Advanced Analysis

Dirichlet's Theorem: Conclusion

1. Proving \(L(1,\chi) \neq 0\) for real characters \(\chi\)

The proof hinges on the analysis of the quantity
\[ S_N := \sum_{nm \leq N} \frac{\chi(n)}{\sqrt{nm}}\]
where \(\chi\) is a real Dirichlet character.

To see why \(L(1,\chi) \neq 0\), we will prove two key facts:
  1. The difference \(S_N - 2 \sqrt{N} L(1,\chi)\) is a bounded function of \(N\).
  2. \(S_N\) is an unbounded function of \(N\).
This proves that \(L(1,\chi)\) cannot be zero because if it were, then \(S_N\) would be both bounded and unbounded as \(N \rightarrow \infty\).

The strategy is to compute the sum by grouping terms in two different ways.
Figure. Illustration of the different groupings of terms which accomplish the two key facts

The proof of the first fact relies on two hard analysis estimates contained in the proposition immediately below. It is accomplished by summing over \(m\) first for small values of \(n\) and summing over \(n\) first when it exceeds the threshold \(\lfloor N^{1/2}\rfloor\). The second key fact is established by grouping terms on the hyperbola \(nm = k\) and then summing over \(k\).

2. Step 1: Analysis of \(S_N\) in terms of \(L\)-functions

Proposition (Asymptotic Analysis of \(S_N\))
There exist constants \(C,C',C''\) such that for each natural number \(N\),
\[{}|S_N - 2 \sqrt{N} L(1,\chi)| \leq C{}\]
\[{}+ C' \sup_M M \left| \sum_{n=M+1}^\infty \frac{\chi(n)}{n} \right|{}\]
\[{}+ C'' \sup_{\substack{M,M' \\ M' > M}} \sqrt{M} \left| \sum_{n=M+1}^{M'} \frac{\chi(n)}{\sqrt{n}} \right|{}\]
Proof
Fix the integer \(M := \lfloor N\rfloor^{1/2}\) and split the sum into pieces based on the size of \(n\):
\[{} S_N{}\]
\[{}= \sum_{n=1}^{M} \frac{\chi(n)}{\sqrt{n}} \sum_{m=1}^{\lfloor N n^{-1} \rfloor} \frac{1}{\sqrt{m}}{}\]
\[{}+ \sum_{n=M+1}^{N} \frac{\chi(n)}{\sqrt{n}} \sum_{m=1}^{\lfloor N n^{-1} \rfloor} \frac{1}{\sqrt{m}}{}\]
\[{}=: I + II{}\]
\[{} I{}\]
\[{}= \sum_{n=1}^M \frac{\chi(n)}{\sqrt{n}} \sum_{m=1}^{\lfloor N n^{-1} \rfloor} \frac{1}{\sqrt{m}}{}\]
\[{}= \sum_{n=1}^M \frac{\chi(n)}{\sqrt{n}} \Big[ 2 \lfloor N n^{-1}\rfloor^{1/2} + c {}\]
\[{}+ O( (N n^{-1})^{-1/2}) \Big]{}\]
\[{}= \sum_{n=1}^M \frac{\chi(n)}{\sqrt{n}} \Big[ 2 ( N n^{-1})^{1/2} + c{}\]
\[{}+ O( (N n^{-1})^{-1/2}) \Big]{}\]
\[{}= 2 \sqrt{N} \sum_{n=1}^M \frac{\chi(n)}{n}{}\]
\[{}+ c \sum_{n=1}^M \frac{\chi(n)}{\sqrt{n}}{}\]
\[{}+ O(N^{-1/2} M),{}\]
\[{}II{}\]
\[{}= \sum_{n=M+1}^{N} \frac{\chi(n)}{\sqrt{n}} \sum_{m=1}^{\lfloor N n^{-1} \rfloor} \frac{1}{\sqrt{m}}{}\]
\[{}= \sum_{m=1}^{\lfloor NM^{-1}\rfloor} \frac{1}{\sqrt{m}} \sum_{\substack{M+1 \leq n \leq N m^{-1}}} \frac{\chi(n)}{\sqrt{n}}.{}\]
In the computations above, we use that \(\lfloor x \rfloor^{1/2} = x^{1/2} + O(x^{-1/2})\) by the Mean Value Theorem.

It follows that
\[{}S_N - 2 \sqrt{N} L(1,\chi) {}\]
\[{}= -2 \sqrt{N} \sum_{n=M+1}^\infty \frac{\chi(n)}{n}{}\]
\[{}+ c \sum_{n=1}^M \frac{\chi(n)}{\sqrt{n}} + O(1) {}\]
\[{}+\sum_{m=1}^{\lfloor NM^{-1}\rfloor} \frac{1}{\sqrt{m}} \sum_{\substack{M+1 \leq n \leq N m^{-1}}} \frac{\chi(n)}{\sqrt{n}}.{}\]
Each of the first three terms on the right-hand side is bounded in magnitude by one of the three expressions on the right-hand side of our desired inequality (when \(C,C'\) and \(C''\) are chosen appropriately). For the fourth term, bounding the magnitude of the inner sum by a constant independent of \(m\) and comparing to an integral gives that
\[{}\left| \sum_{m=1}^{\lfloor NM^{-1}\rfloor} \frac{1}{\sqrt{m}} \sum_{\substack{M+1 \leq n \leq N m^{-1}}} \frac{\chi(n)}{\sqrt{n}} \right|{}\]
\[{}\leq 2 \lfloor N M^{-1} \rfloor^{1/2}{}\]
\[{}\cdot \max_m \left| \sum_{\substack{M+1 \leq n \leq N m^{-1}}} \frac{\chi(n)}{\sqrt{n}} \right|.{}\]
Since \(\lfloor N M^{-1} \rfloor \leq N M^{-1}\) and since \(N < (M+1)^2\), \(\lfloor N M^{-1} \rfloor < (M+1) M^{-1/2} \leq 2 \sqrt{M}\). Thus this fourth term on the right-hand side of our expression for \(S_N - 2 \sqrt{N} L(1,\chi)\) is also bounded above by one of the terms on the right-hand side of the promised inequality (namely, the term with constant \(C''\)).
Lemma
There exist constants \(C'\) and \(C''\) such that for every nontrivial character \(\chi\),
\[{}\left| \sum_{n=M+1}^\infty \frac{\chi(n)}{n} \right|{}\]
\[{}\leq \frac{C'}{M}{}\]
\[{}\left| \sum_{n=M+1}^{M'} \frac{\chi(n)}{\sqrt{n}} \right|{}\]
\[{}\leq \frac{C''}{\sqrt{M}}{}\]
for all \(M\) and \(M' > M\).
Proof
Meta (Key Idea)
Summation by parts does the trick because the partial sums of \(\chi\) always remain bounded.

3. Step 2: Analysis of \(S_N\) by summing on hyperbolas \(nm=k\).

Proposition
\(S_N \rightarrow \infty\) as \(N \rightarrow \infty\).
Proof
Meta (Main Ideas)
Summing along hyperbolas \(nm = k\) gives
\[ S_N = \sum_{k=1}^N \frac{1}{\sqrt{k}} \sum_{n | k} \chi(n). \]
Factor \(k = p_1^{\ell_1} \cdots p_m^{\ell_m}\) and note that for each prime factor \(p_1,\ldots,p_m\), \(\chi(p_i)\) must equal \(-1,0\), or \(1\). The divisors of \(k\) are exactly the integers of the form \(p_1^{\ell_1'} \cdots p_m^{\ell'_m}\) for \(0 \leq \ell'_i \leq \ell_i\). Thus
\[{}\sum_{n | k} \chi(n){}\]
\[{}= \sum_{\ell'_1=0}^{\ell_1} \cdots \sum_{\ell'_m=0}^{\ell_m} (\chi(p_1))^{\ell'_1} \cdots (\chi(p_m))^{\ell'_m}, {}\]
which factors as a product of finite geometric series with common ratios equal to \(\pm 1\). In particular, if \(\chi(p_i) = 0\), then
\[ \sum_{\ell'=0}^{\ell_i} \chi(p_i)^{\ell'} = 1; \]
when \(\chi(p_i) = 1\), we have
\[ \sum_{\ell'=0}^{\ell_i} \chi(p_i)^{\ell'} = \ell_i + 1, \]
and if \(\chi(p_i) = -1\), then
\[ \sum_{\ell'=0}^{\ell_i} \chi(p_i)^{\ell'} = \begin{cases} 0 & \ell_i \text{ odd}, \\ 1 & \ell_i \text{ even}. \end{cases}\]
This implies that \(\sum_{n|k} \chi(n)\) is, in fact, never negative, and it is always at least \(1\) when \(k\) is a perfect square. Thus \(S_N\) diverges at least logarithmically as \(N \rightarrow \infty\).