Primes of Arithmetic Progression
Theorem
Given an integer \(N > 1\) and an integer a coprime to \(N\), there exists infinitely many primes \(p\) such that \(p \equiv a \pmod N\).
Tip
Recall: Proof of infinity of primes via \(\zeta(s)\).
Recall: Legendre Symbol \(\left(\frac{a}{p}\right)\).
\[
\left(\dfrac{a}{p}\right) = \begin{cases} 0 \enspace \textrm{if} \enspace a \equiv 0 \pmod p, \\ 1 \enspace \textrm{if} \enspace p \not \mid a, x^2 \equiv a \pmod p \enspace \textrm{has a solution}, \\ -1 \enspace \textrm{if} \enspace p \not \mid a, x^2 \equiv a \pmod p \enspace \textrm{has no solution}. \end{cases}
\]
Definition
(Dirichlet Character)
A Dirichlet character mod \(N\) is a function \(\chi: \mathbb{Z}/N\mathbb{Z} \rightarrow \{z \in \mathbb{C} \mid \lvert z \rvert = 1\} \cup \{0\}\) such that
(i) \(\chi(a) = 0\) iff \((a, N) \neq 1\).
(ii) \(\chi(ab) = \chi(a)\chi(b)\) for all \(a, b \in \mathbb{Z}\).
If \(\chi(a) = 1\) for all \((a, N) = 1\), then we say \(\chi\) is trivial, denoted by \(\chi_0\).
Lemma
\[
\sum_{a = 0}^{N - 1} \chi(a) = \begin{cases} \varphi(N) \enspace \textrm{if} \enspace \chi = \chi_0, \\ 0 \enspace \textrm{if} \enspace \chi \neq \chi_0. \end{cases}
\]
Lemma
Define \(C_N = \{\textrm{Dirichlet Character mod } N \}\), then \(\lvert C_N \rvert = \varphi(N)\).
Proof
For \(N = p\) prime.
\[
\chi: (\mathbb{Z}/p\mathbb{Z})^{\times} \rightarrow \mathbb{C}, \chi(0) = 0.
\]
Set the generator of \({\mathbb{Z}/p\mathbb{Z}}^{\times}\) to be \(g\). Then \(\chi(g^k) = \chi(g)^k\). So \(\chi\) is uniquely decided by \(\chi(g)\) and \(\chi(1) = 1\). According to Fermat's Little Theorem, \(g^{p-1} \equiv 1 \pmod p\). So \(\chi(g)^{p-1} = 1\). So \(\chi(g)\) is a \(p - 1\)-th root of unity. So \(\chi(g) \in \{e^{\frac{k}{p-1}2\pi \mathrm{i}} \mid 0 \leqslant k \leqslant p-2\}\). So \(\lvert C_p \rvert = p - 1 = \varphi(p)\).
Lemma
Given \(a \in (\mathbb{Z}/N\mathbb{Z})^{\times}, \exists \chi \in C_N\) such that \(\chi(a) \neq 1\).
Lemma
Given \(a\) coprime to \(N\),
\[
\sum_{\chi \in C_N} \chi(a) = \begin{cases} \varphi(N) \enspace \textrm{if} \enspace a = 1, \\ 0 \enspace \textrm{if} \enspace a \neq 1. \end{cases}
\]
Definition
\(L(s, \chi) = \sum_{n = 1}^{\infty} \dfrac{\chi(n)}{n^s}\) is called the Dirichlet \(L\)-function associated to \(\chi\). \(L(s, \chi) = \prod_{p \textrm{ prime}} \left(1 - \chi(p)p^{-s}\right)^{-1}\).
Obsereve that \(L(s, \chi_0) = \prod_{p \textrm{ prime}} \left(1 - \chi_0(p)p^{-s}\right)^{-1} = \prod_{p \textrm{ prime} \atop p \nmid N} \left(1 - p^{-s}\right)^{-1}\), while \(\zeta(s) = \prod_{p \textrm{ prime}} \left(1 - p^{-s}\right)^{-1}\). So \(L(s, \chi_0) \cdot \prod_{p \mid N} \left(1 - p^{-s}\right) = \zeta(s)\). And \(\lim_{s \rightarrow 1^+} L(s, \chi_0) = \infty\).
Prop
\(\lim_{s \rightarrow 1^+} (s - 1) \zeta(s)\) and \(\lim_{s \rightarrow 1^+} (s - 1) L(s, \chi_0)\) both exist.
Proof
\[\begin{align}
\zeta(s) - \frac{1}{s - 1} & = \sum_{n = 1}^{+\infty} \frac{1}{n^s} - \int_{1}^{+\infty} \frac{1}{x^s} \mathrm{d}x \\ & = \sum_{n = 1}^{+\infty} \left(\frac{1}{n^s} - \int_{n}^{n + 1} \frac{1}{x^s} \mathrm{d}x \right) \\
\frac{1}{n^s} - \int_{n}^{n + 1} \frac{1}{x^s} \mathrm{d}x & = \frac{1}{n^s} - \frac{1}{x_0^s}, x_0 \in (n, n + 1) \\ & = \frac{1}{n^s} \left(1 - \frac{1}{(x_0/n)^s}\right) \\ & = \frac{1}{n^s} \cdot s \xi^{-s - 1} \left(\frac{x_0}{n} - 1\right) \\ & \leqslant \frac{1}{n^s} \cdot s \cdot \frac{1}{n} \\ & = \frac{s}{n^{s + 1}}.
\end{align}\]
So \(\lim_{s \rightarrow 1^+} \left(\zeta(s) - \frac{1}{s - 1}\right) = \lim_{s \rightarrow 1^+} C(s)\), then \(\lim_{s \rightarrow 1^+} ((s - 1) \zeta(s) - 1) = \lim_{s \rightarrow 1^+} (s - 1) C(s)\), where \(C(s)\) is a convergent series. So \(\lim_{s \rightarrow 1^+} (s - 1) \zeta(s)\) exists.
Prop
\(\lim_{s \rightarrow 1^+} L(s, \chi)\) exists if \(\chi \neq \chi_0\).
Fact
\(L'(s, \chi)\) exists.
Formal Proof of Dirichlet's Theorem
\[
L(s, \chi) = \prod_{p \textrm{ prime}} \left(1 - \chi(p)p^{-s}\right)^{-1},
\]
so
\[
\ln L(s, \chi) = \sum_{p \textrm{ prime}} -\ln \left(1 - \chi(p)p^{-s}\right)
\]
Use Taylor expansion, \(- \ln (1 - x) = \sum_{n = 1}^{+\infty} \frac{x^n}{n}\), so
\[
\ln L(s, \chi) = \sum_{p \textrm{ prime}} \sum_{m = 1}^{+\infty} \frac{\chi(p)^m}{mp^{ms}}.
\]
Then
\[
- \frac{L'(s, \chi)}{L(s, \chi)} = \sum_{p \textrm{ prime}} \sum_{m = 1}^{+\infty} \frac{\ln p}{\chi(p)^m p^{ms}} = \sum_{p \textrm{ prime}} \frac{\ln p}{\chi(p) p^s} + \sum_{p \textrm{ prime}} \sum_{m = 2}^{+\infty} \frac{\ln p}{\chi(p)^m p^{ms}}.
\]
\[
- \chi(a) \frac{L'(s, \chi)}{L(s, \chi)} = \sum_{p \textrm{ prime}} \chi(a) \frac{\ln p}{\chi(p) p^s} + \sum_{p \textrm{ prime}} \sum_{m = 2}^{+\infty} \chi(a) \frac{\ln p}{\chi(p)^m p^{ms}}.
\]
Sum both sides over all the \(\chi \in C_N\), we get
\[\begin{align}
\textrm{LHS} & = \sum_{\chi \in C_N} - \chi(a) \frac{L'(s, \chi)}{L(s, \chi)} \\
\textrm{RHS} & = \sum_{\chi \in C_N} \sum_{p \textrm{ prime}} \chi(ap^{-1}) \frac{\ln p}{p^s} + \sum_{\chi \in C_N} \sum_{p \textrm{ prime}} \sum_{m = 2}^{+\infty} \chi(a) \frac{\ln p}{\chi(p)^m p^{ms}}. \\ & = \varphi(N) \sum_{p \equiv a \pmod N} \frac{\ln p}{p^s} + \sum_{\chi \in C_N} \sum_{p \textrm{ prime}} \sum_{m = 2}^{+\infty} \chi(a) \frac{\ln p}{\chi(p)^m p^{ms}}.
\end{align}\]
The second term on the RHS
\[
\sum_{\chi \in C_N} \sum_{p \textrm{ prime}} \sum_{m = 2}^{+\infty} \chi(a) \frac{\ln p}{\chi(p)^m p^{ms}} \leqslant \sum_{n = 2}^{+\infty} \frac{\ln n}{n^{2s}}(1 - \frac{1}{n^{-s}})^{-1} < +\infty
\]
\[
\textrm{LHS} = - \frac{L'(s, \chi_0)}{L(s, \chi_0)} - \sum_{\chi \neq \chi_0} \chi(a) \frac{L'(s, \chi)}{L(s, \chi)}.
\]
It remains to prove \(L(1, \chi) \neq 0\) when \(\chi \neq \chi_0\).
Proof when \(\chi^2 \neq \chi_0\).
If \(L(1, \chi) = 0\), then it should have a form like \((s - 1) g(s)\).
\[\begin{align}
\lambda(s) & := L(s, \chi_0)^3 \cdot L(s, \chi)^4 \cdot L(s, \chi^2) \\ & = \prod_{p} \frac{1}{1 - p^{-s}} \cdot \prod_{p} (\frac{1}{1 - \chi(p)p^{-s}})^4 \cdot \prod_{p} \frac{1}{1 - \chi^2(p)p^{-s}} \\ & = \mathrm{exp}\left( \sum_{m, p} \frac{3 + 4 \chi(p)^m + \chi(p^{2m})}{m \cdot p^{ms}}\right)
\end{align}\]
\[
\lvert \lambda(s) \rvert = \lvert \mathrm{exp}\left( \sum_{m, p} \frac{3 + 4 cos \theta_{m, p} + cos 2\theta_{m, p}}{m \cdot p^{ms}}\right) \rvert \Rightarrow \lambda(s) \neq 0 \Rightarrow \lim_{s \to 1^+} \lambda(s) \neq 0.
\]
Where \(\theta_{m, p} = \arg \chi(p)^m\).
Then
\[
\lim_{s \to 1^+} \lambda(s) = \lim_{s \to 1^+} L(s, \chi_0)^3 \cdot L(s, \chi)^4 \cdot L(s, \chi^2) = \lim_{s \to 1^+} (\frac{1}{s - 1})^3 \cdot (s - 1)^4 \cdot L(s, \chi^2) = 0.
\]
Which is a contradiction.