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For a finite Galois extension $L/K$ and a conjugacy class $C$ in $G=\mathrm{Gal}(L/K)$, the unramified primes with Frobenius conjugacy class $C$ have Dirichlet density $|C|/|G|$. In particular,One proof uses Brauer induction, the Artin L-function order at one, and character orthogonality. The trivial Artin L-function is the Dedekind zeta function of $K$. The full theorem also gives the natural density, but this logarithmic estimate alone establishes the Dirichlet density assertion.

A prime ideal of a number field splits completely in a finite extension of degree $n$ when it factors into $n$ distinct prime ideals, each of residue-field degree one. In a Galois extension, an unramified prime ideal splits completely exactly when its Frobenius conjugacy class is the identity class. The Chebotarev density theorem then gives these primes Dirichlet density $1/n$.

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

Let $F/K$ be a finite Galois extension, choose a prime ideal $\mathfrak{q}$ above $\mathfrak{p}$, and write $k_{\mathfrak{q}}$ and $k_{\mathfrak{p}}$ for the residue fields. The decomposition group and inertia group areThe Frobenius automorphism on the finite field $k_{\mathfrak{q}}$ is $x\mapsto x^{N\mathfrak{p}}$. Its inverse image defines a distinguished coset in $D_{\mathfrak{q}}/I_{\mathfrak{q}}$; a Frobenius element is a lift of that coset. At an unramified prime, $I_{\mathfrak{q}}=1$ and the lift is unique. Changing the prime above $\mathfrak{p}$ conjugates it, so a Frobenius conjugacy class is intrinsically attached to $\mathfrak{p}$. These are arithmetic Frobenius conventions.

We prove the Dirichlet density form of the Chebotarev density theorem. Let $L/K$ be a finite Galois extension with Galois group $G$, let $C$ be a conjugacy class in $G$, and let $S_C$ consist of the unramified prime ideals of $K$ whose Frobenius conjugacy class is $C$. The assertion isThe same result holds for any union of conjugacy classes, by addition. The following proof uses exactly the allowed one-dimensional Artin L-function behavior at $1$ and the permitted induction and Euler-product properties.

First determine the order of vanishing at $1$ for an arbitrary character of a representation $\chi$ of $G$. The Brauer induction theorem supplies subgroups $H_j\le G$, linear characters ${\psi }_j$ of $H_j$, and integers $n_j$, possibly negative, such thatPut $K_j=L^{H_j}$. Compatibility of Artin L-functions with induced representations and direct sums gives, initially for $\mathrm{Re}s>1$,Each ${\psi }_j$ is a one-dimensional Galois representation for $L/K_j$, so the stated analytic assumptions apply to every factor. A trivial ${\psi }_j$ contributes a simple pole; a nontrivial ${\psi }_j$ is holomorphic and nonzero at one. Thus the product is meromorphic near one andThe character inner product is $⟨\alpha ,\beta {⟩}_G=|G{|}^{-1}{\sum }_{g\in G}\alpha (g)\overline{\beta (g)}$. Applying Frobenius reciprocity to each induced term givesIn particular, a nontrivial irreducible character has order zero, so its Artin L-function is holomorphic and nonzero near one. For the trivial character, $L_K(1_G,s)={\zeta }_K(s)$, the Dedekind zeta function of the base field $K$, and its pole is simple. This argument establishes the behavior needed at one, not holomorphy everywhere for arbitrary Artin L-functions.

Next convert that order calculation into a prime sum. For an irreducible character $\chi$, choose a unitary representation of dimension $d_{\chi }$ affording it, and choose a Frobenius automorphism $g_{\mathfrak{p}}$ at each unramified prime. The logarithm specified by the absolutely convergent Euler product gives, for real $s>1$,Only finitely many ramified primes occur. Their Euler-factor logarithms $R_{\chi }(s)$ stay bounded near one, since the relevant eigenvalues have modulus one and $(N\mathfrak{p}{)}^{-1}<1$. Also $|\chi (g)|\le d_{\chi }$, so the contribution with $m\ge 2$ is uniformly bounded for $s\ge 1$ byConvergence follows, for example, from the convergent ideal series defining ${\zeta }_K(2)$. ThereforeWriting $a_{\chi }=⟨\chi ,1_G{⟩}_G$, the order computation means $L_K(\chi ,s)=(s-1{)}^{-a_{\chi }}{U}_{\chi }(s)$, where $U_{\chi }$ is holomorphic and nonzero near one. Choose its local logarithm. Along the real interval approaching one, this and the Euler-product logarithm differ only by a fixed multiple of $2\pi i$. ConsequentlyOnly the trivial irreducible character has $a_{\chi }=1$; all the others have $a_{\chi }=0$.

Finally, character orthogonality makes the irreducible characters an orthonormal basis of class functions. For $c\in C$, the coefficient of $\chi$ in the class indicator is $⟨1_C,\chi {⟩}_G=(|C|/|G|)\overline{\chi (c)}$. HenceSum this identity over the unramified primes with weight $(N\mathfrak{p}{)}^{-s}$. The finite character sum and the boxed estimates giveDividing by $\log (1/(s-1))$ proves the asserted Dirichlet density. The same Euler-product calculation with ${\zeta }_K$ gives ${\sum }_{\mathfrak{p}}(N\mathfrak{p}{)}^{-s}=\log (1/(s-1))+O(1)$, so using all primes as the denominator is equivalent. Finitely many ramified primes do not affect either limit. No prime-counting asymptotic or natural-density Tauberian step is required by this proof.

A set $S$ of prime numbers has Dirichlet density $\delta$ if the following limit exists:Equivalently, the denominator can be ${\sum }_p{p}^{-s}$, since that sum is $\log (1/(s-1))+O(1)$. Adding or removing finitely many prime numbers leaves the Dirichlet density unchanged.

Put $\alpha =\sqrt[4]{2}>0$ and $L=\mathbb{Q}(\alpha ,i)$. This is the splitting field of $X^4-2$, whose four roots of a polynomial are $\alpha ,i\alpha ,-\alpha ,-i\alpha$. The polynomial is an Eisenstein polynomial at $2$, so $[\mathbb{Q}(\alpha ):\mathbb{Q}]=4$. Since $\mathbb{Q}(\alpha )$ is contained in the real numbers, it does not contain $i$, giving $[L:\mathbb{Q}]=8$. Thus $L/\mathbb{Q}$ is a Galois extension of degree $8$. Its Galois group is the dihedral group of order $8$: the field automorphisms $r(\alpha )=i\alpha$, $r(i)=i$ and $s(\alpha )=\alpha$, $s(i)=-i$ satisfy $r^4=s^2=1$ and $srs=r^{-1}$.

For an odd prime number $p$, the condition $p\equiv 1(\mathrm{mod}4)$ means that the finite field ${\mathbb{F}}_p$ contains all fourth roots of unity. Under this condition, $2$ is a quartic residue exactly when $X^4-2$ has a root in ${\mathbb{F}}_p$; multiplying that root by $1,i,-1,-i$ gives all four distinct roots. Conversely, complete splitting of $X^4-2$ over ${\mathbb{F}}_p$ gives both a fourth root of $2$ and a primitive fourth root of unity, so it also forces $p\equiv 1(\mathrm{mod}4)$.

The polynomial discriminant of $X^4-2$ is $-2^{11}$. Hence every odd prime number is unramified in $L$, and the root-splitting criterion is equivalent to its Frobenius automorphism acting trivially on all the roots. Since the roots generate $L$, this is equivalent to the Frobenius automorphism being the identity, or to $p$ being a completely split prime of $L/\mathbb{Q}$. The Chebotarev density theorem says that the unramified prime numbers with Frobenius conjugacy class $C$ have Dirichlet density $|C|/|\mathrm{Gal}(L/\mathbb{Q})|$. Here $C=\{1\}$, so