Advanced Analysis

Dirichlet's Theorem: Setup

Theorem (Part 1: Partial Dirichlet's Theorem)
For each natural number \(d \geq 3\), if
\[ \sum_{p \text{ prime}} \frac{\chi(p)}{p^s}\]
remains bounded as \(s \rightarrow 1^+\) for each nontrivial character \(\chi \in \widehat{{\mathbb Z}^*_d}\), then
\[ \sum_{\substack{p \text{ prime} \\ p \equiv r \mod d}} \frac{1}{p} = \infty\]
for all integers \(r\). Consequently, there are infinitely many primes \(p\) congruent to \(r\) modulo \(d\) for each value of \(r\).
Proof
Let \(\delta_r(n) = 1\) if \(n \equiv r \mod d\) and zero otherwise. By regarding \(\delta_r\) as a function on \({\mathbb Z}^d_*\), it can be expressed in terms of characters. If \(\chi_0(n)\) is the trivial character for \({\mathbb Z}^{*}_d\), then
\[ \frac{1}{\# {\mathbb Z}_d^*} \sum_{n \in {\mathbb Z}_d^*} \overline{\chi_0}(n) \delta_r(n) = \frac{1}{\# {\mathbb Z}_d^*}. \]
This means that
\[ \delta_r(n) = \frac{1}{\# {\mathbb Z}_d^*} \chi_0(n) + \sum_{\chi \neq \chi_0} c_\chi \chi(n) \]
for some constants \(c_\chi\). So for each \(s > 1\),
\[{}\sum_{p \text{ prime}} \frac{\delta_r(p)}{p^s}{}\]
\[{}= \frac{1}{\# {\mathbb Z}_d^*} \sum_{p \text{ prime}} \frac{\chi_0(p)}{p^s}{}\]
\[{}+ \sum_{\chi \neq \chi_0} c_\chi \sum_{p \text{ prime}} \frac{\chi(p)}{p^s}.{}\]
Now \(\chi_0(p) = 1\) unless \(p\) and \(d\) are not relatively prime. But for fixed \(d\), at most finitely many primes \(p\) are not relatively prime to \(d\), and it thus follows that
\[ \frac{1}{\# {\mathbb Z}_d^*} \sum_{p \text{ prime}} \frac{\chi_0(p)}{p^s} - \frac{1}{\# {\mathbb Z}_d^*} \sum_{p \text{ prime}} \frac{1}{p^s}\]
differ by some bounded amount for all \(s > 0\). This means that
\[ \sum_{p \text{ prime}} \frac{\delta_r(p)}{p^s} - \frac{1}{\# {\mathbb Z}_d^*} \sum_{p \text{ prime}} \frac{1}{p^s}\]
remains bounded as \(s \rightarrow 1^+\). But we know that the second term is unboundedly large as \(s \rightarrow 1\), so it means the former is unbounded as well. However, if \(\sum_{p \text{ prime}} \delta_r(p) p^{-1}\) were finite, it would dominate the sum with exponent \(p^{-s}\) for \(s > 1\) and everything would have to be bounded as \(s \rightarrow 1^+\).

1. Factorization of \(L\)-functions

Theorem (\(L\)-function Factorization)
For all \(s > 1\),
\[ L(s,\chi) = \prod_{p \text{ prime}} \frac{1}{1 - \chi(p) p^{-s}}. \]
Proof
Meta (Main Ideas)
The proof is similar to the proof for the Zeta function. Fix \(N\) and \(K\) and let \(S_{N,K}\) be the set of natural numbers \(n\) expressible in terms of a prime factorization into primes less than or equal to \(N\) with exponents less than or equal to \(K\). Then
\[ \prod_{\substack{p \text{ prime} \\ p \leq N}} \sum_{\ell=0}^{K} \left( \frac{\chi(p)}{p^s} \right)^\ell = \sum_{n \in S_{N,K}} \frac{\chi(n)}{n^s}\]
Because \(s > 1\), the sum on the right-hand side is absolutely convergent over \(n\) and so in particular when \(K \geq N\), \(S_N\) contains \(\{1,\ldots,N\}\) and so
\[{}\left| \prod_{\substack{p \text{ prime} \\ p \leq N}} \sum_{\ell=0}^{K} \left( \frac{\chi(p)}{p^s} \right)^\ell - \sum_{n=1}^N \frac{\chi(n)}{n^s} \right|{}\]
\[{}\leq \sum_{n=N+1}^\infty \frac{1}{n^s}.{}\]
Passing to the limit \(K \rightarrow \infty\) followed by \(N \rightarrow \infty\) gives the desired conclusion.

2. Nonvanishing of \(L(1,\chi)\)

As was the case for the Zeta function, the relationship between
\[ \sum_{p \text{ prime}} \frac{\chi(p)}{p^s} \text{ and } \prod_{p \text{ prime}} \frac{1}{1 - \chi(p)p^{-s}}\]
will be what allows us to show that the sum remains bounded as \(s \rightarrow 1^+\). The from one to the other is roughly logarithmic, which means that we will need to see that the \(L\)-function \(L(s,\chi)\) has bounded logarithm as \(s \rightarrow 1^+\). Since we already know that it's smooth there (when \(\chi\) is nontrivial), we simply need to establish that \(L(1,\chi) \neq 0\) for each nontrivial character \(\chi\).
Theorem
For all \(s > 1\),
\[ \prod_{\chi} L(s,\chi) > 1 \]
where the product is taken over all characters \(\chi\) of \({\mathbb Z}^*_d\). In particular, the product is always real.
Proof
Because the character group is finite, for any \(s > 1\), we may write
\[ \prod_{\chi} L(s,\chi) = \prod_{p \text{ prime}} \prod_\chi \frac{1}{(1- \chi(p) p^{-s})}\]
We computed the innermost product over \(\chi\) already: we showed that it must equal
\[ \prod_{p \text{ prime}} \frac{1}{(1 - p^{-s n_1(p)})^{n_2(p)}}\]
for integers \(n_1(p)\) and \(n_2(p)\) depending on \(p\). Each term is real and greater than \(1\), so the whole product must be greater than \(1\) as well (with strict inequality because partial products are increasing).
Unfortunately, this observation by itself is not quite enough because the trivial character \(\chi\) is in the product and it diverges:
Proposition
If \(\chi_0\) is the trivial character in \(\widehat{{\mathbb Z}_d^*}\), then there is a constant \(C < \infty\) such that
\[ L(s,\chi_0) \leq 1+\frac{1}{s-1}\]
for all \(s > 1\).
Proof
Meta (Main Idea)
Write \(L(s,\chi_0)\) as a sum and use the integral test.
Corollary
If \(\chi \in \widehat{{\mathbb Z}_d^*}\) is a character which is not real-valued, then \(L(1,\chi) \neq 0\).
Proof
Meta (Main Idea)
If it did vanish, both \(L(x,\chi)\) and \(L(s,\overline{\chi})\) would vanish at \(s = 1\), and since they're both smooth, their product would vanish at least to second order at \(s=1\). Since all nontrivial \(L\)-functions remain bounded and the trivial \(L\)-function diverges at most like \(1/(s-1)\), it would follow that the product of \(L(s,\chi)\) over all \(\chi\) would vanish to at least first order as \(s \rightarrow 1^+\), which isn't possible by the previous theorem.

3. To-Do List

To complete the proof of Dirichlet's Theorem, we need to:
  1. Establish that \(L(1,\chi) \neq 0\) when \(\chi\) is a real nontrivial character.
  2. Establish for all nontrivial characters that \(L(\chi,1) \neq 0\) implies that
    \[ \sum_{p \text{ prime}} \frac{\chi(p)}{p^s}\]
remains bounded as \(s \rightarrow 1^+\).