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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,

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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

2 Answers 2

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).

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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.

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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 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 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 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 at 18:44

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