Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 150 / 4

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

For $\sigma >1$, Orthogonality of Dirichlet characters givesThe principal-character term isPart (a) makes every nonprincipal logarithmic derivative bounded as $\sigma \to 1^{+}$, soIf the nondecreasing Chebyshev function in an arithmetic progressionwere bounded, the Abel summation formula would keep the displayed Dirichlet series bounded near $\sigma =1$. Therefore

There is an absolute $c>0$ such that the product of the Dirichlet L-functions modulo $q$ has no zero inexcept possibly one zero. If it exists, this exceptional zero $\beta$ is real and simple, belongs to a real nonprincipal character ${\chi }_1$, and lies very close to one. Among the primitive characters whose conductors divide $q$, at most one can have such a zero. This is the Classical zero-free region for Dirichlet L-functions.

Let ${\chi }_1$ be the real character associated with the exceptional zero $\beta$. The prime number theorem in an arithmetic progression with an exceptional zero gives, uniformly for $x\ge 2$ and $(a,q)=1$,If no exceptional zero exists, the middle term is omitted. The constant $c>0$ is absolute.

Assume $\beta$ exists and choose a reduced residue class $a$ with ${\chi }_1(a)=-1$; such a class exists because ${\chi }_1$ is nonprincipal. Fix a sufficiently large constant $A=A(ϵ)$ and takeThen $x\ge q^2$ for large $q$. Multiplying the formula from part (d) by $\phi (q)/x$ givesChoose $A$ so large that the error is at most $ϵ/4$ for all sufficiently large $q$. The assumed upper bound then impliesSince $\beta <1$, this givesTaking logarithms,Therefore, with a positive constant depending only on $ϵ$,