MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top


Let $\Gamma$ be a free subgroup of rank 2 in $\mathbb{G}_m^2(\mathbb{Q})$. For all but finitely many primes p we can reduce $\Gamma$ modulo p. Let $S$ be the of primes for which $\Gamma$ does not reduce modulo p, and for any $p$ not in $S$, let $\gamma_p$ be the size of $\Gamma \mod p$. My question is what is known about the function

$f(x)= \sum_{p\not\in S,\ p\leq x}\frac{\log p }{\gamma_p}$

In particular what is the asymptotic behavior of $f$? Is the corresponding infinite series convergent whenever $\Gamma$ is not contained in an algebraic subgroup of $\mathbb{G}_m^2$? Do you know of any references that might be relevant to those questions?

Thanks in advance,

share|cite|improve this question
What exactly is the "exceptional set"? Is it the primes dividing the numerator or denominator of some element of $\Gamma$? – David Loeffler May 30 '11 at 15:43
Yes, that is what I meant. I corrected the question to make the statement clearer. Thank you for for the remark. – Tzanko Matev May 31 '11 at 7:31
up vote 5 down vote accepted

Presumably "exceptional" means primes where either one of the generators of $\Gamma$ is 0 or $\infty$ mod p, or where $\Gamma$ mod $p$ has rank smaller than $2$. The following reference is possibly relevant to your question, although we consider a somewhat different sum. We give an upper bound (that should be fairly sharp) for the sum $$\sum_{p} \frac{\log p}{p\cdot\gamma_p^\epsilon}.$$ In particular, we prove that $$\limsup_{\epsilon\to0} ~~\epsilon \cdot \sum_{p} \frac{\log p}{p\cdot\gamma_p^\epsilon} \le 1+\frac{1}{\text{rank}~\Gamma}.$$ The article is

Murty, M. Ram and Rosen, Michael and Silverman, Joseph H., Variations on a theme of Romanoff, Internat. J. Math. 7 (1996), 373-391 (MR1395936).

share|cite|improve this answer
Joe: For any prime number $p$ (such that the two generators of $\Gamma$ are $p$-adic units), the group $\Gamma\mod p$ is finite, so it always has rank smaller than $2$! – ACL May 31 '11 at 7:02
@ACL: I believe "rank" here means "minimal number of generators". – S. Carnahan May 31 '11 at 8:08
Actually, our result is a little different in that we are looking at subgroups $\Gamma$ of $\mathbb{G}(\mathbb{Q})$. The rank means the free rank, which is the dimension of $\Gamma\otimes\mathbb{Q}$ as a $\mathbb{Q}$-vector space. We also deal with abelian varieties of arbitrary dimension, but looking again at our paper, I see that we didn't do the case of finitely generated subgroups of $\mathbb{G}^d(\mathbb{Q})$ for $d\ge2$. However, the argument that we use will easily generalize, and I suspect that the limsup formula remains the same. – Joe Silverman May 31 '11 at 12:47
@ACL: The rank refers to the free rank of $\Gamma$ as a finitely generated abelian group, not to its rank after being reduced modulo p. – Joe Silverman May 31 '11 at 12:48

I would just like to give a small update for the question. In my thesis I showed that the group $\Gamma\ \mod{p}$ has two generators for almost all primes p. So I would conjecture that on average $\gamma_p \sim p^2$ which would imply that the sum above indeed converges.

share|cite|improve this answer
How is the average understood? You don't mean asymptotic here; that should only happen for a set $p$ of positive density, as in the Artin primitive root conjecture. E.g., if $\Gamma$ is the group generated by $(2,1)$ and $(1,3)$ then $\gamma_p = \mathrm{ord}_p^{\times}{2} \cdot \mathrm{ord}_p^{\times}{3}$. Yes, the sum should converge, and likewise we should have $\sum_{p < X} \log{p} / \mathrm{ord}_p^{\times}{2} = O(\log{X})$ (trivial bound is $O(\sqrt{X}$)), but even on GRH (which is greatly relevant here) these problems seem to be hopelessly difficult. I am curious how the sum came up? – Vesselin Dimitrov Sep 23 '15 at 15:25
I am sorry for being vague. I stopped doing mathematics two years ago and I have forgotten much. I guess I mean "average" in the sense that the sum $\sum_{p} p^{1-\epsilon}/\gamma_p$ converges for any $\epsilon > 0$. – Tzanko Matev Sep 23 '15 at 15:58
If I recall the sum appeared when I was looking at some questions from p-adic transcendence theory. We cannot prove the p-adic four exponentials conjecture, but I tried to show that for a given set of 4 numbers the statement of the conjecture holds for almost all p. One stumbling block was to estimate the sum given above. I don't remember the details. – Tzanko Matev Sep 23 '15 at 16:07
I see. Yes, transcendence is much more mysterious in the $p$-adic world, even with regard to statements having a long known Archimedean avatar (e.g., Leopoldt's conjecture would be the simplest example of such a statement). That's a nice result by the way! (about the rank of the reduction being two for almost all $p$) – Vesselin Dimitrov Sep 23 '15 at 18:44

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.