explicit division and torsion points on superelliptic...

Explicit Division and Torsion Points onSuperelliptic Curves and Jacobians

Vishal Arul’s Thesis Defense

MIT

April 3, 2020

Vishal Arul’s Thesis Defense Explicit Division and Torsion Points 1 of 24

Fruit problem

99.9999% of people cannot solve this! Solve if you are a genius!

++

++

+= 16

Can you find positive whole values for , , and ?


Infinite descent: irrationality of√2

ArithmeticShow that

√2 is irrational,

i.e, that one cannot find twononzero integers a, b suchthat √

2 =a

b.

1 If such a, b existed, theywould both be even

2 Divide a, b by 2 to get asmaller solution

(a2 ,

b2

)3 Can’t divide by 2

forever!

GeometryShow that there is no point(a, b) ∈ (Z \ {0})2 that lieson the (degenerate)hyperbola x2 − 2y2 = 0.

x

y


Rational points on curves

Informally, a curve is a 2-variable polynomial equation

f(x, y) = 0

A rational point on this curve is a solution (a, b) where a, b ∈ Q.

Example: x3 − y2 + 1 = 0Some solutions:

(−1, 0)(0, 1)(0,−1)(2, 3)(2,−3)

x

y


Rational points on conics

Ax2 +Bxy + Cy2 +Dx+ Ey + F = 0

Some examples:

x2 + 2y2 − 1 = 0 2x2 + y − 2 = 0 x2 − 3y2 − 1 = 0

1 Are there any rational points?2 If so, how do we find one?3 Is there a formula that gives all solutions?

See Abhinav Kumar’s Lecture 24 for MIT’s 18.781 for moreinformation.


https://ocw.mit.edu/courses/mathematics/18-781-theory-of-numbers-spring-2012/lecture-notes/MIT18_781S12_lec24.pdf

https://ocw.mit.edu/courses/mathematics/18-781-theory-of-numbers-spring-2012/lecture-notes/MIT18_781S12_lec24.pdf

Example: x2 + y2 = 1

(0, 1) is one rational point. Take any line through (0, 1) ofrational slope; it intersects the circle at exactly one other point,which must be a rational point! This procedure generates allthe rational points.

Example:Line is y = 3

2x+ 1Intersection point is(

−1213 ,−

513

) x

y


Elliptic curvesInformally, elliptic curves are given by an equation f(x, y) = 0where deg f = 3. After a change of coordinates, the equation is

y2 = x3 +Ax+B.

We can add points. The sum of two rational points is anotherrational point.

x

y

P

Q−(P +Q)

P +Q

y2 = x3 − 2x+ 2


Strategy for finding rational points on elliptic curves

By the Mordell-Weil theorem, there exist finitely many rationalpoints that generate all of them.

1 Use a computer to find small rational points.2 Determine whether we have found them all using 2-descent.

(We return to what this means later.)


Back to the fruit problem

Bremner and MacLeod solved this problem (and did more)1

x

y + z+

y

x+ z+

z

x+ y= 16.

Clearing denominators, this becomes a degree 3 equation – anelliptic curve! Computer finds P = (−1729, 1909, 2511) is asolution, but x(P ) is negative. Try multiples of P !

Px = −1729y = 1909z = 2511

2Px = −59704795693360y = 58440761954029z = 60611523515451

3Px = −2860839847498395711385911675329y = 5502830308807460377711231185551z = 3225095062622507332335161309589

4Px = 5595948611224060224364017631176103062582886764573084679y = 1736089310886316841024156986935534509488541328603612320z = −5500750928993484313189193690198645161760377304528912439

5Px = 29306973965939511385616058054230695096641981174315979687608373450722909732659496417749y = 33410032195872509393087670433431148212674791764424544744567912881635223088249073886351z = −26556547643917023101089714889680809129736599315550969939439263851901574038728121206849

1Andrew Bremner and Allan MacLeod. “An unusual cubicrepresentation problem.” Annales Mathematicae et Informaticae. 2014.


Solution to the fruit problem

Calculations are restricted to 120 seconds.Input is limited to 50000 bytes.Running Magma V2.25-4.Seed: 1519260268; Total time: 0.310 seconds; Total memory usage: 85.16MB.

Found a solution after multiplying by 11!

Number of apples is 75695883920707641508654826369959980969484451183645312281679\

4437380752701088755812980091049260660373658301414260453983030970812870540605007\

7357739096869256607369466849049598689320109753370254292760383586908218687016167\

1591261838569615705225865940066075319029896125903861891981127258138299976866161\

7652089849345328389884032389254792615170485647887842866692663123727097675015331\

5515608939686715005617866255912952511

Number of tangerines is 7166369758780814676912316128994352747367541683001253028\

3003079424699216676553002703222405496100897367085440114493618868063648054160182\

0518276882452685812328425046020925426588717716068885887864369991064394215179763\

5162781962169934058777034950490334867644810567745551982537193820787907197051080\

8009030074990221125473470601943040943508285885974430400086910850816628931992243\

9346786021192652347943472958586821777673871

Number of bananas is 3739353473639792037899660845586527588666096869066509865973\

5784952220869157385809042423277508375964340579066162049915895424902178387079529\

1682616970345665316405351511025345400253316500560354614645904106598045889574950\

8195105532469600315226908892502589998301861315135515383031079377766503151960589\

6705877276805455129511173175225722575347604765502284715832294577561038757745387\

491222687245758103143877455222635105109

https://math.mit.edu/~varul/fruitsVishal Arul’s Thesis Defense Explicit Division and Torsion Points 10 of 24

https://math.mit.edu/~varul/fruits

2-descent on elliptic curves

For an elliptic curve E, we want to compute the group ofrational points E(Q). (Assume we already found E(Q)tors.)

1 Use a computer to find small rational points.2 Determine whether we have found them all using 2-descent.

If we are lucky, then we will know when we found enoughrational points to generate E(Q)/2E(Q).

1 Suppose that P is a rational point not expressible as a sumof our generators. By shifting P by a combination of ourgenerators, we can make it divisible by 2.

2 Divide P by 2 to get a smaller rational point (one whichonly has about 25% the digits that P does!)

3 Repeat this until the height of P is small enough. Manuallysearch for points of small height we may have missed.

In fact, Fermat’s infinite descent for X4 − Y 4 = Z2 is equivalentto 2-descent for the elliptic curve y2 = x3 + 4x!


The genus of a curve

We go back to the general case of a curve C given by someequation f(x, y) = 0. Topologically, the complex points of Clook like a g-holed torus, where g is the genus of the curve.2

Conics have genus 0. Elliptic curves have genus 1.

Faltings’ Theorem. When g ≥ 2, there are only finitely manyrational points.Unfortunately, when g ≥ 2, there is no way to add points of C.

2Pictures taken from Wikipedia.Vishal Arul’s Thesis Defense Explicit Division and Torsion Points 12 of 24

The jacobian of a curve

We work in a bigger space J where it is possible to add points.There are two ways to think about elements of J .

A point of J is anunordered collection ofk ≤ g points of the curve C.This interpretation “talks tothe curve” but addition (thegroup structure) is notobvious to write down.

A point of J is a point ofCg/Λ, where Λ ' Z2g is theperiod lattice of C. (As a group,the jacobian is (S1)2g.)This interpretation makesaddition easy to write down(just add in Cg) but it does not“talk to the curve.”

The Abel-Jacobi map goes from the left interpretation to theright interpretation, but the Abel-Jacobi map is not algebraic;it is described in terms of integrals of holomorphic 1-forms.


Odd-degree hyperelliptic curves and their jacobians

An odd-degree hyperelliptic curve C is one of the formy2 = f(x) where d := deg f is odd. Its genus is g = (d− 1)/2.

Mumford representation is an efficient way to represent k ≤ gunordered points of C, i.e, a point of its jacobian. Suppose thepoints are P1, . . . , Pk. The Mumford data is two polynomials(U(x), V (x)) such that

1 The roots of U(x) are x(P1), …, x(Pk).2 For each i, we have y(Pi) = V (x(Pi)).


Division by 2 on odd-degree hyperelliptic curves

Suppose that we want to “divide by 2” on an odd-degreehyperelliptic curve C. Each point P of C will have 22g halves,i.e, points D on the jacobian J such that 2D = P . Zarhin givesthe Mumford representation of each such D.

Example: the genus 2 hyperelliptic curvey2 = x(x+ 1)(x+ 4)(x+ 9)(x+ 16) and P = (0, 0). Then P has22g = 16 halves in J . One of them has Mumford representation

U(x) = x2 − 35x+ 24

V (x) = 300x− 240

which in terms of points is{(1

2(35±

√1129), 5010± 150

√1129

)}.


Division by 1− ζ on superelliptic curves

I generalize Zarhin’s work to the superelliptic situation

yn = f(x)

where d := deg f is coprime to n. Then C has genus(n− 1)(d− 1)/2. The automorphism ζ : (x, y) 7→ (x, ζny) of Cinduces one on its jacobian J , so 1− ζ will be an endomorphismof J . I provide a formula for dividing by 1− ζ. In terms ofCg/Λ, this divides by the complex number 1− e2πi/n.

There is no Mumford representation because there are n pointsof C with the same x-coordinate. Instead, my representation forpoints of J is n equations whose simultaneous vanishing locusgives g points on C. When n = 2, this recovers Zarhin’s formula.


Example of division by 1− ζ

Consider the point (0, 4) on the genus 3 superelliptic curve C:

y3 = (x+ 8)(x+ 1)(x− 1)(x− 8).

Applying the formula to divide (0, 4) by 1− ζ3 gives the threepolynomials

G1,1 = x2 + 20− 5ζ−13 y

G1,2 = −21x− ζ−23 xy

G1,3 = 16 + 5x2 + 4y + y2

Solving the equations G1,1 = G1,2 = G1,3 = 0 on C gives thepoints{

(0, 4ζ3), (√

−20− 105ζ3,−21ζ23 ), (−√−20− 105ζ3,−21ζ23 )

}.


Connections to Anderson–Ihara theory

Anderson–Ihara theory gives a connection between the Galoisaction on πpro-ℓ

1 (P1 \ {s1, . . . , sr}, ⋆) and the Galois action onthe “higher circular units” associated to {s1, . . . , sr}.

Superelliptic curves are branched n-fold coverings of P1, sowhen n = ℓk, Anderson–Ihara theory applies. Using my formulafor 1− ζn, I hope to make this connection explicit (findequations for the higher circular ℓ-units) for small quotients ofthis π1.


Torsion points on curves

A torsion point of C is one whose image in J is torsion; i.e,some nonzero multiple of it is zero. More precisely, P is atorsion point with respect to a basepoint B if the degree zerodivisor class [P −B] is torsion.

When C has genus at least 2, Raynaud’s theorem states that Chas only finitely many torsion points over C (with respect to afixed basepoint). How can we find them all?


Torsion points on the superelliptic Catalan curve

The Catalan problem asks: what are all pairs of consecutiveperfect powers? The Catalan curve Cn,d is yn = xd + 1. Thefollowing are torsion points of Cn,d:

{ ∞︸︷︷︸1-torsion

, (0, ⋆)︸︷︷︸d-torsion

, (⋆, 0)︸︷︷︸n-torsion

}

Any torsion point not in this list will be called an “exceptionaltorsion point.”

Theorem (A.). Assume n, d ≥ 2 coprime withg = (n− 1)(d− 1)/2 ≥ 2. There are no exceptional torsionpoints unless n+ d = 7.

1 C2,5. Exceptional torsion points = (ζi55√4,±

√5). (5-torsion)

2 C4,3. Exceptional torsion points = (2ζi3, ζj4

√3) (12-torsion)


Strategy for showing no other torsion points existSuppose we had an exceptional torsion point P .

1 We can produce more exceptional torsion points byapplying GQ and Aut(Cn,d) to P .

2 If there are enough torsion points, there will be relationsbetween them.

3 If there are low-degree relations, we get a low degree mapto P1, which would contradict the Castelnuovo–Severiinequality: if C has a degree d1 map and a degree d2 mapto P1 where d1 and d2 are coprime, then

genus(C) ≤ (d1 − 1)(d2 − 1).

4 If Castelnuovo–Severi isn’t sufficient, we have anothergeometric tool: Weierstrass weights. The sum of all theWeierstrass weights on a genus g curve must be g3 − g.Torsion points of small order have nonzero weight, so ifthere are too many of them, we get a contradiction.

(This is a sketch when there are no exceptional torsion points.)Vishal Arul’s Thesis Defense Explicit Division and Torsion Points 21 of 24

Galois action on torsionInformally, GQ acts by permuting roots of irreduciblepolynomials. For example, it can move

√2 to −

√2 but cannot

move√2 to

√3.

By proving a new congruence for Jacobi sums, I explicitlycomputed the GQ-action on the p-torsion of Jp,q.

Theorem (A.). The field generated by the p-torsion of Jp,q is

Q

ζpq, p

√√√√√p−1∏t=1

(1− ζtpζiq)︸︷︷︸

cyclotomic unit

(tj): 1 ≤ i ≤ q − 1, 0 ≤ j ≤ p− 3

.

This means that if were to try to write explicit equations forJp,q, the coefficients would lie in this field. The cyclotomic unitsthat show up are related to Iwasawa theory, Vandiver’sconjecture…


Torsion on the generic superelliptic curve

Now consider a “generic” superelliptic curve

yn = (x− a1) · · · (x− ad)

where n, d ≥ 2 are coprime and a1, . . . , ad are indeterminates.Then {∞, (ai, 0)} are torsion points; define an exceptionaltorsion point to be one outside this list. Assume that the genusg = (n− 1)(d− 1)/2 is at least 2.

Theorem (Poonen-Stoll). When n = 2, there are no exceptionaltorsion points.

Theorem (A.). When d ≥ 3, there are no exceptional torsionpoints. When d = 2 and n ≥ 7, there are no exceptional torsionpoints. There are exceptional torsion points when(d, n) = (2, 5), and they were already classified by work ofColeman.


Thank you

Thank you for attending my thesis defense!


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