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Section 2.10 Dessins, integer points on elliptic curves and a proof of the ABC conjecture (Alex)

Subsection 2.10.1 A proof of the ABC theorem (for polynomials)

Last week Angus told us about the incredibly powerful ABC conjecture and its arithmetic consequences (apparently). This week we will prove this conjecture (for polynomials). The proof is very similar to some of the things Angus mentioned, but seeing as I wasn't there its new to me... Following Goldring / Stothers / Parab.

Let \(K\) be algebraically closed of characteristic 0, with \(f \in K\lb x \rb\text{,}\) we can define the radical as before

\begin{equation*} \operatorname{rad}(f) = \prod_{p | f} p \end{equation*}

over the primes/irreducibles dividing \(f\text{,}\) this is the maximal squarefree polynomial dividing \(f\text{.}\) How do we measure the size of a polynomial? Let \(r(f) = \deg \operatorname{rad}(f)\text{,}\) and \(h(f_1, \ldots, f_n) = \max\{\deg f_i\}\text{.}\) This is a complicated way of saying

\begin{equation*} \#\{x \in K : f(x) = 0 \}\text{,} \end{equation*}

but we do so to emphasise the link with ABC.

The result is then

First of all we note that the statement is symmetric in \(e,f,g\text{,}\) so we may arrange that \(h(g) \le h(e,f)\) which implies that \(h(e) = h(f) = h(e,f,g)\text{.}\) The second statement is less obviously invariant but note that \(\phi\) is a Belyi function is equivalent to \(1-\phi\) and \(1/\phi\) being Belyi also and this preserves \(\phi(\infty) \in \{0,1,\infty\}\text{,}\) so rearranging does not change the truth of the second statement either. Let \(\phi = f/g\) so \(\deg(\phi) = \max\{\deg (f), \deg(g)\} = h(e,f,g)\text{,}\) we will denote this by \(h\) now. Apply Riemann-Hurwitz (surprise-surprise)

\begin{equation*} -2 = -2h + \sum_{x\in \PP^1} e_\phi(x) - 1\text{.} \end{equation*}

Let

\begin{equation*} R_y = \sum_{x : f(x) = y} e_\phi(x) - 1 \end{equation*}

be the ramification above \(y\text{,}\) we will consider \(B_0, B_1, B_\infty\text{.}\) These ramification numbers will simply be \(h - \#(\phi\inv(y))\text{.}\) Lets begin with \(R_1\text{,}\) we have \(f(x)/g(x) = 1\) so \(e(x) = 0\) and in fact

\begin{equation*} R_1 = h(e) - r(e) = h - r(e)\text{.} \end{equation*}

For \(R_0\) we have either \(f(x) = 0\) or \(g(x) = \infty\text{.}\) Having \(g(x) = \infty\) means \(x = \infty\) but this cannot really happen as \(h(f) \ge h(g)\text{.}\) So this is really just

\begin{equation*} \sum_{x : f(x) = 0} e_\phi(x) - 1 = h - r(f)\text{.} \end{equation*}

Finally \(\phi(x) = \infty\) only when \(g(x) = 0\) or \(x = \infty\text{.}\) If \(h(f) = h(g)\) then \(\phi(\infty) \ne \infty\) and we have simply

\begin{equation*} R_\infty = h - r(g)\text{.} \end{equation*}

If \(h(g) \lt h(f)\) then we also have \(\phi(\infty) = \infty\) so we pick up an extra preimage and we get instead

\begin{equation*} R_\infty = h - (r(g) + 1)\text{.} \end{equation*}

Back up in Riemann-Hurwitz this comes down (magically?) to

\begin{equation*} -2 = \cancelto{0}{-2 h + h + h} + h - r(e) -r(f) - r(g) + R - \delta_{h(f) \gt h(g)} \end{equation*}

so

\begin{equation*} R = h - r(efg) - 2 + \delta_{h(f) \gt h(g)} \end{equation*}

but of course \(R \ge 0\) so

\begin{equation*} h \ge r(efg) + 1 \end{equation*}

with equality exactly when

\begin{equation*} h = r(efg) + 1 \implies R = 0,\, h(f) \gt h(g)\text{.} \end{equation*}

\(R= 0\) is equivalent to being Belyi.

Subsection 2.10.2 Back to number theory

That was all well and good, but this is a number theory seminar, not a function field analogues of number theory seminar, so let's take it back to why we are all here, solving Diophantine equations.

Let's try and find nontrivial integral points on Mordell curves!

\begin{equation*} E_k\colon y^2 = x^3 + k\text{.} \end{equation*}
Example 2.10.2.
\begin{equation*} 1001^2 = 5009^3 - (5009^3 - 1001^2) \end{equation*}

so I found a large point on

\begin{equation*} y^2 = x^3 - (5009^3 - 1001^2) = x^3 - 125675213728 \end{equation*}

are you not impressed?

Although this point would look slightly non-trivial if I started with the curve \(5009^3\) is roughly \(125675213728\) anyway so you should only be impressed if I find points of height somewhat larger than the coefficients. We should probably ask that

\begin{equation*} |x|^3 \gt |k| \end{equation*}

by some margin at least.

A nice question is then given \(k\) how big can an integer point \((x,y)\) on \(E_k\) be? Bounds are known, e.g. Via work of Baker we get

\begin{equation*} \max(|x|, |y|) \lt e^{10^{10}|k|^{1000}}\text{.} \end{equation*}

Ouch.

If we want to study more realistic bounds we can instead reverse the problem. Can we minimise \(x^3 - y^2\) for integer \(x,y\text{,}\) how close can the square of a large integer and the cube of a large integer be? Euler showed that \(|x^3 - y^2| = 1\) has only 1 (interesting) solution, for example.

Marshall Hall was interested in this, did some nice computations and conjectured:

This is false!

Example 2.10.4. Elkies (who else?).

If

\begin{equation*} x= 5853886516781223,\,y = 447884928428402042307918 \end{equation*}

is a point on

\begin{equation*} y^2 = x^3 - 1641843 \end{equation*}

then

\begin{equation*} \frac{\sqrt{|x|}}{k} = 46.6004943471754\text{.} \end{equation*}

This is far larger than the previous best known, but still remains the record as far as I can tell. It seems Hall's conjecture is unlikely to be true for any fixed constant, but the following of Stark-Trotter is more believable.

If Hall's/Stark-Trotter is true we get a huge improvement on Baker

\begin{equation*} \frac{\sqrt{|x|}}{|k|} \lt 100 \implies x \lt 10^4k^2 \end{equation*}

and hence

\begin{equation*} y^2 = x^3 + k \lt 10^{12}k^6 + k \end{equation*}

giving polynomial bounds on \(x,y\) in terms of \(k\text{.}\)

How might one find such triple \((x,y,k)\) that is extremal? One approach is to try and come up with a parametrisation of nice triples. We can search for polynomials \(X(t),Y(t), K(t)\) and then plug in various integer values for \(t\) and hope for the best. To give ourselves the best chance of succeeding we want \(K(t)\) to be smaller than \(X(t)^3\) and \(Y(t)^2\) for some values of \(t\text{.}\) This leads us to ask for \(K\) to be of smallest degree possible. So how low can we go?

This is the point where we come full circle right, we are searching for

\begin{equation*} X(t)^3 - Y(t)^2 = K(t) \end{equation*}

with degree of \(K\) minimised, so we apply Mason-Stothers to see that, if \(M\) is the degree of the left hand terms we have \(\deg(X) = 2m\) and \(\deg (Y) = 3m\text{,}\) indeed \(h\) in Mason-Stothers is then \(6m\) We also have \(r(X^3) = r(X) \le 2m\) and \(r(X^2) = r(Y) \le 3m\) so together Mason-Stothers gives

\begin{equation*} 6m \lt 2m + 3m + r(K) \end{equation*}

or \(m \lt r(K)\text{.}\) So we have lower-bounded the degree of \(K\) in terms of \(\frac 12 \deg(X)\) for example.

We just proved:

The first part was proved initially by Davenport (in the same year, and journal). The second part had to wait until '81 for Stothers to prove it.

Someone else should probably also ask, how is any of this related to Dessins?

To prove sharpness we have to exhibit for each \(m\) triple of polynomials \(X,Y,K\) of degrees \(2m,3m,m+1\text{.}\) Coming up with polynomial families is hard, drawing stupid pictures is easy, can Dessins aid us here?

Lets back-track, when we proved Mason-Stothers we also said that sharpness was equivalent to \(f/g\) being Belyi, so \(X(t)^3/K(t) = (K(t) + Y(t)^2) / K(t) = Y(t)^2/K(t) + 1\) should be a Belyi map of degree \(6m\) from \(\PP^1\to \PP^1\text{.}\) What does its ramification look like? We should have all preimages of \(0\) degree 3, preimages of \(1\) degree 2, and above infinity \(m + 1\) points of degree \(1\) and the remaining of degree \(6m - (m + 1) = 5m - 1\text{.}\)

How can we draw a Dessin like this? Begin with a tree with all internal vertices degree 3, with \(2m\) vertices, this will have \(2m - 1\) edges, and as it is trivalent by the handshake lemma

\begin{equation*} 3 \#\{\text{internal}\} + \#\{\text{leaves}\} = 4m - 2 \end{equation*}

and

\begin{equation*} \#\{\text{internal}\} + \#\{\text{leaves}\} = 2m \end{equation*}

giving

\begin{equation*} 2\#\{\text{internal}\} = 2m - 2 \end{equation*}
\begin{equation*} \#\{\text{internal}\} = m - 1 \end{equation*}
\begin{equation*} \#\{\text{leaves}\} = m + 1 \end{equation*}

Add loops to the leaves, you now have a clean Dessin as above. It has \(2m- 1 + m + 1 = 3m\) edges. We have a face for every loop of degree 1, and one on the outside of degree \(m+ 1 +2(2m-1) = 5m - 1\) as each internal edge is traversed twice if you walk around the outside. So this works!

Example 2.10.7.

For \(m= 1\)

\begin{equation*} (x^2 + 2)^3 - (x^3 + 3x)^2 = 3x^2 + 8\text{.} \end{equation*}

\(m= 2\)

\begin{equation*} (x^4 - 4x)^3 - (x^6 -6x^3+6)^2 = 8x^3 - 36\text{.} \end{equation*}
Example 2.10.8.

For \(m =5\)

\begin{equation*} X(t) = \frac 19 (t^{10} + 6t^7 +15 t^4 + 12t) \end{equation*}
\begin{equation*} Y(t) = \frac{1}{54} (2t^{15} + 18t^{12} +72 t^9 + 144 t^6 + 135 t^3 + 27) \end{equation*}
\begin{equation*} K(t) = -\frac{1}{108} (3t^{6} + 14t^3 +27) \end{equation*}

and we can let \(t = -3\) to get \(X(-3) = 5234\text{,}\) \(Y(-3) = -378661\) and \(K(-3) = -17\text{,}\) so we have a point

\begin{equation*} (5234, 378661) \in E_{17} \colon y^2 = x^3 + 17 \end{equation*}

letting \(t = \pm 9\) we get

\begin{equation*} |384242766^3 - 7531969451458^2| = 14668 \end{equation*}
\begin{equation*} |390620082^3 - 7720258643465^2| = 14857 \end{equation*}

both of which have

\begin{equation*} \frac{\sqrt{|x|}}{k} \approx 1.33\text{,} \end{equation*}

these get lower as we increase \(t\) though.

We should expect this decrease from this method as if \(\deg X = 2m\) and \(\deg K = m + 1\) then \(\sqrt{X(t)}/K(t)\) grows like \(t^{m}/t^{m+1} = t\inv\text{.}\)

Can we do the same for abc?

Take the Dessin with a deg 1 vertex at infinity, degree 3 at 0 with an edge surrounding 1, we get a Belyi function

\begin{equation*} f(x) = \frac{64x^3}{(x+9)^3 (x+1)},\,f(x) - 1 = -\frac{(x^2 - 18x -27)^2}{(x+9)^3(x+1)} \end{equation*}

plugging in \(x=a/b\) and cross multiplying gives

\begin{equation*} 64a^3 b + (a^2 - 18ab - 27b^2)^2 = (a+9b)^3(a+b) \end{equation*}

which could of course be verified independently, but how would you find this identity without Dessins? Now for \(a = -32, b= 23\) we get

\begin{equation*} - 2^{21} \cdot 23 + 11^2 = -1 \cdot 3^2 \cdot 5^6 \cdot 7^3 \end{equation*}

or

\begin{equation*} 11^2 + 3^2 \cdot 5^6 \cdot 7^3 = 2^{21} \cdot 23 \end{equation*}

This is the second highest quality abc triple known with quality

\begin{equation*} \frac{\log c}{\log R} = 1.62599 \end{equation*}

(the current winner has quality \(1.6299\)).

References.

A semi-random order, maybe starting at the top is nice though. If you have trouble finding something let me know.

  1. Belyi’s theorem and Dessins d’enfant - Koundinya Vajjha https://kodyvajjha.github.io/images/bel.pdf
  2. On Computing Belyi Maps - J. Sijsling, J. Voight
  3. Belyi Functions: Examples, Properties, and Applications - Zvonkin (really nice survey)
  4. On Davenport’s bound for the degree of \(f^3 - g^2\) and Riemann's Existence Theorem - Umberto Zannier
  5. Unifying Themes Suggested by Belyi's Theorem - Wushi Goldring
  6. Polynomial Identities and Hauptmoduln - W. W. Stothers
  7. Elliptic Surfaces and Davenport-Stothers Triples - Tetsuji Shioda
  8. The abc-theorem, Davenport’s inequality and elliptic surfaces - Tetsuji Shioda
  9. It's As Easy As abc - Andrew Granville, Thomas J. Tucker
  10. Polynomial and Fermat-Pell families that attain the Davenport-Mason bound - Noam D. Elkies, Mark Watkins (on Watkins webpage)
  11. Halltripels en kindertekeningen - Hans Montanus (in Dutch, but math is universal right?)
  12. Computational Number Theory and Algebraic Geometry Spring 2012, taught by Noam Elkies, notes by Jason Bland
  13. Davenport-Zannier polynomials over \(\QQ\) - Fedor Pakovich, Alexander K. Zvonkin (a nice extension perhaps?)
  14. Minimum Degree of the Difference of Two Polynomials over Q, and Weighted Plane Trees - Fedor Pakovich, Alexander K. Zvonkin (as above)
  15. The ABC-conjecture for polynomials - Abhishek Parab
  16. On Marshall Hall's Conjecture and Gaps Between Integer Points on Mordell Elliptic Curves - Ryan D'Mello
  17. Neighboring powers - F. Beukers, C. L. Stewart (a more general problem, but nice history and examples)
  18. Rational Points Near Curves and Small Nonzero \(| x^3 - y^2|\) via Lattice Reduction - Elkies
  19. ABC implies Mordell - Elkies
  20. Dessins d'enfant - Jeroen Sijsling (master thesis)
  21. Algorithms and differential relations for Belyi functions - Mark van Hoeij, Raimundas Vidunas.
  22. Belyi functions for hyperbolic hypergeometric-to-Heun transformations - Mark van Hoeij, Raimundas Vidunas (has application to ABC over number fields at the end)
  23. Some remarks on the S-unit equation in function fields - Umberto Zannier
  24. A note on integral points on elliptic curves - Mark Watkins
  25. On Hall’s conjecture - Andrej Dujella (more recent progress)
  26. Hecke Groups, Dessins d'Enfants and the Archimedean Solids - Yang-Hui He, and James Read
  27. Belyi functions for Archimedean solids - Nicolas Magot, Alexander Zvonkin (didn't really use this but it's nice!)