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Diophantine problems in polynomial theory Lee, Paul David 2011-10-25

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Diophantine Problems in Polynomial Theory by Paul David Lee B.Sc. Mathematics, The University of British Columbia, 2009 A THESIS SUBMITTED IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE DEGREE OF MASTER OF SCIENCE in The College of Graduate Studies (Mathematics) THE UNIVERSITY OF BRITISH COLUMBIA (Okanagan) August 2011 c Paul David Lee 2011Abstract Algebraic curves and surfaces are playing an increasing role in mod- ern mathematics. From the well known applications to cryptography, to computer vision and manufacturing, studying these curves is a prevalent problem that is appearing more often. With the advancement of computers, dramatic progress has been made in all branches of algebraic computation. In particular, computer algebra software has made it much easier to  nd rational or integral points on algebraic curves. Computers have also made it easier to obtain rational parametrizations of certain curves and surfaces. Each algebraic curve has an associated genus, essentially a classi cation, that determines its topological structure. Advancements on methods and theory on curves of genus 0, 1 and 2 have been made in recent years. Curves of genus 0 are the only algebraic curves that you can obtain a rational parametrization for. Curves of genus 1 (also known as elliptic curves) have the property that their rational points have a group structure and thus one can call upon the massive  eld of group theory to help with their study. Curves of higher genus (such as genus 2) do not have the background and theory that genus 0 and 1 do but recent advancements in theory have rapidly expanded advancements on the topic. In this thesis, we will  rst outline some methods used to  nd rational and integral points on curves of genus 0, 1, and 2. We will then solve some new problems related to polynomial theory that require  nding the solutions to systems of Diophantine equations. We are required to  nd rational or integral points on algebraic curves to garner the solutions to these systems. iiPreface This thesis was written with the collaboration of my supervisor, Dr. Blair Spearman. Dr. Spearman and myself both contributed to the selection of the topic, the research, and the writing of this thesis. All LaTeX document preparation and coding was done by myself. The contents of Chapter 5 have been published in the International Mathematical Forum under the title A Diophantine System and a Prob- lem on Cubic Fields [16]. The contents of Chapter 6 have been published in Scientiae Mathematicae Japonicae under the title The Factorization of x5 + axm + 1 [17]. iiiTable of Contents Abstract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ii Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iii Table of Contents . . . . . . . . . . . . . . . . . . . . . . . . . . . . iv List of Tables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . vi List of Symbols . . . . . . . . . . . . . . . . . . . . . . . . . . . . . vii Acknowledgments . . . . . . . . . . . . . . . . . . . . . . . . . . . viii 1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1.1 Elementary Number Theory . . . . . . . . . . . . . . . . . . 1 1.2 Algebraic Number Theory . . . . . . . . . . . . . . . . . . . 5 1.3 Group Theory . . . . . . . . . . . . . . . . . . . . . . . . . . 10 1.4 Algebraic Curves . . . . . . . . . . . . . . . . . . . . . . . . . 12 2 Algebraic Curves of Genus 0 . . . . . . . . . . . . . . . . . . . 14 2.1 Integral Points on Genus 0 Curves . . . . . . . . . . . . . . . 15 3 Algebraic Curves of Genus 1 (Elliptic Curves) . . . . . . . 18 3.1 Adding Points on an Elliptic Curve . . . . . . . . . . . . . . 18 3.2 Projective Space and Points at In nity . . . . . . . . . . . . 19 3.3 The Group of Rational Points,  . . . . . . . . . . . . . . . . 20 3.3.1 The Torsion Subgroup of an Elliptic Curve . . . . . . 21 3.3.2 The Rank of an Elliptic Curve . . . . . . . . . . . . . 23 4 Algebraic Curves of Genus 2 . . . . . . . . . . . . . . . . . . . 28 4.1 The Jacobian . . . . . . . . . . . . . . . . . . . . . . . . . . . 28 4.2 Adding Points on the Jacobian . . . . . . . . . . . . . . . . . 30 4.2.1 The Interpolating Polynomial . . . . . . . . . . . . . 30 4.3 Chabauty’s Method . . . . . . . . . . . . . . . . . . . . . . . 34 ivTable of Contents 4.4 Chabauty’s Method in Magma . . . . . . . . . . . . . . . . . 35 4.4.1 Some Examples . . . . . . . . . . . . . . . . . . . . . 36 4.5 A Special Case: The Curve Ck : y2 = x5 + k . . . . . . . . . . 37 5 A Diophantine System and a Problem on Cubic Fields . . 38 5.1 Relevant Lemmas . . . . . . . . . . . . . . . . . . . . . . . . 39 5.2 Proof of Theorem . . . . . . . . . . . . . . . . . . . . . . . . 40 5.3 Another System . . . . . . . . . . . . . . . . . . . . . . . . . 43 6 The Factorization of x5 + axm + 1 . . . . . . . . . . . . . . . . 46 6.1 Some Lemmas on Rational Points . . . . . . . . . . . . . . . 48 6.2 Proof of Theorem . . . . . . . . . . . . . . . . . . . . . . . . 49 7 Conclusion and Future Work . . . . . . . . . . . . . . . . . . 52 7.1 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52 7.2 Future Work . . . . . . . . . . . . . . . . . . . . . . . . . . . 53 Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55 Appendices A Magma Code . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58 A.1 Chapter 4 Magma Code . . . . . . . . . . . . . . . . . . . . . 58 A.1.1 Example 4.3 . . . . . . . . . . . . . . . . . . . . . . . 58 A.1.2 Example 4.4 . . . . . . . . . . . . . . . . . . . . . . . 58 A.2 Chapter 5 Magma Code . . . . . . . . . . . . . . . . . . . . . 59 A.2.1 Lemma 5.4 . . . . . . . . . . . . . . . . . . . . . . . . 59 A.2.2 Proof of Theorem 5.2 . . . . . . . . . . . . . . . . . . 60 A.3 Chapter 6 Magma Code . . . . . . . . . . . . . . . . . . . . . 60 A.3.1 Lemma 6.1 . . . . . . . . . . . . . . . . . . . . . . . . 60 A.3.2 Lemma 6.2 . . . . . . . . . . . . . . . . . . . . . . . . 60 vList of Tables 5.1 Values of m and corresponding solutions (X;Y ) to Y 2 = 12X4 +m . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40 6.1 The factorizations of the quintic trinomial x5 + axm + 1 for corresponding values of a and m . . . . . . . . . . . . . . . . 48 viList of Symbols Z Set of integers Q Set of rationals C Set of complex numbers D(f) Discriminant of a polynomial f(x)  Discriminant of a polynomial or integral basis P2 The projective plane f(X : Y : Z) j X;Y; Z 2 Zg Z[x] The polynomial ring of integers in variable x Q[x] The polynomial ring of rationals in variable y Z[x; y] The polynomial ring of integers in variables x and y Q[x; y] The polynomial ring of rationals in variables x and y deg(f); @f The degree of a polynomial f(x) j ~Epj The number of points on ~E modulo p (plus the point at in nity) of an elliptic curve OK The ring of integers of a number  eld, K U(OK) The group of units of the ring of integers of a number  eld, K R The set of non-zero real numbers Q The set of non-zero rational numbers Q 2 The set of square-free rational numbers Zn The cyclic group of n elements Zn1  Zn2      Znr The direct product of cyclic groups C(Q) The algebraic curve C over the rationals J(Q) The Jacobian of a curve C over the rationals viiAcknowledgments I would like to thank my family and friends, especially my parents, whose constant love and support have made my accomplishments possible. I would also like to thank my supervisor, Dr. Blair Spearman, for his con- stant help and patience throughout my academic career at UBC Okanagan. My chosen career path is largely thanks to his in uence and inspiration. My gratitude also goes out to all of my colleagues and other professors from UBC Okanagan that have been such a great source of friendship and support. viiiChapter 1 Introduction 1.1 Elementary Number Theory We begin with some elementary number theory that will be used through- out this paper. The majority of the material in this section can be found in [21], chapters 3-5. De nition 1.1. The greatest common divisor of two integers a and b, which are not both 0, is the largest integer that divides both a and b. Theorem 1.1. The greatest common divisor of the integers a and b, not both 0, is the least positive integer that is a linear combination of a and b. Corollary 1.1. If a and b are relatively prime integers, then there are in- tegers m and n such that ma+ nb = 1. Theorem 1.2 (The Fundamental Theorem of Arithmetic). Every positive integer greater than 1 can be written uniquely as a product of primes, with the prime factors in the product written in nondecreasing order. Example 1.1. Some examples of integers written as products of primes: 385 = 5  7  11, 10944 = 26  32  19; 15525 = 33  52  23. Integer factorization is a well-known problem and Maple has a command ifactor that will do this. De nition 1.2. A diophantine equation is an equation where integer or rational solutions are sought. De nition 1.3. Let m be a positive integer. If a and b are integers, we say that a is congruent to b modulo m if m j (a b). Example 1.2. Let a = 22, b = 4. Then 22 is congruent to 4 mod 9 since 9 j (22 4) = 18. Theorem 1.3. If a and b are integers, then a  b (mod m) if and only if there is an integer k such that a = b+ km. 11.1. Elementary Number Theory Theorem 1.4 (The Chinese Remainder Theorem). Let m1;m2;    mr be pairwise relatively prime positive integers. Then the system of congruences x  a1 (mod m1); x  a2 (mod m2); : : : x  ar (mod mr); has a unique solution modulo M = m1m2    mr. We now provide some useful tools to  nd solutions of congruences of the form f(x) = 0 (mod m), where f(x) 2 Z[x] with deg(f) > 1. If m has the prime power factorization m = pa11 p a2 2    p ak k , then by the Chinese Remainder Theorem, solving f(x) = 0 (mod m) is equivalent to solving the system of congruences f(x)  0 (mod paii ); i = 1; 2;    ; k: Example 1.3. We try to solve the congruence 2x3 + 7x 4  0 (mod 200). By the Chinese Remainder Theorem, this reduces to  nding the solutions of 2x3 + 7x 4  0 (mod 8) and 2x3 + 7x 4  0 (mod 25) since 200 = 2352 = 8  25. First consider the congruence modulo 8. Assume that x is odd. That is, x = 2n+ 1 for n 2 Z. Then 2x3 + 7x 4 = 2(8n3 + 3n2 + 13n+ 2) + 1 which is an odd integer. However, an odd integer can never be congruent to 0 modulo 8. Therefore we consider the case where x is even. Let x = 2m where m 2 Z. Then 2x3 + 7x 4  14m 4  0 (mod 8) 21.1. Elementary Number Theory which is equivalent to  nding the solution to the congruence 3m  2 (mod 4): Multiplying by 3 we get m  2 (mod 4) which leads to x  4 (mod 8): Now consider the congruence modulo 25. To solve the congruence modulo 25, we can  rst  nd the solutions modulo 5 and work up to modulo 25. Consider the congruence 2x3 + 7x 4  0 (mod 5). By testing x = 0; 1; 2; 3; and 4, we  nd that the solution is x  1 (mod 5). We therefore substitute x = 1 + 5t for t 2 Z into the above congruence yielding 2(1 + 5t)3 + 7(1 + 5t) 4  0 (mod 25): Simplifying this congruence we obtain 15t+ 5  0 (mod 25) and eliminate a factor 5 so that 3t+ 1  0 (mod 5): The solutions to this congruence are t  3 (mod 5). Therefore the solutions modulo 25 are x  1 + 5t  1 + 5  3  16 (mod 25). In summary we now have the two congruences x  4 (mod 8); x  16 (mod 25): Using the Chinese Remainder Theorem, we must solve the congruences 25y1  1 (mod 8) 8y2  1 (mod 25) which lead to the solutions y1  1 (mod 5) and y2  22 (mod 25). Then x  4  25  1 + 16  8  22  2916  116 (mod 200). 31.1. Elementary Number Theory Lemma 1.1. If a; b; and c are positive integers such that gcd(a; b) = 1 and a j bc, then a j c. Proof. Since gcd(a; b) = 1, there exists integers x and y such that ax+ by = 1. Multiplying this equation by c we get acx+ bcy = c. Since a j bc then it is clear that a j acx+ bcy. Therefore a divides the left hand side, and thus must divide the right hand side. In other words, a j c. Lemma 1.2. If p divides a1a2    an, where p is a prime and a1; a2;    ; an are positive integers, then there is an integer i with 1  i  n such that p j ai. Proof. This proof is by induction. The case when n = 1 is trivial. As- sume the result is true for n. Consider the product a1a2    an+1 that is divisible by the prime p. We know that either gcd(p; a1a2    an) = 1 or gcd(p; a1a2    an) = p. If the former is true, then by the previous lemma, p j an+1. On the other hand, if p j a1a2    an, using the induction hypothe- sis, there is an integer i with 1  i  n such that p j ai. Consequently, p j ai for some i with 1  i  n+ 1. We now give some simple results about the squares of integers. Theorem 1.5. Squares of integers are congruent to 0 or 1 modulo 3. Proof. Assume x  0 (mod 3). Then clearly x2  0 (mod 3). Now assume x  1 (mod 3). Then x = 3n + 1 for n 2 Z. Then x2 = (3n + 1)2 = 9n2 + 6n+ 1  1 (mod 3). Lastly, assume x  2 (mod 3). Then x = 3n+ 2 for n 2 Z. Then x2 = (3n+ 2)2 = 9n2 + 12n+ 4  1 (mod 3). Theorem 1.6. Squares of integers are congruent to 0 or 1 modulo 4. Proof. Assume x is even. That is, x = 2n for some n 2 Z. Then x2 = (2n)2 = 4n2  0 (mod 4). Now assume x is odd. That is, x = 2n + 1 for some n 2 Z. Then x2 = (2n+ 1)2 = 4n2 + 4n+ 1  1 (mod 4). Theorem 1.7. Squares of integers are congruent to 0, 1, 4, or 9 modulo 16 Proof. Assume x  0 (mod 4). Then x = 4n for some n 2 Z. Then x2 = (4n)2 = 16n2  0 (mod 16). Now assume x   1 (mod 8). Then x = 8n 1 for some n 2 Z. Then x2 = (8n 1)2 = 64n2 16n+ 1  1 (mod 16). Now assume x   2 (mod 8). Then x = 8n 2 for some n 2 Z. Then x2 = (8n 2)2 = 64n2 16n+4  4 (mod 16). Lastly, assume x   3 (mod 8). Then x = 8n 3 for some n 2 Z. Then x2 = (8n 3)2 = 64n2 48n+9  9 (mod 16). 41.2. Algebraic Number Theory Theorem 1.8. Fourth powers of integers are congruent to 0 or 1 modulo 16. Proof. First, assume x is even. That is, x = 2n for some n 2 Z. Then x4 = (2n)4 = 16n4  0 (mod 16). Now assume x   1 (mod 4). That is, x = 4m 1 for some m 2 Z. Then x4 = 256m4 256m3+96m2 16m+1  1 (mod 16). 1.2 Algebraic Number Theory Some material in this section closely follows online notes from Chapman [8], along with well-known results from algebraic number theory. De nition 1.4. An element  in C is an algebraic number if f( ) = 0 for some monic polynomial in Q[x]. Example 1.4.  = p 2 is an algebraic number since it is a root of f(x) = x2  2. Example 1.5.  = p 3 p 5 is an algebraic number. If we square both sides and subtract 3 we get  2  3 =  p 5: Squaring both sides again we get  4  6 2 + 9 = 5: Which leads to  4  6 2 + 4 = 0. Thus  is a root of the polynomial f(x) = x4  6x2 + 4. De nition 1.5. Let  2 C. If f( ) = 0 for some monic polynomial f(x) 2 Z[x], then  is called an algebraic integer. Since the previous two examples were roots of monic polynomials over the integers, they were in fact algebraic integers. De nition 1.6. Let  be an algebraic number. The monic polynomial f(x) 2 Q[x] of least degree such that f( ) = 0 is called the minimal polyno- mial of  . We now introduced some notation that will describe algebraic numbers and integers. Denote A as the set of algebraic numbers and B the set of 51.2. Algebraic Number Theory algebraic integers. We know that A  B. We now give an example of an element that is an algebraic number, not an algebraic integer (i.e.  2 AnB). Example 1.6. Let  = 1= p 2.  satis es f(x) = x2 1=2 2 Q[x], not Z[x]. f(x) is the minimal polynomial for  . In cases where the minimal polynomial is over the rationals, we can per- form scaling operations so that the polynomial has integer coe cients. This is extremely useful for reducibility and converting from algebraic numbers to algebraic integers. Lemma 1.3 (An algorithm for scaling). [8] Given f(x) 2 Q[x] (monic), say f(x) = xn + an 1x n 1 + an 2x n 2 +    + a1x+ a0: We can substitute x = y=c where c 6= 0; c 2 Q and multiply by cn to clear denominators. We get a new polynomial g(y) = yn + an 1cy n 1 + an 2c 2yn 2 +    + a1c n 1y + a0c n: You can pick c so that g(y) has integer coe cients. Example 1.7. Let f(x) = x2  1=2x  1=3 2 Q[x]. Substituting x = y=6 we get g(y) = 1=36y2  1=12y  1=3. Now multiply by 36 and get h(y) = y2  3y  12 2 Z[x]. Theorem 1.9. [8] Let  2 A. Then there is exactly one monic polynomial f(x) 2 Q[x] of minimum degree with f( ) = 0. This polynomial has the property that if g(x) 2 Q[x] and g( ) = 0 then f(x) j g(x). Proof. Let P be the set of monic polynomials h(x) 2 Q[x] with h( ) = 0. Note that P 6= ; as  2 A. By well-ordering, there is an element of least degree, call it f(x). We have to prove that f(x) is unique and if g( ) = 0, then f(x) j g(x). Suppose f(x) is not unique. Then there exists another monic polynomial in P of least degree, say f2(x). Consider the function f  f2. Then (f  f2)( ) = f( )  f2( ) = 0. However, the degree of f  f2 is less than the degree of f and you could scale f  f2 if necessary so that f  f2 2 P . This contradicts the minimality of f . Therefore, f is unique. Now suppose g(x) 2 Q[x]; g( ) = 0. We want to show that f(x) j g(x). 61.2. Algebraic Number Theory By the division algorithm for polynomials, g(x) = f(x)q(x)+r(x) where either @r(x) < @f(x) or r(x) = 0. Since r( ) = g( )  f( )q( ) = 0  0 = 0 and we can scale r(x) if necessary to be monic, r(x) 2 P . This contradicts the minimality of @f(x) unless r(x) = 0. Therefore, f(x) j g(x). Next we quote some results from basic polynomial theory. Lemma 1.4. Let f(x) be the minimal polynomial of  2 A. Then f(x) is irreducible over Q. Theorem 1.10. [8] If a monic polynomial in Z[x] factors in Q[x], it also factors in Z[x]. Theorem 1.11. [8] Let  2 A with minimal polynomial f(x). Then  2 B , f(x) 2 Z[x]. Theorem 1.12 (Eisenstein’s Criterion). [8] Let f(x) = anxn + an 1xn 2 +    + a1x + a0 2 Z[x]. If there exists a prime number p such that all the following conditions are true:  p j ai for i 6= n,  p - an, and  p2 - a0, then f(x) is irreducible over Q. Theorem 1.13. [8] If  2 A then 1= 2 A ( 6= 0). De nition 1.7 (Number Field). Let  2 A with degree n. We de ne Q( ) = fa0 + a1 +    + an 1 n 1 j ai 2 Q; i = 0; 1;    ; n 1g. Theorem 1.14. [8] For each  xed  2 A;Q( ) is a  eld, a sub eld of A. De nition 1.8. The degree of Q( ) over Q (denoted as [Q( ) : Q]) is the dimension of Q( ) as a vector space over Q. We write it as [Q( ) : Q] = n. De nition 1.9. Let f(x) 2 F [x] be a polynomial over a  eld F . A splitting  eld for f(x) is a  eld extension K of F such that 1. f(x) factors into a product of linear factors in K[x], 71.2. Algebraic Number Theory 2. K is the smallest  eld with this property. De nition 1.10. An irreducible polynomial f(x) 2 F [x] with coe cients in a  eld F is separable if f(x) factors into distinct linear factors over a splitting  eld K of f(x). Theorem 1.15 (Primitive Element Theorem). Let F and K be  elds and let K be a  nite extension of F . Then there exists an element  2 K such that K = F ( ) if and only if there are  nitely many  elds L with F  L  K. Corollary 1.2. Let F be a  eld and let [F ( ;  ) : F ] be  nite and separable. There there exists  2 F ( ;  ) such that F ( ;  ) = F ( ). Let  2 A and let f(x) be the minimal polynomial of  of degree n. Let  =  1;  2;    ;  n be the roots of f(x). These  i are the conjugates of  . Suppose now that we form the number  elds Q( 1);Q( 2);    ;Q( n). Some of these may be equal but assume they are di erent. Even though they are di erent, these number  elds are isomorphic. Let  be the isomorphism mapping Q( i) to Q( j). There are n mapping functions  i : Q( 1)! Q( i) de ned by  i(g( 1)) = g( i). De nition 1.11. Let  2 Q( ). We de ne the norm N( ) and the trace T ( ) by N( ) = nY i=1  i( ) T ( ) = nX i=1  i( ) Theorem 1.16. 1. If  2 Q( ) then N( ); T ( ) 2 Q. 2. If  2 Q( ) \B then N( ); T ( ) 2 Z. De nition 1.12. A unit of a ring R is an element u 2 R such that uv = vu = 1R for some v, where 1R is the multiplicative identity element. De nition 1.13. A quadratic  eld is an algebraic number  eld Q( ) of degree two over Q. De nition 1.14. Let K = Q( ) be a number  eld. De ne its ring of integers as OK = K \B. 81.2. Algebraic Number Theory Theorem 1.17. Let K be a quadratic  eld. Let m be the unique squarefree integer m 6= 0; 1 such that K = Q( p m). Then the set OK is given by OK = 8 >< >: Z + Z p m if m is not of the form 4t+ 1; t 2 Z. Z + Z  1 + p m 2  if m = 4t+ 1; t 2 Z. De nition 1.15. An integral basis of the ring of integers is a basis b1;    bn 2 OK of the Q-vector space K such that each element x 2 OK can be uniquely represented as x = nX i=1 aibi; with ai 2 Z. De nition 1.16. A fundamental unit is a generator for the torsion-free unit group of the ring of integers of a number  eld when that group is an in nite cyclic group. For rings of the form Z[ p n], the fundamental unit has the form x+y p n, where (x; y) is the smallest nontrivial solution to the Pell equation x2  ny2 =  1. Theorem 1.18 (Dirichlet’s Unit Theorem). [8] The group of units of the ring of integers of a number  eld (denoted as U(OK)) is  nitely generated and has rank equal to r = r1+r2 1 where r1 is the number of real embeddings and r2 is the number of conjugate pairs of complex embeddings of the number  eld K. Theorem 1.19. [8] U(OK) is an abelian group under multiplication Lemma 1.5. Suppose that K is a number  eld of degree n and  2 OK . Then  2 U(OK) if and only if N( ) =  1. De nition 1.17. For  ;  2 OK with  6= 0 we say  j  if 9 2 OK such that  =   , or equivalently,   2 OK . Lemma 1.6. Let  ;  2 OK . If  j  then N( ) j N( ) as integers. De nition 1.18. Let  2 OK .  is irreducible if 1.  6= 0, 2.  is not a unit, 3. if  =   with  ;  2 OK , then either  or  is a unit. 91.3. Group Theory De nition 1.19. Let K = Q( ) and let  2 OK . Then  is prime in OK if 1.  6= 0, 2.  is not a unit, 3. if  j   in OK , then  j  or  j  . De nition 1.20. Let K = Q( ) be a number  eld of degree n. Let  1;  2;    ;  n 2 K. We de ne a matrix M( 1;  2;    ;  n) to be the n n matrix whose (j; k) entry is the trace T ( j k). We de ne the discriminant of f 1;  2;    ;  ng by  ( 1;  2;    ;  n) = det(M). Since each T ( j k) 2 Q we have the dis- criminant  ( 1;  2;    ;  n) 2 Q. Lemma 1.7. Let K be a number  eld of degree n and let  1;  2;    ;  n 2 K. Then  ( 1;  2;    ;  n) = det(N)2 where N = N( 1;  2;    ;  n) is the n n matrix whose (j; k) entry is  k( j). Recall that  1;  2;    ;  n were the embeddings of K into C. Consider  i, a rational linear combination of  i. We now give a lemma that describes how the discrimant changes when we consider this linear combination of  . Lemma 1.8. Let K be a number  eld of degree n and  1;  2;    ;  n 2 K. If B = (bjk) is any n n matrix over Q and  j = nX k=1 bjk k then  ( 1;  2;    ;  n) = det(B)2 ( 1;  2;    ;  n). Theorem 1.20. All integral bases of OK have the same discriminant  . 1.3 Group Theory We now review some group theory that will be very relevant to Chap- ter 3 on Elliptic Curves. The material in this section follows Fraleigh [10], chapters I-III. De nition 1.21. A binary operation  on a set S is a function mapping S  S into S. For each (a; b) 2 S  S, we will denote the element  ((a; b)) of S by a  b. 101.3. Group Theory Throughout the paper, + and  will denote regular addition and multi- plication. De nition 1.22. A group hG;  i is a set G, closed under a binary operation  , such that  is associative, G has an identity element, and every element in G has an inverse under  . De nition 1.23. A group G is abelian if its binary operation is commuta- tive. De nition 1.24. If G is a group, then the order jGj of G is the number of elements in G. De nition 1.25. If a subset H of a group G is closed under the binary operation of G and if H with the induced operation from G is itself a group, then H is a subgroup of G. We shall let H  G or G  H denote that H is a subgroup of G, and H < G or G > H shall mean H  G but H 6= G. De nition 1.26. An element a of a group G generates G and is a generator for G if < a >= fan j n 2 Zg = G. A group G is cyclic if there is some element a in G that generates G. Theorem 1.21 (Langrange’s Theorem). Let H  G where G is a  nite group. Then the order of H is a divisor of the order of G. That is, jHj j jGj. Corollary 1.3. Every group of prime order is cyclic. Theorem 1.22. Let G1:G2;    ; Gn be groups. For (a1; a2;    ; an) and (b1; b2;    ; bn) in Qn i=1Gi, de ne (a1; a2;    ; an)(b1; b2;    ; bn) to be the element (a1b1; a2b2;    ; anbn). Then Qn i=1Gi is a group, the direct product of the groups Gi, under this binary operation. In the event that the operation of each Gi is commutative, we sometimes use additive notation and refer to Qn i=1Gi as the direct sum of the groups Gi. The notation Ln i=1Gi is sometimes used in this case in place of Qn i=1Gi, especially with abelian groups under addition. Theorem 1.23 (Fundamental Theorem of Finitely Generated Abelian Groups). Every  nitely generated abelian group is isomorphic to a direct product of cyclic groups in the form Zp1r1  Zp2r2      Zpnrn  Z Z     Z where the pi are primes, not necessarily distinct, and ri 2 N. The direct prodct is unique except for possible rearrangement of the factors. 111.4. Algebraic Curves 1.4 Algebraic Curves In this section, we give some de nitions and results about algebraic curves that will be used throughout the paper. De nition 1.27. An algebraic curve over a  eld K is an equation f(x; y) = 0 where f(x; y) 2 K[x; y]. De nition 1.28. For a curve f(x; y) = 0, we make a substitution x = X=Z; y = Y=Z;X; Y; Z 2 Z to get a curve of the form f(X;Y; Z) = 0. This new curve is known as the homogeneous form. The new space that this conversion creates is called projective space. The projective plane is denoted by P2. We now introduce the concept of a singular point, that is a point with no well-de ned tangent. De nition 1.29. A singularity of an algebraic curve f(x; y) = 0 is a point (x; y) on f(x; y) = 0 such that fx(x; y) = 0 and fy(x; y) = 0. For a curve F (X;Y; Z) = 0 in homogeneous coordinates, we require FX(X;Y; Z) = FY (X;Y; Z) = FZ(X;Y; Z) = 0. Example 1.8. Consider the unit circle x2 +y2 = 1. We then have the curve f(x; y) = x2+y2 1 = 0. Taking partial derivatives we get fx(x; y) = 2x and fy(x; y) = 2y. Setting these both to 0 and solving we get the only solution (x; y) = (0; 0). However, this is not a point on the curve. Therefore, there are no singular points on the unit circle. Example 1.9. Consider the curve y2 = x3 or f(x; y) = y2 x3 = 0. Taking partial derivatives we get fx(x; y) =  3x2 = 0; fy(x; y) = 2y = 0. Solving these equations we get the solution (x; y) = (0; 0), which is on the curve. Considering the homogeneous form F (X;Y; Z) = Y 2Z  X3 = 0 we get the same point (0; 0). Example 1.10. Consider f(x; y) = x4 +x2y2 2x2y xy2 +y2 = 0. Taking partial derivatives and solving we get the solutions (x; y) = (0; 0); ( ; 4=3 2+ 5=3) where  is the root of the polynomial 4a3  8a2 + 10a  5. The point (0; 0) is a singular point since it lies on f(x; y); however, the point ( ; 4=3 2 + 5=3) does not. We must also consider the homogeneous form of f(x; y), F (X;Y; Z) = X4 +X2Y 2 2X2Y Z XY 2Z+Y 2Z2 = 0. Taking partial derivatives and solving we get the points (X : Y : Z) = (0 : 0 : Z) and (0 : Y : 0). The point (0 : 0 : Z) corresponds to the point (x; y) = (0; 0) and the point (0 : Y : 0) is a new singularity, a point at in nity. 121.4. Algebraic Curves We now give some de nitions about curves that will help us de ne a curve’s genus. The majority of this information can be found in [11]. De nition 1.30. A double point is a point at which a curve intersects itself, such as a crunode, a point at which two branches of a curve intersect and each has a distinct tangent. De nition 1.31. A delta invariant measures the number of double points concentrated at a point. It is a non-negative integer. We denote the sum of delta invariants as X P  P where the sum is taken over all singular points P of the complex projective plane curve [29]. De nition 1.32 (Genus). The genus of an irreducible algebraic curve is a non-negative integer. It equals g = (d 1)(d 2)2  X P  P where d is the degree of the curve. In other words, g  (d 1)(d 2)2 . If the curve is nonsingular, then g = (d 1)(d 2)2 . Also, if the curve is of the form y 2 = f(x), where f(x) is a polynomial of degree n, then g = bn 12 c, where b c is the  oor function. Example 1.11. Consider the curve y2 = x3. We  nd that the point (0; 0) is a singularity. Therefore, the genus is g = 1 - (at least one) = 0 since g is nonnegative. Example 1.12. Consider the curve y2 = x7  2x3 + x  3. Then g = b7 12 c = 3. 13Chapter 2 Algebraic Curves of Genus 0 In a topological sense, curves of genus 0 are relatively simple. Every genus 0 curve over Q is birationally equivalent to some conic in the projective plane, P2, given by an equation ax2 + by2 + cz2 = 0; where a; b; c 2 Z are square free and pairwise relatively prime. Let C be an algebraic curve of genus 0. If C does in fact have rational solutions (X;Y ), they can be parameterized to the form (X(t); Y (t)) for some t 2 Q. The method of obtaining these parameterizations can be found in [13]. We will outline the method next. One way to parameterize a curve is using the method of base-point pro- jections. Given a point on the curve, we can  nd the equation of a line that goes through that point with arbitrary slope t. We then intersect this line with the curve C and yield an equation for either x or y. Example 2.1. We begin with a simple example: the unit circle x2 +y2 = 1. By observation, we know that the point (x; y) = (1; 0) is on this curve. Using the point-slope form for the equation of a line with slope t, y  y0 = t(x x0); we get y = t(x 1): Substituting into x2 + y2 = 1 and factoring we get (x 1)((t2 + 1)x t2 + 1) = 0 which yields x = t2  1 t2 + 1 Back substituting, we solve for y and get (x; y) =  t2  1 t2 + 1 ; 2t t2 + 1  : 142.1. Integral Points on Genus 0 Curves Letting say t = 3 we obtain the point (45 ; 3 5) which is indeed a point on the unit circle. Example 2.2. There may be cases however where one cannot parametrize a curve over Q because no such solutions exist. For example, the curve x2 + y2 = 3 has no rational (or integral) solutions. We can show this by  rst trans- forming the problem into projective coordinates by letting x = XZ , y = Y Z , X;Y; Z 2 Z and clearing denominators to obtain X2 + Y 2 = 3Z2: (2.1) where we may assume that the greatest common divisor of X;Y; Z is equal to 1. Working locally mod 3 we see that since squares are either 0 or 1 mod 3, then both X and Y must be congruent to 0 mod 3. Equivalently, we have the two equations X = 3X1 Y = 3Y1 where X1; Y1 2 Z. Substituting back into (2.1) we get 9X21 + 9Y 2 1 = 3Z 2: This implies that Z must be a multiple of 3, contradicting our assumption that the greatest common divisor of X;Y and Z is 1. Therefore, the curve x2 + y2 = 3 has no solutions. 2.1 Integral Points on Genus 0 Curves While  nding rational parametrizations for these curves may come easy (especially with the help of computer algebra systems),  nding the integral points can prove to be quite di cult. We will now give a few de nitions and then outline these steps. De nition 2.1. The resultant of two polynomials P and Q over a  eld K with leading coe cients p and q is de ned as the product res(P;Q) = pdeg(Q)qdeg(P ) Y (x;y):P (x)=0;Q(y)=0 (x y) of the di erences of their roots, where x and y take on values in the algebraic closure of K. 152.1. Integral Points on Genus 0 Curves Note: The resultant can also be found by calculating the determinant of the Sylvester matrix or the Bezout matrix of P and Q [5]. De nition 2.2. A Thue equation is a Diophantine equation of the form f(x; y) = n where f is a homogeneous, irreducible polynomial of at least degree 3 over Q and n is a nonzero rational number. Note: This equation is named after Axel Thue who proved that these equations have  nitely many solutions in integers x and y [27]. After obtaining a rational parametrization for the curve, we substitute t = a=b where a; b 2 Z and clear denominators. This leads to a ratio- nal equation with integer values on both the numerator and denominator. With the help of resultants, we can calculate a common divisor between the numerator and denominator. This leads to a system of Thue equations f(x; y) = m for m 2 Z and f(x; y) 2 Z[x; y] where f(x; y) is the denomina- tor of our rational equation. Depending on the resultant, there could be a signi cant number of values for m for which the equation needs to be solved. The number of cases can usually be reduced to a manageable amount using elementary number theory and then the remaining solved using the Thue command in Magma. While these parametrizations may garner solutions, sometimes solutions may be overlooked because the parametrization used may not yield them. In Poulakis [18] [19], a method was given to  nd the integral points on genus 0 curves. In the paper, Poulakis shows that some points may be found through the singularities of the curve. We give an example from [18] where most of the solutions escape parametrization but can be found by calculating the singularities of the curve. Example 2.3. Consider the curve C : f(X;Y ) = X2Y 3  2XY 3 + X3  3XY 2+3Y 3 = 0. The integer solutions of C are (X;Y ) = (0; 0); (1; 1); ( 3; 1). We start by  nding the singularities of the curve. Using the Maple com- mand, singularities we  nd the singularities are the points (0; 0) and (1; 1). It turns out that these points are solutions to the curve C. Again using Maple, we obtain a parametrization for X in terms of t and then substitute t = ab , a; b 2 Z to get the equation X = 8a3 + 36a2b+ 27b3 8a3 : (2.2) 162.1. Integral Points on Genus 0 Curves Using resultants on the numerator and denominator, we determine that the denominator must divide 2939. That is, 8a3 j 2939; or a3 j 2639; which implies a j 2233: This leads to the equation a =  2i3j where 0  i  2 and 0  j  3. We now reduce the number of cases of the above down from 12. Suppose a is even. Then since the gcd(a; b) = 1, b must be odd. If this is the case, the numerator of (2.2) will be odd and the denominator will be even. This scenario will never produce an integer. So we assume that a is odd. That is, a =  1; 3; 9; 27. Now assume that 9 j a. Since gcd(a; b) = 1, then 3 - b. Looking back at (2.2), we conclude that 36 j 27b3, or 3 j b, which is a contradiction. Therefore a =  1; 3. The only one of these four that leads to a solution is a =  3 so we will follow the steps that lead to the solution. Substituting a =  3 into we get X = 1 3 2 b 1 8 b3 Y = 27b3 + 324b 216 6(9b2 + 72) : Since b was even, we substitute b = 2k; k 2 Z into ( ) and get X = 1 3k  k3 Y = k3 + 3k  1 k2 + 2 : Using resultants on the numerator and denominator of Y , we  nd that k2 + 2 j 3 or in other words, k2 + 2 = ( 1)i  3j for 0  i  1 and 0  j  1. This yields the solution k =  1 ) (a; b) = ( 3; 2)) (X;Y ) = ( 3; 1). 17Chapter 3 Algebraic Curves of Genus 1 (Elliptic Curves) We will now discuss the topic of elliptic curves. The reader can refer to Silverman and Tate [23] for more detail relevant to the material. The most general form of an elliptic curve over Q is ax3 + bx2y+ cxy2 + dy3 + ex2 + fxy + gy2 + hx + iy + j = 0 where a; b; c; d; e; f; g; h; i; j 2 Q. However, elliptic curves usually appear in of the form y2 = x3 + ax2 + bx+ c where a; b; c 2 Q. Any curve of genus 1 with a rational point can be converted into what is known as the Weierstrass normal form y2 = x3 + Ax + B where A;B 2 Q (or Z). One condition for a curve to be elliptic is that it must have distinct roots. For example, y2 = x3  x2 = x2(x  1) is not elliptic. In fact, it is a curve of genus 0 and can be parametrized accordingly. An algorithm for computing this normal form is given in [14]. Maple has a command within the algcurves package, called Weierstrassform that performs this algorithm on a given elliptic curve and gives the isomorphic function (and inverse function) mappings used to convert the curve form from its original form to the normal form. Elliptic curves can be de ned over any algebraic number  eld, but for the duration of this chapter, we will only talk about elliptic curves de ned over Q and their corresponding rational solutions. 3.1 Adding Points on an Elliptic Curve In this section we will outline the method used to add points on an elliptic curve. Consider rational points P = (x1; y1) and Q = (x2; y2) on an elliptic curve E and the line going through both P and Q. We construct a third rational point, denoted by P  Q, the intersection of this line with the curve E. If P = Q then we consider the tangent line through this point. De nition 3.1. The inverse of a point P = (x; y) is  P = (x; y). 183.2. Projective Space and Points at In nity Once this third point of intersection is found, we re ect across the x-axis by  nding its inverse. This method of intersecting and re ecting on points P and Q is denoted in its entirety by P +Q. Adding the two points P and  P , we get P +( P ) = O, the point at in nity of the elliptic curve E. This is a consequence of the line going through P and  P being vertical and only passing through E at most twice. The method of intersecting and re ecting points leads to the following important result. Theorem 3.1. [23] This operation (intersect and re ect) causes the rational points on the elliptic curve to form an abelian group. The identity element of this group is the point at in nity, O. We will now give the formulas for adding points on an elliptic curve. Let E be an elliptic curve de ned by y2 = x3 + ax + b. Let P1 = (x1; y1) and P2 = (x2; y2) be points on E with P1; P2 6= O. De ne P1+P2 = P3 = (x3; y3) as follows: 1. If x1 6= x2, then x3 = m2  x1  x2; y3 = m(x1  x3) y1; where m = y2 y1 x2 x1 . 2. If x1 = x2 but y1 6= y2, then P1 + P2 = O. 3. If P1 = P2 and y1 6= 0, then x3 = m2  2x1; y3 = m(x1  x3) y1; where m = 3x21+a 2y1 . 4. If P1 = P2 and y1 = 0, then P1 + P2 = O. Moreover, de ne P +O = P for all points P on E. 3.2 Projective Space and Points at In nity In this section, we will discuss projective space and points at in nity of an elliptic curve. We begin with the construction of the real projective plane P2R. P 2 R consists of equivalence classes of triples (X;Y; Z) with X;Y; Z not all zero. Two triples (X1; Y1; Z1) and (X2; Y2; Z2) are equivalent if there exists  2 R ( 6= 0) with (X2; Y2; Z2) = ( X1;  Y1;  Z1): Since an equivalence class depends only on ratios, we write our triples (X : Y : Z). Points at in nity occur when Z = 0. Therefore we can  nd 193.3. The Group of Rational Points,  these points by  rst converting the curve into its homogeneous form. That is, f(x; y)! f(X=Z; Y=Z). We can now think of the projective plane as the intersection of the real points of the curve E with the points at in nity. Example 3.1 (Two distinct parallel lines intersect at in nity). We de ne the two lines y = mx+ b1 y = mx+ b2: Converting to homogenous coordinates and clearing denominators we get Y = mX + Zb1 Y = mX + Zb2: Setting Z = 0 we get the two equations Y = mX and Y = mX. Thus the solutions are (X;Y ) = (X;mX). In projective coordinates this is equivalent to (X : Y : Z) = (X : mX : 0). Scaling by 1=X we simplify it to (1 : m : 0). Example 3.2. Consider the elliptic curve y2 = x3+ax2+bx+c 2 Q[x]. We want to  nd the point(s) at in nity of this curve. Converting to homogenous form and clearing denominators we get the equation Y 2Z = X3 + aX2Z + bXZ2 + cZ3: Setting Z = 0 we get the equation 0 = X3 which yields the solution X = 0. Therefore the point at in nity is (0 : Y : 0) which scales to (0 : 1 : 0). 3.3 The Group of Rational Points,  The set of rational points on an elliptic curve, together with the point at in nity, form a group. One of the most interesting results about this group, which we will denote as  , is that it is  nitely generated and abelian. We now state some de nitions and important theorems about  . De nition 3.2. The torsion subgroup of an abelian group is the subgroup consisting of all elements that have  nite order. De nition 3.3. The rank of an elliptic curve is the number of copies of Z in its group of rational points. 203.3. The Group of Rational Points,  Theorem 3.2 (Mordell-Weil Theorem [23]). Let E : y2 = x3+ax2+bx+c 2 Q[x] be an elliptic curve. Let  denote the group of rational points on E. Then   = T  Zr where T is the torsion subgroup and r is the rank. 3.3.1 The Torsion Subgroup of an Elliptic Curve As stated above, the torsion subgroup contains all points of  nite order. We state some theorems and de nitions about the order of a point on an elliptic curve. Note that the point at in nity has order 1 as it is the identity element. De nition 3.4. Let P = (x; y) be a point on an elliptic curve E : y2 = f(x) = x3 + ax2 + bx+ c 2 Z[x; y]. P has order m if mP = P +P +    +P (m times)= O, the point at in nity but m0P 6= O if 1  m0 < m. Theorem 3.3. [23] The rational points of order 2 in  (if they exist) have the form (x; 0) where x 2 Q. Proof. Suppose P = (x; y); (x; y 2 Q) has order 2 in  . Then P 6= O and 2P = O implies that P + P = O or P =  P . This means that P is its own inverse. So P = (x; y) = (x; y) implies y = 0. Example 3.3. Let E : y2 = x3  25x be an elliptic curve. Setting y = 0 we get 0 = x3  25x = x(x 5)(x+ 5). So the points of order 2 are (0; 0) and ( 5; 0). Theorem 3.4. [23] Let P = (x; y) be a point on an elliptic curve E : y2 = f(x) = x3 + ax2 + bx + c, where a; b; c 2 Z. Then P has order 3 if x is a root of 3x4 + 4ax3 + 6bx2 + 12cx+ 4ac b2. We now state a theorem that classi es points of  nite order. Theorem 3.5 (The Theorem of Nagell-Lutz). [23] Let E : y2 = f(x) = x3 + ax2 + bx + c 2 Z[x] be an elliptic curve. Let D be the discriminant of f(x) where D =  4a3c+ a2b2 + 18abc 4b3  27c2: Let P be a rational point on E of  nite order with P = (x; y). Then 1. x; y 2 Z, 2. either y = 0 or y2 j D. 213.3. The Group of Rational Points,  The problem with this theorem is that there may be many candidates for possible points where y2 j D. Nagell-Lutz doesn’t tell you which group the torsion subgroup is isomorphic to, only the points themselves. We now state a de nition and another theorem that will further help to  nd the torsion subgroup. De nition 3.5. Let E : y2 = f(x) = x3 + ax2 + bx+ c be an elliptic curve with a; b; c 2 Z and let D be the discriminant of f(x). Then a bad prime is a prime that divides 2D and a good prime is every prime that is not a bad prime. Theorem 3.6 (Reduction mod p). [23] Let E : y2 = f(x) = x3+ax2+bx+c be an elliptic curve with a; b; c 2 Z and let D be the discriminant of f(x). Let T be the torsion subgroup of  , the group of rational points of E. For any good prime p, let P ! ~P be the reduction map mod p mapping T into ~E (mod p). This mapping  : T ! ~E is an isomorphism. That is, the order of T divides the number of points of ~E (reduced mod p) plus the one point at in nity. Example 3.4. Consider E : y2 = f(x) = x3 + x + 1. The discriminant of f(x) is -31 so that the bad primes are 2 and 31. We  rst reduce modulo 3, solve using the msolve command in Maple, and get a total of 3 solutions plus the point at in nity. Reducing modulo 5, we get a total of 8 solutions plus the point at in nity. Therefore we have jT j j 4 and jT j j 9 which imples that jT j = 1. So then T  = f0g. From now on, we will use the notation j ~Epj for the number of points on E mod p plus the one point at in nity. Example 3.5. Consider E : y2 = f(x) = x3 + 4x = x(x2 + 4). We can see that the point (0; 0) is a point of order 2 on E. From this, we know that jT j > 1. We also have D =  256 so that the bad prime is p = 2. We see that j ~E3j = 4 so jT j j 4 ) jT j = 2 or jT j = 4. We then use Nagell-Lutz to determine if there are any more points than the one point of order 2 that we found. If there is at least one more point, then the order of T must be equal to 4. We now give one more theorem that will help  nd the torsion subgroup. Theorem 3.7 (Mazur’s Theorem). [23] Let E be an elliptic curve. The torsion subgroup T of the group of rational points  is one of the following 15 groups: 223.3. The Group of Rational Points,  1. ZN with 1  N  10 or N = 12 (cyclic group of order N), 2. Z2  Z2N with N = 1; 2; 3; 4 (direct products). We now give some examples that will show how to fully  nd the torsion subgroup of a given elliptic curve. Example 3.6. Consider y2 = f(x) = x3 432x+8208 where discrim(f(x)) =  28  312  11. Reducing modulo 5, we  nd j ~E5j = 5. This implies that jT j = 1 or jT j = 5. Using Nagell-Lutz, we know that y2 = 22i  32j where 0  i  4 and 0  j  6. Running through all the values we yield the solutions (x; y) = ( 12; 108) and (x; y) = (24; 108). Using Magma, we  nd that the order of these points is 5. So we now know that jT j = 5 and the only group of order 5 listed in Mazur’s Theorem is Z5. Thus T  = Z5. Example 3.7. Consider y2 = f(x) = x3  1386747x + 368636886 = (x + 1293)(x 282)(x 1011) where D(f) = 216  320  54  72. By observation, we see that there are 3 points of order 2. From this, we know that the torsion subgroup cannot be cyclic, it must be one of the direct products. Reducing modulo 11, we  nd j ~E11j = 16. Using Nagell-Lutz, we yield a solution (x; y) = (1227; 22680) and using Magma, we  nd that the order of this point is equal to 8. The only direct product in Mazur’s Theorem that this applies to is T  = Z2  Z8. 3.3.2 The Rank of an Elliptic Curve Before we get into rank, we introduce a tool called the height of a ra- tional point. We will use this to prove that  is  nitely generated later on. This concept is used to measure how complicated a rational point is from a number theory perspective. De nition 3.6. Let x = m=n be a rational number in lowest terms. Then the height of x is de ned to be H(x) = maxfjmj; jnjg. Theorem 3.8 (Finiteness Property of Height). [23] The set of all rational numbers whose height is less than some bound is  nite. If P = (x; y) is a rational point on E : Y 2 = X3 + aX2 + bX + c, where a; b; c 2 Q, we de ne the height of P to be H(P ) = H(x): 233.3. The Group of Rational Points,  De nition 3.7. We de ne logarithmic height as h(P ) = ln(H(P )). We now state some lemmas that will be used to prove that  is a  nitely generated group. These lemmas and their proofs can be found in Silverman and Tate [23]. Lemma 3.1. For every real number M , the set fP 2  : h(P )  Mg is  nite. Lemma 3.2. Let P0 be a  xed rational point on a curve C. There is a constant  0, depending on P0 and on a; b; c, so that h(P + P0)  2h(P ) +  0 for all P 2  . Lemma 3.3. There is a constant  , depending on a; b; c, so that h(2P )  4h(P )  for all P 2  . Lemma 3.4. The index ( : 2 ) is  nite. Theorem 3.9 (Descent Theorem). [23] Let  be a commutative group. Sup- pose that there is a function h :  ! [0;1) with the following three properties. 1. For every real number M , the set fP 2  : h(P )  Mg is  nite. 2. For every P0 2  , there is a constant  0 so that h(P + P0)  2h(P ) +  0 for all P 2  . 3. There is a constant  so that h(2P )  4h(P )  for all P 2  . Suppose further that 4. The subgroup 2 has  nite index in  . 243.3. The Group of Rational Points,  Then  is  nitely generated. With this knowledge, we introduce some concepts that will enable us to calculate rank using the algorithm 2-descent by isogeny. Consider the set Q 2 = fr2 = (ab ) 2 j r 2 Q g. Let G = hQ ;  i be a group and let H  G be the subgroup of G where H = hQ 2 ;  i. Then the quotient group with respect to G and H is Q =Q 2 . Elements of this quotient group are r  Q 2 where r is a squarefree rational. We will now outline the algorithm for  nding rank of an elliptic curve of the form E : y2 = x3 + ax2 + bx where a; b 2 Z. Consider the curve  E : y2 = x3+ ax2+ bx where  a =  2a and  b = a2 4b. De ne the squarefree maps  :  ! Q =Q 2 and  :   ! Q =Q 2 by  (P ) = 8 >< >: 1 (mod Q 2 ) if P = O b (mod Q 2 ) if P = (0; 0) x (mod Q 2 ) if P 6= (0; 0); P = (x; y) and  (  P ) = 8 >< >: 1 (mod Q 2 ) if  P =  O  b (mod Q 2 ) if  P = (0; 0)  x (mod Q 2 ) if  P 6= (0; 0);  P = ( x;  y): Furthermore,  (P ) j b and  (  P ) j  b. Each of  ( ) and  (  ) is a  nite subgroup of Q =Q 2 and if r = rank(E) then 2r+2 = j ( )jj (  )j Theorem 3.10. [23] Let E and  E be elliptic curves given by E : y2 = x3 + ax2 + bx;  E : y2 = x3 +  ax2 +  bx where  a =  2a and  b = a2  4b. Let  ;   be the respective groups of rational points on E and  E. Let P = (x; y) be a point on E and  P = ( x;  y) be a point on  E. 1. There is a homomorphism  :  !   de ned by  (P ) = ( ( y 2 x2 ; y(x2 b) x2 ) if P = (x; y); P 6= O; P 6= (0; 0)  O if P = O or P = (0; 0). The kernel of  is fO; (0; 0)g. 253.3. The Group of Rational Points,  2. There is a homomorphism  :   !  de ned by  (  P ) = ( (  y 2 4 x2 ;  y( x2  b) 8 x2 ) if P = ( x;  y); P 6= O; P 6= (0; 0) O if  P =  O or  P = (0; 0). The kernel of  is f  O; (0; 0)g. As a consequence,    (P ) = 2P . Recall  ( ) (respectively  (  )) as the set of square-free values of the x- coordinates of points in  . Furthermore,  ( )  fsquare-free divisors of bg and  (  )  fsquare-free divisors of  bg. Theorem 3.11. [23] For E : y2 = x3 + ax2 + bx, let b1 be a square-free divisor of b. Let b2 = b=b1. Then b1 2  ( ) () 9 integers N;M; e with gcd(M; e) = gcd(N; e) = gcd(b1; e) = gcd(b2;M) = gcd(M;N) = 1 and N2 = b1M 4 + aM2e2 + b2e 4: The above quartic equation is known as a torsor. The above theorem applies for  (  ) by replacing a; b; b1; b2;  ; with their corresponding counterparts, Example 3.8. Let E : y2 = x3 97x be an elliptic curve. We have b =  97. We begin by  nding  ( ). We know  ( )  fsquare-free divisors of bg = f1; 1; 97; 97g. Since  (O) = 1 and  (0; 0) = b =  97 we know  ( )  f1; 97g. We now consider torsors for values of b1. Let b1 = 97. Then the torsor is N2 = 97M4  e4: This equation has solution (N;M; e) = (9; 1; 2) that satis es all 5 gcd con- ditions. Thus 97 2  ( ). By closure, (97)( 97) =  1 is in  ( ) as well. Therefore we have all of the elements for  ( ) = f1; 1; 97; 97g. Now consider  E : y2 = x3 + 388x. We now  nd  ( ). We know  ( )  f1; 1; 2; 2; 97 ; 97; 194; 194g and  ( )  f1; 97g. We  rst consider the torsor N2 =  M4  388e4: This is insolvable as the left hand side is positive while the right hand side is negative. Therefore  1 =2  (  ). By similar reasoning we can also eliminate -2, -97, and -194. We now consider the torsor N2 = 2M4 + 194e4 263.3. The Group of Rational Points,  This torsor has solution (N;M; e) = (14; 1; 1) that satis es the gcd condi- tions. Therefore, 2 2  (  ) and thus 184 2  (  ) as well by closure. Then  (  ) = f1; 2; 97; 194g. We then have 2r+2 = j ( )jj (  )j = 4  4 = 16) r = 2. 27Chapter 4 Algebraic Curves of Genus 2 While curves of genus 1 are known as elliptic curves, any algebraic curve with any genus  1 is known as hyperelliptic. We now explore algebraic curves of genus 2 whose rational points have a corresponding structure sim- ilar to the group of rational points of elliptic curves. Curves of genus 2 generally appear in the form y2 = f(x) where f(x) is a quintic or sextic polynomal over Q with no repeated roots in C. However, just like elliptic curves, curves of genus 2 can appear in a very general form. For example, the curve y2  xy  y = x5  3x2 + 7 is of genus 2. Curves of genus 2 cannot be parametrized like curves of genus 0 nor do their rational points form a group like elliptic curves. One of the most important problems with curves of genus 2 is trying to  nd their rational points. Even though  nding these points can be very di cult or impossible, a very nice result about these curves comes in form of the following theorem: Theorem 4.1 (Faltings’ Theorem). [6] Let C be a curve of genus greater than or equal to 2 over a number  eld K. Then C has only  nitely many rational points. As mentioned above, the rational points on genus 2 curves do not form a group structure. We can, however, consider a parallel structure based from the curve C known as the Jacobian. 4.1 The Jacobian We  rst describe the Jacobian in the language of algebraic geometry used in Freiberg [11]. 284.1. The Jacobian De nition 4.1. The divisor group of a curve C, denoted Div(C), is the free abelian group generated by the points of C. Thus a divisor D 2 Div(C) is a formal sum D = X P2C nP (P ) with nP 2 Z and nP = 0 for all but  nitely many P 2 C. In other words, the points P 2 C form a basis for the divisor group. The degree of D is de ned by deg(D) = X P2C nP : The divisors of degree 0 form a subgroup of Div(C), Div0(C) = fD 2 Div(C) : deg(D) = 0g: We de ne some notation that will be used below. If f 2  k(C) , then f is a nonzero rational function de ned over the curve C with coe cients in  k, the algebraic closure of the  eld k. De nition 4.2. Let f be a rational function. A pole is a point at which f is unde ned. The exponent of this factor determines the multiplicity of this pole. Proposition 4.1. Let C be a smooth curve and f 2  k(C) . Then there are only  nitely many points of C at which f has a pole or a zero. Futher, if f has no poles, then f 2  k. De nition 4.3. Let C be a smooth curve, and let f 2  k(C) . We de ne a map div :  k(C) 7! Div(C) f 7! X P2C ordP (f)(P ): (div(f) is a divisor by Proposition 4.1.) A divisor D 2 Div(C) is called principal if D = div(f) for some f 2  k(C) . De nition 4.4. By Proposition 4.1, the set of principal divisors form a subgroup of Div0(C), denoted Pr(C). We say divisors D1; D2 are linearly equivalent, and write D1  D2 if D1  D2 modulo the subgroup Pr(C). The Picard group of C, Pic(C), is the quotient group Div(C)=Pr(C). We are now ready to give the de nition of the Jacobian. 294.2. Adding Points on the Jacobian De nition 4.5. While Pic(C) = Div(C)=Pr(C), similarily we have Pic0(C) = Div0(C)=Pr(C). This group, Pic0(C), is the subgroup of Pic(C), called the Jacobian of C. This representation of the Jacobian in geometric terms is de nitely non- trivial. Luckily, we can express the Jacobian in a di erent way. Cassels and Flynn [6] describes the Jacobian in the following way: Let C be a curve of genus 2 over Z and let J(Q) be the Jacobian of the curve C. It is easy to  nd the Weierstrass points (x; 0) as they are simply the rational roots of the curve. Therefore we consider the non-Weierstrass points in C(Q). Let P 2 C(Q) be a non-Weierstrass point. Then there exists a corre- spondence P 2 C(Q) ! fP; Pg 2 J(Q): Essentially, the Jacobian can be viewed as pairs of points on C. The Jacobian also has a  nite part, which we will denote as J(Q)tors and an in nite part, which we will denote as D. This leads into an important result analogous to that of elliptic curves. Theorem 4.2. Consider the rational points on C denoted as Pi = (xi; yi). Now consider all pairs of points fPi; Pkg. These pairs of points, with the identity element f1;1g form a  nitely generated abelian group. Now that we have a group structure, we can use it to calculate these pairs of points. Once we have found the structure of the Jacobian, we can work backwards to get the actual rational points on the original curve. 4.2 Adding Points on the Jacobian While adding points directly on a genus 1 curve is quite simple, adding points on the Jacobian of a genus 2 curve is not at all trivial. We consider a curve C : y2 = f(x) = a6x6 +    + a0 where ai 2 Z and f(x) is a quintic or sextic with no repeated roots in C. We want to uniquely represent adding pairs of points fP1; P2g+ fP3; P4g as a single pair of points fP5; P6g. Note that 1 = (0; p a6) denotes the points at in nity on C. 4.2.1 The Interpolating Polynomial We now discuss the interpolating cubic (or quadratic as seen later). As outlined in Freiberg [11], adding points on the Jacobian requires one to  nd 304.2. Adding Points on the Jacobian the intersection of a interpolating cubic y = m(x) with the curve C : y2 = f(x). This intersection leads to the equation f(xi) m 2(xi) = 0; i = 1;    4: These xi are the four of the six roots of the sextic f(x) m2(x): These four roots of this sextic correspond to the original points that are being added together, while the remaining two are the x-coordinates of the new pair of points fP5; P6g. (If one of the Pi =1 , then m(x) is in fact a quadratic and there are  ve points of intersection). Some methods of interpolating include Lagrange Interpolation or Newton Interpolation. We will interpolate by forming a set of linear equations whose solution de nes the coe cients of our interpolating cubic or quadratic. The idea behind this method is given in [11]. For our purposes we will avoid the theory and just stick to outlining the methods for the di erent cases. For all our cases, we start with a general equation of a cubic g = ax3 + bx2 + cx + d and our curve y2 = f(x). Note that yk = p f(xk) and y (n) k =p f (n)(xk). Case 4.1 (Four distinct points). We start with the case that all four points are entirely distinct. In this situation, we have the four equations g(x1) = y1 =) ax 3 1 + bx 2 1 + cx1 + d = y1 g(x2) = y2 =) ax 3 2 + bx 2 2 + cx2 + d = y2 g(x3) = y3 =) ax 3 3 + bx 2 3 + cx3 + d = y3 g(x4) = y4 =) ax 3 4 + bx 2 4 + cx4 + d = y4: Solving these four equations for a; b; c; d gives us the coe cients for our interpolating cubic. Case 4.2 (Three distinct points). If only three points are distinct, then we have to treat it slightly di erently than Case 4.1. Suppose without loss of generality, P1 is the non-distinct point. That is, we are adding say fP1; P1g and fP2; P3g. We then get the four equations g(x1) = y1 =) ax 3 1 + bx 2 1 + cx1 + d = y1 g0(x1) = y 0 1 =) 3x 2 1 + 2bx1 + cx1 = y 0 1 g(x2) = y2 =) ax 3 2 + bx 2 2 + cx2 + d = y2 g(x3) = y3 =) ax 3 3 + bx 2 3 + cx3 + d = y3: 314.2. Adding Points on the Jacobian Case 4.3 (Two distinct points). Say for example we are adding the pairs of points fP1; P1g+ fP1; P2g. Then we get the equations g(x1) = y1 =) ax 3 1 + bx 2 1 + cx1 + d = y1 g0(x1) = y 0 1 =) 3ax 2 1 + 2bx1 + cx1 = y 0 1 g00(x1) = y 00 1 =) 6ax1 + 2b = y 00 1 g(x2) = y2 =) ax 3 2 + bx 2 2 + cx2 + d = y2: If we are adding say fP1; P1g+ fP2; P2g then we get g(x1) = y1 =) ax 3 1 + bx 2 1 + cx1 + d = y1 g0(x1) = y 0 1 =) 3ax 2 1 + 2bx1 + cx1 = y 0 1 g(x2) = y2 =) ax 3 2 + bx 2 2 + cx2 + d = y2 g0(x2) = y 0 2 =) 3ax 2 2 + 2bx2 + cx2 = y 0 2 Case 4.4 (One point). In this case we are adding the pairs of points fP1; P1g+ fP1; P1g. g(x1) = y1 =) ax 3 1 + bx 2 1 + cx1 + d = y1 g0(x1) = y 0 1 =) 3ax 2 1 + 2bx1 + cx1 = y 0 1 g00(x1) = y 00 1 =) 6ax1 + 2b = y 00 1 g000(x1) = y 000 1 =) 6a = y 000 1 Case 4.5 (A point at in nity). Say for example we are adding the points f1; P1g + fP1; P1g. Then instead of a cubic, we start with a quadratic g = ax2 + bx+ c and get the equations g(x1) = y1 =) ax 2 1 + bx1 + c = y1 g0(x1) = y 0 1 =) 2ax1 + b = y 0 1 g00(x1) = y 00 1 =) 2a = y 00 1 : If the three non-in nite points are distinct from each other such as in the previous cases, then the same methodology applies as above except with three linear equations instead of four. We outline this method by using an example from [11]. 324.2. Adding Points on the Jacobian Example 4.1. Consider the curve C : y2 = f(x) = x5  2x de ned over Q. The points on this curve are (0; 0); ( 1; 1), and 1. We want to add the pairs of points f( 1; 1); ( 1; 1)g + f( 1; 1); ( 1; 1)g = f(x5; y5); (x6; y6)g. In this case, our four equations are a( 1)3 + b( 1)2 + c( 1) + d = 1 3a( 1)2 + 2b( 1) + c = 3 2 6a( 1) + 2b =  49 4 6a = 681 8 which leads to  a+ b c+ d 1 = 0 3a 2b+ c 3 2 = 0  6a+ 2b+ 49 4 = 0 6a 681 8 = 0 After solving the system above, we obtain our solution for a; b; c; d and our interpolating cubic is m(x) = 227 16 x3 + 583 16 x2 + 509 16 x+ 169 16 : Factoring the equation f(x) m2(x), we get f(x) m2(x) =  1 256 (51529x2 + 58310x+ 28561)(x+ 1)4: The four points (x =  1) come from the quartic factor. We are interested in the other two points from the quadratic above. Solving the quadratic, we obtain the solutions x5 =  29155 51529 + 132 51529 p 35681 x6 =  29155 51529  132 51529 p 35681: Then y5 =  m(x5) and y6 =  m(x6). 334.3. Chabauty’s Method Example 4.2. We now add the pairs of points f1; ( 1; 1)g+f( 1; 1); ( 1; 1)g. Since one of the points is a point at in nity, we only need to consider three linear equations. They are as follows. a b+ c 1 = 0  2a+ b 3 2 = 0 2a+ 49 4 = 0: The solution to this system leads to the interpolating quadratic m(x) =  49 8 x2  43 4 x 29 8 : We now  nd the di erence f(x) m2(x) and factor to get 1 64 (64x2  2593x 841)(x+ 1)3: Finding the solution to the above quadratic yields x5; x6 = 1 128 (2593 p 6938945) where y5; y6 =  m(x5); m(x6). We now introduce a method of  nding points on a curve C that applies when the rank of the Jacobian is strictly less than the genus of C. 4.3 Chabauty’s Method We begin this section with a relevant theorem to curves of genus 2. Theorem 4.3. [6] Let C be a curve of genus g > 1 de ned over a number  eld K. If the Jacobian has rank less than g then C(K) is  nite. Since we are considering curves C with genus equal to 2, we only need to consider curves with the rank of the Jacobian of 0 and 1. First we consider the case when the rank of the Jacobian is 0. If J(Q) has rank 0, then J(Q) = J(Q)tors. This group J(Q)tors is  nite and the points are easy to  nd. If the rank of J(Q) is 1, then it becomes necessary to  nd a generator for the in nite part of J(Q), denoted by D so that J(Q) =< J(Q)tors;D > : 344.4. Chabauty’s Method in Magma Once this generator is found, then some local calculations are done at some prime p - 2 where C has good reduction. The algorithm then uses a p-adic method involving Strassman’s Theorem to bound the number of rational points. Theorem 4.4 (Strassman’s Theorem). [6] Let  (X) = c0 + c1X +    2 Zp[[X]] satisfy ck ! 0 in Zp. De ne l uniquely by: jcljv  jcj jv for all j  0, and jcljv > jcj jv for all j > l. Then there are at most l values of x 2 Zp such that  (x) = 0 and jxjv  1. Cassels and Flynn [6] outlines the steps of this algorithm as follows: 1. Try to  nd J(Q)tors and J(Q)=2J(Q), and show that the Jacobian has rank 1. 2. Use heights to  nd a generator D such that J(Q) =< J(Q)tors;D >. 3. Find the Chabauty bound with respect to some good prime p. We now provide some interesting results about curves of genus 2 and then talk about using Magma when confronted with a genus 2 curve. Theorem 4.5. [6] Let C be a curve of genus 2 over Q and p  5 be a prime of good reduction. If J(Q) has rank at most 1 and eC is the reduction of C mod p. Then jC(Qj  jeC(Fp)j+ 2 Theorem 4.6. [6] Let C be the curve of genus 2 C : Y 2 = X(X2  1)(X  1= )(X2 + aX + b) ( ; a; b 2 Z): Suppose 32r j  for some r > 0 and 3 - b(1  a + b)(1 + a + b) and that the Jacobian of C has rank at most 1. Then C(Q) contains precisely the points (0; 0); (1; 0); (1= ; 0) and the two rational points at in nity. 4.4 Chabauty’s Method in Magma The previous algorithm has been implemented into Magma and uses a combination of this algorithm, along with the Mordell-Weil sieve [1] to precisely determine the rational points on the curve. There are two di erent cases that require two slightly di erent methods in order to  nd all of the rational points on the curve. 354.4. Chabauty’s Method in Magma Case 4.6 (When the rank of the Jacobian is equal to 0). In this case, known as the trivial case, we can use the Magma command Chabauty0 to enumerate all of the points on the curve. This command only requires the Jacobian of the curve as input. Case 4.7 (When the rank of the Jacobian is equal to 1). This case requires a little more calculation and work. Let C be a hyperelliptic curve and let J(Q) be the Jacobian. In order to use the Chabauty command, we require a point on the Jacobian that is a generator of J(Q)=J(Q)tors. Let Ci = (xi; yi) be the points of C. The generator point can be found by  nding a point fCi; Cjg on the Jacobian for some i; j that has in nite order in J(Q). Once this point is found, we can input it into the Chabauty command to enumerate all of the points. Case 4.8 (When the rank of the Jacobian is  2). This case requires going to Elliptic Chabauty methods that we will omit in this thesis. For more information on these methods, see [4]. 4.4.1 Some Examples We now give some examples outlined in the Magma handbook [1] that use Chabauty’s method to solve. Example 4.3. Consider the curve C : y2 = x6 + 4: Using a height search with the command Points, Magma  nds the points (x; y) = (0; 2) and the two rational points at in nity. The command RankBounds  nds that the rank of the Jacobian is 0. Thus, we can enu- merate all the points on C using the command Chabauty0. The 4 points listed above are reiterated as the only points on C. Example 4.4. Consider the curve C : y2 = x6 + x2 + 2: Using the command Points, Magma  nds the 4 points (x; y) = ( 1; 2) plus the 2 rational points at in nity. Using the command RankBounds we  nd that the rank of the Jacobian is 1. We now need to  nd points on C that give us points of in nite order on the Jacobian J(Q). Now consider the point f(1; 2);1 g. The order of this point is in nite and it can be shown that it generates J(Q)=J(Q)tors, thus being able to use it in the Chabauty command. Using this command, we  nd that the points listed above are the only points on this curve. 364.5. A Special Case: The Curve Ck : y2 = x5 + k 4.5 A Special Case: The Curve Ck : y2 = x5 + k We are now going to talk about the family of genus 2 curves Ck : y2 = x5 + k where k is a tenth-power-free integer considered in a paper by Stoll [25]. Stoll proved a very nice result concerning these curves stated in the following theorem: Theorem 4.7. Let Ck = x5 +k be a genus 2 curve with k a tenth-free-power integer. Let J(Q) be the Jacobian of Ck. Assume rank(J(Q))=1. Then jCkj  7. The bound of 7 is achieved only for k = 324. For all other values of k, the bound is 6 if there exists any Weierstrass points (points of the form (x; 0)) and 5 otherwise. The idea behind this theorem comes from counting the number of ratio- nal points on the curve Ck. Stoll showed that the number of rational points on the curve (denoted by #Ck(Q)) is equal to 2nk + dk where nk is half the number of "nontrivial" points in Ck(Q), i.e.,  nite points with nonvanishing x and y coordinates, and dk = 1; 2; 3, or 4 if k is neither a square nor a  fth power, a  fth power but k 6= 1, a square but k 6= 1, or k = 1, respectively. The bound stated in the theorem is a very nice result since we can use a height search to  nd points on a hyperelliptic curve using Magma. If we have found, say 5 points (if k 6= 324) and there are no points of the form (x; 0) then we know we have found them all. The usefulness of this theorem will become more apparent in Chapter 6. 37Chapter 5 A Diophantine System and a Problem on Cubic Fields In this chapter we will discuss a problem from Kaneko [15] who con- sidered families of cubic  elds and their fundamental units. Kaneko proved that there are  nitely many triples (A;B; b) of integers satisfying the system A2  2B = 3(b2 + 1); B2  2A = 3(b4 + b2 + 1); (5.1) and the following solutions were given. (A;B; b) = ( 1; 1; 0); (3; 3; 0); and (0; 3; 1): The study of this system was related to the following problem about cubic  elds. Let  be a root of the irredicble cubic polynomial f(x) given by f(x) = x3  3x b3; b(6= 0) 2 Z: Let K = Q ( ) be the cubic  eld de ned by f(x) and set " = 1 1 b(  b) : As proved in [15] "(> 1) is the fundamental unit of K for in nitely many values of b: A solution of the above system of equations would yield a value of b for which  was not the fundamental unit of K: In [16], we gave a simple complete solution of this Diophantine system,  nding a total of six solutions. We begin by stating our main result, follow with some relevant lemmas and then  nish with a proof. Theorem 5.1. The Diophantine system (5.1) has the solutions (A;B; b) = (0; 3; 1); ( 1; 1; 0); (3; 3; 0) and (8; 17; 3): 385.1. Relevant Lemmas Our method of proof involves  nding the integral points on a genus 0 curve. A general method for solving this type of problem is given in [18], [19]. As described in Section 2.1, we parametrize the rational points on the curve then reduce the determination of the integral solutions to a  nite set of Thue equations. These can be solved for example with the assistance of Magma [1]. In Section 5.1 we prove some lemmas involving the integral solutions of certain quartic equations and then prove our theorem in Section 5.2. 5.1 Relevant Lemmas Lemma 5.1. If c; d and m are integers with gcd(c; d) = 1 and c4  6c2d2  3d4 = m then m 6 2;3(mod 4) and m 6 2(mod 3): Proof. We can rearrange the given equation to obtain (c2  3d2)2  12d4 = m which yields the pair of congruences (c2  3d2)2  m(mod 3) and (c2  3d2)2  m(mod 4): The solvability of these congruences impose the conditions on m stated in this lemma. Lemma 5.2. If c; d and m are integers with gcd(c; d) = 1 and c4  6c2d2  3d4 = m then either m is odd or 8 k m: Proof. Suppose that m is even. Clearly c and d must both be odd. This forces m to be a multiple of 8: Rearranging the given equation gives (c2  3d2)2  12d4 = m If 16 j m then we deduce the congruence (c2  3d2)2  12d4  12(mod 16) which is insolvable, completing the proof. 395.2. Proof of Theorem Lemma 5.3. If c; d and m are integers with gcd(c; d) = 1 then c4  6c2d2  3d4 =  24 is insolvable. Proof. Solvability requires 4c4  3(c2 + d2)2 =  24: Clearly this implies that 3 j c so that  3(c2 + d2)2   24(mod 9): This in turn implies that (c2 + d2)2  2(mod 3) which is impossible. Lemma 5.4. If m 2 f1; 3; 8; 24g then the quartic equation Y 2 = 12X4 + m has the integral solutions given in the table below. Table 5.1: Values ofm and corresponding solutions (X;Y ) to Y 2 = 12X4+m m = 1 (X;Y ) = (0; 1) m =  3 (X;Y ) = ( 1; 3) m =  8 (X;Y ) = ( 1; 2) m = 24 (X;Y ) = ( 1; 6) Proof. The elliptic curve Y 2 = 12X4 + 1 has rank 0 so the given inte- gral points are easy to determine. The three remaining curves have rank 1. At any rate all of them can be solved using the Magma command IntegralQuarticPoints. The solutions are as listed. 5.2 Proof of Theorem If we solve the  rst equation in (5.1) for B, substitute into the second equation in (5.1) and simplify we obtain A4  6(b2 + 1)A2  8A 3(b2  1)2 = 0: (5.2) 405.2. Proof of Theorem This polynomial equation de nes an algebraic curve of genus 0. We give a parametrization over Q of the rational points on this curve. First suppose that b = 0: Then using (5.1) we obtain the values A =  1; 3; and once again using (5.1) we obtain the solutions (A;B; b) = (3; 3; 0) and ( 1; 1; 0): Now assuming that b 6= 0 we may choose a rational number r such that A = br  1: Substituting this expression for A into (5.2) yields b3((r4  6r2  3)b 4r(r2  3)) = 0: As b 6= 0 we may solve for b giving b = 4r(r2  3) r4  6r2  3 : (5.3) Recalling that A = br  1 we obtain A = 3(r2  1)2 r4  6r2  3 : (5.4) Having obtained this parametric formula for A we derive a set of Thue equations in order to solve the original system. Choosing relatively prime integers c and d 6= 0 so that r = c=d we substitute into (5.4) giving gives A = 3(c2  d2)2 c4  6c2d2  3d4 : (5.5) For convenience we rewrite (5.5) as A = 3F 2 G ; (5.6) where F = (c2  d2) and G = c4  6c2d2  3d4: From the two identities G (c2  5d2)F =  8d4 and G (9c2 + 3d2)F =  8c4 415.2. Proof of Theorem we deduce that gcd(F;G) is a divisor of 8: Thus the gcd of the numerator and denominator of (5.6) is a divisor of 26  3: It follows that in order for (5.6) to yield an integer value for A we must have G j 3F 2 which is only possible if G j 26  3: Thus we deduce that c4  6c2d2  3d4 = m with m =  2e3f ; 0  e  6; 0  f  1: (5.7) Now it remains to consider these 28 Thue equations given by (5.7). We can reduce the number of these equations as follows. By Lemma 5.1 m 6=  1; 2; 2; 3; 4; 6; 6; 8; 16; 32; 64; and by Lemma 5.2 we further deduce that m 6= 4; 12; 12; 16; 32; 48; 48; 64; 96; 96; 192; 192: Lemma 5.3 shows that m 6=  24 so that our list of admissible values of m is reduced to m = 1; 3; 8; 24: Completing the square in (5.7) yields (c2  3d2)2 = 12d4 +m: If m = 1 then Lemma 5.4 gives d = 0 c2  3d2 =  1 which is impossible as d 6= 0: If m =  3 then Lemma 5.4 gives d =  1 c2  3d2 =  3 425.3. Another System which yields integral solutions (c; d) = (0; 1): Using r = c=d = 0; equations (5.3), (5.4) and (5.1) give us the solutions (A;B; b) = ( 1; 1; 0) which was obtained already in the  rst part of this proof. If m =  8 then Lemma 5.4 gives d =  1 c2  3d2 =  2 which yields the integral solutions (c; d) = ( 1; 1): Using r = c=d =  1; equations (5.3), (5.4) and (5.1) give us the solutions (A;B; b) = (0; 3; 1): If m = 24 then Lemma 5.4 gives d =  1 c2  3d2 =  6 which yields the integral solutions (c; d) = ( 3; 1): Using r = c=d =  3; equations (5.3), (5.4) and (5.1) give us the solutions (A;B; b) = (8; 17; 3): This completes the proof.  5.3 Another System In his proof showing that  is the fundamental unit of Q( ), Kaneko [15] considered cases where there exists a unit  0(> 1) of Q( ) such that  =  n0 with some n 2 Z; n > 1. The above system was considered when  =  20. As an exercise, we will now completely solve the system that was considered when  =  30. Solving this system will produce two of the same b values as the previous system. Theorem 5.2. The Diophantine system A3  3AB + 3 = 3(b2 + 1) B3  3AB + 3 = 3(b4 + b2 + 1) has solutions (A;B; b) = (0; 0; 0); ( 3; 6; 3) and (3; 3; 0) 435.3. Another System Proof. Consider the Diophantine system A3  3AB + 3 = 3(b2 + 1) B3  3AB + 3 = 3(b4 + b2 + 1): Eliminating B from the two equations we get A9  9A6b2  54A3b4 + 27b6 + 27A6 = 0: (5.8) This is a curve of genus 4. We then make the substitution b = p c where c 2 Z. This yields the new equation A9  9A6c 54A3c2  27c3  27A6 = 0: (5.9) This new equation (5.9) is now a genus 1 (elliptic) curve. Using the Weierstrassform command in Maple we yield the curve x3  109418989131512359209=4 + y2 = 0: (5.10) Using the command Rank in Magma, we  nd that the rank of this curve is 0. This implies that the curve has  nitely many solutions. From the birational mappings given by Maple between the curves (5.9) and (5.10) we have the relation x =  1594323(7A6  6A5  18A4 + 21cA3  9cA2 + 9c2) A4(A 3)2 : (5.11) Looking at the denominator of (5.11) we see that possible solutions A = 0 and A = 3 may escape the solutions (x; y) of the curve (5.10). Substituting A = 3 and A = 0 into our curve and solving for B and b we get the solutions (A;B; b) = (3; 3; 0) and (A;B; b) = (0; 0; 0) respectively. We now perform some variable changes and scaling to the above Weierstrass cubic to convert it into the pure Weierstrass form. Solving for y2, substituting x =  x and factoring the integers we are now considering y2 = x3 + 342 22 : (5.12) We now divide the equation by 342 and substitute x = 314x and y = 321y and get y2 = x3 + 1 4 (5.13) 445.3. Another System Now multiplying the equation by 43 and substituting x = x=4 and y = y=8 we get y2 = x3 + 16: (5.14) We  nd that the torsion subgroup of this curve is isomorphic to Z3 and the rank is 0. By observation, we see that the curve has solutions (x; y) = (0; 4). These two solutions, plus the point at in nity, make up the neces- sary three points to satisfy the torsion subgroup. After back substituting the two solutions through all the scaling and variable changes, we yield the solutions (A;B; b) = ( 3; 6; 3) and (A;B; b) = (0; 0; 0). 45Chapter 6 The Factorization of x5 + axm + 1 Let f(x) be a polynomial with rational coe cients. The determination of those polynomials f(x) with a speci c form and a prescribed factoriza- tion often leads to interesting Diophantine problems. A general source of information on this type of problem is [22] where Schinzel classi ed many forms of reducible trinomials. In a paper by Rabinowitz [20] the factorization of x5  x + n, for n an integer, into the product of an irreducible quadratic and an irreducible cubic over the rational numbers Q was studied and a  nite number of polynomi- als was determined. The results of this paper are given in the folllowing theorems: Theorem 6.1. [20] The only integral n for which x5 + x + n factors into the product of an irreducible quadratic and an irreducible cubic are n =  1 and n =  6. The factorizations are x5 + x 1 = (x2  x+ 1)(x3  x2  1) x5 + x 6 = (x2  x+ 2)(x3  x2  x 3): Theorem 6.2. [20] The only integral n for which x5 x+n factors into the product of an irreducible quadratic and an irreducible cubic are n =  15; n =  22; 400; and n =  2; 759; 640. The factorizations are x5  x 15 = (x2  x+ 3)(x3  x2  2x 5) x5  x 22400 = (x2  12x+ 55)(x3  12x2 + 89x 408) x5  x 2759640 = (x2  12x+ 377)(x3  12x2  233x 7320): This type of result was extended by Spearman and Williams [24] to polynomials of the form x5  xm + n, for 1  m  4 and is outlined in the following theorems: 46Chapter 6. The Factorization of x5 + axm + 1 Theorem 6.3. [24] The only integers n for which x5 + x2 + n factors into the product of an irreducible quadratic and an irreducible cubic are n =  90; 4; 18; and 11466. The only integers nn for which x5 x2 +n factors into the product of an irreducible quadratic and an irreducible cubic are n =  11466; 18; 4; and 90. Theorem 6.4. [24] The only integers n for which x5  x3 + n factors into the product of an irreducible quadratic and an irreducible cubic are n =  8. There are no integers n for which x5 +x3 +n factors into the product of an irreducible quadratic and an irreducible cubic. Theorem 6.5. [24] The only integers n for which x5  x4 + n factors into the product of an irreducible quadratic and an irreducible cubic are n =  1. In addition, every quintic of the form x5 +  ( 1)kx4 +  F 2k 1F 4 k+1F 4 k+2 and x5 +  ( 1)kx4   F 4k 1F 4 kF 2 k+2 factors into the product of an irreducible quadratic and irreducible cubic where  =  1, k is an integer with k  2, and Fk is the kth Fibonacci number. The purpose of this chapter is to study in a similar manner to [20] and [24] the particular class of quintic polynomials f(x) given by f(x) = x5 + axm + 1; where a is a rational number and 1  m  4. This factorization is an analogous problem whose solution is accessible thanks to recent powerful methods. We shall determine those rational values of a for which f(x) is equal to the product of an irreducible quadratic and an irreducible cubic over Q. In doing so, we take full advantage of a recent theoretical result of Stoll, described in Section 4.5, on rational points on certain genus 2 curves. We also take advantage of the computer algebra system Magma [1]. We note that such a factorization for f(x) = x5 + axm + 1 immediately yields a factorization for the polynomial x5 + ax5 m + 1, by using the reverse poly- nomial x5f (1=x) : We state all of the factorizations of f(x) for m satisfying 1  m  4; for completeness. Finally, a factorization of x5 +axm+1 immediately yields a factorization for x5 + axm  1 if m is odd and x5  axm  1 if m is even by scaling with 476.1. Some Lemmas on Rational Points x!  x: Therefore we only treat the case where the constant term of f(x) is equal to positive 1. Our new result is summarized in the following theorem. Theorem 6.6. Let f(x) = x5 + axm + 1 where a is a rational number and m is an integer with 1  m  4. Then f(x) factors into the product of an irreducible quadratic and an irreducible cubic if and only if a and m assume the values listed in the following table. In each case the factorization is given. Table 6.1: The factorizations of the quintic trinomial x5 + axm + 1 for corresponding values of a and m (a;m) factorization of x5 + axm + 1 (1; 1) x5 + x+ 1 = (x2 + x+ 1)(x3  x2 + 1) ( 11=4; 1) x5  11=4x+ 1 = (x2 + x 1=2)(x3  x2 + 3=2x 2) (1; 4) x5 + x4 + 1 = (x2 + x+ 1)(x3  x+ 1) ( 11=4; 4) x5  11=4x4 + 1 = (x2  2x 2)(x3  3=4x2 + 1=2x 1=2) In Section 6.1 we give some preliminary results concerning rational points on speci c genus 2 curves. In Section 6.2 we give the proof of our theorem. 6.1 Some Lemmas on Rational Points In this section we give two lemmas which determine the set of rational points on two genus 2 curves. We refer the reader to Cassels and Flynn [6] as a reference for these types of algebraic curves. We will use a theorem of Stoll [25] which bounds the number of rational points on Ck : y2 = x5 + k where k is a tenth-power-free integer. These results are discussed in a paper by Bremner [3] as well. This theorem of Stoll states that if the rank of the Jacobian of this curve is at most one then the number of rational points is bounded above by 7 (this bound is achieved only for k = 324). Assuming that Ck has no rational point of the form (x; 0) for k 6= 324, the bound is 5. Magma will be used to determine the rank of the Jacobian for our curves using the command RankBounds. Finding the rank of the Jacobian is an extremely di cult and complicated problem to do by hand so we rely on computers to aid with the calculation. 486.2. Proof of Theorem Additionally we use the fact that if the Jacobian of a genus 2 curve has a rank of zero, then one can enumerate all points in the Jacobian and consequently  nd all rational points on Ck. The Magma command that does this is Chabauty0. Now we analyze the rational points on two relevant genus 2 curves. Lemma 6.1. The only  nite rational points on the genus 2 curve y2 = x5+4 are (0; 2); (2; 6). Proof. We observe the four given points (0; 2); (2; 6) on the curve. Magma con rms, using RankBounds that the rank of the Jacobian of this curve is equal to 1. Consequently the theorem of Stoll applies. The bound in this case, including the point at in nity, is 5 so that all of them are deter- mined. Lemma 6.2. The only  nite rational points on the genus 2 curve y2 = x5 + 256 are (0; 16). Proof. The RankBounds command in Magma con rms that the rank of the Jacobian of the given curve is equal to 0. Chabauty0 shows that the the  nite points on this curve are indeed those listed in the statement of this lemma. 6.2 Proof of Theorem As mentioned in the introduction we need only treat the cases m = 1; 2. Suppose that x5 + ax+ 1 is divisible by a quadratic polynomial x2 + ux+ v where a; u; v 2 Q. Division of these two polynomials leads to x5 + ax+ 1 = (x2 + ux+ v)(x3  x2u+ ( v + u2)x+ 2uv  u3) +(v2  3vu2 + a+ u4)x+ 1 + vu3  2uv2: (6.1) In equation (1) we equate the coe cients of x and 1 in the remainder to zero yieldng the pair of equations v2  3vu2 + a+ u4 = 0; 1 + vu3  2uv2 = 0: (6.2) The second equation in (6.2) shows that u 6= 0 and v 6= 0. Eliminating v from (6.2), using a resultant, produces the equation  11u5 + 1 + 4ua u10 + 3u6a+ 4u2a2 = 0: (6.3) 496.2. Proof of Theorem The discriminant of (6.3), viewed as quadratic equation in a is equal to 25u7(8 + u5): (6.4) If (6.3) has a rational root a then (6.4) must be equal to a square in Q, so that 25u7(8 + u5) = w2; (6.5) for some rational number w: Since u 6= 0; it follows from (6.5) that  2 u ; 2w 5u6  is a point on y2 = x5 + 4: (6.6) From Lemma 6.1, we know that x = 2; so that u = 1: Substituting u = 1 into (6.3) and factoring gives (4a+ 11)(a 1) = 0: The two choices of a = 1; a =  11=4 produce the factorizations given in the table in the theorem. In this case, suppose similarily to the  rst case, that f(x) = x5 +ax2 + 1 is divisible by a quadratic polynomial x2 +ux+v where a; u; v 2 Q. Division of these two polynomials leads to x5 + ax2 + 1 = (x2 + ux+ v)(x3  ux2 + ( v + u2)x+ a+ 2uv  u3) + (v2  ua 3vu2 + u4)x+ 1 + vu3  2uv2  va: (6.7) In equation (6.7) we equate the coe cients of x and 1 in the remainder to zero yielding the pair of equations v2  3vu2  ua+ u4 = 0; (6.8) 1 + vu3  2uv2  va = 0: If there exists a solution to this pair of equations with u = 0, then the  rst equation simpli es to v2 = 0 so that v = 0: It would then follow that the irreducible quadratic factor of f(x) is x2 +ux+v = x2 which violates irreducibility of the quadratic factor. Then since u 6= 0, we may solve the  rst equation in (6.8) to give a = u4 + v2  3u2v u (6.9) 506.2. Proof of Theorem Substituting the value of a given in (6.9) into the second equation in (6.8) yields u v3 + u2v2 u = 0 so that u v3 + u2v2 = 0: (6.10) The existence of a rational solution u to (6.10) requires the discriminant of this quadratic in u to be equal to a square in Q. That is 1 + 4v5 = z2 (6.11) for some rational number z: From (6.11) we see that (x; y) = (4v; 16z) is a rational point on the genus 2 curve y2 = x5 + 256: (6.12) Lemma 6.2 tells us that the only rational solution to (6.12) has x = 0 and since x = 4v we must have v = 0. However this contradicts the assumption that x2 + ux + v is irreducible over Q. Thus f(x) = x5 + ax2 + 1 cannot factor over Q as the product of an irreducible quadratic and an irreducible cubic.  51Chapter 7 Conclusion and Future Work 7.1 Conclusion The purpose of this thesis was to solve some Diophantine problems that required  nding integral or rational points on algebraic curves of genus 0, 1, and 2. We started in Chapters 2, 3, and 4 by outlining the methods used to  nd the points on these curves. We hope that interested readers could use this thesis as a guide to solving algebraic curves and that it would be layed out in a structured, easy to follow manner. In Chapter 5, we solved a Diophantine system related to a problem on cubic  elds based o a family of cubics of the form x3 3x b3. We provided six solutions that yielded  ve di erent values of b for which a given formula was not the fundamental unit of the cubic  eld. Finding these solutions required adapting an algorithm to  nd integral points on genus 0 curves. In Chapter 6, we then considered the factorization of a family of quintic trinomials x5 + axm + 1. This problem took advantage of a recent result from Stoll [25] that bounded the number of rational points on a family of genus 2 curves. Without this result, the problem would most likely remain unsolved as we would not have been able to determine if a computer search yielded all of the points on the curve. We concluded that the family of quintic trinomials only factored for two di erent values of a and gave the factorizations. As shown in this thesis,  nding points on algebraic curves are necessary for solving di erent problems of varying topics of mathematics. An algebraic number theory problem on cubic  elds and a polynomial theory problem on factorization are only a small fraction of problems that require  nding solutions to algebraic curves. The di culty of  nding the points on these curves can range from following a simple algorithm to using complicated methods and nontrivial theory. However, the advancements of computers and theory are making the more di cult problems easier to solve. These problems are extremely interesting and are becoming more preva- lent in mathematics, which make them all the more desirable to research. 527.2. Future Work 7.2 Future Work Possible research that would extend past this thesis are: 1. Can rational points be found on curves of genus 3 or higher? 2. Can the idea of the Jacobian be extended to curves of genus 3 or higher? 3. Are there other unsolved problems that require  nding rational points on genus 2 curves that could take advantage of recent results? 4. Let f(x) be a polynomial with rational coe cient. We say that f(x) is Q derived if f(x) and all of its derivatives have rational roots. An example of such a polynomial is f(x) = x2(x 308)(x 360): An extended study of these polynomials is found in [5]. The current state and description of the research into these problems is given next. Over the rational numbers Q, these polynomials are almost com- pletely classi ed with the notable exception of whether or not there exists a rational derived quartic with distinct roots. In each case of the classi cation, a parametrizing algebraic curve is produced, whose rational points yield the desired polynomials. A variety of algebraic curves is studied including curves of genus 0, 1 and 2. Over algebraic number  elds, much less is known about derived polynomials. The above mentioned reference [5] considers some cases where the coe - cients belong to quadratic  elds. The paper [26] also considers the quadratic  eld case. The main idea on this problem would be to concentrate on complet- ing the quadratic case, with many separate classi cations depending on suitable algebraic curves. For example even treating the polynomial g(x) = x2(x 1)2(x a) and trying to determine the rational numbers a for which g(x) is k derived for some quadratic  eld k requires the study of the rational points on the genus 2 curve y2 = x6  75x4 + 600x2 + 50000: This is a di cult problem by itself. Returning to the rational derived quartic for a moment, introductory calculations show that it 537.2. Future Work may be possible to reduce its solution to an in nite family of genus 2 curves. If any of these curves have a rational point, the problem will be solved in the a rmative. 54Bibliography [1] W. Bosma, J. Cannon, and C. Playoust, The Magma Algebra System I: The User Language, J. Symbolic Comput. 24 (1997), 235-265. [2] B. Buchberger, G.E. Collins, and R. Loos, eds. Computer Algebra: Sym- bolic & Algebraic Computation. New York. Springer-Verlag, 1982 [3] A. Bremner, On the equation Y 2 = X5 + k, Exp. Math. 17:3 (2008), 371-374. [4] N. Bruin, Chabauty methods using elliptic curves, J. reine angew. Math. 562 (2003), 27-49. [5] R. Buchholz and J. MacDougall, When Newton met Diophantus: a study of rational-derived polynomials and their extension to quadratic  elds, J. Number Theory 81:2 (2000), 210{233. [6] J. W. S. Cassels and E. V. Flynn, Prolegomena To A Middlebrow Arith- metic of Curves of Genus 2, Cambridge University Press, 1996. [7] C. Chabauty, Sur les points rationnels des courbes alg ebriques de genre sup erieur  a l’unit e, C. R. Acad. Sci. Paris 212 (1941), 882-885 (French). [8] R. Chapman. Algebraic Number Theory - Summary of Notes, Home page. U. of Exeter. 3 May. 2000 <http://empslocal.ex.ac.uk/people/sta /rjchapma/courses/ant99/notes5.dvi>. [9] H. Cohen. A Course in Computational Algebraic Number Theory, Springer{Verlag, Germany, 1993. [10] J. B. Fraleigh, A First Course in Abstract Algebra, Addison Wesley, Boston, 2003. [11] T. Freiberg, An Outline of the Flynn-Chabauty Method for Curves of Genus 2 with an Application to the Curve C : y2 = x5  2x, Master’s Thesis, University of Queensland, 2006. 55Bibliography [12] M. van Hoeij, Computing Parametrizations of Rational Algebraic Curves, Conf. ISSAC94, July 2022, 1994, Oxford, 1994. [13] M. van Hoeij, Rational parametrizations of algebraic curves using a canonical divisor, J. Symbolic Comput. 23 (1997), 209-227. [14] M. van Hoeij, An algorithm for computing the Weierstrass normal form, ISSAC’95 Proceedings, p. 90-95 (1995). [15] K. Kaneko, Integral bases and fundamental units of certain cubic num- ber  elds, SUT J. Math. 39:2 (2003), 117-124. [16] P. D. Lee and B. K. Spearman, A Diophantine system and a problem on cubic  elds, International Mathematical Forum 6:3 (2011), 141-146. [17] P. D. Lee and B. K. Spearman, The Factorization of x5 + axm + 1, Scientiae Mathematicae Japonicae 73:2-3 (2011), 171-174: e-2011, 77- 80. [18] D. Poulakis and E. Voskos, On the practical solution of genus zero Diophantine equations, J. Symbolic Comput. 30 (2000), 573-582. [19] D. Poulakis and E. Voskos, Solving genus zero Diophantine equations with at most two in nite valuations, J. Symbolic Comput. 33 (2002), 479-491. [20] S. Rabinowitz, The factorization of x5  x+ n, Math. Mag. 61 (1988), 191-193. [21] K. H. Rosen, Elementary Number Theory and its Applications, Addison Wesley, Boston, 2005. [22] A. Schinzel, On reducible trinomials, Dissertationes Math. (Rozprawy Mat.) 329 (1993), 1-83. [23] J. H. Silverman and J. Tate, Rational Points on Elliptic Curves, Springer, New York, 1992. [24] B. K. Spearman and K. S. Williams, The factorization of x5  xa + n, Fibonacci Quart. 36 (1998), 158-170. [25] M. Stoll, On the number of rational squares at  xed distance from a  fth power, Acta Arith. 125:1 (2006), 79-88. 56Bibliography [26] R.J. Stroeker, On Q derived polynomials, Rocky Mountain J. Math. 36:5 (2006), 1705-1713. [27] A. Thue, ber Annherungswerte algebraischer Zahlen, Journal f ur die reine und angewandte Mathematik 135 (1909), 284-305. [28] N. Tzanakis and B. M. M. De Weger, On the practical solution of the Thue equation, J. Number Theory 31 (1989), 99-132. [29] ||||{. Algebraic Curve. Wikipedia. 4 September. 2010. <http://en.wikipedia.org/wiki/Algebraic curve>. 57Appendix A Magma Code This appendix includes relevant Magma code for chapters 4, 5 and 6. A.1 Chapter 4 Magma Code A.1.1 Example 4.3 P<x>:=PolynomialRing(Rationals()); C:=HyperellipticCurve(x^6+4); C; Hyperelliptic Curve de ned by y^2 = x^6 + 4 over Rational Field Points(C:Bound:=1000); f@ (1 : -1 : 0), (1 : 1 : 0), (0 : -2 : 1), (0 : 2 : 1) @g J:=Jacobian(C); RankBounds(J); 0 , 0 \\ gives both a lower and upper bound of 0 Chabauty0(J); f@ (1 : -1 : 0), (1 : 1 : 0), (0 : -2 : 1), (0 : 2 : 1) @g A.1.2 Example 4.4 P<x>:=PolynomialRing(Rationals()); C:=HyperellipticCurve(x^6+x^2+2; C; Hyperelliptic Curve de ned by y^2 = x^6+x^2 + 2 over Rational Field ptsC:=Points(C:Bound:=1000); ptsC; 58A.2. Chapter 5 Magma Code f@ (1 : -1 : 0), (1 : 1 : 0), (-1 : -2 : 1), (-1 : 2 : 1), (1 : -2 : 1), (1 : 2 : 1) @g J:=Jacobian(C); RankBounds(J); 1 , 1 \\ gives both a lower and upper bound of 1 PT1:=J![ptsC[3], ptsC[1]]; \\ The point f(-1,-2), 1 g Order(PT1); 8 PT2:=J![ptsC[5], ptsC[1]]; \\ The point f(1,-2), 1 g Order(PT2); *no output* \\ therefore, PT2 has in nite order Chabauty(PT2); f (1 : -2 : 1), (-1 : -2 : 1), (1 : 2 : 1), (1 : -1 : 0), (1 : 1 : 0), (-1 : 2 : 1) g A.2 Chapter 5 Magma Code A.2.1 Lemma 5.4 IntegralQuarticPoints([12,0,0,0,1]); [ [ 0, 1 ] ] IntegralQuarticPoints([12,0,0,0,-3]); [ [ 1,3 ] [-1,3] ] IntegralQuarticPoints([12,0,0,0,-8]); [ [ 1,2 ] [-1,2] ] IntegralQuarticPoints([12,0,0,0,24]); [ [ 1,6 ] [-1,6] ] 59A.3. Chapter 6 Magma Code A.2.2 Proof of Theorem 5.2 P<x>:=PolynomialRing(Rationals()); E:=EllipticCurve([0,109418989131512359209/4]); E; Elliptic Curve de ned by y^2 = x^3 + 109418989131512359209/4 over Rational Field Rank(E); 0 A.3 Chapter 6 Magma Code A.3.1 Lemma 6.1 P<x>:=PolynomialRing(Rationals()); C:=HyperellipticCurve(x^5+4); C; Hyperelliptic Curve de ned by y^2 = x^5 + 4 over Rational Field RationalPoints(C:Bound:=1000); f@ (1 : 0 : 0), (0 : -2 : 1), (0 : 2 : 1), (2 : -6 : 1), (2 : 6 : 1) @g J:=Jacobian(C); RankBounds(J); 1 , 1 \\ gives both a lower and upper bound of 1 A.3.2 Lemma 6.2 P<x>:=PolynomialRing(Rationals()); C:=HyperellipticCurve(x^5+256); C; Hyperelliptic Curve de ned by y^2 = x^5 + 256 over Rational Field RationalPoints(C:Bound:=1000); f@ (1 : 0 : 0), (0 : -16 : 1), (0 : 16 : 1) @g J:=Jacobian(C); RankBounds(J); 60A.3. Chapter 6 Magma Code 0 , 0 \\ gives both a lower and upper bound of 0 Chabauty0(J); f@ (1 : 0 : 0), (0 : -16 : 1), (0 : 16 : 1) @g 61

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