## Which conjectures only need the Grand Riemann Hypothesis to become genuine theorems?

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Hello,

I've been interested in number theory for several years, and as time goes by, I read more and more articles in which theorems begin with "Assume the Riemann Hypothesis holds." But up to now, I think I've almost never seen any beginning with "Assume the Grand Riemann Hypothesis holds". So, which are those "theorems" that only need the Grand Riemann Hypothesis to become certain results?

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Since this ask for a list, I suggest community wiki mode. – quid Jul 27 2011 at 20:36

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The Wikipedia article (en.wikipedia.org/wiki/Grand_Riemann_hypothesis) has this wonderful remark: "The Siegel zero, conjectured not to exist, is a possible real zero of a Dirichlet L-series, rather near s = 1." – TonyK Jul 27 2011 at 20:42

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A trivial answer to the question: the Grand Riemann Hypothesis only needs to Grand Riemann Hypothesis to proved in order to become a theorem ;) – Peter Humphries Jul 28 2011 at 10:36

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This is a duplicate question. See mathoverflow.net/questions/17209/…. By the way, I am surprised you read "more and more" articles which only assume RH but have almost never seen any assuming GRH. What applications have you been reading? Most of the really juicy applications need GRH. – KConrad Jul 28 2011 at 15:21

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@Ryan, maybe you should edit your question to let everyone know what is meant by "Grand" RH as distinguished from "Extended" RH and "Generalized" RH. For a sense of scope here, most applications of the Generalized RH (meaning for Dedekind zeta-functions of number fields) don't really need it for all number fields, but they do need it for infinitely many number fields. – KConrad Jul 28 2011 at 15:27
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## 4 Answers

I like the phrase "only need the grand Riemann hypothesis"...

One of my favorite results known contingent on this result (rather, the weaker generalized Riemann hypothesis), is that the ring of integers in a number field (EDIT: with infinite unit group) is Euclidean with respect to some Euclidean algorithm if an only if is is a PID. Interestingly, the "amount" of GRH needed here far exceeds that of the field in question. One must assume GRH for an infinite number of extension fields as well.

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@Ramsey: I am not seeing "except for the imaginary quadratic fields $\mathbb{Q}(\sqrt{-19}),\ldots,\mathbb{Q}(\sqrt{-163})$" in your answer.  But it should be there, shouldn't it? – Pete L. Clark Jul 28 2011 at 11:12
I remember reading this a couple of years ago. The condition needed was stronger if I recall correctly, and it was something like: the unit rank is at least 3. – Dror Speiser Jul 28 2011 at 11:31
@Pete:  Of course!  My slip up.  I meant to include "infinite unit group" to make the statement more relevant to the question (oh, and, uh, correct as well).  I'll edit. – Ramsey Jul 28 2011 at 13:26
@Dror: Infinitely many units suffices (conditional on a bunch of GRH's).  This was proven by P.J. Weinberger in "On Euclidean rings of algebraic integers."  – Ramsey Jul 28 2011 at 13:35
Nick, it's par for the course that theorems which assume GRH do so for infinitely many number fields. I agree it's important to be aware when describing this theorem of Weinberger that GRH is used not just for the number field under discussion, but rather for infinitely many of its extensions. However, in light of how typical it is for infinitely many instances of GRH to be required for theorems "under GRH", it's not a surprise that the "amount" of GRH goes beyond the specific number field in question. – KConrad Jul 28 2011 at 15:16
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For the Grand Riemann Hypothesis (RH for zeros of all automorphic $L$-functions), see the (somewhat technical) answer to

http://mathoverflow.net/questions/2826/equivalent-forms-of-the-grand-riemann-hypothesis

I think the Generalized Riemann Hypothesis (RH for zeros of Dirichlet $L$ functions) has the most significant number theoretic consequences. In addition to those listed at

http://en.wikipedia.org/wiki/Generalized_Riemann_hypothesis#Consequences_of_GRH

such as easy primality testing and good bounds on primes in arithmetic progressions, one also gets good lower bounds on class numbers for positive definite binary quadratic forms of discriminant $D$ (or equivalently, rings of integers in complex quadratic fields): for every $\epsilon>0$ there exists an effective constant $C(\epsilon)$ such that the class number $h(d)>C(\epsilon)|D|^{1/2-\epsilon}$.

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The main result

''Assume that the generalized Riemann hypothesis (GRH) for zeta functions of number ﬁelds holds. There exists a deterministic algorithm that on input positive integers $n$ and $k$, together with the factorization of $n$ into prime factors, computes the element $T_n$ of the Hecke algebra $T(1, k)$ in running time polynomial in $k$ and $\log n$.''

of the recent book by Couveignes, Edixhoven, et al. (page 3)

http://www.math.univ-toulouse.fr/~couveig/book.htm

assumes the generalized Riemann hypothesis.

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Consult the chapter entitled Assuming the Riemann Hypothesis and Its Extensions … on pages 61--67 of the recent book The Riemann Hypothesis: A Resource for the Afficionado and Virtuoso Alike by Peter Borwein, Stephen Choi, Brendan Rooney and Andrea Weirathmueller.

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Squarely off-topic, but: is aficionado really the contrasting term to virtuoso? As I understand the meanings of these terms, there is a lot of overlap between the two. – Pete L. Clark Jul 28 2011 at 11:16

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I find it akin to critic and performer. Gerhard "Ask Me About System Design" Paseman, 201.08.11 – Gerhard Paseman Aug 12 2011 at 1:17
"aficionado" implies firstly enthusiasm (affection, maybe akin to "amateur" in the sense of "amare") and secondly perhaps some competence, while "virtuoso" (skill) could be said to reverse the two. Yet I agree they do not directly contrast. – Junkie Aug 12 2011 at 11:47

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Yes, I think they are using fancy words they don't fully understand. Probably they mean: A resource for both the enthusiastic amateur and the professional. – anon Aug 12 2011 at 12:42
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