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My question concerns the least prime (denoted $p(a, q)$) in the arithmetic progression $a \pmod q$ where $a$ and $q$ are coprime. Quite a time ago Linnik demonstrated that $$p(a, q) \ll q^L$$ for some absolute constant $L$. Wiki page for this theorem lists a number of papers that estimate $L$ with the most recent result by Xylouris who proved that $L \leq 5.2$.

It is also known that the Generalized Riemann Hypothesis implies $$p(a, q) \ll (q\log q)^2 \text{,}$$ while in 1978, Heath-Brown conjectured even tighter bound: $$p(a, q) \ll q(\log q)^2 \text{.}$$ I'm wondering whether this last bound, if true (it is still an open problem), implies something non-trivial about $L$-functions?

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I haven't looked at such types of questions, however my first thought is no: I don't see (or haven't seen yet) how the existence of one prime (or a small number of them) would force the Dirichlet L-functions (I guess these are which you meant) to look in a certain way.

As an example, take the explicit formula which counts the number of primes in arithmetic progressions by using zeros of some L-functions.

Now, if you know that the "left side" in the particular equation (counting the primes) is 1 instead of 0, this does not seem to force anything noteworthy for the zeros which appear in the sum of the right side....Some small change in the imaginary parts of a couple of zeros (with very large imaginary part) might be enough to change the total value by 1, which then implies my answer for the special case if you only use the explicit formula. However, most arguments in analytic number theory (where this theorem on the least prime number comes from) tend to give similar behavior.

I hope, you understand the point I am trying to make besides my unclear presentation.

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    $\begingroup$ I think I agree. If you knew there were lots of small primes in every arithmetic progression - essentially the desired asymptotic number with a small error term - then that would probably improve the known zero-free region for Dirichlet $L$-functions, up to a proof of the generalized Riemann hypothesis if the error term were good enough. But just one small prime in each residue class, I'm not sure that would give us any leverage. $\endgroup$ – Greg Martin Mar 13 '12 at 18:41
  • $\begingroup$ That's a good point Greg which I wanted to mention as well and forgot. $\endgroup$ – tyrex Mar 14 '12 at 10:03
  • $\begingroup$ Thanks to both of you: unknown and Greg. I think I understand. Your arguments sound convincing. $\endgroup$ – kdr Mar 25 '12 at 21:26

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