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Use the Fourier transform convention $\hat{m}(t)=\int e^{-itx}dm(x)$. The complex measure is finite because ${\sum }_n|f(n)|n^{-\sigma }<\infty$. The Möbius divisor-sum identity, proved in the preceding solution, allows multiplication of absolutely convergent Dirichlet series and givesHere $\zeta$ is the Riemann zeta function. We prove the estimates in the intended range $1<\sigma \le 2$. A uniform estimate of the printed form for all $\sigma >1$ would be false: at $t=0$, $F(\sigma )\to 1$ as $\sigma \to \infty$, while its proposed upper bound tends to zero. A version valid for every $\sigma >1$ replaces $(\sigma -1{)}^{-3/4}$ by $1+(\sigma -1{)}^{-3/4}$.

For completeness, unique prime factorization and absolute convergence give $\zeta (s)={\prod }_p(1-p^{-s}{)}^{-1}$ andas the logarithm defined by that Euler product. The elementary nonnegative trigonometric polynomialtherefore proves the three-four-one inequality for Euler productsWe also need elementary uniform bounds for derivatives of the Riemann zeta function. For an integer $H\ge 1$, summation by parts givesThis continues the function to $\Re s>0$ with just the pole at one. For $1\le \sigma \le 2$, $|t|\ge 1$, choose $H=\lceil 2+|t|\rceil$. Differentiating $j$ times, the finite Dirichlet polynomial is bounded by ${\sum }_{n\le H}(\log n{)}^j/n{\ll }_j{\log }^{j+1}(2+|t|)$. Derivatives of the second term have the same bound, since $|s-1|\ge 1$. In the last term, differentiated integrands are bounded by $u^{-\sigma -1}(\log u{)}^j$, whose integral is $O_j(H^{-\sigma }{\log }^j(2H))$; multiplication by $|s|\ll H$ is harmless. It follows thatAs $\zeta (\sigma )\le 1+(\sigma -1{)}^{-1}$, the three-four-one inequality for Euler products gives, for $|t|\ge 1$,To handle bounded $t$, first note that $\zeta$ cannot vanish at $1+i{t}_0$ for $t_0\ne 0$: a zero of order $m\ge 1$ would make the three-four-one product $O((\sigma -1{)}^{4m-3})\to 0$, contradicting its lower bound one. At $s=1$, the reciprocal $F$ extends holomorphically with a simple zero, since the pole of $\zeta$ has residue one. Thus $F$ and all its derivatives are bounded on the compact rectangle $1\le \sigma \le 2$, $|t|\le 1$. This proves the requested reciprocal zeta bounds near the line one:

For the derivative estimate, repeatedly differentiate $F=1/\zeta$. Every term of $F^{(k)}$ is a constant timesThe preceding derivative bounds and reciprocal estimate bound this by $O_k((\sigma -1{)}^{-3(k+1)/4}{\log }^{3k+1}(2+|t|))$. Compact-height bounds supply the remaining range. Since ${\partial }_t^{k}F(\sigma +it)=i^k{F}^{(k)}(\sigma +it)$, this has exactly the requested dependence, with an explicit logarithmic exponent linear in $k$ for $k\ge 1$.

Here is the unsmoothing a logarithmically weighted sum argument with the normalization repaired. For real $v$,Indeed the Fourier transform of $u{e}^{-\sigma u}{1}_{u\ge 0}$ is $(\sigma +it{)}^{-2}$, so Fourier inversion gives the identity, also at $v=0$ by continuity. In a vertical contour integral the coefficient is $1/(2\pi i)$ and the differential is $ds=idt$. The printed real-$t$ integral has $dt$ with $1/(2\pi i)$ and is missing this factor of $i$.

Set $a_m=\mu (m)(\log m{)}^k$, taking $(\log 1{)}^0=1$ when $k=0$. Its Dirichlet series is $(-1{)}^k{F}^{(k)}(s)$. The logarithmically smoothed Perron formula, justified by absolute convergence, givesFor $x\ge 3$ choose $\sigma =1+1/\log x$. The integral of ${\log }^{3k+1}(2+|t|)/|\sigma +it{|}^2$ is finite, with a constant depending only on $k$, and $x^{\sigma }=ex$. Thus, with $\beta =3(k+1)/4$,For $0<h\le 1$, subtraction gives the exact finite-difference identityThe last sum is $O_k((x{h}^2+h)(\log x{)}^k)$, since there are $O(xh+1)$ terms. ConsequentlyFor $k\ge 3$, choose $h=(\log x{)}^{(\beta -k)/2}$ to get exponent $(\beta +k)/2=7k/8+3/8$. For $0\le k<3$, the trivial bound $O_k(x(\log x{)}^k)$ is already no larger than that estimate. The extra $(\log x{)}^k$ term is absorbed for fixed $k$. HenceThis also yields the cancellation needed for the Prime number theorem. Choose a fixed $k>3$, discard $m\le \sqrt{x}$ at cost $O(\sqrt{x})$ for the unweighted sum, and use partial summation with $(\log m{)}^{-k}$ on the rest. The displayed bound gives $M(x)=O_k(x(\log x{)}^{-k/8+3/8})=o(x)$; the preceding solution then gives $\psi (x)\sim x$.

For $\Re s>1$, the Euler products areThe Dirichlet character is completely multiplicative, and unique prime factorization expands each finite product into the terms of the corresponding Dirichlet series supported on those primes. Absolute convergence permits passage to all primes. More strongly, the absolutely convergent Euler product logarithm givessince the sum of absolute values is bounded by ${\sum }_{n\ge 2}{\sum }_{m\ge 1}{n}^{-m\Re s}/m<\infty$.

We prove nonvanishing of a nonprincipal Dirichlet L-function at one, including real Dirichlet characters. If $\chi$ is nonprincipal modulo $q$, periodicity and the given Orthogonality of Dirichlet characters bound $A_{\chi }(x)={\sum }_{n\le x}\chi (n)$ by $q$. Partial summation giveswhich is holomorphic for $\Re s>0$ by locally uniform convergence, including after differentiation. The principal Dirichlet character instead haswith a simple pole at one.

For a nonreal Dirichlet character, consider $P(s)={\prod }_{\chi \mathrm{mod}q}L(s,\chi )$. For real $\sigma >1$, its Euler product logarithm isThe given character inversion with $a=1$ makes each inner sum either $\phi (q)$ or zero, so $P(\sigma )\ge 1$. If $L(1,\chi )=0$ for a nonreal Dirichlet character, then $L(1,\overline{\chi })=0$ as well, since their values at real $s$ are conjugate. These are two distinct factors. All nonprincipal factors are holomorphic at one, and only the principal factor has a simple pole. Their two zeros would force $P(\sigma )\to 0$, a contradiction.

For a real nonprincipal Dirichlet character, suppose $L(1,\chi )=0$. Then $G(s)=\zeta (s)L(s,\chi )$ is holomorphic throughout $\Re s>0$: the supplied continuation of the Riemann zeta function has no other pole there, and the zero cancels the pole at one. On $\Re s>1$ its Dirichlet series has coefficientsTo check the sign explicitly, for a prime power $p^j$ the local sum is $j+1$ if $\chi (p)=1$, is $1$ for even $j$ and $0$ for odd $j$ if $\chi (p)=-1$, and is $1$ if $\chi (p)=0$. Multiplication over prime factors proves nonnegativity and $g_{m^2}\ge 1$. Also $g_n\le \tau (n)$, so termwise derivatives at $s=2$ are justified by absolute convergence:The disc $|s-2|<2$ lies in $\Re s>0$. Its Taylor series therefore converges at $s=1/2$, and Tonelli theorem givesThis contradicts holomorphy at $1/2$. It is the positive-coefficient argument behind the Landau theorem for a Dirichlet series with nonnegative coefficients, proved here rather than invoked. Thus $L(1,\chi )\ne 0$ for every nonprincipal Dirichlet character.

It remains to detect the primes. For real $\sigma >1$, the Euler product logarithm satisfiesuniformly as $\sigma \downarrow 1$: terms with $m\ge 2$ are bounded in absolute value by ${\sum }_{n\ge 2}{\sum }_{m\ge 2}{n}^{-m}/m<\infty$. For a nonprincipal Dirichlet character, a holomorphic logarithm of $L$ exists near one by the nonvanishing just proved. On a short real interval it differs from the Euler product logarithm by a fixed multiple of $2\pi i$, so the prime sum stays bounded. For the principal Dirichlet character, the simple pole instead givesApplying the supplied Orthogonality of Dirichlet characters for a reduced residue class $a$ now yieldsThe right side tends to infinity, whereas a finite set of primes would give a bounded left side. Therefore every residue class coprime to $q$ contains infinitely many primes, proving the Dirichlet theorem on primes in arithmetic progressions.

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

For $\sigma =\mathrm{Re}s>1$, the Dirichlet series $\zeta (s)={\sum }_{n\ge 1}{n}^{-s}$ is absolutely convergent. A finite product over primes $p\le P$ can be expanded into absolutely convergent geometric series. By unique prime factorization,The omitted integers are all greater than $P$, so the difference from $\zeta (s)$ has modulus at most ${\sum }_{n>P}{n}^{-\sigma }\to 0$. ThusThis defines the infinite Euler product as the limit of finite products. Its absolute convergence also follows from ${\sum }_p{p}^{-\sigma }<\infty$, with $\log (1-p^{-s}{)}^{-1}=p^{-s}+O(p^{-2\sigma })$ for sufficiently large $p$.

For the unit-modulus prime values in this question, the Granville-Soundararajan distance isThe second identity uses $|f(p)|=|g(p)|=1$. Thus it is the Euclidean distance between the finite weighted vectors $(f(p)/\sqrt{2p}{)}_{p\le X}$ and $(g(p)/\sqrt{2p}{)}_{p\le X}$. Nonnegativity, symmetry and the triangle inequality follow from that norm representation, and it vanishes exactly when the two prime-value vectors agree. This proves the prime-restriction metric for pretentious distance.

On whole arithmetic functions at fixed $X$ it is a pseudometric. For example $f=1$ and $g={\mu }^2$ have identical values at every prime and zero distance for every $X$, but different values at four. Thus the literal identity-of-indiscernibles assertion needs the prime-restriction quotient.

Since $\mu (p)=-1$,For completeness, partial summation and the prime-counting form of the Prime number theorem giveTo verify the constant and the error, replace $\pi$ by $\mathrm{Li}+E$ in the first expression. Differentiating $\mathrm{Li}(u)/u$ shows its two contributions combine to $\log \log X$ plus a constant. The error integral converges absolutely because $|E(u)|/u^2\ll e^{-c\sqrt{\log u}}/u$, and $E(X)/X\to 0$. This proves the Mertens second theorem in the form needed here. In particular the pretentious distance between one and the Möbius function satisfies

For the last request, retain the earlier prime normalization $|f(p)|=1$; the proof also works with the weaker assumption $|f(p)|\le 1$. Without a bound on prime values, the original definition and the absolute convergence of the displayed measure need not apply to an unrestricted multiplicative arithmetic function. For example $f(n)=\mu (n{)}^{2}n$ is multiplicative and squarefree-supported, but for $X>e$ the displayed measure has infinite positive mass: its prime terms alone are ${\sum }_p{p}^{-1/\log X}$, which diverges by the Prime number theorem.

Put $L=\log X$, $\sigma =1+1/L$, and use the Fourier transform convention $\hat{\nu }(t)=\int e^{-itu}d\nu (u)$. Since $f$ is supported on squarefree integers, the defining property of a multiplicative arithmetic function gives the absolutely convergent Euler productThe bound on prime values gives $|f(n)|\le 1$ for squarefree $n$, so the total variation norm of a measure is bounded by $\zeta (\sigma )$; thus $\nu$ is a finite measure. Expanding the logarithm of the modulus, with an absolute $O(1)$ remainder since ${\sum }_p{p}^{-2\sigma }\ll 1$, givesAll constants here are uniform in $t$. Moreoverusing the pole of the Riemann zeta function. The nonnegative deficits $1-\mathrm{Re}(f(p)p^{-it})$ giveIndeed, the replacement of $p^{-\sigma }$ by $1/p$ for $p\le X$ costs at mostby the Mertens first theorem, while discarding the $p>X$ terms costs nothing in this lower bound. Thus the large Fourier coefficient criterion for pretentiousness isIn particular,The comparison function is the Archimedean character $n^{it}$ with this sign because our Fourier transform uses $e^{-itu}$.

For $\mathrm{Re}s>1$, define the Riemann zeta function by the absolutely convergent Dirichlet seriesThe Euler product follows from unique prime factorization and absolute convergence. The counting-function integral givesBecause the fractional part is bounded, the last integral, and its derivatives with respect to $s$ on compact subsets, converge uniformly for $\mathrm{Re}s>0$. It is a holomorphic function there. This proves the Meromorphic continuation of the Riemann zeta function to the right half-plane, with a simple pole at one of residue one and no other singularity in that half-plane.

The Gamma function isThe Gamma function recurrence extends it meromorphically; its poles are at the nonpositive integers and it has no zeros. The functional equation of the Riemann zeta function can be stated asor, equivalently,These are identities of meromorphic functions; the second also extends the Riemann zeta function to the remaining half-plane. The critical strip is $0<\mathrm{Re}s<1$; its boundary lines will also be treated below, rather than included among possible exceptional zeros.

For $\mathrm{Re}s>1$, the absolutely convergent reciprocal Euler product gives $\zeta (s)\ne 0$. For $\mathrm{Re}s<0$, the factors $2^s$, ${\pi }^{s-1}$, $\Gamma (1-s)$ and $\zeta (1-s)$ in the second functional equation of the Riemann zeta function are finite and nonzero. Hence the only zeros there are the zeros of the sine factor:Each is simple. At zero, the sine zero cancels the pole of $\zeta (1-s)$; using its residue one gives $\zeta (0)=-1/2$, not zero. Nonvanishing on the rest of $\mathrm{Re}s=0$ follows from the boundary-line proof below and the functional equation of the Riemann zeta function. Together these facts prove that the trivial zeros of the Riemann zeta function are the only zeros outside the open critical strip.

Here is a Jensen disk proof of the zeta zero-count bound which avoids any unproved left-half-plane growth estimate. Set $F(s)=(s-1)\zeta (s)$, a holomorphic function throughout $\mathrm{Re}s>0$, including at one. The integral continuation formula givesFor each integer $j$, apply Jensen's formula in the disk with centre $2+ij$, outer radius $R=7/4$ and inner radius $r=8/5$. The outer disk stays in $\mathrm{Re}s\ge 1/4$, so its maximum modulus is $O((|j|+2{)}^2)$. At its centre,since $|1/\zeta (2+ij)|\le {\sum }_n|\mu (n)|n^{-2}\le \zeta (2)$. Therefore the number of zeros in its inner disk, counted with multiplicity, is at mostThe rectangle $1/2\le \mathrm{Re}s\le 1$, $|\mathrm{Im}s-j|\le 1/2$ fits in the inner disk because its farthest point has distance $\sqrt{(3/2{)}^2+(1/2{)}^2}=\sqrt{5/2}<8/5$. Summing over $O(T)$ such disks counts $O(T\log T)$ zeros in the right half of the critical strip up to height $T$. The first functional equation of the Riemann zeta function bijects zeros, with multiplicities, in the left half with their reflections $s\mapsto 1-s$ in the right half; its Gamma function factors are finite and nonzero in the strip. ThusThe pole of zeta function at one does not count as a zero: $F(1)=1$.

Finally, for real $\sigma >1$ and $t\ne 0$, the Euler-product logarithms and the nonnegative trigonometric polynomialgive the three-four-one product proof of zeta boundary nonvanishing:If $\zeta (1+it)$ had a zero of order $a\ge 1$, its factor would be $O((\sigma -1{)}^{4a})$ as $\sigma \downarrow 1$. The real zeta function factor has pole order three, and the factor at $1+2it$ stays bounded because $t\ne 0$. The product would tend to zero as $O((\sigma -1{)}^{4a-3})$, contradicting that it is at least one. ThereforeAt $t=0$ there is a pole, not a zero. Applying the sine-form functional equation of the Riemann zeta function at $s=it\ne 0$ now proves nonvanishing on the remaining imaginary axis, completing the earlier assertion about all zeros outside the critical strip.

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.