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Assume that every nontrivial one-dimensional Artin L-function over the intermediate fields is holomorphic and nonzero at one, while the trivial one is the base Dedekind zeta function with a simple pole. By Brauer induction, $\chi ={\sum }_j{n}_j{\mathrm{Ind}}_{H_j}^G{\psi }_j$ with integers $n_j$ and linear characters ${\psi }_j$. Compatibility of Artin L-functions with induced representations expresses $L_K(\chi ,s)$ as the corresponding product of $L_{L^{H_j}}({\psi }_j,s{)}^{n_j}$. Its order at one is $-{\sum }_j{n}_j⟨{\psi }_j,1_{H_j}⟩$. Frobenius reciprocity turns this sum into $-⟨\chi ,1_G⟩$. Hence every nontrivial irreducible character gives order zero and a holomorphic nonzero function near one. This local conclusion does not assert global Artin holomorphy.

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.

Let $E$ be a finite normal closure of the number field $K$. The permutation representation of its Galois group on the embeddings of $K$ into $E$ has one trivial constituent, because the action is transitive. At a prime, the inertia group invariant basis consists of sums over inertia orbits of embeddings; Frobenius automorphism permutes them in cycles of lengths equal to the residue degrees of the primes of $K$. A cycle of length $f$ contributes $\det (1-T\rho (\mathrm{Frob}))=1-T^f$. Hence the local factors equal those of the Dedekind zeta function, including at ramified primes. If nontrivial irreducible Artin L-functions are holomorphic and nonzero at one, this factorization and the simple pole of the Riemann zeta function show that ${\zeta }_K$ has a simple pole there.

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 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.

For $\mathrm{Re}s>1$, the Dedekind zeta function of $K$ isUnique factorization of ideals gives its Euler productover the nonzero prime ideals of ${\mathcal{O}}_K$.

For each rational prime $p$, defineThese are the three cases in splitting of rational primes in a quadratic field.

Set $x=p^{-s}$. In the split case the local factor of the Dedekind zeta function is $(1-x{)}^{-2}$; in the inert case it is $(1-x^2{)}^{-1}$ because the prime ideal has norm $p^2$; and in the ramified case it is $(1-x{)}^{-1}$. On the other hand, the product of the local factors of ${\zeta }_{\mathbb{Q}}(s)$ and $L(\chi ,s)$ iswhich gives exactly those three expressions when $\chi (p)=1,-1,0$. Multiplying over all rational primes proves the quadratic Dedekind zeta factorizationformally.

The Dedekind zeta function of $K$ isfor $\mathrm{Re}s>1$.