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  1. Artin's conjecture on primitive roots over the integers states that a given integer $0\ne h\in \mathbb{Z}$ that is neither a square number nor $-1$ is a primitive root modulo infinitely many primes $p$, and the number of primes is $c(h)\pi(x)$ where $c(h)>0$ is Artin's corrected constant, see [Theorem, 2].

  2. Artin's conjecture for polynomials and rational functions over finite fields $\mathbb{F}_{q}$ states that a given polynomial $a(z)\in \mathbb{F}_q[z]$ that is neither a square number nor constant is a primitive root modulo infinitely many finite fields $\mathbb{F}_{q^n}$, see [1, page 6].

Equivalently, there are infinitely many irreducible polynomial $f(z)\in \mathbb{F}_q[z]$ of degree $\text{deg} \;f(z)=n$ as $n\to\infty$ such that $\mathbb{F}_{q^n}\cong \mathbb{F}_q[z]/f(z)$ and the fixed polynomial nonconstant $a(z)\ne b(z)^2$ generates the multiplicative group $\mathbb{F}_{q^n}^{\times}$.

Question 1. Equation (3.5) in [1] displayed here, $$ \sum_{\theta\in \mathbb{F}_{q^n}} \frac{\varphi(q^n-1)}{q^n-1}\left\{1+\sum_{\substack{d\mid q^n-1\\d>1}}\frac{\mu(d)}{\varphi(d)}\sum_{\text{ord}\chi=d}\chi(a(\theta))\right\}, $$is not counting finite fields $\mathbb{F}_{q^n}$ generated by a fixed polynomial $a(z)$ as $n \to \infty$, it counts the number generators in a fixed finite field $\mathbb{F}_{q^n}$, which is $\varphi(q^n-1)+O(q^{n/2})$.

Equivalently, there are $\varphi(q^n-1)/n+O(q^{n/2})$ primitive polynomial of degree $n$ over $\mathbb{F}_{q}$. Each of these polynomials generates exactly \underline{$1$ finite field}: all these fields are the same finite field $\mathbb{F}_{q^n}$, so it does not prove the conjecture for infinitely many finite fields. Moreover, it is the same result for primitive polynomials proved by Carlitz, see [3]. Is this correct?

Question 2. Based on the authors' argument, the polynomial $a(z)=c_1z^2+c_0\in \mathbb{F}_{q}[z]$ generates every finite field $\mathbb{F}_{q^n}$ as $n\to\infty$. This follows from their proof.

But, counterexamples are easy to construct (Sage):

$a(z)=2z^2-1\in\mathbb{F}_{5}[z]$ has order $5^2-1=24$ in $\mathbb{F}_{5^2}\cong \mathbb{F}_{5}[x]/(x^2+x+1)$, it generates the multiplicative group.

But, it has order $62<5^3-1=124$ in $\mathbb{F}_{5^3}\cong \mathbb{F}_{5}[x]/(x^3+x+1)$, so it does not generate the multiplicative group. It has order $6<5^4-1$ in $\mathbb{F}_{5^4}\cong \mathbb{F}_{5}[x]/(x^4+x^2+2)$, and so on. Is this correct?

Question 3. The statement or description of Artin's conjecture for polynomials and rational functions over finite fields $\mathbb{F}_{q}$ \underline{is about the number of finite fields} $$\mathbb{F}_{q^2},\quad \mathbb{F}_{q^3},\quad \mathbb{F}_{q^4},\quad\ldots,\quad \mathbb{F}_{q^n},\quad \mathbb{F}_{q^{n+1}}, \quad\mathbb{F}_{q^{n+2}},\quad\ldots, \quad \mathbb{F}_{q^x}, $$ such that a fixed polynomial $a(z)\ne b^2(z)$ generates the multiplicative group. The quantitative version should be something of the form $$(*)\quad \#\{n\leq x: \langle a(z)\rangle =\mathbb{F}_{q^n}^{\times}\}\stackrel{???}{=} \delta(a,q)x+o(x),$$ where $\delta(a,q)>0$ is 'Artin's constant' for function fields. This problem seems to be more difficult than the original Artin's conjecture. Even a weak lower bound $$\#\{n\leq x: \langle a(z)\rangle =\mathbb{F}_{q^n}^{\times}\}\stackrel{???}{\gg} \log \log x,$$

appears to be difficult to prove. Paper [1] does not address this question because it is proving something else. Is this correct?

Edit. I respect the extensive works accomplished by these two authors and previous authors, including Bilharz, but I do not agree that this is a proof, it should be something like (*).

Reply to Professor Conrad: Thank your for your informative and detailed answer. The paper itself does not have a description as you have stated, the formulation given here was taken from the paper [1] and other papers have similar error, in fact the MR's reviewer of the paper commented on these things. True, in odd characteristic $q\ne 2^k$, the polynomial $a(z)\ne b(z)^d$ for all $d\mid q-1$. Changing the polynomial does not changes the main idea.

The disagreements seems to be about different versions Artin conjecture in function fields and numbers fields.

One version on the isomorphic class of finite field of fixed norm or size $q^n$.

Another version on the sequence of finite fields of norm or size $q^n$ as $n\to\infty$, see the discussion in [6].

Several publications on Artin conjecture in numbers fields defines it exactly the same as (*) is defined, \textbf{each norm counts as one field}. For example

  1. Cohen's Theorem 1.1 in the paper [5] is about primitive root over number fields, it implicitly uses (*), which is the correct version, not in isomprphic class of prime/irreducible ideals as in [1].

  2. Troupe's Theorem 1.3 in the paper [7] is about primitive root over function field, it implicitly uses (*), which is the correct version, not in isomprphic class of finite fields as in [1].

Thus, the version (*) does make sense, and it has the correct density, also has a Dirichlet density, see [6]. All the papers on Artin's conjecture in function fields have failed to cite [3]. Carliz proved the same result as in [1], about 30 years after Artin primitive root conjecture over the integers, he was a leading mathematician, but he did not called it Artin's conjecture in function fields.

As in the case of the Selberg class for L-functions and the Riemann hypothesis, there should be a Artin primitive root conjecture class to classify the fields, rings, modules or variety. for example,

  1. The field $\mathcal{K}$ is an infinite field.
  2. The element $\mathfrak{a}\ne \mathfrak{b}^d$ is not a $d$-power for a finite number of integers $d\geq2$.
  3. The subset primes $$\mathcal{A}=\{N(\mathfrak{p})\leq x: <\mathfrak{a}>=\left (\mathcal{K}/\mathfrak{p}\right )^{\times}\}$$ has nonzero density in the set of primes $\mathcal{P}$.
  4. Other criteria.

Reply to Gorodetsky's comment. The density computed by Bilharz is the number of primitive polynomials/generators, not the number of distinct finite fields, compare this to Carlitz's result in [4]. These polynomial generates exactly one finite field, see [4, Section 19.3] for discussion. In the original conjecture, the density primes with a fixed primitive root $c(a)$ is dependent on $a$, it is not fixed, it varies, this well known, see [2]. Likewise, the density of number of finite fields with a fixed generator is dependent on the polynomial $a(z)$ and $q$.

The authors should consider extending the Hooley method's to prove Artin's conjecture over function field over finite fields. It should count the number of finite fields, they have accumulated lot of results, and they should complete the proof of (*).

Note: Very sorry but I can't use the Mathoverflow reply/editing function. This Network blocks almost all the features of your apps, most of the Mathoverflow's features are disable.

References

[1] Kim, Seoyoung; Ram Murty, M. Artin's primitive root conjecture for function fields revisited. Finite Fields Appl. 67 (2020), 101713, 15 pp. MR4125887.

[2] Hooley, Christopher, On Artin's conjecture. J. Reine Angew. Math. 225 (1967), 209-220. MR0207630.

[3] L. Carlitz, Primitive roots in a finite field, Trans. Amer. Math. Soc. 73 (1952), 373-382.

[4] Victor Shoup, A computational introduction to number theory and algebra, Second edition, Cambridge University Press, 2005.

[5] Joseph Cohen, Primitive Roots in Quadratic Fields II, arXiv:math/0501120.

[6] H.W. Lenstra, Jr. On Artin's Conjecture and Euclid's Algorithm in Global Fields, Inventiones math. 42, 201-224 (1977).

[7] Lee Troupe. Bounded gaps between prime polynomials with a given primitive root. arXiv:1503.06634.

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  • $\begingroup$ Regarding Q2: I believe the authors prove that for every sufficiently large $n$, there exists an irreducible polynomial $p$ of degree $n$ such that $a(x)$ generates $(\mathbb{F}_q[x]/(p(x)))^{\times}$ (this is the function field analogue of Artin's conjecture), so I do not see how your counterexamples are valid. You simply exhibited a finite set of degrees (2 and 3) and a finite set of $p(x)$ of these degrees for which $a(x)$ is not a generator. This is trivial to find by taking irreducible factors of $a(x)^d - 1$. $\endgroup$ Commented Mar 13, 2023 at 19:33
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    $\begingroup$ Regarding Q3: What do you mean by $a(x)$ being a generator, without specifying $x$ as an element of a finite field instead of an indeterminate? It is natural to let $S_n = \{ p(x): \deg(p)=n, \, p \text{ irreducible}, a(x)\text{ generates } (\mathbb{F}_q[x]/p(x))^{\times}\}$ and study $|S_n|$. The authors show that $|S_n| \sim \phi(q^n-1)/n$, so you can say that Artin's constant is $\sim \phi(q^n-1)/q^n$ as $n \to \infty$. This is independent of $a(x)$ in the same way Artin's constant is independent of $a$ in the integer setting. $\endgroup$ Commented Mar 13, 2023 at 20:13
  • $\begingroup$ Of course if $a(x)$ is an $l$th power for some $l\ge 2$ the answer is more complicated. This is treated e.g. in Chapter 10 of Rosen's book "Number Theory in Function Fields", where an evaluation of the Dirichlet density of $\cup_n S_n$ is given in general; if $a$ is not an $l$th power for any $l \ge 2$ it is $\sum_{(m,q)=1} \mu(m)/(mf(m))$ where $f(m)$ is the order of $q$ mod $m$. In view of the results on $|S_n|$ I believe this density should coincide with $\lim_{s \to 1^+}\sum_{n \ge 1} \phi(q^n-1) q^{-ns}/n / (-\log(1-q^{1-s}))$. This Dirichlet density estimate is due to Bilharz. $\endgroup$ Commented Mar 13, 2023 at 20:21
  • $\begingroup$ Please use the Contact Us form to have your accounts merged, to regain full control over your posts. $\endgroup$ Commented Mar 22, 2023 at 10:02
  • $\begingroup$ You write "all these fields are the same finite field $\mathbf F_{q^n}$", which confuses isomorphism and equality. In $\mathbf F_q[x]$, irreducibles $\pi(x)$ of degree $n$ lead to finite fields $\mathbf F_q[x]/(\pi(x))$ that are all isomorphic, but they are not equal. How $a(x)$ reduces mod $\pi(x)$ depends on $\pi(x)$! Consider $K=\mathbf F_5[x]/(x^2+2)$ and $L=\mathbf F_5[x]/(x^2+x+2)$. These fields have size 25, but $x$ doesn't generate $K^\times$ (its order mod $x^2+2$ is $8$) and $x$ generates $L^\times$ (its order mod $x^2+x+2$ is $24$). That's why your $(*)$ makes no sense. $\endgroup$ Commented Mar 24, 2023 at 14:23

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Your formulation of Artin’s primitive root conjecture over $\mathbf F_q[x]$ is incorrect: you need to avoid not just that $a(x)$ is a square (for odd $q$), but also that $a(x)$ is a $d$th power where $d>1$ and $d\mid (q-1)$: for every irreducible $\pi(x)$, the field $\mathbf F_q[x]/(\pi(x))$ has a multiplicative group of order $q^n-1$, so if $a(x)$ is a $d$th power in $\mathbf F_q[x]$ where $d \mid (q-1)$ and $\pi(x) \nmid a(x)$ then $a(x) \bmod \pi(x)$ will have order dividing $(q^n-1)/d < q^n-1$.

For example, if $a(x)$ is a cube in $\mathbf F_7[x]$ then it does not generate the units in $\mathbf F_7[x]/(\pi(x))$ for irreducible $\pi(x)$ not dividing $a(x)$ since $a(x)^{(7^n - 1)/3} \equiv 1 \bmod \pi(x)$.

A correct statement and proof of the Artin primitive root conjecture in $\mathbf F_q[x]$ is in Chapter 10 of Michael Rosen’s book Number Theory in Function Fields. He works in the setting of a general function field $K$ over a finite field $\mathbf F_q$, but you can focus on the case $K = \mathbf F_q(x)$. (To avoid some technicalities he assumes $a(x)$ is what he calls geometric on p. 152, which essentially means for each prime $\ell$ dividing $q-1$ that $a(x)$ is not a constant multiple of an $\ell$th power.)

Update: you wrote that on page 6 of your reference [1], the formulation of the Artin primitive root conjecture for $a(x)$ in $\mathbf F_q[x]$ is that $a(x)$ "is neither a square number nor constant". I looked at that paper here and on page 6 it does not say what you wrote. They impose more constraints on nonconstant $a(x)$ than just not being a square.

You are unhappy with the way densities of primes are defined in $\mathbf F_q[x]$. A comparison of different ways to defined densities in that context is in a paper by Christian Ballot: Competing prime asymptotic densities in $\mathbf F_q[X]$: a discussion, Enseign. Math. 54 (2008), 303–327.

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