What's the idea of Dirichlet’s Theorem on Arithmetic Progressions proof?

Question

Dirichlet’s Theorem on Arithmetic Progressions says that if $a, m$ are natural numbers such that $gcd (a,m) = 1$, then there are infinitely many prime numbers in the arithmetic progression $a + km, k \in \mathbb{N}$. I would like to know what is the idea of the proof, and the sketch of the proof, to try to understand it. I understand the Dirichlet L-functions, to obtain the set of complex values such that the Dirichlet L-functions becomes zero, the Dirichlet character, but I heard that "Dirichlet's theorem is proved by showing that the value of the Dirichlet L-function (of a non-trivial character) at 1 is nonzero", how is that? Can someone explain how a arithmetic progression is related with a L-function?

Answer 1

The sum of the reciprocals of the primes diverges, $\sum_p\frac{1}{p}=\infty$. This can be seen using the Euler product for the Riemann zeta function (say $s>1$ real), $\zeta(s)=\prod_p\frac{1}{1-p^{-s}}$, $$ \log\left(\zeta(s)\right)=\sum_p-\log(1-p^{-s})=\sum_{p,n}\frac{p^{-ns}}{n}=\sum_p\frac{1}{p^s}+O(1), $$ and letting $s\to1^+$.

Dirichlet extends this by replacing $\zeta(s)$ with $L(s,\chi)=\prod_p\frac{1}{1-\chi(p)p^{-s}}$, $$ \log(L(s,\chi))=-\sum_p\log(1-\chi(p)p^{-s})=\sum_{n,p}\frac{\chi(p)}{np^{ns}}=\sum_p\frac{\chi(p)}{p^s}+O(1), $$ and using properties of characters to select the progression (i.e. average over all characters and "shift" $1\bmod q$ to $a\bmod q$) $$ \frac{1}{\phi(q)}\sum_{\chi}\bar{\chi}(a)\log(L(s,\chi))=\sum_pp^{-s}\sum_{\chi}\chi(pa^{-1})+O(1)=\sum_{p\equiv a(q)}\frac{1}{p^s}+O(1). $$ The proof goes through (letting $s\to1^+$) as long as $L(1,\chi)\neq0$ for all $\chi$. There is a pole for the principal character $\chi_0$, and the other $L(1,\chi)$ values are finite, but we get "$\infty-\infty$" on the LHS if some of these values are zero.

The "relate arithmetic progressions to characters" question is basically orthogonality (some algebra) $$ \begin{align*} \sum_{\chi}\chi(a)&= \left\{ \begin{array}{cc} 1 & \chi=\chi_0\\ 0 & \text{else},\\ \end{array} \right. \\ \sum_{a\in(\mathbb{Z}/q\mathbb{Z})^{\times}}\chi(a)&= \left\{ \begin{array}{cc} 1 & a\equiv1(q)\\ 0 & \text{else}.\\ \end{array} \right. \end{align*} $$ If you want to see some proofs, try Davenport's Multiplicative Number Theory or these somewhat concise notes I wrote once upon a time (with a few proofs of non-vanishing of $L(1,\chi)$).

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To more directly address the concerns of the question, the algebraic trick to select an arithmetic progression using characters is essentially the sum $$ \frac{1}{\phi(q)}\sum_{\chi\bmod q}\chi(pa^{-1})= \left\{ \begin{array}{cc} 1 & p\equiv a\bmod q\\ 0 & \text{else},\\ \end{array} \right. $$ which is a direct consequence of the orthogonality relations above.

For example, lets consider $q=3$. There are two characters modulo 3, $$ \chi_0 = (0,1,1)=(\chi_0(0), \chi_0(1),\chi_0(2)), \quad \psi = (0,1,-1)=(\psi(0), \psi(1),\psi(2)), $$ (listing the values on $0,1,2 \bmod 3$). If you average these, they "interfere" to give $$ \frac{1}{\phi(q)}\sum_{\chi\bmod q}\chi=\frac{1}{2}(0,2,0)=(0,1,0) $$ which picks out the residue class $1\bmod 3$. To pick out the residue class $2\bmod 3$ we can "shift the indices" by considering the translated characters ($x\mapsto 2^{-1}x=2x$ working $\bmod 3$) $$ \chi_0(2x) = (0,1,1)=(\chi_0(0), \chi_0(2),\chi_0(4)), \quad \psi(2x) = (0,-1,1) =(\psi(0), \psi(2),\psi(4)). $$ Averaging these instead, and noting $\chi(2x)=\chi(2)\chi(x)$, we get $$ \frac{1}{\phi(q)}\sum_{\chi\bmod q}\overline{\chi}(a)\chi=\frac{1}{2}(1\cdot(0,1,1)+(-1)\cdot(0,1,-1))=(0,0,1), $$ which picks out $2\bmod 3$.

[Note that $2\cdot2=4\equiv1\bmod 3$, i.e. $2=2^{-1}\bmod 3$, and $\chi(a^{-1})=\overline{\chi}(a)$ by multiplicativity if the bar or $a^{-1}$ is confusing.]