Graduate Texts in Mathematics 97 Editorial Board F W Gehring P R Halmos (Managing Editor) C C Moore Neal Koblitz Introduction to Elliptic Curves and Modular Forms Springer-Verlag New York Berlin Heidelberg Tokyo Neal Koblitz Department of Mathematics University of Washington Seattle, W A 98195 U.S.A Editorial Board P R Halmos F W Gehring c C Managing Editor Indiana University Department of Mathematics Bloomington, Indiana 47405 U.S.A University of Michigan Department of Mathematics Ann Arbor, Michigan 48109 U.S.A University of California Department of Mathematics Berkeley, California 94720 U.S.A Moore AMS Subject Classifications: 10-01, IODI2, IOH08, 10HIO, 12-01, 14H45 Library of Congress Cataloging in Publication Data Koblitz, Neal Introduction to elliptic curves and modular forms (Graduate texts in mathematics; 97) Bibliography: p Includes index Curves, Elliptic Forms, Modular Numbers Theory of I Title II Series QA567.K63 1984 516.3'5 84-10517 With 24 Illustrations © 1984 by Springer-Verlag New York Inc Softcover reprint of the hardcover 1st edition 1984 All rights reserved No part of this book may be translated or reproduced in any form without written permission from Springer-Verlag, 175 Fifth Avenue, New York, New York 10010, U.S.A Typeset by Asco Trade Typesetting Ltd., Hong Kong 9876 54 32 I ISBN-13: 978-1-4684-0257-5 e-ISBN-13: 978-1-4684-0255-1 DOl: 10.1007/978-1-4684-0255-1 Preface This textbook covers the basic properties of elliptic curves and modular forms, with emphasis on certain connections with number theory The ancient "congruent number problem" is the central motivating example for most of the book My purpose is to make the subject accessible to those who find it hard to read more advanced or more algebraically oriented treatments At the same time I want to introduce topics which are at the forefront of current research Down-to-earth examples are given in the text and exercises, with the aim of making the material readable and interesting to mathematicians in fields far removed from the subject of the book With numerous exercises (and answers) included, the textbook is also intended for graduate students who have completed the standard first-year courses in real and complex analysis and algebra Such students would learn applications of techniques from those courses, thereby solidifying their understanding of some basic tools used throughout mathematics Graduate students wanting to work in number theory or algebraic geometry would get a motivational, example-oriented introduction In addition, advanced undergraduates could use the book for independent study projects, senior theses, and seminar work This book grew out of lecture notes for a course I gave at the University of Washington in 1981-1982, and from a series of lectures at the Hanoi Mathematical Institute in April, 1983 I would like to thank the auditors of both courses for their interest and suggestions My special gratitude is due to Gary Nelson for his thorough reading of the manuscript and his detailed comments and corrections I would also like to thank Professors J Buhler, B Mazur, B H Gross, and Huynh Mui for their interest, advice and encouragement VI Preface The frontispiece was drawn by Professor A T Fomenko of Moscow State University to illustrate the theme of this book It depicts the family of elliptic curves (tori) that arises in the congruent number problem The elliptic curve corresponding to a natural number n has branch points at 0,00, nand -no In the drawing we see how the elliptic curves interlock and deform as the branch points ± n go to infinity Note: References are given in the form [Author year]; in case of multiple works by the same author in the same year, we use a, b, after the date to indicate the order in which they are listed in the Bibliography Seattle, Washington NEAL KOBLITZ Contents CHAPTER I From Congruent Numbers to Elliptic Curves Congruent numbers A certain cubic equation Elliptic curves Doubly periodic functions The field of elliptic functions Elliptic curves in Weierstrass form The addition law Points of finite order Points over finite fields, and the congruent number problem 14 18 22 29 36 43 CHAPTER II The Hasse-Weil L-Function of an Elliptic Curve 51 51 56 The congruence zeta-function The zeta-function of En Varying the prime p The prototype: the Riemann zeta-function The Hasse-Weil L-function and its functional equation The critical value 64 70 79 90 CHAPTER III Modular forms SL£Z.) and its congruence subgroups Modular forms for SL 2(Z.) 98 98 108 Vlll Modular fOnTIS for congruence subgroups Transformation formula for the theta-function The modular interpretation, and Hecke operators Contents 124 147 153 CHAPTER IV Modular Forms of Half Integer Weight 176 177 185 202 Definitions and examples Eisenstein series of half integer weight for f 0(4) Hecke operators on forms of half integer weight The theorems of Shimura, Waldspurger, Tunnell, and the congruent number problem 212 Answers, Hints, and References for Selected Exercises 223 Bibliography 240 Index 245 CHAPTER I From Congruent Numbers to Elliptic Curves The theory of elliptic curves and modular forms is one subject where the most diverse branches of mathematics come together: complex analysis, algebraic geometry, representation theory, number theory While our point of view will be number theoretic, we shall find ourselves using the type of techniques that one learns in basic courses in complex variables, real variables, and algebra A well-known feature of number theory is the abundance of conjectures and theorems whose statements are accessible to high school students but whose proofs either are unknown or, in some cases, are the culmination of decades of research using some of the most powerful tools of twentieth century mathematics We shall motivate our choice of topics by one such theorem: an elegant characterization of so-called "congruent numbers" that was recently proved by J Tunnell [Tunnell 1983] A few of the proofs of necessary results go beyond our scope, but many of the ingredients in the proof of Tunnell's theorem will be developed in complete detail Tunnell's theorem gives an almost complete answer to an ancient problem: find a simple test to determine whether or not a given integer n is the area of some right triangle all of whose sides are rational numbers A natural number n is called "congruent" if there exists a right triangle with all three sides rational and area n For example, is the area of the 3-4-5 right triangle, and so is a congruent number Right triangles whose sides are integers X, Y, Z (a "Pythagorean triple") were studied in ancient Greece by Pythagoras, Euclid, Diophantus, and others Their central discovery was that there is an easy way to generate all such triangles Namely, take any two positive integers a and b with a > b, draw the line in the uv-plane through the point (-1,0) with slope bfa Let (u, v) be the second point of intersection of this line with the unit circle (see Fig I.1) It is not hard to show that 235 Answers, Hints, and References for Selected Exercises Figure A.3 SC \ - - l SB Figure AA (b) Lf(s) = IIp[(l - p-s)(1 - pk-l-S)]-I = IIil - (h-I {p)p-S + pk-I-2')-I (c) L f (s) = IIp[(l - X(p)p-')(l - X(p)pk-l-S)]-I )( = IIil - X{P)Uk- 1(p)p-S + X(p)2pk-l-2')-I 17 (a) Any point r-equivalent to i must be r o(4)-equivalent to one of the points IJ.jli, where r = j r o(4) In this way, find that i, (2 + i)/5EInt F' are r-equivalent to i; wand {5 + i,J3)/14EInt F' are r-equivalent to w; the two r o(4)-equivalent boundary points (-1 + i)/2 and {3 + i)/10 are r-equivalent to i; and no boundary points are r-equivalent to w (b) Follow the proof of Proposition 8, but with slightly more involved computations Note that the three "corners" (-3 + iy'3)/4, (1 + iy'3)/4, (9 + iy'3)/28 are all r o(4)-equivalent, and the sum of the angles at the three corners is 360° To illustrate the elements in the integration around the cusps and along the circular arcs, let us compute (2nO- JI'(z)dzlf(z) over the contour ABCDEF pictured in Fig A-3, where we suppose thatf(z) has no zeros or poles on the contour (but may have a zero or pole at the cusp 0), and we ultimately want the limit as e - O Here BC is a circular arc of radius e centered at o The element IJ I = (! ~) E ro(4) takes the arc from A to to the arc from D to O The image of B is very close to C, and so we can use the integral from D to IJ.IB to approximate the integral from D to C First, we use S to take the arc BC to a horizontal line between Re z = -3 and Re z = (see Fig AA) By (2.24), we have UJ=llJ f c I'(z) dz= B fez) fSC I'(z) dz dz-k fe SB fez) B z But the latter integral approaches as e - (since the angle of the arc BC approaches zero), while the first integral on the right approaches -vo(f) (as in the proof of Proposition 8, but we use the map q4 = e21tiz/4 into the unit disc) Next, by 236 Answers, Hints, and References for Selected Exercises f (2.24), we have alB I'(z) f.D I'(z) fB f'(z) ~z+ ~z~ ~z dz= -k fB -dzf Bf'(z) A fez) c fez) A fez) f o - -k a A z dz + 1/4 = f fez) alA -k A l /4 z + 1/4 dz (-2+iv'3V4 ;' Similarly, using Ci = =D Ero(4) to take the arc DE to the arc FE, we find that the sum of the integrals over those two arcs is equal to -k times the integral of from (2 + i-J3)/28 to i These two integrals are evaluated by taking In z between the limits of integration, where the branch is determined by following the contour As a result, we find that the sum of the two integrals is - k times the logarithm of d: 1/4 ( - + i-J3)/4 (2 1/4 + i-J3)/28 = -1 ' where, if we keep track of the contour, we see that we must take In( -1) = -ni Thus, (2ni) -1 times the sum of these two integrals is equal to k/2 For a systematic treatment of formulas for the sum of orders of zero of a modular form for a congruence subgroup, see [Shimura 1971, Chapter 2] (c) The only zero of is a simple zero at the cusp -1; the only zero of F is a simple zero at 00 (d) If fEM2(ro(4)), apply part (b) to the elementf - f(oo) - 16f( -i) FEM2(ro(4», which is zero at -1 and 00, and hence is the zero function (e)-(f) See the proof of Proposition (g) Look at the value at each cusp of f= a0 12 + b0 F + c0 F2 + dF3 :f(oo) = a,/( -i) = d/16 3, so a = d = 0; then f(O) = (-£ - 6C4) ( -i4)( -i), so c = -16b (h) Letf(z) = 1]12(2z) Since f(Z)2 = ~(2Z)EM12(ro(2)), we immediately have vanishing at the cusps Let Ci = (_~ ?) By writing 2Ci.Z = -1/(1 - 1/2z), show thatfl[Ci.]6 = -f, and so f¢M6(ro(2» However, ro(4) is generated by Tand Ci Next, since S6(ro(4» = (:(0 F - 160 F2), to check the last equality in part (h), it suffices to check equality of the coefficient of q 18 (b) If fE M1 (4, X), by subtracting off a multiple of we may suppose thatf(oo) = 0; then M2(4, 1)3f2 = afq2 + But a multiple zero at 00 would contradict Problem l7(b), unlessfis the zero function (c) First show that, for k;:: odd, Sk(4, X) = 0-2Sk+1(4, 1); then, by Problem 17(i), dim Sk(4, l) = [(k - 3)/2] (where [ ] is the greatest integer function) (d) Use Proposition 20, Problem 17 (with r' = ro(2), Ci = (6 ?)), and part (c) §III.4 Show that.J -i(cz + d)/.J -i( -cz - d) = i- sgnd (it suffices to check for z = 0), and then divide into four cases, depending on the sign of c and d (b)-(c) Use the relation 0(Ci.{3z) = o (Ci.{3z) o ({3z) 0(z) o ({3z) 0(z) §III.5 (a) By (5.27) and (5.28), T.UI[Ci.N]k) = n(k/2)-1 ~a.JI[Ci.NO'a(g mk' Let Ci.a.b = O'nCi.NO'.(g ~)Ci.l/ E~n(N, {I}, Z), and show that the Ci.a.b are in distinct cosets of r (N) in ~n(N, {l}, Z), and hence form a set of representatives; so, by (5.27), 237 Answers, Hints, and References for Selected Exercises (T"f)I[O(NJk = n(k/2)-1 La,dl[O(a,bO(NJk = n(k/2)-1 L/I[anO(Naa(g ~)Jk = T"UI[O(NJk), because II [anJk =f (b) T2F= 0, T20 = + 16F, so {F, iF+ 14 } is a normalized eigenbasis, (c) (T2F) I[0(4J2 = =1= T2(FI[0(4J2)' (d) Any eigenvector of T2 or [0(4Jk must be an eigenvector of Tn; this includes F, iF + 14 , , and so all of M (ro(4)); by Proposition 40 applied to F, we have An = a (n) (n odd), (e) Let 4(z) = Lanqn For n odd, an = a (n)a = 8a 1(n) by Proposition 40 For n = 2no twice an odd number, compare coefficients of qno in T20 = + 16Fto get an = an + 16a1(no) = 24a 1(no) For n = 2nl divisible by 4, compare coefficients of qnl in 20 = + 16Fto get an = an + = 24a1(n O) by induction The first part follows from (5.19) with n = 0; a ~ounterexample for other cusps is in Problem 2(b), since + 16F does not vanish at s = -1/2 (a) Both T"/and T:Jare equal to n(k/2)-1 L/I[aa(o ~)Jk over the same a, b, d, because g.c.d.(n, M) = Note that = M- k/2II[(~ mk' Thus, Mk/ 2T;g = n: 0, 111 = otherwise Thus, 238 Answers, Hints, and References for Selected Exercises -Jz)) pyp-l = «-tb -~N), ~1 Nl/4(%)e;;l.ja( -1/Nz) + b (- iN 1/4 = ((-tb -~N), ~1 (%)e;;l.j - Nbz + a) (b) We have Yl = ((-tb -~N), (-:b)e;;-l.j -Nbz + a) Since ad - be = and 41e, we have a == d mod 4, and so ea = ed Thus, it remains to compare ( -:b) with ~1 (%) We suppose e ~ (an analogous argument gives the same result if e < 0) Then (~b)(!'If) = U~.t) = 1, and ( .1) = (-,l ), so that (-:b) = (-%b) Now (-%b)(%) = med"d) = sgn a = (~)~1' Thus, )\ = pyp-l((b ~), (~» Again Yl and pyp-l are equal if N is a perfect square; otherwise there exist y for which the two differ by (1, -I) At 00, P = 1, h = I, t = At 0, take p = ((~ -b), -Jz), so p-l = «_~ b), -i-Jz), and we need ro(4)3p«b ~), t)p-l = «~I/),t.jz+h)«_~ b), -i-Jz) = ((!h ~), -it.j-l/z + h·-Jz) = (eil ~), -it.jhz - 1) This is in ro(4) if h = 4, t = Finally, at the cusp -t take rx = (-1 ~), p = ((-1 ~),.j - 2z + I), so p-l = j2z + 1) Since a(b Drx- E ro(4), we can take h = To find t, compute p«b D, t)p-l = «_! -D, t.j -4z - I), which isj((_! -D, z) provided that t = i (a) sk/z(ro(4)) = if k < For k ~ 9, it consists of elements of the form 0F(16F - 4)P(0, F), where P is a polynomial of pure weight (k - 9)/2 Thus, for k ~ 5, dim sk/z(ro(4» = [(k - 5)/4] (b) Since t = I at the cusps 00 and 0, there are always those two regular cusps Since t = i at the cusp -t, that cusp is k-regular if and only if ik = 1, i.e., 41k But + [k/4J - [(k - 5)/4J = if k ~ 5, 4{k and = if k ~ 5, 41k, as can be verified by checking for k = 5, 6, 7, and then using induction to go from k to k + (c) 0F(0 - 16F)(0 - 2F) (a) If d' == d mod Nand N/4 is odd, then (~) = (lift-) = (_1)(N/4-1)(d- 1lIZ(N14) = (_IYN/4-1)(d'-I)fZU;4) = (:') If N/4 is even, then d' == dmod 8, and the proof works the same way, with the additional observation that (~) = (J ) (b) The proof that the cusp condition holds for II [p Jk/Z is just like the analogous part of the proof of Proposition 17 in §IIL3 Now let y = (~ ~)Ero(N), and let Yl = (-iN -~N) By Problem above, we have pyp-l = (1, XN(d))Yl' Thus, (f1[pJkIZ)I[yJk/z = (f1[pyp- Jk/z)l[pJk/z = x~(d)(fI[YIJkIZ)I[pJk/z = X~(d)x(a)II[pJk/z' But since ad == mod N, we have x(a) = X(d) Thus, (fl[p JkIZ) I[yJk/z = xx~(d)II[p Jk/z, as desired 0(00) = 1, 0(-1/2) = 0, 0(0) = (l - i)/2 m ôm, ĐIV.2 Ek/z(oo) = 1, Ek/z(O) = Ek/z(-1/2) = 0; Fk/2(0) = (i.J2)-k; Fk/2(00) = Fk/z(-1/2) = o If one uses (2.16) and (2.19) rather than (2.17) and (2.20), then the solutions rx and f3 to the resulting x equations involve X, which depends on l By Proposition 8, it suffices to show this for I squarefree In that case use (2.16) and (2.19) to evaluate the l-th coefficient of Ek/z + (1 + i")rk/2Fk/z' Hk/2 = W - 22) + W - 2)q + ; W - 2) = if and only irA ~ is odd Use Problem above and Problem of §IV.l to find a and b so that k - aEklZ - bF;"lz vanishes at the cusps a = I, b = (1 + i)krkIZ = E51Z - (1 + i)/.J2F5IZ, = E712 + (I - i)/.J2F7IZ · §IV.3 The computation is almost identical to that in Problem in §IV.1 (c) By the lemma in Proposition 43 in §III.5, right coset representatives for r (N) modulo rl(N)n rx- 1r 1(N)rx, where rx = (~ :v), are rxb = (~ :v), s b