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This question is partly inspired by David Stork's recent question about the enigmatic complexity of number theory. Are there algebraic systems which are similar enough to the integers that one can formulate a "Riemann hypothesis" but for which the Riemann hypothesis is false? One motivation for constructing such things would be to illustrate the barriers to proving the Riemann hypothesis, and another one would be to illustrate how "delicate" the Riemann hypothesis is (i.e., that it's not something that automatically follows from very general considerations).

I've run across various zeta functions over the years, but I seem to recall that either the Riemann hypothesis is probably/provably true, or the zeta function is too unlike the classical zeta function to yield much insight.

More generally, what happens if we replace "Riemann hypothesis" with some other famous theorem or conjecture of number theory that seems to be "delicate"? Can we construct interesting systems where the result fails to hold?

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    $\begingroup$ Conjecturally, every L-function in the Selberg class en.wikipedia.org/wiki/Selberg_class obeys the Riemann hypothesis. Assuming Selberg's conjecture, one would have to give up the Euler product, the functional equation, the Ramanujan conjecture, and/or analytic continuation in order to obtain what you are asking for. (For instance, the Beurling integer example below keeps the Euler product and Ramanujan conjecture, but sacrifices analytic continuation and the functional equation.) $\endgroup$ – Terry Tao Oct 10 '17 at 16:18
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    $\begingroup$ An analytic (not algebraic) way to think about how 'delicate' the Riemann Hypothesis is, is to perturb the zeros so they flow according to the (backward) heat equation. The conjecture is that any infinitesimal perturbation destroys RH. See this question for details mathoverflow.net/questions/115447/… $\endgroup$ – Stopple Oct 10 '17 at 16:37
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    $\begingroup$ A trivial remark: if RH is not provable from a certain subtheory $T$ of Peano arithmetic, then (by Gödel's completeness theorem) there is a model of $T$ ("fake integers") for which RH fails. • In case $T$ is open induction (i.e., Peano with induction limited to quantifier-free formulæ), explicit models are known and some things are known about their primes, so maybe something can be said. Formulating RH over such weak theories might cause unexpected problems, however. $\endgroup$ – Gro-Tsen Oct 10 '17 at 17:06
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    $\begingroup$ @David, there's a big difference between "every iterated map of such-and-such a form terminates" and "I've tested several thousand and all terminate". Do you, or don't you have (or know of) a proof of the assertion in your first comment? $\endgroup$ – Gerry Myerson Oct 10 '17 at 22:22
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    $\begingroup$ @JoseBrox: The best overall introduction is The ultimate challenge: The 3x + 1 problem, Jeffrey C. Lagarias (ed.), AMS (2000), and its Chapter 2, "The 3x+13x+1 problem and its generalizations" (Lagarias). One relevant result is in John Conway "Unpredictable iterations," Proc. 1972 Number Theory Conference, U. Colorado, pp 49–52 (1972). See too number-theoretic analyses on termination in H. Möller, "Über Hasses Verallgemeinerung des Syracuse-Algorithmus," Acta Arithmetric 34: 219–226 (1978). $\endgroup$ – David G. Stork Oct 11 '17 at 1:23
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One way of making "fake integers" explicit is a Beurling generalized number system, which is the multiplicative semigroup $Z$ generated by a (multi)set $P$ of real numbers exceeding $1$; lots of research has been done on the relationship between the counting function of $P$ (the Beurling primes) and the counting function of $Z$ itself. In this context, it is certainly known that the Riemann hypothesis can fail; see for example this paper of Diamond, Montgomery, and Vorhauer.

If this is not "similar enough to the integers", then I think you should more clearly define what you mean by that phrase.

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    $\begingroup$ However, if I understand correctly from this paper, if the integer counting function of a discrete Beurling system satisfies $N_Z(X) = \kappa X + O(X^{\frac{1}{2} - \epsilon})$ for some $\epsilon > 0$, which implies that the Beurling zeta function can be extended past the $\mathrm{Re}(s) = 1/2$ line, then it could very well be that all the zeros lie on that line. This is especially interesting as the Dedekind systems (of norms of ideals in a number field) are all conjectured to satisfy this regularity condition, by a generalization of the conjectured truth on the circle problem. $\endgroup$ – Vesselin Dimitrov Oct 10 '17 at 23:00
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    $\begingroup$ I wish I could upvote this answer more than once! Is there a paper available online that discusses Beurling primes in a relatively elementary way? Thanks. $\endgroup$ – Yaakov Baruch Oct 11 '17 at 14:21
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    $\begingroup$ @YaakovBaruch, on this link there are a few more elementary articles by Devin Platt on generalised Beurling integers: devinplatt.com/math.html (found via this MO-question mathoverflow.net/questions/185193/…). $\endgroup$ – Agno Oct 11 '17 at 15:41
  • $\begingroup$ @Agno, thank you for these interesting links. $\endgroup$ – Yaakov Baruch Oct 12 '17 at 18:25
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Moving away from the Riemann hypothesis, some questions in additive prime number theory (e.g. twin primes conjecture or even Goldbach conjecture) are considerably more delicate than others (e.g. arithmetic progressions in primes or odd Goldbach conjecture) because in the former case it is easy to "redact" the primes by removing a small number of primes (e.g. all twin primes), reweight the surviving primes by the appropriate renormalising factor, and end up with a set of "fake" primes that is statistically indistinguishable from the original set of primes, except that a conjecture that was once believed to be true is now false. This all but rules out a large number of existing proof techniques for such problems, such as sieve theory or the circle method, unless these techniques are somehow combined with a new method that can distinguish the true primes from their fake counterparts. I discuss this in Section 2 of https://terrytao.wordpress.com/2012/05/20/heuristic-limitations-of-the-circle-method/ .

In a similar vein, one can replace the natural numbers by a fake version of the natural numbers in which products of an even number of primes are counted with double weight and products of an odd number of primes are counted with zero weight, or vice versa. Assuming a standard conjecture in analytic number theory (the Mobius pseudorandomness conjecture), these fake natural numbers are also statistically indistinguishable from the true natural numbers by all the tests for which we can hope to provably compute the statistics for the true natural numbers. On the other hand, any problem subject to the "parity barrier", such as the twin prime conjecture, can end up having the "wrong" answer when one replaces the natural numbers with this fake version (or some variant thereof), leading to the famous "parity problem". I discuss this for instance at https://terrytao.wordpress.com/2014/11/21/a-general-parity-problem-obstruction/

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    $\begingroup$ It seems, judging from the abstract, that M. Ram Murty and another author published in the December 2017 issue of the Journal of Number Theory a conjecture which, together with GEH conjecture, allows to overcome the parity barrier. $\endgroup$ – Sylvain JULIEN Oct 11 '17 at 20:48
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    $\begingroup$ This looks to be a "bilinear" conjecture, which is the sort of additional input that is known to be able to defeat the parity barrier (e.g., this was how the Friedlander-Iwaniec theorem was proven, using some deep tools outside of sieve theory to verify the bilinear axiom). Unfortunately this bilinear conjecture is in turn subject to the parity barrier insofar as trying to prove it, so the problem hasn't actually been circumvented in this case yet. $\endgroup$ – Terry Tao Oct 11 '17 at 21:52
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Also, in addition to Beurling's ideas, there is Landau's example of $\zeta(2s)\cdot \zeta(2s-1)$, which has Euler product, meromorphic continuation and functional equation, but no zeros at all on the critical line (by Hadamard and de la Vallee-Poussin). If one objects that this is artificial, a response might be that this is simply the globally split case of zeta function of a quaternion algebra, which, up to finitely-many Euler factors is again Landau's example. The finitely-many exceptional Euler factors product on-the-line zeros of $1-p^{s-1/2}$ for ramified primes. The collection of such zeros, for any finite (even-cardinality...) collection of ramified primes is $O(T)$ to height $T$, which is a negligible proportion of the $O(T\log T)$ zeros to that height.

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    $\begingroup$ I guess in this case it is the Ramanujan conjecture which is sacrificed amongst Selberg's four options (the $\zeta(2s-1)$ term creates coefficients as large as $\sqrt{n}$). $\endgroup$ – Terry Tao Oct 11 '17 at 16:59
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    $\begingroup$ @reuns, zeta functions of semi-simple algebras appear in Weil's "Basic Number Theory" as a slightly disguised way to talk about group cohomology for classfield purposes, and he treats their analytic properties in the style of Iwasawa-Tate. Also, I recap similar computations in math.umn.edu/~garrett/m/v/eis_std_periods.pdf $\endgroup$ – paul garrett Oct 11 '17 at 18:03
  • $\begingroup$ I can write $\zeta(s)\zeta(s-1) = \sum_{\Lambda' \in S_\Lambda} (\Lambda: \Lambda')^{-s} = \prod_p (1+\sum_{\Lambda' \in S_{\Lambda,p}} (\Lambda: \Lambda')^{-s})$ where $S_\Lambda$ are the sub-lattices of $\Lambda = \mathbb{Z}^2$ and $S_{\Lambda,p}$ those of index a power of $p$, but the additive structure is not clear. Otherwise there is $\zeta_{P^1(\mathbb{Q})}(s)=\zeta_{\mathbb{Q}(x,y)_{proj}}(s) = \prod_{v} \frac{1}{1-|\varpi|_v^{-s}}=\zeta(s)\zeta(s-1)$ where the product is over the valuations or places and $|\varpi|_v$ is the cardinality of the residue field. $\endgroup$ – reuns Oct 11 '17 at 18:41
  • $\begingroup$ In that case I think (as for those over finite fields) there is a Riemann hypothesis for zeta functions of varieties that the zeros are on $n$ vertical lines. $\endgroup$ – reuns Oct 11 '17 at 18:45
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    $\begingroup$ @TerryTao part of the Euler product condition is also not satisfied. The Euler product itself exists, but the $p$-th Euler factor is $\exp(\sum_{k \geq 0} (p^k+1)/p^{2ks})$, so the coefficient of $1/p^{2ks}$ is not $O(p^{2k\theta})$ for some $\theta < 1/2$. $\endgroup$ – KConrad Apr 16 at 14:37

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