Selberg (1949) “An elementary proof of Dirichlet’s theorem about primes in an arithmetic progression”

Consider the sum

$\displaystyle S_{l}(x)=\sum_{p \le x \atop p \equiv l ~ \bmod ~k} \frac{\log p}{p}$

We want to prove lower bound on ${S_l(x) \ge c_k \log x}$

Consider the sieve weights

$\displaystyle \lambda_{d}=\lambda_{d, x}=\mu(d) \log ^{2} \frac{x}{d}$

which have the property

$\displaystyle \theta_{n}=\theta_{n, x}=\sum_{d \mid n} \lambda_{d}=\left\{\begin{array}{ll} \log ^{2} x, & \mbox { for } n=1 \\ \log p \cdot \log x^{2} / p, & \mbox { for } n=p^{\alpha}, \quad \alpha \geq 1, \\ 2 \log p \log q, & \mbox { for } n=p^{\alpha} q^{\beta}, \quad \alpha \geq 1, \quad \beta \geq 1 \\ 0, & \mbox { for all other } n . \end{array}\right.$

The contribution from prime powers is small and we get

$\displaystyle \sum_{n \leq x \atop n \equiv l(k)} \theta_{n} = \sum_{p \leq x \atop p \equiv \bar{l}(k)} \log^{2} p+\sum_{p q \leq x \atop p q=\bar{i}(k)} \log p \log q+O(x)$

On the other hand opening up the sum over the divisors and switching sums we get,

$\displaystyle \sum_{n \leq x \atop n=l(k)} \theta_{n}=\sum_{d \leq x \atop(d, k)=1} \lambda_{d} \sum_{d \mid n \atop{ n \leq x \atop n= l(k)}} 1=\frac{x}{k} \sum_{d \leq x \atop(d, k)=1} \frac{\lambda_{d}}{d}+O\left(\sum_{d \leq x}\left|\lambda_{d}\right|\right)=\frac{x}{k} \sum_{d \leq x \atop(d, k)=1} \frac{\lambda_{d}}{d}+O(x)$ So we have

$\displaystyle \sum_{p \leq x \atop p \equiv \bar{l}(k)} \log ^{2} p+\sum_{p q \leq x \atop p q=l(k)} \log p \log p=\frac{x}{k}\sum_{ d \le x\atop(d,l)=1} \frac{\lambda_{d}}{d}+O(x)$

The contribution from ${(d,l)>}$ is ${O(x)}$ , therefore

$\displaystyle \sum_{p \leq x \atop p \equiv \bar{l}(k)} \log ^{2} p+\sum_{p q \leq x \atop p q \equiv l(k)} \log p \log q =\frac{1}{\varphi\left(k\right)}\left (\sum_{p \leq x} \log ^{2} p+\sum_{p q \leq x} \log p \log q\right)+O(x)$

$\displaystyle \sum_{p \leqq x} \frac{\log p}{p}=\log x+O(1)$ and partial summation with some smooth functions, we get the following

$\displaystyle \sum_{p \leq x \atop p \equiv \bar{l}(k)} \frac{\log ^{2} p}{p}+\sum_{p q \leq x \atop p q=l(k)} \frac{\log p \log q}{p q} = \left(\sum_{p \leq x} \frac{\log ^{2} p}{p}+\sum_{p{q} \leq x}{ }_{p} \frac{\log p \log q}{p q}\right)+O(\log x)=\frac{1}{\varphi(k)} \log ^{2} x+O(\log x) \quad \quad (A)$

$\displaystyle \sum_{p \leq x \atop p \equiv(k)} \frac{\log ^{3} p}{p}+\sum_{p q \leq \underline{x} \atop p q=\bar{l}(k)} \frac{\log p \log q}{p q} \log p q = \frac{2}{3 \varphi(k)} \log ^{3} x+O\left(\log ^{2} x\right)$

$\displaystyle \sum_{p \leq x \atop p \equiv l(k)} \frac{\log ^{2} p}{p} \leqq \frac{1}{\varphi(k)} \log ^{2} x+O(\log x), \quad \sum_{p \leq x \atop p=l(k)} \frac{\log p}{p} \leqq \frac{2}{\varphi(k)} \log x+O(\log \log x) \quad \quad (B)$

(The factor ${2}$ is really important here, it cannot be improved with basic sieve estimates as above.)

Writing

$\displaystyle \sum_{p q \leq x \atop p q=\bar{l}(k)} \frac{\log p \log ^{2} q}{p q}=\sum_{p<x \atop p\not \mid k} \frac{\log p}{p} \sum_{q \leq x \mid p \atop q \equiv l \bar{p}~(k)} \frac{\log ^{2} q}{q}$

and plugging in the above expression ${(A)}$ for $\sum_{q \leq x \mid p \atop q \equiv l \bar{p}~(k)} \frac{\log ^{2} q}{q}$ , we get

$\displaystyle \sum_{p q \leqq \underline{z} \atop p q=l(k)} \frac{\log p \log ^{2} q}{p q}= -\sum_{p q r \leq \atop p q r=l(k)} \frac{\log p \log q \log r}{p q r}+\frac{1}{3 \varphi(k)} \log ^{3} x+O\left(\log ^{2} x\right)$

$\displaystyle \sum_{p \leq x \atop p \equiv l(k)} \frac{\log ^{3} p}{p}=2 \sum_{p q r \leq z \atop p q r=\bar{l}(k)} \frac{\log p \log q \log r}{p q r}+O\left(\log ^{2} x\right)$

So we have

$\displaystyle \boxed{\log^2{x} \cdot S_{l}(x) \geqq 2 \sum_{m m ' m^{\prime\prime} \equiv l(k)} S_{m}\left(x^{1 / 3}\right) S_{m'}\left(x^{1 / 3}\right) S_{m^{\prime\prime}}\left(x^{1 / 3}\right) -O(\log^2 x)} \quad \quad (*)$

$\displaystyle \sum_{p q \leq z \atop p q=\bar{l}(k)} \frac{\log p \log q}{p q} \leqq \sum_{p \leqq x^{1 / 3}} \sum_{q\le x^{1/3} \atop p q=l(k)} \frac{\log p \log q}{p q} + 2 \sum_{x^{1 / 3}<p \leqq x} \frac{\log p}{p} \sum_{q \leqq x / p \atop q=l \bar{p} ~(k)} \frac{\log q}{q}$

$\displaystyle \le \sum_{p \leqq x^{1 / 3}} \sum_{q\le x^{1/3} \atop p q=l(k)} \frac{\log p \log q}{p q} +\frac{8}{9 \varphi(k)} \log ^{2} x+O(\log x \log \log x)$

$\displaystyle \sum_{p \leq z \atop p=\bar{l}(k)} \frac{\log ^{2} p}{p} \geqq \frac{1}{9 \varphi(k)} \log ^{2} x-\sum_{p \leqq x^{1 / 3} \atop p q=l(\bar{k})} \sum_{q \leq x^{1 / 3}} \frac{\log p \log q}{p q}+O(\log x \log \log x)$

$\displaystyle \boxed {\log x \cdot S_{l}(x)>\frac{1}{10 \varphi(k)} \log ^{2} x-\sum_{m m^{\prime} \equiv l(k)} S_{m}\left(x^{1 / 3}\right) S_{m}\left(x^{1 / 3}\right)} \quad \quad (**)$

Till now we did some elementary manipulations and estimates, and these equations are consistent with having primes concentrated in a few residue classes. In fact, as we will see later we have to somehow argue that they don’t all lie in the set ${\chi(p)=-1}$ for the quadratic character. So the main problem of ${L(1, \chi)}$ vanishing is still not really addressed. It’s equivalent to showing lower bounds on $\sum_{p \leq x \atop \chi(p)=1} \frac{\log p}{p}$ , that is we don’t want ${\chi(p)}$ to behave like ${-1}$ for most primes.

Selberg achieves it uses the following lemma.

Main lemma: For every real, non-principal character ${\chi \bmod k}$ , for large enough ${x}$ , we have

$\displaystyle \sum_{p \leq x \atop \chi(p)=1} \frac{\log p}{p}>\frac{1}{9} \log x$

$\displaystyle \sum_{\substack{p \leq x \\ \left(\frac{D} {p} \right)=1}} \frac{\log p}{p}>\frac{1}{9} \log x \text { for } x\gg 1$

There has to be some Class Number Formula kind of ideas to related the ${L(1, \chi)}$ and the arithmetic of ${\mathbb Q(\sqrt{D})}$ (or binary quadratic forms of discriminant ${D}$ ) (In fact, all such ideas relate the information about point counting/size/archimedean behaviour of the forms to their multiplicative behaviour ie how they behave modulo primes.

Selberg consider the product

$\displaystyle P=\prod_{|u| \leq \sqrt{x / 2} \atop { |v| \leqq \sqrt{x / 2 D}}}\left|u^{2}-D v^{2}\right|$

The product excludes term with ${u=v=0}$ . It’s easy to see that ${\log P = \frac{2x (\log x +O(1)) }{\sqrt{D}}}$ . This is the archimedean part.

On the other hand, the contribution of a prime (in factorization) to the product depends on whether ${\left(\frac{D} {p} \right)=1}$ or ${\left(\frac{D} {p} \right)=-1}$

$\displaystyle u^{2}-D v^{2} \equiv 0(p)$

If ${\left(\frac{D} {p} \right)=-1}$ , we need ${u}$ and ${v}$ to be divisible by ${p}$ and the exponent is ${O\left(\frac{x}{p^{2}}\right)}$ exponent. On the other hand for ${\left(\frac{D} {p} \right)=-1}$ , each solution ${(u, v)}$ satisfies ${u v_{o} \pm u_{o} v \equiv 0(p)}$ for ${(u_0, v_0)}$ any fixed solution. Thus we get ${\frac{4 x}{p \sqrt{D}}+O\left(\sqrt{\frac{x}{D}}\right)}$ for the contributions to the exponent of ${p}$ in this case.

Putting both the cases together we get,

$\displaystyle \log P \le 8 \frac{x}{\sqrt{D}} \sum_{\substack{p \leq x \\ \left(\frac{D} {p} \right)=1}} \ \frac{\log p}{p}+O(x)$

and comparing with the lower bound, we attain

$\displaystyle \sum_{p \leq x \atop \chi(p)=1} \frac{\log p}{p} > c \log x.$

Finishing the proof:

Thus we have ${S_m}$ is large for a a lot of residues ${m}$ because we get all primes by adding ${S_m}$ over all the residues ${m}$ . It’s large on average over residues with ${\chi(m)=1}$ from the main lemma.

We want to say ${S_l}$ is large. If it were small, then using ${(**)}$ , we get that there is a pair ${(m,m')}$ such that ${S_m(x^{1/3})S_{m'}(x^{1/3})}$ is large, and in fact each of the quantities ${S_m, S_{m'}}$ is large because by ${(B)}$ we know neither of them can be too large alone. Using this information with can construct ${m_1, m_2, m_3}$ such that ${m_1m_2m_3 = l \bmod k}$ with ${S_{m_i}}$ large.
(Proof: If there are more than ${\varphi(k)}$ ${S_{m_i}}$ which are large, we can see that at least one pair of ${m_r m_s}$ and ${\ \bar m_t}$ should be equal. If there are exactly ${\varphi(k)}$ number of ${S{m_i}}$ large, ${m_i \bar m_j}$ forms a subgroup of index 2 inside ${\mathbb Z/l}$ and ${\chi(m_i) }$ is either ${1}$ for all ${i}$ or ${-1}$ for all ${i}$ . The main lemma produced ${m}$ with ${\chi(m)=1}$ and ${S_m}$ large, hence we see that all of the ${m_i}$ satisfy ${\chi(m_i)=1}$ , and ${S_m}$ is large precisely for the subgroup of ${m}$ such that ${\chi(m)=1}$ .This show that ${l =mm' \bmod k}$ also lies in that subgroup and hence ${m_1=1, m_2=1, m_3=l}$ works)

Once we have ${m_1, m_2, m_3}$ using ${(*)}$ we get

$\displaystyle {\log^2{x} \cdot S_{l}(x) \geqq 2 S_{m_1}\left(x^{1 / 3}\right) S_{m_2}\left(x^{1 / 3}\right) S_{m_3}\left(x^{1 / 3}\right) -O(\log^2 x)}$ is large.

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