Explicit Division and Torsion Points on Superelliptic 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! Can you fjnd positive whole values for , , and ? Vishal Arul’s Thesis Defense Explicit Division and Torsion Points 2 of 24 + + = 16 + + +
√ Infjnite descent: irrationality of would both be even 3 Explicit Division and Torsion Points Vishal Arul’s Thesis Defense on the (degenerate) Show that there is no point Geometry forever! smaller solution of 24 i.e, that one cannot fjnd two that Arithmetic Show that 2 √ 2 is irrational, ( a, b ) ∈ ( Z \ { 0 } ) 2 that lies nonzero integers a, b such hyperbola x 2 − 2 y 2 = 0 . √ 2 = a b . y 1 If such a , b existed, they 2 Divide a , b by 2 to get a � a � 2 , b 2 x 3 Can’t divide by 2
Rational points on curves Informally, a curve is a 2-variable polynomial equation 4 Explicit Division and Torsion Points Vishal Arul’s Thesis Defense of 24 Some solutions: f ( x, y ) = 0 A rational point on this curve is a solution ( a, b ) where a, b ∈ Q . y Example: x 3 − y 2 + 1 = 0 ( − 1 , 0) (0 , 1) x (0 , − 1) (2 , 3) (2 , − 3)
Rational points on conics Some examples: See Abhinav Kumar’s Lecture 24 for MIT’s 18.781 for more information. Vishal Arul’s Thesis Defense Explicit Division and Torsion Points 5 of 24 Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0 x 2 − 3 y 2 − 1 = 0 x 2 + 2 y 2 − 1 = 0 2 x 2 + y − 2 = 0 1 Are there any rational points? 2 If so, how do we fjnd one? 3 Is there a formula that gives all solutions?
of 24 Vishal Arul’s Thesis Defense rational slope; it intersects the circle at exactly one other point, which must be a rational point! This procedure generates all the rational points. Example: 6 Explicit Division and Torsion Points Intersection point is Example: x 2 + y 2 = 1 (0 , 1) is one rational point. Take any line through (0 , 1) of y Line is y = 3 2 x + 1 x � � − 12 13 , − 5 13
Elliptic curves Vishal Arul’s Thesis Defense 7 Explicit Division and Torsion Points of 24 We can add points. The sum of two rational points is another rational point. Informally, elliptic curves are given by an equation f ( x, y ) = 0 where deg f = 3 . After a change of coordinates, the equation is y 2 = x 3 + Ax + B. y Q − ( P + Q ) P x P + Q y 2 = x 3 − 2 x + 2
Strategy for fjnding rational points on elliptic curves By the Mordell-Weil theorem, there exist fjnitely many rational points that generate all of them. (We return to what this means later.) Vishal Arul’s Thesis Defense Explicit Division and Torsion Points 8 of 24 1 Use a computer to fjnd small rational points. 2 Determine whether we have found them all using 2-descent.
Back to the fruit problem Bremner and MacLeod solved this problem (and did more) 1 9 Explicit Division and Torsion Points Vishal Arul’s Thesis Defense representation problem.” Annales Mathematicae et Informaticae. 2014. 1 Andrew Bremner and Allan MacLeod. “An unusual cubic of 24 Clearing denominators, this becomes a degree 3 equation – an x y z y + z + x + z + x + y = 16 . elliptic curve! Computer fjnds P = ( − 1729 , 1909 , 2511) is a solution, but x ( P ) is negative. Try multiples of P ! x = − 1729 P y = 1909 z = 2511 x = − 59704795693360 2 P y = 58440761954029 z = 60611523515451 x = − 2860839847498395711385911675329 3 P y = 5502830308807460377711231185551 z = 3225095062622507332335161309589 x = 5595948611224060224364017631176103062582886764573084679 4 P y = 1736089310886316841024156986935534509488541328603612320 z = − 5500750928993484313189193690198645161760377304528912439 x = 29306973965939511385616058054230695096641981174315979687608373450722909732659496417749 5 P y = 33410032195872509393087670433431148212674791764424544744567912881635223088249073886351 z = − 26556547643917023101089714889680809129736599315550969939439263851901574038728121206849
Solution to the fruit problem https://math.mit.edu/~varul/fruits 10 Explicit Division and Torsion Points Vishal Arul’s Thesis Defense of 24 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 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.
2-descent on elliptic curves generators, we can make it divisible by 2. 11 Explicit Division and Torsion Points Vishal Arul’s Thesis Defense search for points of small height we may have missed. of 24 If we are lucky, then we will know when we found enough For an elliptic curve E , we want to compute the group of rational points E ( Q ) . (Assume we already found E ( Q ) tors .) 1 Use a computer to fjnd small rational points. 2 Determine whether we have found them all using 2-descent. rational points to generate E ( Q )/2 E ( Q ) . 1 Suppose that P is a rational point not expressible as a sum of our generators. By shifting P by a combination of our 2 Divide P by 2 to get a smaller rational point (one which only has about 25 % the digits that P does!) 3 Repeat this until the height of P is small enough. Manually In fact, Fermat’s infjnite descent for X 4 − Y 4 = Z 2 is equivalent to 2-descent for the elliptic curve y 2 = x 3 + 4 x !
The genus of a curve Conics have genus 0. Elliptic curves have genus 1. rational points. 2 Pictures taken from Wikipedia. Vishal Arul’s Thesis Defense Explicit Division and Torsion Points 12 of 24 We go back to the general case of a curve C given by some equation f ( x, y ) = 0 . Topologically, the complex points of C look like a g -holed torus, where g is the genus of the curve. 2 Faltings’ Theorem. When g ≥ 2 , there are only fjnitely many Unfortunately, when g ≥ 2 , there is no way to add points of C .
The jacobian of a curve This interpretation makes 13 Explicit Division and Torsion Points Vishal Arul’s Thesis Defense it is described in terms of integrals of holomorphic 1-forms. right interpretation, but the Abel-Jacobi map is not algebraic; The Abel-Jacobi map goes from the left interpretation to the “talk to the curve.” addition easy to write down of 24 group structure) is not obvious to write down. the curve” but addition (the This interpretation “talks to unordered collection of 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 an A point of J is a point of C g /Λ , where Λ ≃ Z 2 g is the k ≤ g points of the curve C . period lattice of C . (As a group, the jacobian is ( S 1 ) 2 g .) (just add in C g ) but it does not
Odd-degree hyperelliptic curves and their jacobians Vishal Arul’s Thesis Defense Explicit Division and Torsion Points 14 of 24 An odd-degree hyperelliptic curve C is one of the form y 2 = f ( x ) where d := deg f is odd. Its genus is g = ( d − 1)/2 . Mumford representation is an effjcient way to represent k ≤ g unordered points of C , i.e, a point of its jacobian. Suppose the points are P 1 , . . . , P k . The Mumford data is two polynomials ( U ( x ) , V ( x )) such that 1 The roots of U ( x ) are x ( P 1 ) , …, x ( P k ) . 2 For each i , we have y ( P i ) = V ( x ( P i )) .
Division by 2 on odd-degree hyperelliptic curves which in terms of points is 15 Explicit Division and Torsion Points Vishal Arul’s Thesis Defense Suppose that we want to “divide by 2” on an odd-degree of 24 Example: the genus 2 hyperelliptic curve hyperelliptic curve C . Each point P of C will have 2 2 g halves, i.e, points D on the jacobian J such that 2 D = P . Zarhin gives the Mumford representation of each such D . y 2 = x ( x + 1)( x + 4)( x + 9)( x + 16) and P = (0 , 0) . Then P has 2 2 g = 16 halves in J . One of them has Mumford representation U ( x ) = x 2 − 35 x + 24 V ( x ) = 300 x − 240 �� 1 �� √ √ 2(35 ± 1129) , 5010 ± 150 1129 .
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