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For every nonprincipal Dirichlet character, $L(1,\chi )\ne 0$. A nonreal character is covered by Euler product positivity for L-function nonvanishing at height zero, since its squared character is nonprincipal.

For a real character, suppose $L(1,\chi )=0$. The simple zeta pole would cancel, making $F(s)=\zeta (s)L(s,\chi )$ an entire function. Its nonnegative zeta-times-real-L coefficients satisfy $a_n\ge 0$ and $a_{m^2}\ge 1$. Termwise derivatives at two give $(-1{)}^k{F}^{(k)}(2)={\sum }_n{a}_n(\log n{)}^k{n}^{-2}$. The Taylor series of this entire function at two converges at zero. Every term there is nonnegative, and Tonelli theorem identifies its sum asThe last series diverges because all square coefficients are at least one. This contradiction proves the claim. It is the positive-coefficient mechanism behind the more general Landau theorem for a Dirichlet series with nonnegative coefficients.

For a positive integer $N$, a Dirichlet character $\psi$ modulo $N$ is a group homomorphism $(\mathbb{Z}/N\mathbb{Z}{)}^{\times }\to {\mathbb{C}}^{\times }$, extended periodically to the integers and set equal to zero on integers not coprime to $N$. Its nonzero values are roots of unity. For $\mathrm{Re}s>1$, its Dirichlet L-function is the absolutely convergent Dirichlet seriesComplete multiplicativity of $\psi$ proves the Euler product by expanding each factor as a geometric series and applying unique integer factorization; absolute convergence justifies these rearrangements.

Suppose $\psi$ is a nonprincipal Dirichlet character. Choose a unit $u$ with $\psi (u)\ne 1$. Multiplication by $u$ permutes a complete residue system, so ${\sum }_{a=1}^N\psi (a)=\psi (u){\sum }_{a=1}^N\psi (a)$, and the sum is zero. Consequently $A(x)={\sum }_{n\le x}\psi (n)$ is bounded: complete periods contribute zero and the remaining incomplete period has at most $N$ terms. The Abel summation formula gives, initially for $\mathrm{Re}s>1$,For $\mathrm{Re}s>0$ the integral converges, locally uniformly together with its derivatives in $s$, because on a compact set with $\mathrm{Re}s\ge \delta >0$ the differentiated integrands are bounded by constants times $x^{-1-\delta }(1+\log x{)}^k$. Thus holomorphy of a Dirichlet series from bounded partial sums proves

For a nonreal $\psi$, the character ${\psi }^2$ is also nonprincipal. The supplied logarithmic Euler product yields, for real $\sigma >1$,At a prime dividing $N$ the numerator is $3$. At every other prime put $\psi (p{)}^k=e^{i\theta }$; then the numerator is $3+4\cos \theta +\cos 2\theta =2(1+\cos \theta {)}^2\ge 0$. This Euler product positivity for L-function nonvanishing proves that the product inside the logarithm is at least one. If $L_N(\psi ,1)=0$, its holomorphy would give $L_N(\psi ,\sigma )=O(\sigma -1)$. Meanwhile $L_N({\psi }^2,\sigma )$ is bounded near one and the Riemann zeta function has a simple pole there. The product would therefore be $O(\sigma -1)$ and tend to zero, a contradiction.

For a real nonprincipal $\psi$, suppose instead that $L_N(\psi ,1)=0$. The product $H(s)=\zeta (s)L_N(\psi ,s)$ would then be holomorphic on the whole half-plane $\mathrm{Re}s>0$, because the zero cancels the only zeta pole there. In $\mathrm{Re}s>1$ it has coefficientsThese are nonnegative zeta-times-real-L coefficients. They are multiplicative, and their prime-power values are $k+1$ if $\psi (p)=1$, one or zero according as $k$ is even or odd if $\psi (p)=-1$, and one if $\psi (p)=0$. In particular $b_n\ge 0$ and $b_{m^2}\ge 1$ for every $m$.

The Taylor series of the assumed holomorphic $H$ about $2$ converges in the disk $|s-2|<2$. Termwise differentiation of its absolutely convergent Dirichlet series at $2$ gives $(-1{)}^k{H}^{(k)}(2)={\sum }_n{b}_n(\log n{)}^k/n^2$. Evaluate the Taylor series at $s=1/2$ and apply the Tonelli theorem to the resulting nonnegative terms:This contradicts holomorphy at $1/2$. The argument is the positive-coefficient mechanism of the Landau theorem for a Dirichlet series with nonnegative coefficients, with the needed step proved explicitly. Combining the real and nonreal cases gives the nonvanishing of a nonprincipal Dirichlet L-function at one:

Now let $a$ be coprime to $N$. The Dirichlet theorem on primes in arithmetic progressions asserts that infinitely many prime numbers satisfy $p\equiv a(\mathrm{mod}N)$. To prove it, separate the first-power terms in the given logarithmic Euler product:For $\sigma \ge 1$, the remainder is bounded uniformly byFor a nonprincipal character, holomorphy and nonvanishing near one give a local holomorphic logarithm. Along a short real interval to the right of one, this logarithm and the supplied continuous logarithm differ by a fixed integer multiple of $2\pi i$. Hence the prime-character sum near one is bounded:For the principal Dirichlet character ${\psi }_0$,The latter follows from the simple zeta pole and the bounded remainder. Finally Orthogonality of Dirichlet characters givesHere $\phi$ is the Euler totient function. The right-hand side tends to infinity as $\sigma \downarrow 1$, whereas a finite set of primes would give a bounded left-hand side. Therefore

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.

Let $\hat{G}$ be the group of all Dirichlet characters modulo $q$, where $G=(\mathbb{Z}/q\mathbb{Z}{)}^{\times }$, and considerFor $p\nmid q$, let $f_p$ be the order of $p$ in $G$. The values $\chi (p)$ run through the $f_p$th roots of unity, each $|G|/f_p$ times, so the local factor isFor $p\mid q$ the local factor is one. Thus the Dirichlet series for $F$ has nonnegative coefficients.

The principal-character factor has a simple pole at $s=1$, while every nonprincipal Dirichlet L-function is entire. If some nonprincipal $L(1,\chi )$ vanished, its zero would cancel that pole and make $F$ entire. The Landau theorem for a Dirichlet series with nonnegative coefficients would then force the Dirichlet series of $F$ to converge for every real $s$. This is impossible: its coefficient at $m^{|G|}$ is at least one for every $m$ coprime to $q$, as is clear from the local factors. Hence