Do we have an upper bound for Artin’s conjecture on primitive roots?

Question

Let $a$ be an appropriate integer and $\pi_a (x)$ denote the number of prime $p$ such that $a$ is a primitive root modulo $p$. Do we have an upper bound of $\pi_a(x)$ such as $\pi_a(x) \ll x/\log x$? If you know, could you tell me any books or theses which refer such result? Thanks.

Answer 1

Let $$C=\prod_{q\text{ prime}}\left(1-\frac1{q(q-1)}\right)\approx 0.3739558\ldots.$$ It's possible to show unconditionally that $$\limsup_{x\to\infty}\frac{\pi_a(x)}{\pi(x)}\leq C.$$ Consider any subset $Q$ of the primes, and let $\pi_{a,Q}(x)$ be the number of primes less than $x$ modulo for which there exists some $q\in Q$ with $q\mid p-1$ and for which $a$ is a $q$th power modulo $p$. If such a $q$ exists, then $a^{(p-1)/q}=1\pmod p$ and so $a$ is not a primitive root modulo $p$, which means $\pi_{a,Q}(x)\geq \pi_a(x)$, and thus $$\limsup_{x\to\infty}\frac{\pi_a(x)}{\pi(x)}\leq \limsup_{x\to\infty}\frac{\pi_{a,Q}(x)}{\pi(x)}.$$ We claim $$\lim\frac{\pi_{a,Q}(x)}{\pi(x)}=\prod_{q\in Q}\left(1-\frac1{q(q-1)}\right).$$ For a set $R$ of primes, let $\tilde\pi_{a,R}(x)$ be the number of primes $p$ less than $x$ for which, for all $q\in R$, $q\mid p-1$ and $a$ is a $q$th power modulo $p$. Since $$\pi_{a,Q}(x)=\sum_{R\subset Q}(-1)^{|R|}\tilde\pi_{a,R}(x)$$ by the inclusion-exclusion principle, it suffices to show that $$\lim_{x\to\infty}\frac{\pi_{a,R}(x)}{\pi(x)}=\prod_{q\in R}\frac1{q(q-1)}.$$ We see that $$q\mid p-1\text{ and }a\in\{\text{$q$th powers}\pmod p\}\Longleftrightarrow x^q-1\text{ and }x^q-a\text{ split completely }\pmod p.$$ Indeed, $\Longrightarrow$ follows from the fact that $q\mid p-1$ if and only if $x^q-1$ splits completely modulo $p$, and so if $x^q=a$ has some solution modulo $p$ it has $q$ solutions. $\Longleftarrow$ is clear, since $x^q-1$ splitting completely implies $q\mid p-1$ and $x^q-a$ splitting completely implies the existence of some root. To this end, consider the extension $K$ of $\mathbb Q$ formed by adjoining the $q$th roots of unity for all $q\in R$ and a $q$th root of $a$ for each $q\in R$. This extension is Galois, as the splitting field of $$\prod_{q\in R}(x^q-1)(x^q-a),$$ and its degree should be $\prod_{q\in R}(q-1)q$, as adjoining a primitive root modulo $q$ multiplies the degree by $q-1$ and adjoining $\sqrt[q](a)$ multiplies the degree by $q$. So, Chebotarev's density theorem implies that the natural density of primes $p$ for which $K$ splits completely modulo $p$ is $$\prod_{q\in R}\frac1{q(q-1)},$$ as desired.

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Here's a rough heuristic argument that $\lim_{x\to\infty}\tfrac{\pi_a(x)}{\pi(x)}$ should be $C$, if $a$ is not a perfect power. (This argument actually breaks down for some other $a$ too, which is not surprising given how loose it is, but hopefully it's convincing enough.)

Given a prime $p$, the probability that any randomly chosen $1\leq b\leq p-1$ is a primitive root modulo $p$ is $\varphi(p-1)$, where $\varphi$ is the Euler totient function. If we expect $a\in(\mathbb Z/p\mathbb Z)^\times$ to be "random," we'd expect $$\pi_a(x)\approx\sum_{p<x}\frac{\varphi(p-1)}{p-1}.$$ If $$\frac1{\pi(x)}\sum_{p<x}\varphi(p-1)\approx cx$$ for some constant $c$, then we'd expect $$\pi_a(x)\approx 2c\pi(x).$$ For arbitrary integers, the average value of $\varphi$,i.e. $$\lim_{n\to\infty}\frac{\sum_{m=1}^n\varphi(m)}{n},$$ is $\frac3{\pi^2}n$. There is a nice heuristic argument for this that says that the numerator is the number of pairs $(k,m)$ with $1\leq k\leq m\leq n$ and $\gcd(k,m)=1$, and the probability that $\gcd(k,m)=1$ for "randomly chosen" $k$ and $m$ is $$\left(1-\frac1{2^2}\right)\left(1-\frac1{3^2}\right)\left(1-\frac1{5^2}\right)\cdots=\frac1{\zeta(2)}=\frac6{\pi^2}:$$ the probability that $k$ and $m$ don't share a factor of $2$ is $1-\tfrac1{2^2}$, the probability that they don't share a factor of $3$ is $1-\tfrac1{3^2}$, et cetera, and these probabilities "should be independent." (This heuristic can be rigorized). Applying the same heuristic to $$\frac1{\pi(x)}\sum_{p<x}\varphi(p-1)=\frac{\#\{1\leq k<p\leq n\colon \gcd(k,p-1)=1\}}{\pi(x)}$$ gives that a randomly chosen pair $(k,p-1)$ has a probability $$C=\prod_{q\text{ prime}}\left(1-\frac1{q(q-1)}\right)$$ of being relatively prime: for each prime $q$, the probability that it divides $k$ is $1/q$ and the probability that it divides $p-1$ is $\tfrac1{q-1}$, since $p-1$ can be in any residue class modulo $q$ except for $-1$. (The error here when $p=q$ is negligible). Since there should be about $\frac{x\pi(x)}2$ such pairs $(k,p)$, this gives that we'd expect $$\frac{\pi_a(x)}{\pi(x)}\approx C.$$ In some accounts of Artin's conjecture, e.g. the one on Wikipedia, the proposition that a constant fraction of primes should have $a$ as a primitive root is part of the statement of Artin's conjecture (not just the infinitude of such primes).