Artin Primitive Roots Conjecture (Hooley’s Conditional Proof)

Given a integer $a$ , consider the set $S=\{1, a, a^2, \cdots, a^n, \cdots \}$ . For how many values of prime $p$ , the set exhausts all the non-zero residues. That is for how many primes $p$ , the given integer $a$ primitive element modulo $p$ ?

There are some obvious examples where there are a very few/no such primes. For instance take $a=\pm 1 :$ $a=1$ only generates one residue class and primitive only for $p=2$ . $a=-1$ generates at most two residues classes and is primitive only for $p=2, 3$ .
If $a=n^2$ is a perfect square, the residues generated will all be square residues and cannot exhaust $(\mathbb Z/p)^{*} .$ Similar logic applies to perfect powers, but other powers $a^k .$ residues can sometimes exhaust the whole group – for instance if $(k, p-1)=1. .$

So the squares and $-1 .$ seem to be the only issues and we have the following conjecture:

Conjecture. If $a \neq-1 .$ or a perfect square, then there is a constant $C(a)>0 .$ such that

$\displaystyle \left|\left \{p \leq x: a \text{ is a primitive root modulo } p\right\} \right| \sim C(a) \frac{x}{\log x} .$

For $a=2$ ,

$\displaystyle C_{\text {Artin }}=\prod_{p \text { prime }}\left(1-\frac{1}{p(p-1)}\right)=0.3739558136 \ldots$

That is the set

$\displaystyle P(2) =\left\{3,5,11,13,19,29,37,53,59,61,67,83,101,107,131,139,149,163,173,179,181,197, \ldots\right\}$

of primes with primitive element $2$ has density $\approx 0.3739..$ among all the primes.

Below we will discuss a conditional proof under GRH. The problem is still open, but the state of the art is Heath-Brown proof that at least one of 2, 3, or 5 is a primitive root modulo infinitely many primes $p$

Hooley’s Proof conditional on GRH:

Let us fix $a=2$ . Now $2$ is a primitive root modulo $p$ iff

$\displaystyle 2^{d} \not \equiv 1 \mod p$ for $d|p-1, d\neq p-1$ . That is we dont want

$\displaystyle 2^{\frac{p-1}{q}} \equiv 1 \mod p$ to hold for any prime $q.$

We first estimate $S_1$ the set of primes $p$ for which there are no obstructions like above for $q < z$ . We also estimate $S_2$ , the primes for which there is an obstruction $z<q<x$ , then our required count for primes $p$ where 2 is primitive is given $|S_1| +O(|S_2|)$ because those elements of $S_1$ that shouldn’t be included in our count are all elements of $S_2.$

By inclusion-exclusion we write $S_1$ as sifted sum where we remove primes for which the obstruction holds at different primes $q$

$\displaystyle |S_1| =\sum_{d|(n, \displaystyle\prod_{q<y})} \mu(d) \left|\{p \le x: q| p-1, 2^{\frac{p-1}{q}} \equiv 1 \mod p \text{ for all } q|d \}\right|$

Now each of the term inside can be seen as the number of splitting primes in the compositum of Kummer extensions $\displaystyle K_{q}=\mathbb{Q}\left(2^{1 / q}, \zeta_{q}\right)$ for $q|d$ . So by GRH (for the Dedekind zeta function of this compositum), we get the count split primes

$\displaystyle =\frac{1}{d\varphi(d)} \text{Li}(x)+O\left(x^{1/2} \log (d x)\right)$

Plugging this in we get

$\displaystyle |S_1| = \sum_{d|(n, \displaystyle \prod_{q<y} } \mu(d) \left(\frac{\text{Li}(x)}{d \varphi(d)}+O\left(x^{1 / 2} \log (d x)\right)\right)$

$\displaystyle \sim C(2) \frac{x}{\log x}$

by taking $y \le \log x$ and observing that

$\displaystyle =\sum_{d | \displaystyle \prod_{q<y} } \frac{1}{d \varphi(d)}=\prod_{q <y}\left(1-\frac{1}{q(q-1)}\right) \sim C(2)=\prod_{q}\left(1-\frac{1}{q(q-1)}\right) .$

To estimate $S_2$ the primes that have a obstruction $z<q<x$ , we split the range of primes into three parts small, medium and large ( $C\log x \le q \le \frac{\sqrt x}{\log x^2}$ , $\frac{\sqrt x}{\log^2 x} \le q \le \sqrt x \log x$ , and $\sqrt x \log x \le x \le x$

Again for each of the primes $q$ , these counts are exactly split primes in the Kummer extension $K_q$ .

For small primes, summing over all the $q$ gives a bound $\ll \frac{x}{(\log x)^{2}}$ .

For medium primes we just use the condition $p =1 \mod q$ and estimate the primes in arithmetic progressions by Brun-Titchmarsh

$\displaystyle \pi(x ; q, 1) \leq 2 \frac{x}{\varphi(q) \log (x / q)}$

gives an admissible bound like $\ll \frac{x \log \log x}{(\log x)^{2}} .$

Finally for the large primes, use the fact that $\displaystyle 2^{\frac{p-1}{q}} \equiv 1 \mod p$ , to bound the number of primes $p$ by the number of prime factors of all the numbers $\displaystyle 2^n-1$ , where $\displaystyle n =\frac{p-1}{q} \le M=\frac{\sqrt x}{\log x}$