# Is the ABC conjecture known to imply the Riemann hypothesis?

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.

I looked through Goldfeld's paper 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.

What is the known realationship between ABC, BSD, and GRH? Are any two known to imply the third?

-
I suspect the graduate student in question was confusing the Riemann Hypothesis with Fermat's Last Theorem, for which ABC gives an asymptotic affirmative answer. –  S. Carnahan Mar 29 '10 at 5:44
You could be right. I only doubt that because the two of us got into number theory well after Fermat's Last Theorem was proven. I suppose it's third-hand folklore a this point, but I'd still like to know what is known about connections between the big three conjectures if anyone can help me out. –  Jamie Weigandt Mar 29 '10 at 6:04

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?

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.

(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.)

Addendum: 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.

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 Modified Szpiro Conjecture. 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.

-
Re: your addendum: Well, the Dirichlet formula for class numbers is not exactly "likely", is it? :) –  Mariano Suárez-Alvarez Mar 29 '10 at 15:09
@MSA: No, it certainly isn't. As I was getting at above, there's something a bit small-minded in saying, "No, I don't think that X [believed to be true] can be shown to imply Y [believed to be true]." One of the coolest things about mathematics is that apparently very unlikely things turn out to be true/provable rather often. In every case, you can surely find people who naysayed up until the moment of proof. So all I'm doing is giving my intuition as a generic mathematician working in the field. People know better than to take that very seriously, I hope. –  Pete L. Clark Mar 29 '10 at 19:03
Thanks very much for sharing that intuition Pete! Based on the number of people who've seen this and not responded with a positive answer, we can safely say that ABC is not "known" to imply RH, or ERH, or GRH. This reminds me of a paper that was once written about the Erdos Number graph, which proved by its own existence that said graph had a certain property. –  Jamie Weigandt Mar 31 '10 at 21:57

Granville and Tucker's piece in the AMS Notices 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.

-
In the end, though, Scott is probably right. –  Qiaochu Yuan Mar 29 '10 at 6:02
I had heard that there was a mistake in the proof of ABC => no Landau-Siegel zero. Does anyone know the status of it? –  Idoneal Mar 29 '10 at 19:12

Here is a link to many consequences of the ABC-conjecture.

-
Be prepared for a trip back in time, to the internet of the early 90s, if you dare to follow that link! –  Steven Gubkin Apr 30 '10 at 15:58

Perhaps my friend had seen this entry from Terry Tao's blog. Apparently Shou-Wu Zhang 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.

That is to say:

$$\text{GRH} + \text{Motive Generalization of BSD} + \delta \implies \text{ABC}.$$

Where $\delta$ is something a bit vague, but there is optimism that it can be formulated and proven "in the near future."

-
Perhaps, it's better to replace $\epsilon$ with $\delta$? That would also avoid undesirable connotations like $\textrm{Taniyama-Shimura} + \epsilon \implies$ FLT. –  Victor Protsak Aug 26 '10 at 6:20
Done, that's for the advice. Although it's not unreasonable to point out the similarity. As GRH and Beilinson-Bloch are deep statements about L-functions, somewhat like modularity. Also, ABC is the natural generalization of BSD. But likely $\delta$ will look nothing like $\epsilon$ did. –  Jamie Weigandt Aug 26 '10 at 18:38
I guess the last BSD was meant to be FLT. –  Peter Scholze Aug 26 '10 at 19:19
Yeah. Sorry. I wrote my comment very quickly before class. The first couple sentences were also supposed to be "Done! Thanks for the advice." –  Jamie Weigandt Aug 26 '10 at 20:27