Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 28 / 5 / Solution

Fix integers $q\ge 1$ and $a$ with $(a,q)=1$. We prove the Dirichlet theorem on primes in arithmetic progressions by showing that the prime Dirichlet sum in this residue class diverges as its real exponent decreases to one.

First justify the required Dirichlet L-function facts. If $\chi$ is a nonprincipal Dirichlet character modulo $q$, its sum over a full period is zero: choose a unit $b$ with $\chi (b)\ne 1$, and multiplication by $b$ permutes the residue classes, forcing the sum to equal $\chi (b)$ times itself. Therefore the partial sums $C_{\chi }(u)={\sum }_{n\le u}\chi (n)$ are bounded by $q$. Partial summation giveswhich supplies holomorphic continuation near one, with locally uniform convergence. For the principal Dirichlet character ${\chi }_0$,has a simple pole at one, with residue $\phi (q)/q$.

The assumption in the question gives nonvanishing at one for real nonprincipal characters. We must also prove it for nonreal characters. For a nonreal $\chi$, its square ${\chi }^2$ is nonprincipal. For real $\sigma >1$, logarithms of the Euler products giveFor a prime not dividing $q$, its logarithmic coefficient is $3+4\mathrm{Re}\chi (p{)}^k+\mathrm{Re}\chi (p{)}^{2k}=2(1+\cos \theta {)}^2\ge 0$, where $\chi (p{)}^k=e^{i\theta }$. Primes dividing $q$ contribute only the positive zeta term. If $L(1,\chi )$ vanished to order $m\ge 1$, the product would be $O((\sigma -1{)}^{4m-3})$ and tend to zero: the squared-character factor is bounded because ${\chi }^2$ is nonprincipal. This is a contradiction. HenceThis step uses the given real-character hypothesis and the Euler product positivity for L-function nonvanishing for all remaining characters.

In $\sigma >1$ choose the logarithm furnished by the absolutely convergent Euler product. Its prime-power expansion givesThe error from powers $k\ge 2$ is uniformly bounded as $\sigma \downarrow 1$, by ${\sum }_p{\sum }_{k\ge 2}1/(k{p}^{k\sigma })<\infty$. For nonprincipal $\chi$, nonvanishing and holomorphic continuation at one provide a bounded logarithm in a neighborhood of one. The Euler-product branch differs from that branch by a constant integral multiple of $2\pi i$ on the final real interval, so it too is bounded. ThusFor the principal character, its simple pole instead gives

The Orthogonality of Dirichlet characters gives, for primes not dividing $q$,Multiply by $p^{-\sigma }$ and sum. Absolute convergence allows the interchange for $\sigma >1$, soThe right side tends to infinity. If there were only finitely many primes in the progression, the left side would have a finite limit at one. This contradiction proves infinitely many primes occur in every reduced residue class. For $q=1$ or $2$ the character argument simply has no nonprincipal terms and gives the same conclusion.

For the final deduction the modulus $q$ is fixed. Let $A(t)=\pi (t;q,a)$. The given consequence of the Siegel–Walfisz theorem is $A(t)\sim t/(\phi (q)\log t)$. Partial summation, including the prime $2$ through the lower endpoint $2^{-}$ when appropriate, givesThe boundary is $O_q(1/\log x)$. Write $A(t)=t(1+\epsilon (t))/(\phi (q)\log t)$ with $\epsilon (t)\to 0$. The main integral is $(\log \log x-\log \log 2)/\phi (q)$. The error is $o(\log \log x)$: given $\eta >0$, choose $T$ such that $|\epsilon (t)|\le \eta$ for $t\ge T$. Its integral from $2$ to $T$ is a fixed constant, and its absolute value from $T$ to $x$ is at most $\eta \log \log x/\phi (q)$ plus a fixed constant. Divide by $\log \log x$ and then let $\eta \downarrow 0$. We obtain the reciprocal primes in a fixed arithmetic progression asymptoticThe argument uses the supplied prime-counting asymptotic in this last part; the preceding infinitude proof did not assume the Prime number theorem in arithmetic progressions.