Is the ABC conjecture known to imply the Riemann hypothesis? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T09:20:06Z http://mathoverflow.net/feeds/question/19692 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/19692/is-the-abc-conjecture-known-to-imply-the-riemann-hypothesis Is the ABC conjecture known to imply the Riemann hypothesis? Jamie Weigandt 2010-03-29T05:41:27Z 2010-08-26T18:34:43Z <p>I once heard from a graduate student that the ABC conjecture implies the Riemann hypothesis. I can't find a reference for this, but given the department the student is from I tend to believe he might know about these things.</p> <p>I looked through <a href="http://www.math.columbia.edu/~goldfeld/ABC-Conjecture.pdf" rel="nofollow">Goldfeld's paper</a> which shows that certain bounds on Shafarevich-Tate groups plus the Generalized Riemann Hypothesis for L-functions associated to certain modular forms imply a form of the ABC conjecture. The article mentions nothing about the opposite implication.</p> <blockquote> <p>What is the known realationship between ABC, BSD, and GRH? Are any two known to imply the third?</p> </blockquote> http://mathoverflow.net/questions/19692/is-the-abc-conjecture-known-to-imply-the-riemann-hypothesis/19693#19693 Answer by Qiaochu Yuan for Is the ABC conjecture known to imply the Riemann hypothesis? Qiaochu Yuan 2010-03-29T05:49:48Z 2010-03-29T05:49:48Z <p>Granville and Tucker's <a href="http://www.ams.org/notices/200210/fea-granville.pdf" rel="nofollow">piece in the AMS Notices</a> from 2002 doesn't contain a mention of such a result, so if something like this is true it would have been recent. It does mention that ABC implies that certain L-functions have no Siegel zeroes, which is weaker than, but related to, GRH.</p> http://mathoverflow.net/questions/19692/is-the-abc-conjecture-known-to-imply-the-riemann-hypothesis/19694#19694 Answer by Pete L. Clark for Is the ABC conjecture known to imply the Riemann hypothesis? Pete L. Clark 2010-03-29T06:20:25Z 2010-03-31T22:45:04Z <p>I am pretty sure that the answer to the question is no: no two of those big conjectures are known to imply the third. But I feel somewhat sheepish giving this as an answer: what evidence can I bring forth to support this, and if nothing, why should you believe me?</p> <p>The only thing I can think of is that in the function field case, ABC and GRH are fully established, but only parts of BSD are known.</p> <p>(Maybe I should also admit that I didn't know anything about the connection between ABC and bounds on Shafarevich-Tate groups of elliptic curves in terms of the conductor until I glanced just now at the paper of Goldfeld the OP linked to. The fact that you can build examples of large Sha from triples of integers with large ABC exponent is amazing to me.)</p> <p><b>Addendum</b>: I feel especially confident that ABC and GRH do not imply BSD, at least not the part of BSD that asserts finiteness of Shafarevich-Tate groups. The first two conjectures are essentially analytic in nature, whereas the finiteness of Sha is deeply arithmetic. It seems extremely unlikely.</p> <p>Moreover, ABC is really hard, in the sense that for all of the results of the form "X implies ABC" that I've ever seen, X includes a statement which is ABC-like in the sense that it gives a uniform bound on one arithmetic quantity in terms of another. For example, ABC is known to be of a similar flavor to the Szpiro Conjecture (and implies it), but so far as I know it is only known to be implied by a more-explicitly-ABC-like <a href="http://modular.fas.harvard.edu/mcs/archive/Fall2001/notes/12-10-01/12-10-01/12-10-01.html" rel="nofollow">Modified Szpiro Conjecture</a>. Admittedly bounding Sha in terms of the conductor, as in Goldfeld's work, is only vaguely ABC-like, but to an arithmetic geometer like me these bounds still feel very "analytic"; I can't see any connection at all between this and BSD. So I doubt that GRH (let me say ERH, so that I more or less know what I'm talking about -- i.e., Dedekind zeta functions) plus BSD is known to imply ABC.</p> http://mathoverflow.net/questions/19692/is-the-abc-conjecture-known-to-imply-the-riemann-hypothesis/23047#23047 Answer by Dirk Basson for Is the ABC conjecture known to imply the Riemann hypothesis? Dirk Basson 2010-04-29T22:09:06Z 2010-04-29T22:09:06Z <p><a href="http://www.math.unicaen.fr/~nitaj/abc.html" rel="nofollow">Here</a> is a link to many consequences of the ABC-conjecture.</p> http://mathoverflow.net/questions/19692/is-the-abc-conjecture-known-to-imply-the-riemann-hypothesis/36725#36725 Answer by Jamie Weigandt for Is the ABC conjecture known to imply the Riemann hypothesis? Jamie Weigandt 2010-08-26T05:13:05Z 2010-08-26T18:34:43Z <p>Perhaps my friend had seen <a href="http://terrytao.wordpress.com/2007/05/04/distinguished-lecture-series-iii-shou-wu-zhang-%E2%80%9Ctriple-l-series-and-effective-mordell-conjecture%E2%80%9D/%20" rel="nofollow">this entry</a> from Terry Tao's blog. Apparently <a href="http://en.wikipedia.org/wiki/Shou-Wu_Zhang" rel="nofollow">Shou-Wu Zhang</a> raised the possibility that the ABC conjecture could be conditionally proven subject to strong enough versions of the generalized Riemann hypothesis and the Beilinson-Bloch conjecuture.</p> <p>That is to say:</p> <blockquote> <p>$$\text{GRH} + \text{Motive Generalization of BSD} + \delta \implies \text{ABC}.$$</p> </blockquote> <p>Where $\delta$ is something a bit vague, but there is optimism that it can be formulated and proven "in the near future."</p>