# Arithmetic Progressions for Primes

\begin{equation*}

L(s,\chi) = \sum\limits_{n=1}^{+ \infty} \frac{\chi (n)}{n^s}

\end{equation*}

for $s \in \mathbb{C}$, and also the zeta function

\begin{equation*}

\zeta_m (s) = \prod\limits_{\chi \in \widehat{(\mathbb{Z}/m \mathbb{Z})^*}} L(s , \chi )

\end{equation*}

One can prove that $\zeta_m$ has a simple pole at $s=1$.

**Density of a subset of prime numbers**

A fairly precise information about the density of prime numbers among all integers is given by the following estimation, when $s$ tends to 1 :

\begin{equation*}

\sum\limits_{p \in \mathbb{P}} \frac{1}{p^s} \sim \log \frac{1}{s-1}

\end{equation*}

to be compared with

\begin{equation*}

\sum\limits_{n \in \mathbb{N}^*} \frac{1}{n^s} \sim \frac{1}{s-1}

\end{equation*}

The simple fact that the two series diverge for $s=1$ and converge for $s>1$ is striking ! In fact, taking the logarithm of the second relation gives the first one :

\begin{equation*}

\log \sum\limits_{n \in \mathbb{N}^*} \frac{1}{n^s} = \log \prod\limits_{p \in \mathbb{P}} \frac{1}{1-\frac{1}{p^s}} = \sum\limits_{p \in \mathbb{P}} \sum\limits_{k \in \mathbb{N}^*} \frac{1}{k \times p^{ks}} = \sum\limits_{p \in \mathbb{P}} \frac{1}{p^s} + \psi(s)

\end{equation*}

where it is easy to see that $\psi(s)$ remains bounded when $s \rightarrow 1$ (show this !).

Now if we take any subset $A$ of the set of prime numbers, we can define its density as

\begin{equation*}

d(A) = \lim\limits_{s \rightarrow 1} \frac{\sum\limits_{p \in A} \frac{1}{p^s} }{\log \frac{1}{s-1}}.

\end{equation*}

With this at hand we can state the theorem: let $m \in \mathbb{N}^*$ and $a \in \mathbb{Z}$ such that $m$ and $a$ are relatively prime ; then the set $P_{m,a}$ of primes $p$ such that $p \equiv a$ mod $m$ has density $\frac{1}{\phi(m)}$. This is truly fantastic : the first thing to notice is that $P_{m,a}$ is infinite, and more than that, its density doesn't depend on $a$, meaning that the primes are equally distributed between the different classes modulo $m$, and this is true for any $m$.

How can we see that the theorem above is true ? What we want to study is

\begin{equation*}

\sum\limits_{p \in P_{m,a}} \frac{1}{p^s}

\end{equation*}

We would like to use the machinery of $L$-functions and characters. Thus the trick is to translate the condition $p \equiv a$ mod $m$ using the characters. In $(\mathbb{Z}/m \mathbb{Z})^*$, this condition reads $a^{-1} p=1$. But the number 1 has a very neat characterization with the characters : if $p \equiv a$ mod $m$,

\begin{equation*}

\sum\limits_{\chi} \chi (a^{-1} p) = \phi (m)

\end{equation*}

and else the sum vanishes. Thus

\begin{equation*}

\sum\limits_{p \in P_{m,a}} \frac{1}{p^s} = \sum\limits_{p \nmid m} \frac{1}{\phi (m)} \sum\limits_{\chi} \chi (a^{-1} p) \frac{1}{p^s} = \frac{1}{\phi (m)} \sum\limits_{\chi} \chi (a^{-1}) \left[ \sum\limits_{p \nmid m} \frac{\chi(p)}{p^s} \right]

\end{equation*}

The square bracket gives the wanted $\log \frac{1}{s-1}$ for $\chi=1$, and what remains to be shown is that for $\chi \neq 1$, it is bounded. It is slightly more technical, but can be done using similar arguments.

**Other notions of density**

The density defined above was the *analytic density*, it measures roughly "how the sum of the inverses diverge". This is a useful tool to estimate the "size" of a set of integers : for $\mathbb{N}^*$, the divergence is $\frac{1}{s-1}$, for primes, it is $ \log \frac{1}{s-1}$ and for primes in $P_{m,a}$ it is $\frac{1}{\phi (m)} \log \frac{1}{s-1}$.

An other idea to compare the sizes of sets of integer is to truncate them at some fixed value, say $N$, and to compare their finite cardinal. If we let $N$ go to infinity, the ratio of the two cardinals may tend to a limit, which we call the *natural density*. One can prove that, if $A$ has natural density $k$, then the analytic density of $A$ exists and is equal to $k$. The converse does not hold in general.