Granville and Stark (Invent. Math. 139 (2000), 509-523) proved that a uniform version of the abc conjecture for number fields eliminates Siegel zeros for $L$-functions associated with quadratic characters of negative discriminant. Does the recently announced proof by Mochizuki of the abc conjecture cover this statement?

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    $\begingroup$ Has the answer to this question changed in light of Dimitrov's work? $\endgroup$ Commented Aug 3, 2016 at 18:25
  • $\begingroup$ @PaceNielsen: Good question, to which I don't know the answer. Actually, I am not familiar with Dimitrov's work. $\endgroup$
    – GH from MO
    Commented Aug 3, 2016 at 20:55
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    $\begingroup$ It seems that this result has been announced by Mochizuki in a recent Oberwolfach workshop: kurims.kyoto-u.ac.jp/~motizuki/… $\endgroup$ Commented Jan 22 at 20:35
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    $\begingroup$ @Nils I don't think that report of Mochizuki's announces the result, rather it describes allegedly stronger versions of IUT that would imply it, as being under development. $\endgroup$
    – David Roberts
    Commented Apr 3 at 10:04

1 Answer 1


I don't think so. Mochizuki claims to have proved a diophantine result for points of bounded degree, while you need a uniform form of the ABC conjecture for the application that you mention.

EDIT: About your question below, on the version of the ABC conjecture claimed in Mochizuki's work, it is clearly stated in Theorem A of the 4th paper. Anyways, for the benefit of the people that might read this question, I will state in very elementary terms a corollary of Theorem A in the following context: X is the projective line with the usual projective coordinates [x:y], and D is the divisor $[0:1] + [1:0] + [1:1]$ which makes the curve U=X\D hyperbolic (the degree of the canonical divisor $\omega$ of X in this case is -2 and the degree of D is 3, hence the degree of $\omega(D)$ is 1>0). Ok, here is the corollary (the notation is explained below):

Statement: Let $d$ be a positive integer and let $\epsilon>0$. There is a constant $C>0$ depending only on $d$ and $\epsilon$ such that the following is true: If $A,B$ are non-zero algebraic numbers with $A+B=1$, and if the degree over Q of the number field $K=Q(A)$ is at most d, then we have $H(A,B,1) < C(\Delta_K N_K(A,B,1))^{1+\epsilon}.$

Notation: Here I am using the same definition of $\Delta_K$, H(a,b,c) and $N_K(a,b,c)$ as in the paper on Siegel zeros of the question (this notation is explained in the first page of the paper). Well, if you check the reference you'll see that actually there is one difference: the paper uses N(a,b,c), not $N_K(a,b,c)$. However, in the above statement it is crucial that we must compute N(A,B,1) using the number filed K=Q(A), that's why I added this subscript.

I hope that the readers can see the difference between this version and the uniform ABC conjecture for the paper on Siegel zeros: the fact that here the constant C also depends on d, not only $\epsilon$.

A last trivial remark. To get the classical ABC conjecture with coprime integers a+b=c you take A=a/c, B=b/c and hence K=Q which makes $\Delta_K=1$, and N(A,B,1)=rad(abc).

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    $\begingroup$ Can you give more detail, please? For example, which version of the abc conjecture follows from Mochizuki's work? Also, is the constant depending on $X$ and $d$ in his Theorem A effective? $\endgroup$
    – GH from MO
    Commented Sep 5, 2012 at 16:53
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    $\begingroup$ I don't know if the claimed result is effective or not, I started to read the papers just about a week ago and they are certainly hard. However, the first main Diophantine consequence claimed in the 4th paper is for a somewhat restricted class of curves which is nonetheless "sufficiently general". Then the author reduces the general case to this sufficiently general case (sec. 2), and it is remarkable that this reduction step is performed keeping track of explicit constants. Does this indicate that the ultimate goal is an effective result? no idea, I guess that we have to read, not speculate. $\endgroup$
    – Pasten
    Commented Sep 6, 2012 at 1:47
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    $\begingroup$ @Pasten: Thank you. I will leave this question open to collect more information. $\endgroup$
    – GH from MO
    Commented Sep 6, 2012 at 20:59
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    $\begingroup$ @GH: Perhaps the word "clearly" should be omitted in my post. Once I heard that the word "clearly" should be omitted in all the mathematical literature: if something is "so clear" then it is pointless to say that it is clear, while on the other hand if we use the word "clearly" to hide an argument that we don't want to write then we should perhaps be honest and at least give some hint. In this case, what I meant is that Theorem A alone (modulo notation) gives the main result without having to prove further propositions before using it. In any case, sorry about using the word "clearly". $\endgroup$
    – Pasten
    Commented Sep 10, 2012 at 2:58
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    $\begingroup$ @Pasten: Thanks for the clarification regarding "clearly"! $\endgroup$
    – GH from MO
    Commented Sep 10, 2012 at 19:51

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