Let $L$ be the splitting field of the monic irreducible polynomial over $\mathbb{Q}$, and let $\Omega$ be its $n$ roots. The Galois group $G$ acts transitively on $\Omega$. There is a derangement in a transitive group action: count the pairs $(g,\alpha )$ satisfying $g\alpha =\alpha$. The orbit-stabilizer theorem givesIf every $g$ fixed at least one root, the identity, which fixes $n\ge 2$, would make this sum at least $|G|+n-1$, a contradiction. Therefore some $g$ fixes no root, and neither does any conjugate of $g$.

For completeness, relate this group action to reduction without assuming the desired splitting result. All roots are algebraic integers, since the polynomial is monic. Exclude the finitely many primes ramifying in $L$ or dividing the nonzero polynomial discriminant. At a prime $\mathfrak{q}$ over any remaining $p$, the roots reduce to distinct elements of $k_{\mathfrak{q}}$ and give all roots of the reduced polynomial. Arithmetic Frobenius permutes them by $b\mapsto b^p$. A reduced root belongs to ${\mathbb{F}}_p$ exactly when it is fixed by this map. Because reduction is injective on the original root set, this happens exactly when the Frobenius permutation fixes an original root. This proves the Frobenius permutation and roots modulo a prime criterion.

The Chebotarev density theorem assigns the conjugacy class of our fixed-point-free $g$ the positive Dirichlet density $|C_g|/|G|$. All but finitely many of these primes have rootless reduction, so there are infinitely many. More precisely, applying the same argument to the union of all fixed-point-free classes gives