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

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?

• You'd need to provide much more context. EG How much do you know? Are you familiar with complex analysis? analytical number theory? Commented Jun 2, 2023 at 15:17
• I would say yes, but I wish an answer as easy as possible. Commented Jun 2, 2023 at 18:34

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 $$L(1,\chi)$$).

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.]

• What I really do not understand is how do you relate a Dirichlet L-function with a congruence, let say a+kb are the numbers that are congruent to a mod b, right? How L-functions relate to this, and therefore what is the step that makes the proof done so that the progression has infinite primes. Commented Jun 2, 2023 at 18:34