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If Artin's conjecture on primitive roots is true, then 2 generates $(\mathbb{Z}/p\mathbb{Z})^\times$ for infinitely many primes $p$. Can one at least show that $(\mathbb{Z}/p\mathbb{Z})^\times$ is generated by 2 and 3 for infinitely many primes $p$?

edited Jan 28 2011 at 2:55

Zev Chonoles
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asked Jan 28 2011 at 1:22

Hej
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Hej, I've changed $\mathbb{Z}_p^*$ to $(\mathbb{Z}/p\mathbb{Z})^\times$, because the generally accepted meaning of $\mathbb{Z}_p$ is the $p$-adic integers (though I don't think anyone was confused about what you meant), and because after changing that aspect, the asterisks gave me trouble with the markdown. – Zev Chonoles Jan 28 2011 at 2:07

+1 by the way - neat question. – Zev Chonoles Jan 28 2011 at 2:12

I am going to guess "probably not" - if you can find a prime $p$ and positive integers $j,k$ with $2^{j} \equiv 3^{k} \equiv 1 \, \mathrm{mod} \, p$ and $j*k < p-1$, then you've found a counterexample. – David Hansen Jan 28 2011 at 2:31

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Or indeed $\text{lcm}(j,k)<p-1$. But that would only be one counterexample - is there an intuitive reason why we should expect it to occur for all but finitely many primes? – Zev Chonoles Jan 28 2011 at 2:35

@Hej: although your question is quite clear, I think it's natural that people are misreading it (as you mentioned in a comment below). Namely, a primitive root mod p is a generator of (Z/pZ)^*, so it is natural to (mis)read your question as: "Are there infinitely many primes p such that 2 or 3 is a primitive root mod p?" especially if you know about the result of Heath-Brown which is "slightly weaker" than this. You are weakening APR in a somewhat unexpected direction by allowing more than one generator. (Anyway, good question!) – Pete L. Clark Jan 28 2011 at 4:05

Joe Silverman · Answer 1 · 2011-01-28 03:21:20Z UTC

This is an interesting question. More generally people have considered the following. Let $\Gamma$ be a subgroup of $\mathbf{Q}^*$ generated by $r$ primes. What can one say about $$ N_\Gamma(X)=|\{p < X : \Gamma \bmod p \textrm{ generates } (\mathbf{Z}/p \mathbf{Z})^{\times}\}|. $$

There is also a natural elliptic analogue. Thus let $E/\mathbf{Q}$ be an elliptic curve and let $\Gamma\subset E(\mathbf{Q})$ be a subgroup of rank $r$. Then we can consider $$ N_\Gamma(X)=|\{p < X : \Gamma \bmod p \textrm{ generates } E(\mathbf{Z}/p \mathbf{Z})\}|. $$ Gupta and Murty give a number of results, both conditional and unconditional, in their paper Primitive points on elliptic curves, Compositio Math. 58 (1986), 13–44. For example, if $r\ge6$ and $E$ has complex multiplication, then they prove unconditionally that $N_\Gamma(X)\gg X/(\log X)^2$. In the non-CM case, assuming the GRH, they prove that if $r\ge18$, then $N_\Gamma(X)\gg X/(\log X)$.

It would be interesting to investigate similar questions on higher dimensional algebraic groups, either abelian varieties of dimension $\ge2$, or even on $(\mathbf{Q})^{\times}\times(\mathbf{Q})^{\times}\times\cdots\times(\mathbf{Q})^{\times}$.

GH · Answer 2 · 2011-01-28 01:55:45Z UTC

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I think the best approximation is due to Heath-Brown (Quart. J. Math. Oxford Ser. 37, 27-38.): for infinitely many primes p, one of 2,3,5 is a primitive root mod p.

Actually Heath-Brown's theorem works for any three primes in place of 2,3,5. You can find his paper online here (praise Google).

edited Jan 28 2011 at 1:55

answered Jan 28 2011 at 1:50

GH
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There is also a nice presentation by Murty, "Artin's conjecture for primitive roots", the Mathematical Intelligencer, vol 10 (4) (1988), 59-67. – Andres Caicedo Jan 28 2011 at 1:56

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Thank you. I was aware of that result. Maybe my question was not clear. So what I am asking is whether the group generated by 2 and 3 in $\mathbb{Z}_p^*$ is all of $\mathbb{Z}_p^*$. Equivalently, is it true that for infinitely many primes $p$, every number $m$ is congruent to some $2^a3^b$ mod $p$, with $a,b$ natural. – Hej Jan 28 2011 at 1:58

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However, $n_1,\ldots,n_k\in\mathbb{Z}$ could still generate $\mathbb{Z}/p\mathbb{Z}^\times$ even if none of them is a primitive root mod $p$. It seems like Hej's question might not require such hard results. – Zev Chonoles Jan 28 2011 at 2:00

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Heath-Brown's proof goes by first showing that given three primes, they will generate $(\mathbb{Z}/p)^*$ for infinitely many $p$ and current technology does not allow us to reduce the number of primes to two. His method was inspired by an analogous result of Gupta and Murty on elliptic curves, where you ask if a given set of $m$ rational points (satisfying some obvious necessary condition) generates the group mod $p$ for infinitely many $p$ and this can be proved as long as $m$ is big enough (5 for CM and 8 for non-CM? I forget). The conjecture would be $m=1$, but current tech is not enough. – Felipe Voloch Jan 28 2011 at 2:45

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@Felipe: Amusing, I think we were composing our answers at the same time, with more-or-less the same material. – Joe Silverman Jan 28 2011 at 3:03

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Pedro Martins Rodrigues · Answer 3 · 2011-01-28 14:59:21Z UTC

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This is just a comment, but I'm unable to do it, so I post it as an answer:

$2^{11} \equiv 3^{11} \equiv 11 \, \mod 23$.

And a similar situation happens with other primes:

$47, 71, 73, 97, 167, 191, 193, 239, 241, 263, 307, 311, 313, ...$

although not always with $ord_{p}(2)=ord_{p}(3)=\frac{p-1}{2}$.

edited Jan 28 2011 at 14:59

answered Jan 28 2011 at 14:43

Pedro Martins Rodrigues
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	All well and good, but the question is asking for a proof that there are infinitely many primes for which this doesn't happen. – Zev Chonoles Jan 28 2011 at 15:17
	Also you mean $2^{11}\equiv 3^{11}\equiv1\bmod 23$ (not 11). – Zev Chonoles Jan 28 2011 at 15:19

Primitive roots

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