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I was wondering if there are any well-known results or hunches about whether the non-trivial zeroes of Riemann-zeta (or zeta/L-functions in general) are algebraic or not.

flag	asked Jul 17 2010 at 23:11 Nick Salter 1,110●6●17

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Seeing John's answer, I would agree that it is a nice question. – Wadim Zudilin Jul 18 2010 at 2:57

Micah Milinovich · Answer 1 · 2010-07-18 01:50:14Z UTC

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There is a paper by A. E. Ingham, "On two conjectures in the theory of numbers", Amer. J. Math. 64 (1942), 313-319, where he shows that if the ordinates of the non-trivial zeros of the Riemann zeta-function are linearly independent over $\mathbb{Q}$ then Merten's conjecture is false. This is, of course, weaker than the Rubinstein-Sarnak conjecture, but related and much earlier.

answered Jul 18 2010 at 1:50

Micah Milinovich
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John, this is a very surprising (to me) result and the paper is quite influential (for example, the methods where used in [C.B. Haselgrove, Mathematika 5 (1958) 141--145; dx.doi.org/10.1112/S0025579300001480] for a disproof of a conjecture of Pólya. Here is the (jstor) link to Ingham's paper for those who are interested in: dx.doi.org/10.2307/2371685. I learn from this answer more than from David's (which represents the standard observations). – Wadim Zudilin Jul 18 2010 at 2:55

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After some digging I've learnt that the Mertens conjecture (that always $|\sum_{k=1}^n\mu(k)| \le \sqrt n$ holds for all $n$) was only recently shown to fail for some $n$ no smaller than $10^{12}$ and probably larger than $10^{30}$ [J. Havel, Gamma: Exploring Euler’s Constant, Princeton University Press, Princeton, 2003, pp. 207–-210]. – Wadim Zudilin Jul 18 2010 at 4:03

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@Wadim: I agree, this answer is much more useful. :) – David Hansen Jul 18 2010 at 20:21

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It is been a while since I have read the paper, but I think Odlyzko & Riele's disproof of the Merten's conjecture (Crelle 357 (1985), 138-160) relies heavily on Ingham's ideas. – Micah Milinovich Jul 19 2010 at 0:30

Here is their paper: dtc.umn.edu/~odlyzko/doc/arch/… – Micah Milinovich Jul 19 2010 at 0:35

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David Hansen · Answer 2 · 2010-07-17 23:23:26Z UTC

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Every non-trivial zero of every L-function, besides possible zeros at $s=1/2$, is conjectured to be of the form $s=1/2+i\gamma$ with $\gamma$ real (GRH) and transcendental. I learned this from (for example) the Rubinstein-Sarnak paper on Chebyshev biases, but they were not the first to enunciate it.

answered Jul 17 2010 at 23:23

David Hansen
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David, thanks for your selfcriticism. I would add that it is expected that the ordinates of Riemann's zeroes are algebraically independent. But this is definitely out of reach because there are no constructions known even for rational approximations to a single zero. – Wadim Zudilin Jul 18 2010 at 23:43

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Indeed, I don't think it has even been proved that there is a single zero of the Riemann zeta function where the imaginary part is irrational. – Greg Martin Jul 27 2010 at 7:41

What is known about the transcendence of zeroes of Riemann zeta?

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