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In [M24] it is asserted that "considering $abc$ triples of the form $(1,p^n,1+p^n)$ for a prime number $p$ and an arbitrarily large positive integer $n$, one can verify that ABC/Szpiro inequalities can never be obtained as a result of summing up local inequalities at each prime of a number field."

Is this true, and what are references for this, or could you provide a sketch of the proof?

(Please note that I am disregarding all the controversial context and content of [M24], and focusing only on this mathematical statement).

[M24] S.Mochizuki, Report on the recent series of preprints by K.Joshi, march 24, 2024 https://www.kurims.kyoto-u.ac.jp/~motizuki/Report%20on%20a%20certain%20series%20of%20preprints%20(2024-03).pdf

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    $\begingroup$ The idea seems to be that for a given large $n$, for each $p$ one has an ABC inequality with a certain constant coming from the triple $(1,p^n,1+p^n)$, and that one can construct some other triple such that its ABC inequality cannot be obtained by summing those of its prime factors, the large $n$ presumably inversely related to the $\epsilon$ in the usual ABC statement, or something of the sort, but I cannot work out the argument. $\endgroup$
    – Jon23
    Commented Mar 26 at 10:03
  • $\begingroup$ @LuisFerroni Do you have a link? $\endgroup$
    – Nico
    Commented Mar 30 at 19:50
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    $\begingroup$ @Nico See the last line of Will Sawin's answer. $\endgroup$
    – Timothy Chow
    Commented Mar 31 at 14:19

5 Answers 5

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For concreteness, consider the $abc$-inequality with $\epsilon=1$, i.e. that there is some integer $K$ such that for all coprime integers $a,b,c$ with $a+b=c$, one has

$$abc\leq K \prod_{p|abc} p^6.$$

Mochizuki makes the true and very simple observation that this can never be proved by proving it "one prime at a time", i.e. proving that for every prime $p$, it divides the left-hand side at most as often as the right-hand side, up to some error term $K_p$ depending only on $p$ (and not on $a,b,c$). Indeed, the right-hand side is only divisible $6$ times by any large enough $p$, while the left-hand side can become arbitrarily divisible, as the example $(1,p^n,1+p^n)$ pointed out by Mochizuki shows.

This observation is relevant to the discussion of Joshi's manuscript. Indeed, the key result is Theorem 6.10.1 in Joshi's paper, giving upper and lower bounds on a certain volume. Both inequalities are proved by summing up local inequalities, i.e. inequalities at each prime $p$. For the first one, see the proof of Theorem 9.11.1 of Joshi's previous paper. For the second one, see the proof of Theorem 6.10.1. I believe the actual mistake is Proposition 6.10.7, whose proof simply refers to Mochizuki's work. However, if one notes that the proof of Theorem 9.11.1 also proves a corresponding lower bound in the context of Proposition 6.10.7, one sees that these two inequalities are contradictory, using the template of the simple argument above.

Let me stress that this argument is extremely robust: No matter the details of the inequalities around (and possibly other auxiliary hypotheses in place), the $abc$ inequality can not be obtained in this way. But both the proofs of Theorem 9.11.1 and Theorem 6.10.1 clearly indicate that they are obtained in this way. For all I can see, this means that Joshi's proof strategy is irreparably flawed.

I sent an e-mail to Joshi, asking for clarification, on the day he posted his manuscript to the arXiv. While he did kindly answer my e-mail, I have not yet received clarification on this point.

Let me end by saying that I wished that Mochizuki had focused his manuscript only on this mathematical point, and that I feel deeply sorry for Joshi for the rest of the manuscript (and more generally, I feel deep embarassment as a member of the mathematical community).

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    $\begingroup$ Thank you very much for this clear and detailed answer! $\endgroup$
    – Jon23
    Commented Mar 29 at 20:02
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    $\begingroup$ Even if they cannot prove the full $abc$ conjecture, would results along the lines of the $abc$ inequality but "one prime at a time" still be interesting and new? $\endgroup$
    – Sam Hopkins
    Commented Mar 29 at 21:44
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    $\begingroup$ As the argument shows, there are no "one prime at a time" versions of the $abc$ conjecture: It's a purely global phenomenon! $\endgroup$
    – Peter Scholze
    Commented Mar 29 at 21:50
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    $\begingroup$ Does the mistake in Proposition 6.10.7 also invalidate Mochizuki's original proof of Theorem 1.10 in IUTT IV, thus invalidating Mochizuki's original proof of the abc conjecture? $\endgroup$
    – Madeleine Birchfield
    Commented Mar 30 at 2:31
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    $\begingroup$ @SamHopkins In fact the complete relationship between the $p$-adic valuations $v_p(a), v_p(b), v_p(c)$ is known: The set of triples of integers that occur as $v_p(a), v_p(b),v_p(c)$ for nonzero rational numbers with $a+b=c$ is exactly the set of triples of integers whose minimum value is attained at least twice, unless $p=2$ in which case it is the set of integers whose minimum value is attained exactly twice. So nothing new can be said about (only) these quantities. (Of course, while not new, the inequalities following from this relation are very interesting and foundational to number theory) $\endgroup$
    – Will Sawin
    Commented Mar 30 at 23:29
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This is from my recent email to Peter Scholze in response to his post above:

My response to Peter Scholze:

I read your comments on MathOverFlow. Thank you for your kind words.

Somehow, once again, we are in a similar situation as before, when both you and Mochizuki disagreed with my work. But I have clearly understood the intent of Mochizuki's framework and strategy for proof and amplified it in my work by establishing the Theory of Arithmetic Teichmuller Spaces. My [Preprint: Construction II(1/2)] deals primarily with the global aspects of the theory.

I have received many questions, including yours, about my papers in the last two weeks and I am working on answering them. Personally, I do not want to rush into providing instant answers, because we are dealing with complex mathematical content and half-baked answers could lead to further confusion. Once I have an answer to my satisfaction I will email you and also post a response on MathOverFlow.

I appreciate your comments. They help me identify points which need further clarification in my preprints and provide me an opportunity to sort things out. So thank you.

Added 30th April 2024: My response to Mochizuki's comments and Scholze's comments is here and here.

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    $\begingroup$ @DavidRoberts As an outsider to this topic, I don’t think I only speak for myself when I say the following: for all the mudslinging surrounding this topic, your efforts at aiding communication, combined with Joshi’s thoroughly professional responses, is like a breath of fresh air. Thank you! $\endgroup$
    – Carl-Fredrik Nyberg Brodda
    Commented May 1 at 1:47
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    $\begingroup$ I believe the claim "Hence under this assumption, for all embeddings $\mathbb Q(j)\to \mathbb C$, the absolute value $|j|_\mathbb C$ is bounded (from above and from below). Since $j\in U_d(\overline{\mathbb Q})$ means that the degree of $j$ is also bounded above and below, this implies that on each compactly bounded subset there are only finitely many values $j\in U_d(\overline{\mathbb Q})$ one is dealing with" is wrong. $\endgroup$
    – Will Sawin
    Commented May 2 at 2:04
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    $\begingroup$ Taking $d=1$, the claim is that there are finitely many rational numbers with absolute value bounded above and below. This is certainly not true since the rational numbers are dense in $\mathbb R$. One needs to bound absolute values at every single place to get a finiteness statement, or bound them above and below at all places but one. The compactly bounded set involves only finitely many places. $\endgroup$
    – Will Sawin
    Commented May 2 at 2:08
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    $\begingroup$ Similarly, with regards to the counterexamples, it's true for the literal counterexample stated, but a slightly different example like $(p^n, p^n+1, 2p^n+1)$ will have $j$ invariant in a compactly bounded subset since the $j$ invariant is a rational function of $a/c$ which for this sequence converges to $1/2$ as $n\to\infty$ rather than diverging. If $1/2$ is outside the relevant compactly bounded subset then just take a different sequence. $\endgroup$
    – Will Sawin
    Commented May 2 at 2:10
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    $\begingroup$ Thanks to Peter Scholze (by email) and Will Sawin for pointing this out. My discussion of Mochizuki's example is incorrect. Both the above linked files have been updated. $\endgroup$
    – Kirti Joshi
    Commented May 2 at 16:39
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Mochizuki's claim is right in spirit but not stated completely rigorously. I want to explain (1) what part is not completely rigorous and (2) why this is not problematic for the analysis of any particular claimed proof of ABC.

Peter Scholze's answer demonstrates that the most straightforward approach to prove ABC by adding up local inequalities at each place fails.

However, one can try to prove ABC by adding up local inequalities at each place in a slightly more complicated way. You can add to one side of the expression $\log ( abc) \leq \log ( K_\epsilon \operatorname{rad}(abc)^{3(1+\epsilon)})$ some local contributions that are known to sum to zero.

The main way to do this uses the fact that the product of the absolute values of a rational at all places (including the archimedean place) is 1. You can pick any rational number depending on $a,b,c$ and add the log of the $p$-adic absolute value to one side of the local inequality at $p$. This modified local inequality, if proven at each place, would still imply ABC.

It is not difficult to do this in such a way that it avoids this particular counterexample, for example by including the log of the absolute value of $b/c$. However doing this will immediately run into a slightly different counterexample.

The claim that "ABC can never be obtained as a result of summing up local inequalities at each prime of a number field" expresses the heuristic that any way of expressing the ABC inequality as a sum of local inequalities will contradict either this counterexample as a very similar one.

Let me try to give some further justification for this heuristic: The issue is not just that the local inequality fails in this case but it fails arbitrarily badly. As $n$ goes to $\infty$, the $p$-adic valuation of $b$ grows but none of the other terms in the formula (locally to $p$) change at all. If we add a term to change it, the change needs to be pretty drastic - it will essentially have to cancel the $p$-adic valuation of $b$. Similar counterexamples show the term also has to cancel the $p$-adic valuation of $a$ and the $p$-adic valuation of $c$. So we've canceled everything on one side of the inequality and are left with nothing nontrivial to bound, and a bound on nothing also cannot prove ABC.

As far as I know, this heuristic has not been made into a rigorous statement by precisely defining a notion of "proof by summing local inequalities" and showing it is impossible. (Such a proof-of-impossibility has sometimes been achieved in computer science, using ideas like relativization and the natural proofs barrier.) If this has been made rigorous, the proof that such proofs are impossible would be at least a little more complicated than this one-line counterexample. I think it's exaggerating to claim that this example "verifies" the impossibility, as Mochizuki does.

However, this is not problematic for applying the heuristic in any particular case. One just has to examine the local inequalities claimed, plug in values of this type, and look for a counterexample, as Peter Scholze briefly sketches how to do in his answer. (Mochizuki has also added a sketch of this in his updated document.)

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    $\begingroup$ +1. It occurs to me that your discussion here may help people understand better the debate between Scholze and Dupuy on Woit's blog. In both cases, it seems that the claim that "no argument of this type can work" falls epsilon short of a rigorous theorem (akin to the barriers in computational complexity that you mention). But since the burden of proof is on the person claiming to be able to prove abc, the "epsilon" is sort of beside the point. The point is that there's a major gap which, at mininum, requires a major idea to fix. $\endgroup$
    – Timothy Chow
    Commented Mar 30 at 21:43
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This is not an answer to the question but too long for a comment. It is interesting to ask if one can prove the ABC inequality just for the tuple $(1,p^n,1+p^n)$ directly for fixed $p$. ABC says $$\frac{\log(p^n(1+p^n))}{\log(p)+\log(\text{rad}(1+p^n))} < 3+\epsilon,$$ with possibly finitely many exceptions. The numerator is about $2n \log (p)$. If $1+p^n$ is square free, the denominator is $>n \log(p)$ so that we get a bound $\le 2$. To falsify the ABC inequality, $1+p^n$ must have large prime factors $q_n$ dividing it to power at least two for infinitely many $n$. But by theorem 1.1 of https://arxiv.org/abs/2112.04173 , any such $q_n$ must satisfy $$2 \le \nu_{q_n}(1+p^n)=\nu_{q_n}(n)+\nu_{q_n}(p^{q_n-1}-1).$$ If we ignore the $\nu_{q_n}(n)$ which should normally be zero, this implies that the infinitely many primes $q_n$ must be Wieferich primes to the base $p$, something we probably can't prove impossible because nothing much seem to be known about Wieferich primes. So it seems that we can't even prove the inequality just for these tuples without assuming ABC ?

So the role of ABC here is to rule out these situations which is in line with the known fact that ABC implies the existence of infinitely many non-Wieferich prime for a fixed base.

Actually it is precisely looking at this ABC tuple that lead us to look at primes that can divide $2^n-1$ to order at least two that led us to the identity above, see the first section of the cited article. The identity holds for most prime in a divisibility sequence.

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    $\begingroup$ Thanks for the interesting comment!Do you know in which papers these triples where first considered, those of Masser or Osterlé maybe? $\endgroup$
    – Jon23
    Commented Mar 29 at 8:55
  • $\begingroup$ I am not sure but I guess this is the natural first example to look at. $\endgroup$
    – CHUAKS
    Commented Apr 8 at 14:59
  • $\begingroup$ It's known ABC implies there are infinitely many non-Wieferich primes to any base $a$. The role of ABC here can be replaced by assuming that there are only finitely many Wieferich primes with Wieferich order $\nu_p(a^{p-1}-1) \ge 3$ (which is expected by statistical reasoning). A result of Granville state that every $a^n-1,n>6$ has primitive prime divisor $p$ which divide it to odd order, which implies $p \not | n=ord_p(a) \le p-1$. So $\nu_p(a^n-1)=\nu_p(a^{p-1}-1)$ is odd. $\endgroup$
    – CHUAKS
    Commented May 2 at 13:24
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In the context of the conversations here and especially Yaakov Baruch's question above, the following report will be useful. Scholze has previewed this document and we have discussed its contents; it has also been sent to Mochizuki for preview. No response from him yet.

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    $\begingroup$ I am sorry David, MO flagged me as a new user again--I had logged into my account for posting this. (Kirti Joshi) $\endgroup$
    – Kirti Joshi
    Commented Jun 29 at 5:28
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    $\begingroup$ Just for people wondering, mods are aware of this issue. It should be cleared up in the near future. $\endgroup$
    – David Roberts ♦
    Commented Jun 29 at 23:13

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