Dirichlet Characters L Functions and Primes in Progressions

If $m\equiv n(\mathrm{m}\mathrm{o}\mathrm{d}q)$, then $m$ and $n$ determine the same residue class in $\mathbb{Z}\mathrm{/}q\mathbb{Z}$. Hence $(m,q)=1$ iff $(n,q)=1$, because both conditions say exactly that this common class is a unit. When they are units, [L1] gives $\chi (m)=\bar{\chi }(\bar{m})=\bar{\chi }(\bar{n})=\chi (n)$; when they are nonunits, [L1] gives $\chi (m)=\chi (n)=0$.

Step 1.1 is exactly representative-independence, and applying it to $m=n+q$ gives $\chi (n+q)=\chi (n)$ for every integer $n$, so $\chi$ is $q$-periodic. The final clause of [L1] says $\chi (n)=0$ precisely on the nonunit residue classes, equivalently exactly when $(n,q)>1$.

Assume first that $\chi$ comes from a Dirichlet character modulo $q$. Periodicity and the support condition are exactly [L2]. If $(m,q)>1$ or $(n,q)>1$, then both $\chi (mn)$ and $\chi (m)\chi (n)$ are $0$ by [L2]. If both are coprime to $q$, then [L1] gives $\chi (mn)=\bar{\chi }(\bar{m}\bar{n})=\bar{\chi }(\bar{m})\bar{\chi }(\bar{n})=\chi (m)\chi (n)$. Also $\chi (1)=\bar{\chi }(\bar{1})=1$ because every homomorphism sends the identity to the identity.

Conversely, assume properties 1-4. If $(n,q)=1$, define $\bar{\chi }(\bar{n}):=\chi (n)$. This is well defined because property 1 makes $\chi$ constant on residue classes modulo $q$, and property 3 shows that only unit classes receive nonzero values. For unit classes $\bar{m},\bar{n}$, property 2 gives $\bar{\chi }(\bar{m}\bar{n})=\chi (mn)=\chi (m)\chi (n)=\bar{\chi }(\bar{m})\bar{\chi }(\bar{n})$, so $\bar{\chi }$ is a homomorphism $(\mathbb{Z}\mathrm{/}q\mathbb{Z}{)}^{\times }\to {\mathbb{C}}^{\times }$; property 4 makes it unital. Extending this homomorphism by zero recovers the original $\chi$ by property 3.

Step 1.1 proves necessity and step 1.2 proves sufficiency.

If $(n,q)>1$, then [L2] gives $\chi (n)=0$. Assume now that $(n,q)=1$. By [A1], the unit class $\bar{n}$ has some positive order $m$, so ${\bar{n}}^m=\bar{1}$. Applying the homomorphism of [L1] gives $\chi (n{)}^m=\bar{\chi }(\bar{n}{)}^m=\bar{\chi }(\bar{1})=1$. Thus $\chi (n)$ is a root of unity, hence nonzero.

For a nonzero complex number on the unit circle, complex conjugation equals reciprocal. Since step 1.1 gives $\chi (n{)}^m=1$, the value $\chi (n)$ lies on the unit circle, so $\overline{\chi (n)}=\chi (n{)}^{-1}$. Together with the nonunit case from step 1.1, this proves the statement.

By [L1], every irreducible complex character of $G$ has degree $1$, and then [L4] shows that their number is $\mathrm{\mid }G\mathrm{\mid }=\phi (q)$ because $\mathrm{\mid }G\mathrm{\mid }={\sum }_i{1}^2$. Thus the irreducible complex characters of $G$ are exactly the Dirichlet characters modulo $q$. Since $G$ is abelian, every conjugacy class is a singleton and every centralizer is all of $G$.

Applying [L2] to the character group of $G$ gives $\frac{1}{\mathrm{\mid }G\mathrm{\mid }}{\sum }_{a\in G}\chi (a)\overline{\psi (a)}={\delta }_{\chi \psi }$, which is exactly the second displayed formula because $\mathrm{\mid }G\mathrm{\mid }=\phi (q)$. Applying [L3] to the same irreducible character list and using step 1.1 turns the centralizer size into $\mathrm{\mid }G\mathrm{\mid }=\phi (q)$ and conjugacy into literal equality of elements, which yields the first displayed formula.

If $(n,q)>1$, then every Dirichlet character has $\chi (n)=0$, so the sum is $0$, and this matches the fact that $n$ cannot be congruent to the reduced class $a$. If $(n,q)=1$, then both $a$ and $n$ are unit classes modulo $q$ and [L1] applies with $u=n$ and $v=a$.

In the unit case from step 1.1, [L1] gives ${\sum }_{\chi }\overline{\chi (a)}\chi (n)=\phi (q)$ exactly when $n\equiv a(\mathrm{m}\mathrm{o}\mathrm{d}q)$, and $0$ otherwise. Dividing by $\phi (q)$ yields the indicator formula.

Since $\chi \mathrm{\ne }{\chi }_0$, [L1] shows that some unit $a$ modulo $q$ satisfies $\chi (a)\mathrm{\ne }1$. Let $S:={\sum }_{r\in R}\chi (r)$. Multiplication by the unit $a$ permutes the residue classes modulo $q$, so $aR$ is again a complete residue system modulo $q$.

Reindex over $aR$ and use multiplicativity on units: $S={\sum }_{r\in R}\chi (ar)=\chi (a){\sum }_{r\in R}\chi (r)=\chi (a)S$. Because $\chi (a)\mathrm{\ne }1$, this gives $(\chi (a)-1)S=0$, hence $S=0$.

Write $\lfloor x\rfloor =mq+r$ with integers $m\ge 0$ and $0\le r<q$. By [L1], the sum over $1\le n\le x$ is the sum over $1\le n\le \lfloor x\rfloor$, which breaks into $m$ complete blocks of length $q$ and one terminal block of length $r$.

Every complete block contributes $0$ by [L2]. Hence ${\sum }_{1\le n\le x}\chi (n)={\sum }_{mq<n\le mq+r}\chi (n)$. The terminal block has at most $q-1$ terms, and every term has modulus at most $1$ because character values are either $0$ or roots of unity. Therefore the absolute value of the sum is at most $q-1<q$, so certainly at most $q$.

A Dirichlet character is completely multiplicative on $\mathbb{Z}$, because it is multiplicative on unit classes and both sides vanish when a nonunit factor is present. Hence [L2] applied to $f=\chi$ and [L1] give the Euler product formula on $\mathrm{Re}s>1$.

Write $\sigma =\mathrm{Re}s>1$. Then $\mathrm{\mid }\chi (p)p^{-s}\mathrm{\mid }\le p^{-\sigma }$, so ${\sum }_p\mathrm{\mid }\chi (p)p^{-s}\mathrm{\mid }$ converges. Therefore ${\sum }_{p,m\ge 1}\chi (p{)}^m\mathrm{/}(m{p}^{ms})$ converges absolutely, and $-\log (1-\chi (p)p^{-s})$ is the absolutely convergent local series. Summing over $p$ shows that the Euler product of step 1.1 is the exponential of a convergent complex series, so it cannot vanish.

By [L2], $L(s,{\chi }_0)={\prod }_p(1-{\chi }_0(p)p^{-s}{)}^{-1}$. Using [L1], the Euler factors are $1$ at primes dividing $q$ and $(1-p^{-s}{)}^{-1}$ at all other primes, so $L(s,{\chi }_0)={\prod }_{p\nmid q}(1-p^{-s}{)}^{-1}=({\prod }_p(1-p^{-s}{)}^{-1}){\prod }_{p\mid q}(1-p^{-s})=\zeta (s){\prod }_{p\mid q}(1-p^{-s})$.

The finite factor ${\prod }_{p\mid q}(1-p^{-s})$ is holomorphic at $s=1$ and has value ${\prod }_{p\mid q}(1-p^{-1})$. Multiplying this with the residue-one pole from [L3] gives a simple pole of $L(s,{\chi }_0)$ at $1$ with residue ${\prod }_{p\mid q}(1-p^{-1})$. Finally, $\phi (q)=q{\prod }_{p\mid q}(1-p^{-1})$ by the standard totient product, so the residue is $\phi (q)\mathrm{/}q$.

Apply [L2] to the coefficients $a_n=\chi (n)$ with $\theta =0$. By [L1], the summatory function is bounded, so for every $\mathrm{Re}s>0$ one has $L(s,\chi )=s{\int }_1^{\mathrm{\infty }}A(x)x^{-s-1}dx$, and the integral converges absolutely and locally uniformly on each half-plane $\mathrm{Re}s\ge \epsilon >0$ because $A(x)=O(1)$ and $x^{-s-1}=O(x^{-\epsilon -1})$.

A locally uniformly convergent parameter integral of holomorphic integrands is holomorphic in the parameter. Hence the right-hand side of step 1.1 defines a holomorphic function on $\mathrm{Re}s>0$, and on the smaller half-plane $\mathrm{Re}s>1$ it agrees with the defining Dirichlet series [L3]. Therefore $L(s,\chi )$ is holomorphic on $\mathrm{Re}s>0$.

Suppose first that $F(1+i{t}_0)=0$ for some real $t_0\mathrm{\ne }0$. For $\sigma >1$, absolute convergence gives $\log \mathrm{\mid }F(\sigma +iu)\mathrm{\mid }={\sum }_{n\ge 2}{b}_n{n}^{-\sigma }\cos (u\log n)$, so $\log \text{\hspace{0.17em}⁣}(\mathrm{\mid }F(\sigma ){\mathrm{\mid }}^3\mathrm{\mid }F(\sigma +i{t}_0){\mathrm{\mid }}^4\mathrm{\mid }F(\sigma +2i{t}_0)\mathrm{\mid })={\sum }_{n\ge 2}{b}_n{n}^{-\sigma }(3+4\cos {\theta }_n+\cos (2{\theta }_n))$ with ${\theta }_n=t_0\log n$. Since $3+4\cos \theta +\cos (2\theta )=2(1+\cos \theta {)}^2\ge 0$, the product on the left is at least $1$.

Because $F$ is meromorphic with at most a simple pole at $1$, the factor $F(\sigma )$ grows like $O((\sigma -1{)}^{-1})$ as $\sigma \downarrow 1$, while $F(\sigma +2i{t}_0)$ stays bounded and the zero at $1+i{t}_0$ forces $F(\sigma +i{t}_0)=O(\sigma -1)$. Therefore the product from step 1.1 is $O(\sigma -1)\to 0$, contradicting the lower bound $1$. The Landau statement [L1] concerns singularity of the logarithmic series at its own abscissa, so it does not by itself exclude a zero of $F$ at $1$. Instead, if $F(1)=0$, then $F$ is holomorphic at $1$ and $F(\sigma )\to 0$ as $\sigma \downarrow 1$, whereas the assumed logarithmic identity at $t=0$ gives $\log \mathrm{\mid }F(\sigma )\mathrm{\mid }={\sum }_{n\ge 2}{b}_n{n}^{-\sigma }\ge 0$ and hence $\mathrm{\mid }F(\sigma )\mathrm{\mid }\ge 1$ for every $\sigma >1$. This is another contradiction. Thus no zero occurs on the line $\mathrm{Re}s=1$.

For $\mathrm{Re}s>1$, [L2] gives $\log F_q(s)={\sum }_{\chi <mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>q}{\sum }_{p,m\ge 1}\chi (p{)}^m\mathrm{/}(m{p}^{ms})={\sum }_{p,m\ge 1}(m{p}^{ms}{)}^{-1}{\sum }_{\chi <mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>q}\chi (p^m)$. If $p\mid q$, then every term is $0$. If $p\nmid q$, then [L1] applied to the unit class of $p^m$ shows that the inner character sum is $\phi (q)$ when $p^m\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}q)$ and $0$ otherwise. Hence the logarithmic coefficients of $F_q$ are nonnegative.

By [L3] and [L4], the product $F_q$ is meromorphic on a neighbourhood of $\mathrm{Re}s\ge 1$, with at most a simple pole at $1$ and no other singularities on the boundary line. Step 1.1 therefore places $F_q$ under [L5], so $F_q$ has no zero on $\mathrm{Re}s=1$. This proves both the nonvanishing claim and the stated meromorphic control at $s=1$.

Suppose $L(1+it,\chi )=0$. Then the finite product over all characters modulo $q$ also vanishes at $1+it$, because one factor is zero there.

This contradicts [L1]. Therefore $L(1+it,\chi )\mathrm{\ne }0$ for every real $t\mathrm{\ne }0$.

Suppose $L(1,\chi )=0$. Then [A1] gives $L(1,\bar{\chi })=0$ as well. Because $\chi$ is nonreal, the characters $\chi$ and $\bar{\chi }$ are distinct, so these are two different vanishing factors in the full finite product at $s=1$.

In the full product over all characters, [L2] contributes order $-1$ at $s=1$, while step 1.1 contributes at least $+1$ from each of the distinct factors $L(s,\chi )$ and $L(s,\bar{\chi })$. Every remaining nonprincipal factor is holomorphic at $1$ by [L3], so it contributes order at least $0$. Hence the total product has order at least $+1$ at $1$, meaning a zero there. This contradicts [L1]. Therefore $L(1,\chi )\mathrm{\ne }0$.

Suppose $L(1,\chi )=0$, and set $G(s):=L(s,{\chi }_0)L(s,\chi )$. For $\mathrm{Re}s>1$, facts [L2] and [L3] give $G(s)={\prod }_{\chi (p)=1}(1-p^{-s}{)}^{-2}{\prod }_{\chi (p)=-1}(1-p^{-2s}{)}^{-1},$ because primes dividing $q$ contribute the trivial local factor $1$. Expanding the geometric series shows that $G(s)={\sum }_{n\ge 1}{a}_n{n}^{-s}$ with $a_n\ge 0$ for every $n$. Moreover, if $(m,q)=1$, then the square coefficient $a_{m^2}$ is positive: each local factor above has a positive coefficient at every even exponent occurring in $m^2$.

Step 1.1 implies ${\sum }_{n\ge 1}{a}_n{n}^{-1\mathrm{/}2}\ge {\sum }_{\begin{array}{c}m\ge 1\\ (m,q)=1\end{array}}\frac{a_{m^2}}{m}\ge {\sum }_{\begin{array}{c}m\ge 1\\ (m,q)=1\end{array}}\frac{1}{m}=\mathrm{\infty },$ so the abscissa of convergence ${\sigma }_c$ of $\sum a_n{n}^{-s}$ satisfies ${\sigma }_c\ge 1\mathrm{/}2$. On the other hand, [L1] and [L4] show that $G$ is holomorphic on the whole half-plane $\mathrm{Re}s>0$: the only possible singularity there is the simple pole of $L(s,{\chi }_0)$ at $1$, and the assumption of step 1.1 cancels it. Since $G$ is represented by a Dirichlet series with nonnegative coefficients, [L5] forbids any positive abscissa of convergence. Thus ${\sigma }_c\le 0$, contradicting ${\sigma }_c\ge 1\mathrm{/}2$. Therefore $L(1,\chi )\mathrm{\ne }0$.

Every Dirichlet character is either real or nonreal. If $\chi$ is nonreal, [L1] applies; if $\chi$ is real, then because it is also nonprincipal [L2] applies.

In both cases $L(1,\chi )\mathrm{\ne }0$, so the theorem follows.

Apply [L2] to the coefficients $a_n={\mathbf{1}}_A(n)$. Then for $s>1$ one has ${\sum }_{n\in A}{n}^{-s}=s{\int }_1^{\mathrm{\infty }}A(x)x^{-s-1}dx$. Now write $A(x)=\delta x+E(x)$ with $E(x)=o(x)$ by [L1].

Substituting into step 1.1 gives ${\sum }_{n\in A}{n}^{-s}=\delta s{\int }_1^{\mathrm{\infty }}{x}^{-s}dx+s{\int }_1^{\mathrm{\infty }}E(x)x^{-s-1}dx=\delta s\mathrm{/}(s-1)+s{\int }_1^{\mathrm{\infty }}E(x)x^{-s-1}dx$. It therefore suffices to show that $(s-1){\int }_1^{\mathrm{\infty }}E(x)x^{-s-1}dx\to 0$ as $s\downarrow 1$.

Fix $\epsilon >0$. Because $E(x)=o(x)$, choose $X\ge 1$ so that $\mathrm{\mid }E(x)\mathrm{\mid }\le \epsilon x$ for every $x\ge X$. For $1<s\le 2$, the quantity $\mid (s-1)s{\int }_1^{\mathrm{\infty }}E(x)x^{-s-1}dx\mid$ is at most $(s-1)s{\int }_1^X\mathrm{\mid }E(x)\mathrm{\mid }{x}^{-s-1}dx+\epsilon s(s-1){\int }_X^{\mathrm{\infty }}{x}^{-s}dx$. The first term tends to $0$ because the integral over $[1,X]$ is bounded, and the second term is at most $2\epsilon X^{1-s}\le 2\epsilon$. Since $\epsilon$ is arbitrary, the whole expression tends to $0$. Hence $(s-1){\sum }_{n\in A}{n}^{-s}\to \delta$, which is exactly the Dirichlet-density statement from Natural and Dirichlet density.