Definition 7.7.1.
A character \(\chi\) is called real if its values are in the real numbers.
Section 7.7 Proof of Dirichlet’s theorem, part 2
In this section we’ll finish the proof of Dirichlet’s theorem. What is left is to show that \(L(1,\chi) \neq 0\) for every character \(\chi\text{.}\) We already know it for the principal character \(\chi_0\) (in fact \(L(1,\chi_0)=\infty\)), so we just have to show it for nonprincipal characters. This will happen in two steps: First we’ll show it for real nonprincipal characters. The Dirichlet hyperbola method makes an appearance here. Second, we’ll show it for nonreal nonprincipal characters, by a clever trick which hinges on complex conjugation.
We fix a positive integer \(q\text{.}\) For this part of the proof, the integer \(a\) (as in, the primes such that \(p \equiv a \pmod{q}\)) doesn’t play a role.
Subsection 7.7.1 Real nonprincipal characters
Since the values of \(\chi\) are roots of unity (or zero), the only real values are \(\pm 1, 0\text{.}\) For example (we won’t use these examples in the proof):
- The principal character is real.
- If \(q\) is prime, the Legendre symbol \(\chi(n) = \left(\frac{n}{q}\right)\) is a real character.
- As an exercise, one can show that modulo a prime \(q\text{,}\) the principal character and the Legendre symbol are the only real characters.
Theorem 7.7.2.
Let \(\chi\) be a real nonprincipal Dirichlet character modulo \(q\text{.}\) Then \(L(1,\chi) \neq 0\text{.}\)Proof.
We know that \(\chi\) is completely multiplicative.
Claim 7.7.3. Step 1.
Let \(r(n) = \sum_{d \mid n} \chi(d)\text{.}\) Then \(r(n) \geq 0\) for all \(n \in \bbN\) and \(r(n) \geq 1\) if \(n\) is a square.
Proof.
Since \(\chi\) is (completely) multiplicative and \(r = \chi * 1\) (Dirichlet convolution), it follows that \(r\) is multiplicative (but not necessarily completely multiplicative).
Let \(p\) be a prime and \(a \geq 1\text{.}\) Then
\begin{align}
r(p^a) \amp = \sum_{d \mid p^a} \chi(d) \tag{7.7.1}\\
\amp = \chi(1) + \chi(p) + \chi(p^2) + \dotsb + \chi(p^a) \tag{7.7.2}\\
\amp = \chi(1) + \chi(p) + \chi(p)^2 + \dotsb + \chi(p)^a \tag{7.7.3}\\
\amp = \begin{cases}
a+1, \amp \text{if } \chi(p) = 1; \\
1, \amp \text{if } \chi(p) = 0; \\
1, \amp \text{if } \chi(p) = -1 \text{ and } a=2k; \\
0, \amp \text{if } \chi(p) = -1 \text{ and } a=2k+1.
\end{cases} \tag{7.7.4}
\end{align}
Let \(n = p_1^{a_1} \dotsm p_k^{a_k}\text{.}\) Then \(r(n) = r(p_1^{a_1}) \dotsm r(p_k^{a_k})\text{.}\) We saw that every \(r(p^a) \geq 0\text{,}\) hence \(r(n) \geq 0\text{.}\) If \(n\) is a perfect square, then every power \(a_i\) is even, so each \(r(p_i^{a_i}) \geq 1\text{.}\) It follows that \(r(n) \geq 1\text{.}\)
Claim 7.7.4. Step 2.
Let \(A(x) = \sum_{n \leq x} \frac{r(n)}{\sqrt{n}}\text{.}\) Then \(A(x) \to \infty\) as \(x \to \infty\text{.}\)
Proof.
We have
\begin{align*}
A(x) \amp = \sum_{n \leq x} \frac{r(n)}{\sqrt{n}} \\
\amp \geq \sum_{n \leq x, n = m^2} \frac{1}{\sqrt{n}} \\
\amp = \sum_{m \leq \sqrt{x}} \frac{1}{m}
\end{align*}
This is a partial sum of the harmonic series; as \(x \to \infty\) it diverges.
Claim 7.7.5. Step 3.
\(A(x) = 2\sqrt{x} L(1,\chi) + O(1) \text{.}\)
Proof.
We will use the Dirichlet hyperbola method.
We have
\begin{align}
A(x) \amp = \sum_{n \leq x} \frac{r(n)}{\sqrt{n}} \tag{7.7.5}\\
\amp = \sum_{n \leq x} \frac{1}{\sqrt{n}} \sum_{d \mid n} \chi(d) \tag{7.7.6}\\
\amp = \sum_{dk \leq x} \frac{\chi(d)}{\sqrt{dk}} \tag{7.7.7}\\
\amp = \sum_{dk \leq x} \frac{\chi(d)}{\sqrt{d}} \cdot \frac{1}{\sqrt{k}} \tag{7.7.8}
\end{align}
This sum is over lattice points \((d,k)\) where \(d,k\) are postive integers such that \(dk \leq x\text{,}\) i.e., lying on or underneath the hyperbola \(dk = x\) in the \(dk\)-plane. We separate the region under the hyperbola into three parts as follows: the first part is defined by \(d \leq \sqrt{x}\text{,}\) the second is defined by \(k \leq \sqrt{x}\text{,}\) and the third is the intersection of the first two, so \(d, k \leq \sqrt{x}\) (both). Visually, if \(d\) is the horizontal coordinate and \(k\) is the vertical coordinate, then the region under the hyperbola with \(d \leq \sqrt{x}\) is the “upward” branch, the region under the hyperbola with \(k \leq \sqrt{x}\) is the “rightward” branch, and the region with both \(d,k \leq \sqrt{x}\) is a square. Define \(S_1,S_2,S_3\) as the sums over lattice points in these regions:
\begin{equation}
S_1 = \sum_{dk \leq x, d \leq \sqrt{x}} \frac{\chi(d)}{\sqrt{d}} \frac{1}{\sqrt{k}},
\qquad
S_2 = \sum_{dk \leq x, k \leq \sqrt{x}} \frac{\chi(d)}{\sqrt{d}} \frac{1}{\sqrt{k}},
\qquad
S_3 = \sum_{dk \leq x, d \leq \sqrt{x}, k \leq \sqrt{x}} \frac{\chi(d)}{\sqrt{d}} \frac{1}{\sqrt{k}}.\tag{7.7.9}
\end{equation}
We have
\begin{equation}
A(x) = S_1 + S_2 - S_3.\tag{7.7.10}
\end{equation}
Now our goal is to approximate the \(S_i\text{.}\) Looking ahead, we can expect to encounter certain sums, so it will be helpful to recall the following approximations:
\begin{equation}
\sum_{k \leq y} \frac{1}{k^s} = \frac{y^{-s+1}}{-s+1} + O(1)\tag{7.7.11}
\end{equation}
and
\begin{equation}
\sum_{d \leq z} \frac{\chi(d)}{d^s} = L(s,\chi) + O\left(\frac{1}{z^s}\right)\text{,}\tag{7.7.12}
\end{equation}
both of these holding for all \(0 \lt s \lt 1\text{.}\) (The textbook mentions a more accurate approximation for the sum of \(1/k^s\) that can be found by using Euler summation.) We will use them for \(s = \frac{1}{2}\text{.}\) In order to avoid headaches about fractions, here are explicit statements:
\begin{equation}
\sum_{k \leq y} \frac{1}{\sqrt{k}} = \frac{y^{1/2}}{1/2} + O(1) = 2\sqrt{y} + O(1)\tag{7.7.13}
\end{equation}
and
\begin{equation}
\sum_{d \leq z} \frac{\chi(d)}{\sqrt{d}} = L(1/2,\chi) + O\left(\frac{1}{\sqrt{z}}\right)\text{.}\tag{7.7.14}
\end{equation}
We will apply these with various \(y\)’s and \(z\)’s; warning, sometimes we will use, for example, \(y = \sqrt{x}\) in which case \(\sqrt{y} = x^{1/4}\) (similar for \(z\)).
Now, here are the approximations of the \(S_i\text{.}\) The easiest is \(S_3\text{:}\)
\begin{align}
S_3 \amp = \left( \sum_{d \leq \sqrt{x}} \frac{\chi(d)}{\sqrt{d}} \right)
\left( \sum_{k \leq \sqrt{x}} \frac{1}{\sqrt{k}} \right) \tag{7.7.15}\\
\amp = (L(1/2,\chi) + O(1/x^{1/4})) (2x^{1/4} + O(1)) \tag{7.7.16}\\
\amp = 2x^{1/4} L(1/2,\chi) + O(1) \tag{7.7.17}
\end{align}
Next, we approximate \(S_1\text{:}\)
\begin{align}
S_1 \amp = \sum_{d \leq \sqrt{x}} \frac{\chi(d)}{\sqrt{d}} \sum_{k \leq \frac{x}{d}} \frac{1}{\sqrt{k}} \tag{7.7.18}\\
\amp = \sum_{d \leq \sqrt{x}} \frac{\chi(d)}{\sqrt{d}} \left(2\sqrt{x/d} + O(1) \right) \tag{7.7.19}\\
\amp = 2\sqrt{x} \sum_{d \leq \sqrt{x}} \frac{\chi(d)}{d} + O(1) \sum_{d \leq \sqrt{x}} \frac{\chi(d)}{\sqrt{d}} \tag{7.7.20}\\
\amp = 2\sqrt{x} (L(1,\chi) + O(1/\sqrt{x})) + O(1)(L(1/2,\chi) + O(1/x^{1/4})) \tag{7.7.21}\\
\amp = 2 \sqrt{x} L(1,\chi) + O(1) \tag{7.7.22}
\end{align}
(For some reason, the textbook uses more accurate approximations here and does extra work, to arrive at the same result in the end.) Finally, \(S_2\) is similar to \(S_1\text{:}\)
\begin{align}
S_2 \amp = \sum_{k \leq \sqrt{x}} \frac{1}{\sqrt{k}} \sum_{d \leq \frac{x}{k}} \frac{\chi(d)}{\sqrt{d}} \tag{7.7.23}\\
\amp = \sum_{k \leq \sqrt{x}} \frac{1}{\sqrt{k}} (L(1/2,\chi) + O(1/\sqrt{x/k})) \tag{7.7.24}\\
\amp = L(1/2,\chi) \sum_{k \leq \sqrt{x}} \frac{1}{\sqrt{k}} + O(1/\sqrt{x}) \sum_{k \leq \sqrt{x}} \frac{1}{\sqrt{k}} O(\sqrt{k}) \tag{7.7.25}\\
\amp = L(1/2,\chi) (2x^{1/4} + O(1)) + O(1/\sqrt{x}) O(\sqrt{x}) \tag{7.7.26}\\
\amp = 2x^{1/4} L(1/2,\chi) + O(1) \tag{7.7.27}
\end{align}
Putting it all together:
\begin{align}
A(x) \amp = S_1 + S_2 - S_3 \tag{7.7.28}\\
\amp = (2 \sqrt{x} L(1,\chi) + O(1)) + (2 x^{1/4} L(1/2,\chi) + O(1)) - (2x^{1/4}L(1/2,\chi) + O(1)) \tag{7.7.29}\\
\amp = 2 \sqrt{x} L(1,\chi) + O(1) \tag{7.7.30}
\end{align}
which proves the claim.
Claim 7.7.6. Conclusion.
\(L(1,\chi) \neq 0\text{.}\)
Proof.
If \(L(1,\chi)=0\) then \(A(x) = O(1)\text{,}\) contradicting that \(A(x) \to \infty\) as \(x \to \infty\text{.}\)
Subsection 7.7.2 Nonreal nonprincipal characters
Theorem 7.7.7.
Let \(\chi\) be any nonprincipal Dirichlet character modulo \(q\text{.}\) Then \(L(1,\chi) \neq 0\text{.}\)Proof.
Let \(N\) be the number of Dirichlet characters \(\chi\) modulo \(q\) with \(L(1,\chi) = 0\text{.}\) Our goal is to show that \(N = 0\text{.}\)
Recall that:
- For the principal character \(\chi_0\text{,}\) \(\sum_{n \leq x} \frac{\chi_0(n)\Lambda(n)}{n} = \log x + O(1)\text{.}\)
- For \(\chi \neq \chi_0\text{,}\) if \(L(1,\chi) \neq 0\text{,}\) then \(\sum_{n \leq x} \frac{\chi(n)\Lambda(n)}{n} = O(1)\text{.}\)
- For \(\chi \neq \chi_0\text{,}\) if \(L(1,\chi) = 0\text{,}\) then \(\sum_{n \leq x} \frac{\chi(n)\Lambda(n)}{n} = -\log x + O(1)\text{.}\)
Adding this up for all characters modulo \(q\text{,}\) we get
\begin{equation}
\sum_{\chi \mod q} \sum_{n \leq x} \frac{\chi(n)\Lambda(n)}{n} = (1-N)\log x + O(1)\tag{7.7.31}
\end{equation}
We get \(+1\log x\) from the principal character, and \(-N\log x\) from the nonprincipal characters with \(L(1,\chi)=0\text{.}\)
We can use the orthogonality relations to simplify the left hand side of the equation:
\begin{align}
\sum_{\chi \mod q} \sum_{n \leq x} \frac{\chi(n)\Lambda(n)}{n}
\amp = \sum_{n \leq x} \left( \sum_{\chi \mod q} \chi(n) \right) \frac{\Lambda(n)}{n} \tag{7.7.32}\\
\amp = \phi(q) \sum_{n \leq x, n \equiv 1 \pmod{q}} \frac{\Lambda(n)}{n} \tag{7.7.33}\\
\amp \geq 0 \tag{7.7.34}
\end{align}
We used that \(\sum_{\chi \mod q} \chi(n)\) is equal to \(\phi(q)\) if \(n \equiv 1 \pmod{q}\text{,}\) or \(0\) otherwise. And we used that each term \(\frac{\Lambda(n)}{n} \geq 0\text{.}\)
The result is that
\begin{equation*}
(1-N)\log x + O(1) \geq 0
\end{equation*}
Well, if \(N \gt 1\) this is impossible, the left hand side will go to \(-\infty\text{.}\) Therefore \(N \leq 1\text{.}\)
Finally we rule out the possibility that \(N=1\text{.}\) If \(N=1\) it means that there is a nonprincipal character \(\chi\) with \(L(1,\chi) = 0\text{.}\) We know that \(\chi\) must be nonreal, i.e., \(\chi(a) \notin \bbR\) for some \(a\text{.}\)
But then \(\overline{\chi}\) is also a character, distinct from \(\chi\) (\(\overline{\chi} \neq \chi\)), and we have
\begin{equation*}
L(1,\overline{\chi}) = \sum \frac{\overline{\chi}(n)}{n} = \overline{\sum \frac{\chi(n)}{n}}
= \overline{L(1,\chi)} = 0
\end{equation*}
which would force \(N \geq 2\text{.}\)
This contradiction shows that \(N=1\) is impossible. So it must be \(N=0\text{.}\) There are no Dirichlet characters \(\chi\) with \(L(1,\chi)=0\text{.}\)
Subsection 7.7.3 Conclusion: Proof of Dirichlet’s theorem
Theorem 7.7.8. Dirichlet’s theorem.
Let \(q,a\) be positive integers with \((a,q)=1\text{.}\) Then
\begin{equation*}
\sum_{p \leq x, p \equiv a \pmod{q}} \frac{\log p}{p} = \frac{1}{\phi(q)} \log x + O(1)\text{.}
\end{equation*}
In particular, there are infinitely many primes congruent to \(a\) modulo \(q\text{.}\)
Proof.
We have
\begin{align}
\sum_{p \leq x, p \equiv a \pmod{q}} \frac{\log p}{p}
\amp = \sum_{n \leq x, n \equiv a \pmod{q}} \frac{\Lambda(n)}{n} + O(1) \tag{7.7.35}\\
\amp = \frac{1}{\phi(q)} \sum_{\chi \mod q} \overline{\chi}(a) \sum_{n \leq x} \frac{\chi(n)\Lambda(n)}{n} + O(1) \tag{7.7.36}\\
\amp = \frac{1}{\phi(q)} \log x + \frac{1}{\phi(q)} \sum_{\chi \mod q, \chi \neq \chi_0} \sum_{n \leq x} \frac{\chi(n)\Lambda(n)}{n} + O(1) \tag{7.7.37}\\
\amp = \frac{1}{\phi(q)} \log x + \frac{1}{\phi(q)} \sum_{\chi \mod q, \chi \neq \chi_0} O(1) + O(1) \tag{7.7.38}\\
\amp = \frac{1}{\phi(q)} \log x + O(1) \tag{7.7.39}
\end{align}
as claimed.
