UBC Theses and Dissertations

UBC Theses Logo

UBC Theses and Dissertations

An algebraic view of discrete geometry De Zeeuw, Frank 2011

You don't seem to have a PDF reader installed, try download the pdf

Item Metadata

Download

Media
ubc_2011_fall_dezeeuw_frank.pdf [ 316.84kB ]
Metadata
JSON: 1.0072373.json
JSON-LD: 1.0072373+ld.json
RDF/XML (Pretty): 1.0072373.xml
RDF/JSON: 1.0072373+rdf.json
Turtle: 1.0072373+rdf-turtle.txt
N-Triples: 1.0072373+rdf-ntriples.txt
Original Record: 1.0072373 +original-record.json
Full Text
1.0072373.txt
Citation
1.0072373.ris

Full Text

An Algebraic View of Discrete Geometry by Frank de Zeeuw Doctoraal (B.Sc. and M.Sc.) in Mathematics, Rijksuniversiteit Groningen, 2006 B.Sc. in Arti al Intelligence, Rijksuniversiteit Groningen, 2006 Master of Advanced Study (MASt) in Mathematics, University of Cambridge, 2007 A THESIS SUBMITTED IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE DEGREE OF DOCTOR OF PHILOSOPHY in The Faculty of Graduate Studies (Mathematics) THE UNIVERSITY OF BRITISH COLUMBIA (Vancouver) October 2011 c Frank de Zeeuw 2011Abstract This thesis includes three papers and one expository chapter as background for one of the papers. These papers have in common that they combine algebra with discrete geometry, mostly by using algebraic tools to prove statements from discrete geometry. Algebraic curves and number theory also recur throughout the proofs and results. In Chapter 1, we will detail these common threads. In Chapter 2, we prove that an in nite set of points in R2 such that all pairwise distances are rational cannot be contained in an algebraic curve, except if that curve is a line or a circle, in which case at most 4 resp. 3 points of the set can be outside the line or circle. In the proof we use the classi cation of curves by their genus, and Faltings’ Theorem. In Chapter 3, we informally present an elementary method for computing the genus of a planar algebraic curve, illustrating some of the techniques in Chapter 2. In Chapter 4, we prove a bound on the number of unit distances that can occur between points of a  nite set in R2, under the restriction that the line segments corresponding to these distances make a rational angle with the horizontal axis. In the proof we use graph theory and an algebraic theorem of Mann. In Chapter 5, we give an upper bound on the length of a simultaneous arithmetic progression (a two-dimensional generalization of an arithmetic progression) on an elliptic curve, as well as for more general curves. We give a simple proof using a theorem of Jarn  k, and another proof using the Crossing Inequality and some bounds from elementary algebraic geometry, which gives better explicit bounds. iiPreface This thesis is based on 3 papers, of which two have been published and one has been accepted for publication. The results are reproduced in this thesis with permission from the coauthors and journals. All work involved in these publications was shared in equal parts between the authors. Chapter 2 was published as: J. Solymosi and F. de Zeeuw, On a Question of Erd}os and Ulam, Discrete and Computational Geometry 43 (2010), 393{401. Chapter 3 is a writeup for a course taught at UBC by Jozsef Solymosi on Additive Combinatorics. Chapter 4 was accepted for publication: R. Schwartz, J. Solymosi, and F. de Zeeuw, Rational distances with rational angles, accepted for publication in Mathematika; arXiv:1008.3671v2, [math.CO], 2011. Chapter 5 was published as: R. Schwartz, J. Solymosi, and F. de Zeeuw, Simultaneous arithmetic progressions on algebraic curves, International Journal of Number Theory 7 (2011), 921{931. The mathematical content of these papers is unchanged in this thesis, though the exposition has undergone some changes. iiiTable of Contents Abstract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ii Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iii Table of Contents . . . . . . . . . . . . . . . . . . . . . . . . . . . . iv Acknowledgements . . . . . . . . . . . . . . . . . . . . . . . . . . . vi 1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1.1 Outline . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1.2 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1.3 Tools . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 2 Rational sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 2.1 Background . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 2.2 Main results . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 2.3 Proof of Theorem 2.2.1 . . . . . . . . . . . . . . . . . . . . . 11 2.4 Proof of Theorem 2.2.2 . . . . . . . . . . . . . . . . . . . . . 17 3 Computing the genus . . . . . . . . . . . . . . . . . . . . . . . 18 3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 3.2 Basics from algebraic geometry . . . . . . . . . . . . . . . . . 22 3.3 Computing the genus, Part I . . . . . . . . . . . . . . . . . . 23 3.4 Blowing up singularities . . . . . . . . . . . . . . . . . . . . . 25 3.5 Computing the genus, Part II . . . . . . . . . . . . . . . . . 28 4 Rational distances with rational angles . . . . . . . . . . . . 31 4.1 Background . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 4.2 Main results and proof sketch . . . . . . . . . . . . . . . . . 32 4.3 Mann’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 33 4.4 Proofs of main theorems . . . . . . . . . . . . . . . . . . . . 36 4.5 Lower bounds . . . . . . . . . . . . . . . . . . . . . . . . . . 40 ivTable of Contents 5 Simultaneous arithmetic progressions . . . . . . . . . . . . . 42 5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . 42 5.2 First proof for algebraic curves . . . . . . . . . . . . . . . . . 43 5.3 Second proof for algebraic curves . . . . . . . . . . . . . . . . 45 5.4 SAPs on elliptic curves . . . . . . . . . . . . . . . . . . . . . 48 Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51 vAcknowledgements I would like to thank my advisor Jozsef Solymosi, fellow student and co- author Ryan Schwartz, my supervisory committee, other UBC faculty, sta and students, my family and friends, and Julia, for all their guidance, coop- eration, support and entertainment. I would also like to thank Kalle Karu, Jirka Matou sek, Trevor Wooley and three anonymous referees for help with the individual papers. viChapter 1 Introduction 1.1 Outline This thesis is based on 3 papers. What these papers have in common is that they combine discrete geometry with algebra (or algebraic geometry), with a touch of number theory. In Chapters 2 and 4, the problems are from discrete geometry (recognizable from the fact that Erd}os coined them), and the tools are, more or less, from algebra. In both cases, we translate the geometric problem into algebraic equations, and apply some algebraic result, in the form of a bound on the number of solutions to the equations. Chapter 5 is di erent: its problem is algebraic (a statement about alge- braic curves), and our main tool is discrete (graph theory). In Chapter 3, we present an elementary method for computing the genus of an algebraic curve, as background for Chapter 2. All three papers also involve some number theory: the problems revolve around the rationality or integrality of numbers, and the tools partly belong to number theory (in particular the theorems of Faltings, Mann, and Jarn  k). In the rest of this introduction, we describe the three problems that form the basis of this thesis in Section 1.2, and in Section 1.3 we list the tools that we will apply to these problems. 1.2 Problems 1.2.1 Rational distance sets We de ne a rational distance set to be a set S  R2 such that the distance between any two points in S is a rational number. We are interested in the existence of in nite rational distance sets on algebraic curves. This notion was introduced by Erd}os in 1945, when he proved with Anning [4] that any in nite set with all distances integral must be contained in a line. For a rational distance set, the same is not true, because one can  nd in nite rational distance sets contained in a circle. In fact, rational distance sets can be dense on lines or circles. In 1960 Ulam wrote [53] that 11.2. Problems he believed that an in nite rational distance set cannot be dense in the plane, but no progress has been made on this question to date. Erd}os conjectured [23] that an in nite rational distance set must be very restricted, but wrote that it was probably a very deep problem. We prove the following. Theorem 1.2.1. A rational distance set S can have only  nitely many points in common with an algebraic curve de ned over R, unless the curve has a component which is a line or a circle. If S has in nitely many points on a line or on a circle, then all but at most 4 resp. 3 points of S are on the line or on the circle. Our proof will use the genus of algebraic curves and Faltings’ Theorem (see Section 1.3.2). Roughly, the idea of the proof is that a curve with an in nite rational distance set on it lets us construct a new curve with higher genus, and with a set of rational points on it corresponding to the points in the rational distance set. Faltings’ Theorem then tells us that this set of points must be  nite, hence the rational distance set could not have been in nite. 1.2.2 Unit distances A famous problem of Erd}os concerns the maximal number of unit distances among n points in the plane; we will denote this number by u(n). Erd}os [18] showed that u(n) > n1+ c log logn . The best known upper bound is u(n) < cn4=3, due to Spencer, Szemer edi and Trotter [51]. We will approach this problem under the restriction that we only consider unit distances that have rational angle, by which we mean that the line through the pair of points makes a rational angle in degrees with the x-axis (or equivalently, its angle in radians divided by  is rational). This leads to the following result. Theorem 1.2.2. Suppose we have n points in R2. Then the number of pairs of points with unit distance and rational angle is at most n1+6= p logn. In the proof we view each unit distance line segment between two of the points as a complex number, which is then a root of unity thanks to the rational angle condition. Given a large set with many unit distances with rational angles, we deduce linear equations with many solutions in roots of unity. Comparing with a uniform upper bound on the number of such solutions, derived from Mann’s Theorem (see Section 1.3.1), then leads to our bound. In fact, this proof works for more general situations, leading to the fol- lowing three theorems. 21.2. Problems Theorem 1.2.3. Suppose we have n points in R2, no three of which are on a line. Then the number of pairs of points with rational distance and rational angle is at most n1+6= p logn. Theorem 1.2.4. Suppose we have n points in R2, with no more than n on a line, where 0 <  < 1=2. Then the number of pairs of points with rational distance and rational angle is at most n1+ +6= p logn. Theorem 1.2.5. Suppose we have n points in R2, with no more than n on a line, where 1=2    1. Then the number of pairs of points with rational distance and rational angle is at most 4n1+ . 1.2.3 Simultaneous arithmetic progressions on curves There are interesting problems in number theory related to arithmetic pro- gressions on elliptic curves (see Section 1.3.2 and Chapter 3 for more about elliptic curves). An example of such an open problem is to  nd the maxi- mum number (if it exists) of rational points on an elliptic curve such that their x-coordinates are in arithmetic progression [8]; these are for instance related to 3 3 magic squares with square entries [7]. In [30], Garcia-Selfa and Tornero looked instead for \simultaneous" arith- metic progressions on elliptic curves, which are de ned as follows. De nition. A simultaneous arithmetic progression (SAP) of length k con- sists of points (xi; y (i)) in R2, where xi = a1 + i  d1 and yi = a2 + i  d2 for i = 0; 1; : : : ; k 1 are arithmetic progressions, and  is a permutation of the numbers 0; 1; : : : ; k  1. Garcia-Selfa and Tornero gave examples of elliptic curves over Q that contain an SAP of length 6, and showed that no SAP of length 7 exists on an elliptic curve. However, it did not seem computationally feasible to extend their methods to longer SAPs. The  nal open problem they suggested in their paper is  nding a universal bound for the length of SAP’s on elliptic curves over Q. We prove the following (unlikely to be optimal) bound to this e ect. Theorem 1.2.6. Consider an elliptic curve over a sub eld of R given by y2 + axy+ by = x3 + cx2 + dx+ e. Then the length of an SAP on this curve is  4319. In our proof of this bound we construct a graph out of translates of the curve in a grid. Then we derive an algebraic upper bound on the number of intersections, from B ezout’s Theorem (see Section 1.3.3), and compare it 31.3. Tools with a combinatorial lower bound on the number of crossings in the graph, obtained using the Crossing Inequality (see Section 1.3.3). We then extend this proof to algebraic curves of any degree. It turns out that the structure of an SAP is not crucial; our proof only uses the fact that an SAP has k points on a k  k grid. This led to the following more general theorem. Theorem 1.2.7. If f is a real plane algebraic curve of degree d  2 with no linear factor, and f contains at least k points from a k  k grid, then there are absolute constants c and m such that k  cdm. The proof outlined above gives m = 7 for an arbitrary curve, and m = 4 when the curve is irreducible. We also give an alternative proof of this theorem, using Jarn  k’s Theorem (see Section 1.3.3), which gives an upper bound on the number of integer grid points on a curve. This is a more direct proof, but the resulting bound is weaker: m = 9 in general and m = 6 for irreducible curves. 1.3 Tools 1.3.1 Linear equations The simplest type of equation that we might derive from our geometric problems are linear equations kX i=1 aixi = 0; ai 2 C: Of course, without restrictions on the values that the variables can take, there is not much to say about the solutions. Under suitable restrictions, however, it is possible to obtain bounds on the number of solutions, that are even uniform, in the sense that they do not depend on the coe cients. The grandest example of this is the following theorem of Evertse, Schlickewei and Schmidt [26], where the solutions are restricted to a multiplicative subgroup of C of  nite rank. It is a corollary of the Subspace Theorem proved by Schmidt [45]. Theorem 1.3.1. Let  be a subgroup of C of  nite rank r, and let ai 2 C . Then the equation Pk i=1 aixi = 1 has at most S(k; r) = e (6k)5k(r+1) non- degenerate solutions. The proof of this theorem is very di cult, and we did not use the theorem in this thesis. But we mention it because it did lead us to consider a special 41.3. Tools case which has an elementary proof. We take the group in C to be  nite, which means it must consist of roots of unity. In that case, the following result was proved by Mann in 1965 [41]. Theorem 1.3.2 (Mann). The number of k-tuples f igki=1 of roots of unity, with  1 = 1, that satisfy Pk i=1  i = 0, but no shorter such equation, is bounded by a constant C(k). Because the proof of this theorem uses only basic algebra, we were able to modify it to obtain an extension which exactly suits our needs in the proof of Theorem 1.2.2. 1.3.2 Algebraic curves We will also encounter equations of higher degree, which will lead us to consider algebraic curves. For these we will not have the same kind of explicit bound as for linear equations, but much can still be said about their solution sets, in particular for points with rational coordinates. To explain this, we need to outline the classi cation of algebraic curves by genus. By a curve Cf we mean the zero set in R2 of a polynomial f 2 R[x; y]: Cf = Cf (R) = f(x; y) 2 R2 : f(x; y) = 0g: The curve is irreducible if f is irreducible. In our problems we will also consider the set of rational points Cf (Q) = f(x; y) 2 Q2 : f(x; y) = 0g of a curve Cf . We will frequently denote the curve by its polynomial f . Note that in this thesis we are restricting ourselves to plane curves, except in Section 2.3, where we use curves in R3. The genus g is an important invariant of an irreducible curve, and we have devoted Chapter 3 to it. It has several equivalent de nitions, but they are not elementary, and it would take us too far a eld to introduce them here. Computationally, we can use the following formula in terms of the degree d of the curve: g = (d 1)(d 2) 2  X P  P ; where the sum ranges over the singularities P , points of the curve where both derivatives of f vanish (possibly including projective points). The  P  0 are nonnegative invariants associated to singularities. In Chapter 3 we will specify what the  P are, and how one can compute them in an elementary 51.3. Tools way. But we can already see that the genus of a nonsingular curve is simply given by g = (d 1)(d 2)2 . It turns out that the genus of a curve C is related to the structure of the set of rational points C(Q) on the curve, as follows: g = 0 () the curve is rational g = 1 () the curve is an elliptic curve g  2 =) C(Q) is  nite In Chapter 3 we will go into more detail about this classi cation. For now, the important thing is that curves with g = 0 or g = 1 have a well- understood structure (we will explain what rational and elliptic curves are in Chapter 3; we will meet elliptic curves again in Chapter 5). Note that the linear equations from Section 1.3.1 are curves with genus 0, but there we consider solutions taken from a group of  nite rank, rather than Q, which allows us to say more about their size. When g  2, the structure of C(Q) is not so easy to understand, but thanks to the following theorem of Faltings [27], we do know that C(Q) is  nite: Theorem 1.3.3 (Faltings). An irreducible curve of genus g  2, de ned over a number  eld, contains only  nitely many rational points. The proof is very complicated, but the statement is relatively easy to apply, and we will use it as a black box. A related conjecture that should be mentioned here (although we have not used it in this thesis), is the Uniformity Conjecture from [11]. It says that for every number  eld K and integer g  2 there exists an integer B(K; g) such that no smooth curve of genus g de ned over K has more than B(K; g) rational points. If this is true, it could perhaps be used to improve our  niteness result in Chapter 2. 61.3. Tools 1.3.3 Points on curves The tools that we will use in Chapter 5 can be roughly characterized as deal- ing with points on curves: intersection points of algebraic curves, crossings of planar graphs, and integer grid points on convex curves. The  rst of these is a standard theorem from algebraic geometry, B ezout’s Theorem. It will provide the upper bound that we use in the proof of The- orem 1.2.6. Theorem 1.3.4 (B ezout). Suppose F and G are curves of degree m and n. If F and G do not have a common factor, then they intersect in at most mn points. The second tool is a well-known theorem from graph theory, bounding the crossing number of a graph. We will apply it to obtain the lower bound used in the proof of Theorem 1.2.6. Given a simple graph G, we de ne the crossing number cr(G) as the minimum number of pairs of crossing edges in a planar drawing of G. Theorem 1.3.5 (Crossing Inequality). Suppose G is a simple graph with n vertices and e edges. If e > 7:5n then cr(G)  e3 33:75n2 : The third tool is a theorem of Jarn  k [35], which will give us an alternative proof of Theorem 1.2.7. Theorem 1.3.6 (Jarn  k). If f is a strictly convex di erentiable curve of length N , then the number of integer points on f is less than  N2=3 for some constant  . 7Chapter 2 Rational sets 2.1 Background Recall that in the introduction we de ned a rational distance set to be a set S  R2 such that the distance between any two elements is a rational number. For briefness, in this chapter we will refer to rational distance sets as rational sets. We are interested in the existence of in nite rational sets on plane algebraic curves. On any line, one can easily  nd an in nite rational set that is in fact dense. It is also possible to  nd an everywhere dense rational subset of the unit circle (see Section 2.2). On the other hand, it is not known if there is a rational set with 8 points in general position, i.e. no 3 on a line, no 4 on a circle. In 1945, Anning and Erd}os proved that any in nite integral set, i.e. where all distances are integers, must be contained in a line [4]. Problems related to rational and integral sets became one of Erd}os’ favorite subjects in combinatorial geometry [19, 20, 21, 22, 23, 25]. In 1945, when Ulam heard Erd}os’ simple proof [17] of his theorem with Anning, he said that he believed there is no everywhere dense rational set in the plane (see Problem III.5. in [53], and also [24]). Erd}os conjectured that an in nite rational set must be very restricted, but that it was probably a very deep problem [23, 24]. Not much progress has been made on Ulam’s question. There were attempts to  nd rational sets on parabolas [10, 12], and there were some results on integral sets, in particular bounds were found on the diameter of integral sets [33, 49]. Recently Kreisel and Kurz found an integral set with 7 points in general position [38]. In this thesis, we prove that lines and circles are the only irreducible algebraic curves that contain in nite rational sets. Our main tool is Faltings’ Theorem [27]. We will also show that if a rational set S has in nitely many points on a line or on a circle, then all but at most 4 resp. 3 points of S are on the line or on the circle. This answers questions of Guy, Problem D20 in [31], and Pach, Section 5.11 in [6]. 82.2. Main results 2.2 Main results 2.2.1 Main Theorems Our main result is the following. Theorem 2.2.1. A rational set in the plane can have only  nitely many points in common with an algebraic curve de ned over R, unless the curve has a component which is a line or a circle. The two special cases, line and circle, are treated in the next theorem. Theorem 2.2.2. If a rational set S has in nitely many points on a line or on a circle, then all but at most 4 resp. 3 points of S are on the line or on the circle. These numbers are best possible: there are in nite rational sets with all but 4 points on a line, and there are in nite rational sets with all but 3 points on a circle. A construction of Hu [34, 43] gives an in nite rational set with all but 4 points on a line. The circle case follows from the line case by applying an inversion (see Section 2.2.2) with rational radius and center one of the 4 points not on the line. This will give an in nite rational set on the circle, with 3 points o it (the fourth point, the one that was used as center, is lost because the inversion sends it to in nity). We can formulate our Theorem 2.2.1 in a di erent way by using the term curve-general position: we say that a point set S of R2 is in curve-general position if no algebraic curve of degree d contains more than d(d + 3)=2 points of S. Note that d(d+3)=2 is the number of points in general position that determine a unique curve of degree d (see Section 5.2 in [29]). Corollary 2.2.3. If S is an in nite rational set in general position, then there is an in nite S0  S such that S0 is in curve-general position. Proof. We will construct S0 with a greedy approach: we consecutively add as many points of S as possible while maintaining curve-general position. Let S5 consist of any  ve points in S, and let T5 be the set of  nitely many points on the unique conic through those 5 points. Continue recursively; at step n, add a point from SnTn 1 to Sn 1 to get Sn. For each d such that d(d + 3)=2  n, let Tn be the union of Tn 1 and the set of points of S that are on a curve of degree d through any d(d+ 3)=2 points in Sn. Since each Tn is  nite by Theorem 2.2.1, we can always add another point. Then the in nite union of the sets Sn is an in nite subset of S with the required property. 92.2. Main results 2.2.2 Two lemmas Rationality of distances in R2 is clearly preserved by translations, rotations and uniform scaling ((x; y) 7! ( x;  y) with  2 Q). More surprisingly, rational sets are preserved under certain central inversions. This will be an important tool in our proof below. Lemma 2.2.4. If we apply inversion to a rational set S, with center a point x 2 S and rational radius, then the image of Snfxg is a rational set. Proof. We may assume the center to be the origin and the radius to be 1. The properties of inversion are most easily seen in complex notation, where the map is z 7! 1=z (ignoring a re ection). Suppose we have two points z1, z2 with rational distances jz1j, jz2j from the origin, and with jz2  z1j rational. Then the distance between their images is also rational:     1 z1  1 z2     =     z2  z1 z1z2     = jz2  z1j jz1jjz2j : A priori, points in a rational set could take any form. But after moving two of the points to  xed rational points by translating, rotating, and scaling, the points in the set are almost rational. This fact is well-known among researchers working with integer sets, and as far as we know, it was  rst proved by Kemnitz [36]. Note that the lemma implies that any curve of degree d that contains at least d(d + 3)=2 points from a rational set containing (0; 0) and (1; 0) is de ned over Q( p k), for some k 2 N. Lemma 2.2.5. For any rational set S containing (0; 0) and (1; 0) there is a square-free integer k such that any point in S is of the form (r1; r2 p k); r1; r2 2 Q: Proof. Suppose we have a rational set containing (0; 0) and (1; 0), and a third point (x; y). Then x2 + y2 2 Q and (x  1)2 + y2 = x2 + y2  2x + 1 2 Q, hence by subtracting we have  2x+ 1 2 Q, so x 2 Q. Combining that with x2 + y2 2 Q gives y = s p k for s 2 Q and a square-free integer k. It remains to show that k is the same for di erent points. Let (r1; s1 p k) and (r2; s2 p l) be two points of our rational set. Then their distance, (r1  r2) 2 + (s1 p k  s2 p l)2 = (r1  r2) 2 + s21k + s 2 2l  2s1s2 p kl; is rational, which implies that p kl 2 Q. If both k and l are square-free integers, that is only possible if k = l. 102.3. Proof of Theorem 2.2.1 2.3 Proof of Theorem 2.2.1 2.3.1 Outline Our proof relies on the following theorem of Faltings [27]. Theorem 2.3.1 (Faltings). An irreducible curve of genus  2, de ned over a number  eld K, contains only  nitely many K-rational points. See Chapter 3 or [48] for de nitions. As in the Introduction, in this chapter by curve (de ned over a  eld K  R) we usually mean the zero set in R2 of a polynomial in two variables with coe cients from K. But when we consider the genus of a curve, we are actually talking about the projective variety de ned by the polynomial. In the proof we will also brie y encounter curves in R3; for de nitions we again refer to [48], especially for their genus, which is not covered in Chapter 3. Before we start to outline the proof, note that Theorem 2.2.1 allows re- ducible curves, but if the statement holds for irreducible curves, it follows for reducible curves. So in our proof we need only consider irreducible curves, and in the rest of this chapter we will assume that all curves are irreducible. For a curve C of genus  2, de ned over R, we can prove our theorem very quickly. Suppose we have an in nite rational set S on such a C. We can move two points in S to (0; 0) and (0; 1), so that by Lemma 2.2.5 the elements of S are of the form (r1; r2 p k). Then by the remark right before Lemma 2.2.5, the curve is de ned over the number  eld Q( p k). By Faltings’ Theorem, S must be  nite. In the following sections we will treat the cases with genus 0 or 1. The main idea of the proof is that if we have an in nite rational set S on a curve C1 of genus 0 or 1, which is not a line or a circle, then we can construct a curve C2 of genus  2 with in nitely many rational points, contradicting Faltings’ Theorem. But the cases split into two groups, for which the details of the proofs are very di erent. For a curve C1 with genus 1, or with genus 0 and degree  4, a point (r1; r2 p k) from the rational set on C1 will give a point (r1; r2 p k; r3) on a curve C2 in R3. Then using the Riemann-Hurwitz formula we can show that the genus of C2 is  2. For curves with genus 0 and degree < 4, this does not work, and we need a di erent approach. First we apply a central inversion: by Lemma 2.2.4 this preserves the rational set, but it transforms most curves into curves with degree  4, so that the previous proof applies. There are two exceptions: 112.3. Proof of Theorem 2.2.1  rst of all, lines and circles (as we would hope!), and second, a set of curves with d = 3. The curves in this second group have such speci c polynomials that we can use them to explicitly construct hyperelliptic curves with in-  nitely many rational points (similar to the constructions in Section 3.1.2). That  nishes the proof. 2.3.2 Curves of genus 1 Let C1 : f(x; y) = 0 be an irreducible algebraic curve of genus g1 = 1, and suppose that there is an in nite set S on C1 with pairwise rational distances. Assume that the points (0; 0) and (1; 0) are on C1 and in S, and that (0; 0) is not a singularity of C1. Below we will be allowed to make any other assumptions on C1 that we can achieve by translating, rotating or scaling it, as long as we also satisfy the assumptions above. In particular, we can use any of the in nitely many rotations about the origin that put a di erent point of S on the x-axis. We wish to show that the intersection curve C2 of the surfaces X : f(x; y) = 0; Y : x2 + y2 = z2; has genus g2  2. Graphically, Y is a cone around the z-axis, and X is a "cylinder" above (and below) C1, if C1 is viewed as a curve in the z = 0 plane. The intersection curve C2 then consists of the points on the cone that are directly above or below C1. De ne the map  : C2 ! C1 by (x; y; z) 7! (x; y), i.e. the restriction to C2 of the projection from the cone Y to the z = 0 plane. The preimage of a point (x; y) consists of two points (x; y; p x2 + y2), except when x2 + y2 = 0, which in C2 happens on the two lines x+ iy = 0 and x iy = 0. We can determine (or at least bound from below) the genus of C2 using the Riemann-Hurwitz formula applied to  (see [48], Ch. I, for the formula and for the de nitions of the degree of a map and the rami cation index eP of  at a point), 2g2  2  deg   (2g1  2) + X P2C2 (eP  1): This formula is usually stated with equality for smooth curves, but we are allowing C1 and C2 to have singularities. To justify this, observe that the map  corresponds to a map e : eC1 ! eC2 between the normalizations of the curves, for which Riemann-Hurwitz holds. The normalizations have the 122.3. Proof of Theorem 2.2.1 same genera as the original curves, and e has the same degree. Furthermore a rami cation point of  away from any singularities gives a rami cation point of e . It is enough for us to have this inequality, but there could be more rami cation points for e , above where the singularities used to be. Applying this formula with g1 = 1, deg  = 2, we have g2  1 + 1 2 X P2C2 (eP  1); so to get g2  2, we only need to show that  contains some point P with eP  2, i.e. a rami cation point. The potential rami cation points are above the intersection points of C1 with the lines x  iy = 0, of which there are 2d by B ezout’s Theorem (Theorem 5.3.4), counting with multiplicities. Such an intersection point P can only fail to have a rami cation point above it if the curve has a singularity at P , or if the curve is tangent to the line there. We will show that there are only  nitely many lines through the origin on which one of those two things happens. Then certainly one of the in nitely many rotations of C1 that we allowed above will give an intersection point of C1 with x iy = 0 that has a rami cation point above it. The intersection of a line y = ax with f(x; y) = 0 is given by pa(x) = f(x; ax) = 0, and if the point of intersection is a singularity or a point of tangency, then pa(x) has a multiple root. We can detect such multiple roots by taking the discriminant of pa(x), which will be a polynomial in a that vanishes if and only if pa(x) has a multiple root. Hence for all but  nitely many values of a the line y = ax has d simple intersection points with f(x; y) = 0. So indeed there is an allowed rotation after which  is certain to have a rami cation point. 2.3.3 Curves of genus 0, d  4 Let C1 : f(x; y) = 0 be an algebraic curve of genus g1 = 0, and again assume that it passes through the origin, but does not have a singularity there. The Riemann-Hurwitz formula with the same map  as above gives g2   1 + 1 2 X P2C2 (eP  1); so to get g2  2 it su ces to show that there are at least 5 rami cation points. As above, we can ensure that the lines x  iy each have d simple points of intersection. Discounting the intersection point of the two lines, 132.3. Proof of Theorem 2.2.1 this gives 2d  2 rami cation points. Hence if the degree of f is d  4 we are done. 2.3.4 Curves of genus 0, d = 2; 3. Let f(x; y) = 0 be an irreducible algebraic curve of genus g1 = 0 and degree d = 2 or d = 3, and again assume that it passes through the origin, but does not have a singularity there. Consider applying inversion with the origin as center to the curve. This is a birational transformation, so does not change the genus. Therefore, when inversion increases the degree of f to above 4, we are done by the previous section. Algebraically, inversion in the circle around the origin with radius 1 is given by (x; y) 7!  x x2 + y2 ; y x2 + y2  ; and since this map is its own inverse, the curve f(x; y) = 0 is sent to the curve C3 : (x 2 + y2)k  f  x x2 + y2 ; y x2 + y2  = 0; where k  d is the lowest integer that makes this a polynomial. This curve is irreducible if and only if the original curve is irreducible. Since f does not have a singularity at the origin, it has a linear term ax + by with a, b not both zero. After inversion this gives a highest degree term (ax+ by)(x2 + y2)k 1: If d = 3 and k = 3, then the curve C3 has degree 2k  1 = 5, and we are done. This does not work if d = 3 and k = 2, which happens if x2 + y2 divides the leading terms of f . We will treat this case in a completely di erent way in the next section. If d = 2, then applying inversion will give a curve of degree 3, unless its leading terms are a multiple of x2 + y2. But that exactly means that the curve is a circle! So we treat this case by going to the d = 3 case, leaving only circles behind, as we should expect. Note that the curves with d = 3 resulting from this inversion have x2 + y2 dividing their leading terms, so are of the type that is handled in the next section. 142.3. Proof of Theorem 2.2.1 2.3.5 Curves of genus 0, d = 3, k = 2. Since f has genus 0 and degree 3, it must have a singularity (see Chapter 3). The singularity need not be in our rational set, but it is always a rational point, so we can move it to the origin, while maintaining the almost-rational form of the points in our rational set. Then f must have the form (ax+ by)(x2 + y2) + cx2 + dy2 + exy: Note that this is exactly what we get if we apply inversion to a quadratic that is not a circle and goes through the origin. In fact, we can ensure that (1; 0) is on the curve again, so that a+ c = 0. Then if we divide by c, f is of the form ( x+ by)(x2 + y2) + x2 + dy2 + exy: We can parametrize this curve using lines x = ty, giving the parametrization y(t) = t2 + et+ d (t b)(t2 + 1) =: p(t) q(t) ; x(t) = t  y(t): If we let tj be a value of t that gives one of the points from our rational set, it follows that for in nitely many t, (y(t) y(tj)) 2 +(x(t) y(tj)) 2 =  p(t) q(t)  p(tj) q(tj)  2 +  t  p(t) q(t)  tj  p(tj) q(tj)  2 is a square. Then we can multiply by q(t)2q(tj)2 to get in nitely many squares of the form (p(t)q(tj) p(tj)q(t)) 2 + (tp(t)q(tj) tjp(tj)q(t)) 2 : This polynomial has degree 6 in t. It has a factor (t  tj)2, and a factor t2 + 1, since taking t =  i gives (using q( i) = 0) (p( i)q(tj)) 2 + ( i  p( i)q(tj)) 2 = 0: Factoring these out, we get a quadratic polynomial Qj(t) in t. Its leading coe cient is (t2j + 1)((1 + (e+ b) 2)t2j + 2(bd b+ de)tj + d 2 + b2): and its constant term is (t2j + 1)((d 2 + b2)t2j + 2(b 2e+ db d2b)tj + b 2e2 + b2d2 + d2 + 2ebd); 152.3. Proof of Theorem 2.2.1 These polynomials in tj are not identically zero (if b and d were both 0, then f would be reducible), hence we can pick many tj so that they are not zero. Then in turn Qj(t) is a proper quadratic polynomial, and since it is essentially a distance function in the real plane, it cannot have real roots other than the t = tj that we have already factored out, so it has two distinct imaginary roots. Therefore an in nite rational set gives in nitely many solutions to equa- tions of the form z2j = (t 2 + 1)  Qj(t): Multiplying three of these together, and moving (t2 + 1)2 into the square on the left, we get in nitely many solutions to z2 = (t2 + 1)Q1(t)Q2(t)Q3(t): If there are no multiple roots on the right, then this is a hyperelliptic curve of degree 8, so it has genus 3, hence cannot have in nitely many solutions, a contradiction. The one thing we need to check is that we can choose the tj so that the Qj do not have roots in common. We need some notation: write Qj(t) = c2(tj)t 2 + c1(tj)t+ c0(tj); where c2(tj) = (1 + (e+ b) 2)t2j + 2(bd+ de b)tj + d 2 + b2 c1(tj) = 2(bd+ de b)t 2 j + 2(b 2 + d2  bed bd be d)tj + 2(bd+ b 2e bd2) c0(tj) = (d 2 + b2)t2j + 2(b 2e+ db d2b)tj + b 2e2 + b2d2 + d2 + 2ebd: Suppose that for in nitely many tj the polynomial Qj(t) has the same roots x1 and x2. Then for each of those in nitely many tj we have c1(tj) =  (x1 + x2)  c2(tj); c0(tj) = x1  x2  c2(tj); which implies that the corresponding coe cients of these polynomials in tj are equal. If we look at the coe cients of the linear tj terms, we see that  x1  x2 = 2(b2 + d2  bed bd be d) 2(bd b+ de) =  b be+ d d2 bd+ de b ; x1  x2 = 2(b2e+ db d2b) 2(bd+ de b) = b  be+ d d2 bd+ de b : Here we can read o that the roots are x1 = b and x2 = be+d d 2 bd+de b , which is a contradiction, since the roots had to be imaginary (or since plugging them into Qj(t) does not always give zero). 162.4. Proof of Theorem 2.2.2 2.4 Proof of Theorem 2.2.2 We will prove that if a rational set has in nitely many points on a line, then it can have at most 4 points o the line. The corresponding statement for 3 points o a circle then follows by applying an inversion. More precisely, suppose we have a rational set S with in nitely many points on a circle C and at least 4 points o that circle. Assume that the origin is one of the points in S \ C, and apply inversion with the origin as center, and with some rational radius. That turns C into a line L, and we get a rational set with in nitely many points on L, and 4 other points. Moreover, the new origin can be added to S, so that we get 5 points o the line, contradicting what we will prove below. To see that the new origin has rational distance to all points in S, observe that in complex notation the distances jzj to the old origin were rational for all z 2 S, and that the distances to the new origin are 1=jzj. To prove the statement for a line, our main tool will again be Faltings’ Theorem, but now applied to a hyperelliptic curve y2 = 6Y i=1 (x  i); which has genus 2 if and only if the  i are distinct (see Chapter 3). Suppose we have a rational set S with in nitely many points on a line, say the x-axis, and 5 or more points o that line. Then we can assume that 3 of those points are above the x-axis, say (a1; b1), (a2; b2), and (a3; b3). Note that we are taking 3 points on one side of the line, because we want to avoid having one point a re ection of another. If we had, say, (a2; b2) = (a1; b1), the argument below would break down. Take a point (x; 0) of S on the x-axis, with x 6= 0; a1; a2; a3. Then (x a1) 2 + b21; (x a2) 2 + b22; and (x a3) 2 + b23 are rational squares, so that we get a rational point (x; y) on the curve y2 = ((x a1) 2 + b21)((x a2) 2 + b22)((x a2) 2 + b22): This is a curve of genus 2, since the roots on the right hand side distinct: they are x = ai q  b2i for i = 1; 2; 3, which are distinct by the assumptions on the points (ai; bi). Therefore the curve has genus 2, and cannot contain in nitely many rational points, contradicting the fact that S has in nitely many points on the line. 17Chapter 3 Computing the genus 3.1 Introduction In this chapter we will informally describe an elementary method for com- puting the genus from the equation for any planar algebraic curve with- out multiple components. This can often be done with the simple formula g = (d 1)(d 2)=2, but this only holds for nonsingular curves. For singular curves, the corresponding formula is g = (d 1)(d 2) 2  X P  P : The trouble here is in the delta-invariant  P : it is zero for nonsingular points, but it is not always easy to determine for singular points. This formula is mentioned in many books on algebraic geometry (for instance [32], p. 393), but usually no elementary method for computing  P in hard cases is given. One exception is the wonderful book [1] (the formula is on p. 148), which was the main source for this chapter. A speci c target we aim for is to compute the genus of hyperelliptic curves (curves of the form y2 = f(x), where f has no multiple roots) in this way, which we do in Section 3.5.2; these are important in Chapter 2 (in particular in Sections 2.3.5 and 2.4) and in the examples from additive combinatorics that we will give in Section 3.1.2. It should be noted that the exposition here is intentionally informal, so de nitions and claims may not be fully rigorous. In particular, we will not formally de ne the genus, although the formula above could be taken as a de nition, together with the method that we will describe for computing the  P . We will also not give proofs for general claims, but we will work out the examples in detail. To be more precise would take up far more space, and be redundant; our goal here is to give a swift and practical introduction (the chapter is based on a writeup for a graduate course, entitled "Genus for Dummies"). 183.1. Introduction 3.1.1 De nitions To repeat, by a curve we will mean the zero set in R2 of a polynomial f 2 R[x; y]: Cf = Cf (R) = f(x; y) 2 R2 : f(x; y) = 0g: For number-theoretic questions we mostly consider the set of rational points Cf (Q) = f(x; y) 2 Q2 : f(x; y) = 0g of a curve Cf . In this chapter, we will assume all curves to be irreducible. An important invariant of a curve is its genus, a nonnegative integer g. For nonsingular curves, one can think of it as the number of holes in the surface that we get when we view the curve as a subset of R4. There are many ways to de ne and interpret the genus for possibly singular curves, for instance using di erential forms, the Hilbert polynomial, or cohomology. We will try to give a view of the genus of a (possibly singular) curve that is more elementary and doesn’t use any of these di cult words. What’s more, this will give an algorithm for computing the genus given any ir- reducible polynomial that is straightforward and usually doable on paper. In particular, this approach works for singular curves without leaving the plane, whereas with the more advanced methods one will have to  nd an equivalent nonsingular curve, which may live in higher dimensions. On the other hand, the more advanced methods will often be more insightful, and generalize better to non-planar curves and higher-dimensional varieties. For number theory, the interest in the genus lies in the following classi-  cation: g = 0 () rational curves g = 1 () elliptic curves g  2 =) C(Q) is  nite So when a curve has genus 0, it is rational, i.e. birationally equivalent to a projective line, hence we can parametrize it by rational functions, which makes the curve easy to understand, if not boring. Curves with genus 1 are elliptic curves, which have a group structure on C(Q), and with that we can understand its set of points very well. Elliptic curves occur as objects of study in Chapter 5, but in this thesis we will not use any elliptic curve techniques (except that in the next section we mention their outcome as an illustration). When the genus is 2 or higher, it is not so easy to understand the rational points, but one does know that there are only  nitely many, which is what Faltings’ Theorem (Theorem 2.3.1) says. 193.1. Introduction An important class of curves is that of hyperelliptic curves, which have the form y2 = ’(x); ’ 2 Q[x] no multiple roots and monic: Their form makes them convenient to handle, and there is an easy formula for their genus: deg’ = 2k + 1 or 2k + 2 =) g = k: We used this formula in Sections 2.3.5 and 2.4. We will prove it in Section 3.5.2. 3.1.2 An application: progressions in powers Here is an example of a problem from additive combinatorics where being able to compute the genus of singular curves comes in handy. These are not new results or techniques (see for instance [13] and [14] for a similar approach). We will not fully prove the conclusions, especially since we do not want to include elliptic curve methods in this thesis; we will merely indicate where elliptic curve methods could be applied. But these calculations should illustrate why this approach is useful for discrete problems; in fact they were the inspiration for our work in Chapter 2 (in particular see Sections 2.3.5 and 2.4). Suppose we have an arithmetic progression of length 3 (a 3AP) that consists of squares (in the integers). Then we can write them in the form x, x+ d, x+ 2d, with d 6= 0, so that since they are squares, multiplying them all together gives a square, and we have a solution to y2 = x(x+ d)(x+ 2d): As we saw in Section 3.3.1, this curve has genus g = 1, so it is an elliptic curve, and we could  nd out a lot about its solutions. This is merely a  rst idea, and not too fruitful, but there are many variants. For instance, given a 4-term AP of squares x2, x2 + d, x2 + 2d, x2 + 3d, we have a solution to y2 = (x2 + d)(x2 + 2d)(x2 + 3d); which is a hyperelliptic curve with genus 2 (as we will prove in 3.5.2), hence we immediately know that there are only  nitely many 4APs with given common di erence in the squares. 203.1. Introduction That’s still not too exciting, because with a bit more ingenuity we could show that there are no 4APs whatsoever in the squares, as was  rst proven by Euler (see [15]) using very di erent methods. Let x21, x 2 2, x 2 3, x 2 4 form a 4AP. That means that x22  x 2 1 = x 2 3  x 2 2 = x 2 4  x 2 3 ) x21 = 2x 2 2  x 2 3; x 2 4 = 2x 2 3  x 2 2 )  x1 x3  2 = 2  x2 x3  2  1;  x4 x3  2 = 2  x2 x3  2 ; hence we have a rational point on the curve y2 = (2x2  1)(2 x2): This is an elliptic curve (as we will see in Section 3.5.2; note that the formula (d 1)(d 2)=2 would give the wrong genus). Basic elliptic curve methods would tell us that this curve has only a few rational points, and these only give trivial 4APs, proving that there are no 4APs in the squares (see [14], where this is done via a di erent curve). We can also apply this approach to progressions of higher powers; we only need to look at 3APs. If xk1, x k 2 = x k 1 + d, x k 3 = x k 1 + 2d is a 3AP of kth powers, then we have 2xk2  2x k 1 = 2d = x k 3  x k 1 ) 2x k 2 = x k 1 + x k 3 ) 2 =  x1 x2  k +  x3 x2  k ; so we get a rational point on the curve xk + yk = 2: We’ll see in Section 3.3.1 that for k = 3, this curve has genus 1, so we could determine the solutions. For k  4, we have genus  2, hence there are only  nitely many 3APs of such powers. That also implies there are no arbitrarily long progressions for higher powers. Actually, using Wiles-type methods it has been shown [16] that xk+yk = 2zk has no nontrivial integer solutions for k  3, which implies that there are no 3APs of higher powers at all. 213.2. Basics from algebraic geometry 3.2 Basics from algebraic geometry We will informally introduce a few notions from algebraic geometry that are necessary here. We follow the notation of [48]. 3.2.1 Projective curves The most natural setting for algebraic curves is actually not the a ne plane C2, but the projective plane P2(C), which is de ned by taking C3, identifying two points if they are scalar multiples of each other, and removing (0; 0; 0). As notation for an equivalence class of points we will write [x : y : z], where (x; y; z) is some representative from the class. The points [x : y : 1] will be considered as the points from the "a ne" plane, and the points [x : 1 : 0] and [1 : 0 : 0] make up the projective line at in nity. Given a curve f 2 C[x; y] with deg f = d, we can view it projectively by homogenizing the polynomial: F (X;Y; Z) = Zd  f(X=Z; Y=Z); then the curve is given projectively as the points [X : Y : Z] 2 P2(C) where F (X;Y; Z) = 0. If we want to take a closer look at how f behaves around a point at in nity like P = [0 : 1 : 0], we put g(x; z) = F (x; 1; z), and consider g = 0 around P = (0; 0) in the xz-plane. 3.2.2 Morphisms and rational maps We’ll need to know what a rational map is to be able to make sense of the important statement "genus is a birational invariant of a curve". We say that a map f : C1 ! C2; (x; y) 7! (f1(x; y); f2(x; y)) is a morphism if f1 and f2 are polynomials from Q[x; y], and an isomorphism if it has an inverse which is also a morphism. Basically, such a map is a rational map if f1 and f2 are not polynomials but rational functions from Q(x; y), though that doesn’t quite make sense because a rational function might have poles. So a rational map is only a partial map, de ned on all but  nitely many points. A birational equivalence is then a rational map with an inverse that’s also a rational map. Note that we are leaving out a number of subtleties, e.g. that because these maps are only de ned at points satisfying some equation, what looks like a rational function might actually be a polynomial. The thing that 223.3. Computing the genus, Part I matters here is that birational equivalences do not change the genus. This can be seen pretty easily from the di erential form concept of genus, but is harder to prove in an elementary way. 3.2.3 Singularities The a ne curve given by f has a singularity at (a; b) if f(a; b) = 0 and the partial derivatives fx(a; b) = 0 and fy(a; b) = 0. Projectively, F has a singularity at [A : B : C] if F (A;B;C) = 0, FX(A;B;C) = 0, FY (A;B;C) = 0, and FZ(A;B;C) = 0 (this might seem like one more condition, but a homogeneous polynomial satis es XFX + Y FY + ZFZ = deg(F )  F ). For example, the curves y2 = x3 and y2 = x2(x+ 1) both have a singularity at (0; 0). A hyperelliptic curve y2 = ’(x) has no a ne singularity, since it would have to have y = 0 and x a root of ’0, but by de nition ’ and ’0 have no root in common. However, if d = deg’ > 3, it does have a singularity when Z = 0: since its homogenization is F = Y 2Zd 2 Zd’(X=Z), we get X = 0 from FX = 0, and then FY = FZ = 0 follow, hence it has the singularity [0 : 1 : 0] at in nity. 3.2.4 Multiplicity of a singularity With a singularity we can associate a number m that says how "bad" it is. Assume the singularity is (0; 0) and write f = P aijxiyj . Then having a singularity at (0; 0) is the same as having a00 = a01 = a10 = 0. The multiplicity is de ned by m = minfk : i+ j = k; aij 6= 0g; i.e. the lowest total order of a nonzero term. For instance, x4  y4 + 3y7x2 has a singularity with multiplicity m = 4 at (0; 0). 3.3 Computing the genus, Part I 3.3.1 Easy Cases Now we can explain how to compute the genus in the easier cases. First of all, if the curve is nonsingular, there is a simple formula for the genus in terms of the degree d of the curve: g = (d 1)(d 2) 2 : 233.3. Computing the genus, Part I So for instance a curve of the form y2 = x3 + ax+ b, with no multiple roots on the right, has no singularities, hence has genus (3 1)(2 1)2 = 1, so is an elliptic curve. As a quick consequence, we see that there are no nonsingular plane curves of genus 2, since (d 1)(d 2)=2 never equals 2; the same holds for other values not assumed by (d 1)(d 2)=2. For the curves xk + yk = 2 that we obtained for 3APs of kth powers in 3.1.2, there are no singularities, so we can apply the formula above to see that the genus is (k  1)(k  2)=2, which gives the values that we used in 3.1.2. We can extend this formula to curves with only singularities that have all tangents distinct. More precisely, if we move the singularity to the origin and it has multiplicity m, and we write f = X i+j=m aijx iyj + X i+j>m aijx iyj = mY k=1 ( kx+  ky) + X i+j>m aijx iyj ; then the tangents at the origin are the lines  kx+ ky = 0. If these tangents are all distinct lines, then the genus of f is given by g = (d 1)(d 2) 2  X P mP (mP  1) 2 : Here mP is the multiplicity of the curve at the point P , which is 1 when P is on the curve but not a singularity, and 0 when P is not on the curve, so this sum really only runs over the singularities. For instance, if f = y2  x3  x2, then its only singularity is the origin, where it has distinct tangents x  y = 0 and m = 2, hence its genus is (3 1)(2 1) 2  2 1 2 = 0. 3.3.2 Not all cases are easy Here is an example of a curve for which the above formula fails: y2 = x4  x5: It has a singularity with m = 2 at the origin (but not with distinct tangents), and it has a singularity at [0 : 1 : 0], where its equation is z3 = x4z  x5, so m = 3. Then the formula above would give g = 4 32  2 1 2  3 2 2 = 6 1 3 = 2. However, we can see that the genus must be 0: we can rewrite the equation to y = x2 p 1 x, hence x = cos2  , y = sin  cos4  is a parametrization. Since we can rationally parametrize cos and sin by cos  = 1 t 2 1+t2 , sin  = 2t 1+t2 , we can rationally parametrize this curve, hence its genus must be zero. We will see in the next section how to get this genus right. 243.4. Blowing up singularities 3.4 Blowing up singularities 3.4.1 Local blowup To extend the formula above, we need to "blow up" the singularities. The idea is to untangle the singularity by separating the branches of the curve according to their tangents. For simplicity, we will illustrate this process for the example C : y2 = x2(x+ 1); by blowing up its singularity at the origin. We will end up with a birational equivalence C 00 ! C from a nonsingular curve C 00, but we will need several steps to construct it. Fortunately, afterwards we will see that the equation for C 00 is quite easy to calculate. De ne the "blowup surface" B = f(x; y; t) : y = xtg  C3; with the corresponding projection  : B ! C2, (x; y; t) 7! (x; y). Then the inverse images of points under this map are pretty simple:   1(x; y) =  f(x; y; yx)g if x 6= 0 f(0; 0; t) : t 2 Cg if x = 0 : So  is bijective away from the origin, and collapses the line L = f(0; 0; t) : t 2 Cg to the origin. The inverse image of C under  is   1(C) = f(x; y; t) : y = xt; y2 = x2(x+ 1)g = f(x; xt; t) : x2t2 = x2(x+ 1)g = f(x; xt; t) : x2 = 0g [ f(x; xt; t) : t2 = x+ 1g = L [ C 0; where C 0 = f(x; y; t) : y = xt; t2 = x+ 1g is a new curve in C3. Now the restricted map  : C 0 ! C is a birational equivalence, which is bijective away from the line L, and sends the two points (0; 0; 1) on L to the origin. We say that these points "lie above" the singularity (0; 0) of C. 253.4. Blowing up singularities We can simplify this map a bit. We project C 0 onto a curve in xt-space with  : C 0 ! C 00, (x; y; t) 7! (x; t), where C 00 = f(x; t) : t2 = x+ 1g: This projection has a well de ned inverse   1 : (x; t) 7! (x; xt; t). Then the blowup of C, with center the origin, is the map     1 : C 00 ! C; (x; t) 7! (x; xt): This blowup is a birational map from a nonsingular curve to our original curve, which is bijective away from the origin, and the inverse image of the singularity at the origin consists of the two nonsingular points (0; 1) and (0; 1). If we took a closer look, we would see that these points correspond to the two tangent directions the original curve had there; the blowup has "untangled" the singularity. In practice, we do not have to consider all these maps explicitly: given the curve y2 = x2(x+ 1), we plug in y = xt to get t2x2 = x2(x+ 1); then cancel out x2 (which corresponds to removing the line x2 = 0) to get t2 = x + 1, which is the blown-up curve. Since this curve is nonsingular, its genus is (2 1)(2 2) 2 = 0, hence our original curve also has genus 0. To show this for another example, let’s blow up the singularity at the origin of the cusp y2 = x3. We plug in y = xt to get x2t2 = x3, then remove the new line x2 = 0 to obtain the blowup t2 = x, which is nonsingular and birational to y2 = x3. Since t2 = x has genus 0, the genus of y2 = x3 must also be 0. In general, the points lying above the blown-up singularity are the points of C 00 that are on the t-axis. Note that it is possible for these to be singularities as well. For example, if we start with y2 = x5, then blowing up gives t2 = x3, which still has a singularity at the origin. But another blowup (with for instance t = xu) then gives the birational nonsingular curve u2 = x with genus 0, hence y2 = x5 has genus 0 as well. More generally, for y2 = x2k+1, blowing up k times will show that the genus is 0. In fact, one can show that any singularity can be resolved in  nitely many steps this way ([1], p.137). Finally, note that the curve C should not have a vertical tangent, since this would correspond to a point on C 00 with t =1. But this can be easily avoided by applying a rational rotation. 263.4. Blowing up singularities 3.4.2 Resolutions The above kind of blowup is not quite satisfactory, because it does not behave well on the line at in nity; in fact, it might create new singularities. For instance, the curve x4 + y3 + y2 x2 has a singularity at the origin, and no others, including at in nity. Blowing up like above gives x2 +xt3 + t2 1, which is nonsingular in the a ne plane, but has a new singularity at in nity, at [1 : 0 : 0]. This can be avoided, by doing projective blowups and gluing them to- gether. We won’t do this, but the result is a global blowup: a projective birational map which creates no new singularities, and "improves" at least one of the singularities, by which we mean that on an a ne neighborhood around that singularity it is just a local blowup like above. The main result is then that any curve C has a resolution C  C1      Cn; where each map is a global blowup, and Cn is nonsingular ([1], p.137, or [32], p. 391). So one way to  nd the genus of any curve would be to  nd this resolution, and then just compute (d 1)(d 2)2 for Cn. However,  nding that resolution is not so easy. Luckily there is a shortcut: locally, on an a ne neighborhood of a singularity, these global blowups act just like local blowups. So if we can  nd the "contribution" to the genus of every such local blowup, putting all of those together we can compute the genus of C, without having to do all the global blowups. In other words, for each singularity P , we repeatedly apply local blowups, until we have resolved it, i.e. the singularity is replaced by nonsingular points. We refer to the singularities Q that appear during this process as lying above P . Note that this needn’t be a linear process like the resolution above: blowing up a singularity might split it into two new singularities (or more), and then one has to continue by blowing up each of those. But again one can prove that this process will terminate in  nitely many steps. For each such singularity Q lying above P , take its multiplicity mQ, and compute  P = X Q mQ(mQ  1) 2 : As it turns out, this quantity is what we have to subtract from the degree 273.5. Computing the genus, Part II formula to get the correct genus: g = (d 1)(d 2) 2  X P  P : A di erent way of looking at it is to think of these Q as points "in nitely near" P , and then the genus formula is g = (d 1)(d 2) 2  X Q mQ(mQ  1) 2 ; where the sum runs over all in nitely near points Q. 3.5 Computing the genus, Part II 3.5.1 Examples Let’s consider y2 = x4 x5 from Section 3.3.2 again. We showed that it has a singularity with m = 2 at the origin, and one with m = 3 at [0 : 1 : 0]. First let’s blow up the origin. Putting y = xt gives x2t2 = x4 x5, hence the new curve is t2 = x2 x3. This has a singularity at the origin with m = 2 (and we should check that there are no others on the t-axis), and there the tangent lines are t x = 0, so we don’t have to blow up again, because there wouldn’t be any new singularities. So the origin has  = 2 12 + 2 1 2 = 2. For the singularity at in nity, the local equation is z3 = x4z  x5, and blowing up with z = xt gives t3 = x2t x2, which has a singularity at (0; 0) with m = 2. Blowing up again with t = xs, we get xs3 = xs  1, which has no singularities on the s-axis, so we are done. Hence the singularity at in nity has  = 3 22 + 2 1 2 = 4. Now the genus is (5 1)(5 2) 2  2 4 = 0; as we expected. Apparently, when we miscalculated this genus to be 2 in Section 3.3.2, we missed out on the two in nitely near points, for which we should have subtracted 1 each. 283.5. Computing the genus, Part II For another example, consider the four-leaved clover (x2 + y2)3 = x2y2. It has a singularity at the origin, and at [1 : i : 0] and [1 :  i : 0]. Those two at in nity can easily be seen to have  = 1. The singularity with m = 4 at the origin is more complicated. Since it has vertical and horizontal tangents, we will have to rotate the curve before we can blow up. Put for instance x = u + v, y = u  v, so that x2 + y2 = 2(u2 + v2), and xy = u2  v2, so that the curve becomes 8(u2 + v2)3 = (u2  v2)2: Now its tangents at the origin are u v = 0, each double. We blow up with v = tu, which gives the new curve 2u2(1 + t2)3 = (1 t2)2: We only care about its points on the t-axis, which are at t =  1, both singular with m = 2. To analyze them we need to move them to the origin, so put for instance t = w + 1, then we get 2u2(w2 + 2w + 2)3 = w2(w + 2)2: This has lowest terms 16u2  4w2, so here the curve has distinct tangents, and we do not need to blow up any further. Same for t =  1. Hence the origin of the clover has  = 4 32 + 2 1 2 + 2 1 2 = 8. Hence the curve has genus (6 1)(6 2)=2 1 1 8 = 0. That makes sense, as this is a polar curve (2r = sin 2 ), hence has a parametrization. Note that both these examples happen to have genus 0, which is of course not representative. But they show how complicated singularities can get, and we can check our genus calculation by  nding a parametrization. 293.5. Computing the genus, Part II 3.5.2 Hyperelliptic curves Finally we will prove the following formula for the genus g of a hyperelliptic curve y2 = ’(x), where ’ has no multiple roots and is monic: deg(’) = 2k + 1 or 2k + 2 =) g = k: We only need to consider d  3. First we assume that deg’ is even, so that we can write the equation as y2  x2k+2  h(x) = 0, deg h = l  2k + 1. Then homogenizing and setting y = 1 gives z2k  x2k+2  z  H(x; z) = 0; where H(x; z) = z2k+1 lh(x=z) is homogeneous and degH = 2k + 1. Then the singularity is at (0; 0) and has multiplicity m = 2k. Applying the blowup z = xt and factoring out x2k gives t2k  x2  x2t  H0 = 0; where H0 = H(x; xt)=x is a polynomial since H is homogeneous. The only singularity on the t-axis is at (0; 0), and has m = 2. Next we apply x = st and factor out t2 to get t2k 2  s2  s2t  H0 = 0: Again we have a singularity with m = 2. Repeating this, we see that we get k in nitely near singularities with m = 2. Hence the genus is g = (2k + 1)2k 2  2k(2k  1) 2  k  2  1 2 = k ((2k + 1) (2k  1) 1) = k; as desired. For deg’ = 2k + 1 the calculation is similar. From z2k 1  x2k+1  z  H(x; z) = 0 with m = 2k  1 we get t2k 1  x2  x2t  H0 = 0 with a singularity with m = 2. Repeating, we get k  1 such singularities with m = 2, so that the genus is g = 2k(2k  1) 2  (2k  1)(2k  2) 2  (k  1)  2  1 2 = (2k  1) (k  (k  1)) (k  1) = k: 30Chapter 4 Rational distances with rational angles 4.1 Background A famous problem of Erd}os from 1946 [18] concerns the maximum number of unit distances among n points in the plane; we will denote this number by u(n). He showed that u(n) > n1+c= log logn, using a p n p n piece of a scaled integer lattice, and conjectured that this was the true magnitude. The best known upper bound is u(n) < cn4=3,  rst proved by Spencer, Szemer edi and Trotter in 1984 [51]. This bound has several other proofs, the simplest of which was the proof by Sz ekely [52], using a lower bound for the crossing number of graphs (the very same that we use in Section 5.3). A recent result of Matou sek [42] shows that the number of unit distances is bounded above by cn log n log log n for most norms. As a general reference for work done on the unit distances problem, see [6]. We will show that the upper bound n1+6= p logn holds if we only consider unit distances that have rational angle, by which we mean that the line through the pair of points makes a rational angle in degrees with the x-axis (or equivalently, its angle in radians, divided by  , is rational). Under this restriction, we can use an algebraic theorem of Henry Mann [41], Theorem 4.3.1, to get a uniform bound on the number of paths between two  xed vertices in the unit distance graph, which will lead to a contradiction if there are too many unit distances with rational angle between the points. In fact, our proof also shows that the bound n1+6= p logn holds for the number of rational distances with rational angles, if we have no three points on a line. The lower bound, n1+c= log logn, of Erd}os does not apply in this case as we are restricted to rational angles. But a construction of Erd}os and Purdy [25] gives a superlinear lower bound for unit (and hence rational) distances with rational angles (see Section 4.5). If instead we allow up to n points on a line where 1=2    1, the number of rational distances with rational angles is bounded by 4n1+ . This 314.2. Main results and proof sketch bound is tight up to a constant factor with the lower bound now coming from an n1   n square grid. If we allow up to n points on a line where 0 <  < 1=2, the number of rational distances with rational angles is bounded above by n1+ +6= p logn. We get a lower bound of cn1+ from n1  horizontal lines each containing n rational points so that no three points on di erent lines are collinear (see Section 4.5). In Section 4.2 we will state our main results and give an outline of the proof. Section 4.3 contains the algebraic tools that we will use, including, for completeness, a proof of Mann’s Theorem. In Section 4.4 we use the bounds obtained from Mann’s Theorem and some graph theory to prove our main results. In Section 4.5 we give lower bounds for the main results. 4.2 Main results and proof sketch We will say that a pair of points in R2 has rational angle if the line segment between them, viewed as a complex number z = re i , has  2 Q. Our  rst result is the following. Theorem 4.2.1. Given n points in R2, the number of pairs of points with unit distance and rational angle is at most n1+6= p logn. Roughly speaking, our proof goes as follows. Given n points in the plane, we construct a graph with the points as vertices, and as edges the unit line segments that have rational angle. We can represent these unit line segments as complex numbers, which must be roots of unity because of the rational angle condition. Then if this graph has many edges, it should have many cycles of a given length k, and each such cycle would give a solution to the equation kX i=1  i = 0; with  i a root of unity. Using Mann’s Theorem, we could give a uniform bound on the number of such solutions, depending only on k (under the non-degeneracy condition that no subsum vanishes). If the number of non- degenerate cycles goes to in nity with n, this would give a contradiction. However, dealing with cycles of arbitrary length is not so easy, so instead in our proof we count non-degenerate paths of length k between two  xed vertices, which correspond to solutions of the equation kX i=1  i = a; 324.3. Mann’s Theorem where a 2 C; a 6= 0, corresponds to the line segment between the two points. We have extended Mann’s Theorem to this type of equation, giving a similar upper bound and proving our result. In fact, in our proof it turns out that it is not necessary for the lengths to be 1, but that they only need to be rational. This is because our extension of Mann’s Theorem also works for equations of the type kX i=1 ai i = a; where ai 2 Q and a 2 C; a 6= 0. This leads to the following results (the  rst supersedes 4.2.1; we have stated both because 4.2.1 answers our initial question). Theorem 4.2.2. Suppose we have n points in R2, no three of which are on a line. Then the number of pairs of points with rational distance and rational angle is at most n1+6= p logn. Theorem 4.2.3. Suppose we have n points in R2, with no more than n on a line, where 0 <  < 1=2. Then the number of pairs of points with rational distance and rational angle is at most n1+ +6= p logn. Theorem 4.2.4. Suppose we have n points in R2, with no more than n on a line, where 1=2    1. Then the number of pairs of points with rational distance and rational angle is at most 4n1+ . The constants 6 and 4 in these theorems are not optimal, but they are the smallest integers that followed directly from our proof. 4.3 Mann’s Theorem For completeness we provide a proof of Mann’s Theorem. We then prove the extension that we will need to prove the main result in the next section. Theorem 4.3.1 (Mann). Suppose we have kX i=1 ai i = 0; with ai 2 Q, the  i roots of unity, and no subrelations X i2I ai i = 0 where ; 6= I ( [k]. Then ( i= j) m = 1 334.3. Mann’s Theorem for all i; j, with m = Y p k p prime p. Proof. We can assume that  1 = 1 and a1 = 1, so that we have 1 +Pk i=2 ai i = 0. We take a minimal m such that  m i = 1 for each i. We will show that m must be squarefree, and that a prime p that divides m must satisfy p  k. Together these facts prove the theorem. Let p be a prime dividing m. Write m = pj  m with (p;m ) = 1, and use that to factor each  i as follows:  i =   i    i ; with  a primitive pjth root of unity so  p j = 1; (  i ) pj 1m = 1; 0   i  p 1: Now reorganize the equation as follows: 0 = 1 + kX i=2 ai i = 1 + p 1X l=0  ‘ ‘ = f( ); where the coe cients are of the form  ‘ = X i2I‘ ai  i 2 Q(  2 ; : : : ;   k) = K; with I‘ = fi 2 [k] :  i = ‘g. So f is a polynomial over the  eld K of degree  p  1 and f( ) = 0. The polynomial f isn’t identically zero, since that would give a subrelation containing strictly fewer than k terms. To see this, observe that we must have  i  1 for at least one i, otherwise  m=p i = 1 for each i, contradicting the minimality of m. But we can compute the degree of  over K to be degK( ) =  (m)  (pj 1m ) =  (pj)  (pj 1) =  p 1 if j = 1 p if j > 1: This is a contradiction unless j = 1, which proves that m is squarefree. Knowing that m is squarefree, we have m = p  m with (p;m ) = 1, and  i =   i    i ;  p = 1; (  i ) m = 1; 0   i  p 1: Still f( ) = 0 for f(x) a polynomial over K, not identically zero. But we know ([39], Ch. VI.3) that the minimal irreducible polynomial of  over K is F (x) = xp 1 + xp 2 +    + x+ 1, hence we must have f(x) = cF (x) for some c 2 K. In particular, f has p terms, which implies that our original relation had at least p terms, so k  p. 344.3. Mann’s Theorem In our proof in the next section we will not use Mann’s Theorem itself, but the following modi ed version and its corollary. Theorem 4.3.2. Suppose we have kX i=1 ai i = a; kX j=1 a j  j = a; with a 2 C; a 6= 0, ai 2 Q, roots of unity  i, and no subrelations X i2I ai i = 0 or X j2J a j  j = 0 where ; 6= I ( [k] and ; 6= J ( [k]. Then for any   j there is a  i such that    j = i  m = 1 with m = Y p 2k p prime p. Proof. We have P ai i = a = P a j  j , which gives the single equation kX i=1 ai i  kX j=1 a j  j = 0: (4.1) Mann’s Theorem does not apply immediately, because there might be sub- relations. But we can break the equation up into minimal subrelations X i2I‘ ai i  X j2I ‘ a j  j = 0; (4.2) where each I‘ 6= ;, I ‘ 6= ;, and there are no further subrelations. Given   j , there is such a minimal subrelation of length  2k in which it occurs, and which must also contain some  i. Applying Mann’s Theorem to this equation gives    j = i  m = 1 with m = Y p 2k p prime p. Note that in the above proof we require a 6= 0. If a = 0 and there is no proper subrelation as in (4.2) then (4.1) still has the subrelations kX i=1 ai i = 0; kX j=1 a j  j = 0; 354.4. Proofs of main theorems so we cannot use Mann’s Theorem to get a relation between a  i and   j . For a 2 C; a 6= 0; k 2 Z; k > 0 we de ne Zka to be the set of k-tuples of roots of unity ( 1; : : : ;  k) for which there are ai 2 Q such that Pk i=1 ai i = a with no subrelations, i.e.: Zka = f( 1; : : : ;  k) j 9ai 2 Q : kX i=1 ai i = a; X i2I ai i 6= 0 for ; 6= I  [k]g: Corollary 4.3.3. Given a 2 C with a 6= 0, we have jZka j  (k  C(k)) k, where C(k) = Y p 2k p prime p. Proof. Fix an element ( 1; : : : ;  k) 2 Zka and let m = C(k) and Mi =   m i for 1  i  k. Then for   j in any element of Z k a , we have an i such that Mi    j  m = 1. In other words   j is a solution of Mix m = 1. Each of these k equations has m = C(k) solutions, hence there are at most k  m = k  C(k) choices for each   j . 4.4 Proofs of main theorems 4.4.1 Preparation We are now in a position to prove the main results. Suppose we have a graph G = G(V;E) on v(G) = n vertices and e(G) = cn1+ edges. We will denote the minimum degree in G by  (G). The following lemma assures us that we can remove low-degree vertices from our graph without greatly a ecting the number of edges. Lemma 4.4.1. Let G be as above. Then G contains a subgraph H, with e(H)  (c=2)n1+ edges and v(H)  (c=2)n vertices, such that the mini- mum degree  (H)  (c=2)n . Proof. We iteratively remove vertices from G of degree less than (c=2)n . Then the resulting subgraph H has  (H)  (c=2)n , and we removed fewer than (c=2)n1+ edges, so H contains more than (c=2)n1+ edges. 364.4. Proofs of main theorems Suppose we are given a path Pk = p0p1 : : : pk on k edges in this graph. We will denote by   !pipj the complex number representing the vector between the points in R2 corresponding to the vertices pi; pj . We call the path irredundant if X i2I    !pipi+1 6= 0 for any ; 6= I  f0; 1; : : : ; k  1g. 4.4.2 Proof of Theorem 4.2.2 Let G be the graph with the n points in the plane as vertices and the rational distances with rational angles between pairs of points as edges. Suppose there are n1+f(n) such distances for some positive function f . Then e(G)  n1+f(n). We will count the number of irredundant paths Pk in G, for a  xed k that we will choose later. By Lemma 4.4.1 we can assume that e(G)  (1=2)n1+f(n); v(G)  (1=2)nf(n) and  (G)  (1=2)nf(n). The number of irredundant paths Pk starting at any vertex v is at least N = k 1Y ‘=0 ( (G) 2‘ + 1); since, if we have constructed a subpath P‘ of Pk, then at most 2‘  1 of the at least  (G) continuations are forbidden. Thus the total number of irredundant paths Pk is at least nN 2  (n=2) k 1Y ‘=0 ((1=2)nf(n)  2‘ + 1)  nkf(n)+1 22k+1 if 2k  (1=2)nf(n), which is true as long as k < f(n) log n= log 2. It follows that there are two vertices v and w with at least N n  (1=n) k 1Y ‘=0 ((1=2)nf(n)  2‘ + 1)  nkf(n) 1 4k irredundant paths Pk between them. We will call the set of these paths Pvw, so that we have jPvwj  nkf(n) 1=4k. Given Pk 2 Pvw, Pk = p0p1 : : : pk, consider the k-tuple ( 1; : : : ;  k) where  i is the root of unity in the direction from pi 1 to pi, i.e.  i =    !pi 1pi=j    !pi 1pij. Note that ( 1; : : : ;  k) 2 Zka , because Pk is irredundant. Since there are no 374.4. Proofs of main theorems three points on a line, this process gives an injective map from Pvw to Zka . Hence jPvwj  (k  C(k))k by Corollary 4.3.3, and we get nkf(n) 1 4k  (k  C(k))k =) nkf(n) 1  (4k  C(k))k: This gives (kf(n) 1) log n  k log(4k  C(k)) =) f(n)  log(4k) + log(C(k)) log n + 1 k : The term log(C(k)) is the log of the product of the primes less than or equal to 2k. This is a well known number-theoretic function called the Chebyshev function and denoted by #, speci cally #(2k) = log(C(k)). We use the bound #(x) < 4x log 2 < 3x (for a proof see [3]). This gives f(n) < log(4k) + 6k log n + 1 k < 7 log n k + 1 k : Let k be an integer such that f(n) log n=18 < k < f(n) log n=14 (possible since otherwise f(n) = O(1= log n) giving nf(n) = O(1)). Then the condition that k < f(n) log n= log 2 is clearly satis ed, and we get f(n) < 7 log n  f(n) log n 14 + 18 f(n) log n =) f(n) < 6 p log n : This completes the proof. 4.4.3 Proof of Theorem 4.2.1 Theorem 4.2.1 follows from the same proof as 4.2.2: the condition that no three points are on a line was used to show that the map from Pvw to Zka is injective, which now follows from the fact that the edges in the graph all have unit length. 4.4.4 Proof of Theorem 4.2.3 Consider a path Pk = p0p1 : : : pk. If the distance from pi 1 to pi is less than the distance from pi 1 to any vertex on the line connecting pi 1 and pi and not in Pi 1 = p0p1 : : : pi 1 then Pk is called a shortest path. This proof is almost the same as the proof of Theorem 4.2.2 except that instead of considering all irredundant paths Pk, we only consider shortest 384.4. Proofs of main theorems irredundant paths. Suppose there are n1+ +f(n) edges in the rational dis- tance graph. Since there are at most n points on a line, we get that from any vertex v there are at least N = k 1Y ‘=0   (G) n  2‘ + 1   nkf(n) 4k shortest irredundant paths Pk, if k < f(n) log n= log 2. For any two ver- tices v; w let Pv;w be the set of shortest irredundant paths Pk between v and w. Then there are two vertices v; w such that the number of shortest irredundant paths between v and w is at least jPvwj  nkf(n) 1 4k : By Mann’s Theorem, since we are looking at shortest irredundant paths, jPvwj  (k  C(k))k: Let k be an integer such that f(n) log n=18 < k < f(n) log n=14. Then nkf(n) 1 4k  (k  C(k))k =) f(n) < 6 p log n : That completes the proof. 4.4.5 Proof of Theorem 4.2.4 Assume we have a con guration of n points with at most n on a line, 1=2    1, and n1+ +f(n) rational distances with rational angles, for some positive function f(n). The graph G on these points has e(G) = n1+ +f(n). By Lemma 4.4.1 we can assume that e(G)  n1+ +f(n)=2, v(G)  n +f(n)=2 and  (G)  n +f(n)=2. We now count irredundant paths P2 of length 2. Note that an irredundant path on two edges is just a noncollinear path. For any vertex v, since we have at most n points on a line, v is the midpoint of at least N =  (G)( (G) n )  n2( +f(n)) 8 paths P2 if f(n)  log 4= log n (if f(n) < log 4= log n then nf(n) < 4, com- pleting the proof.) Thus there are two vertices v and w with at least (1=8)n2( +f(n)) 1 noncollinear paths P2 between them. 394.5. Lower bounds But by Corollary 4.3.3 there is a constant number of directions from each of v and w. Since we are looking at noncollinear paths P2, the direction from v and the direction from w uniquely determine the midpoint for a path P2. Thus there are at most (k  C(k))k = 144 noncollinear paths P2 between v and w, since k = 2. Putting the upper and lower bounds together we get n2( +f(n)) 1  2732. This gives f(n)  7 log 2 + 2 log 3 2 log n + 1 2    7 log 2 + 2 log 3 2 log n < 4 log n ; since   1=2. But this gives nf(n) < 4, completing the proof. 4.5 Lower bounds In this section we give lower bounds for the theorems given in Section 4.2. We will be a little more informal, since these are known constructions. The bounds in Theorems 4.2.1 and 4.2.2 are not far from optimal as the following construction of Erd}os and Purdy [25] shows. Suppose we have n points, no three on a line, with the maximum possible number of unit distances with rational angles; we call this number f(n). Consider these points as a set fz1; : : : ; zng of complex numbers. For a 2 C with jaj = 1 and a 6= zi zj for any i 6= j, the set fz1; : : : ; zn; z1+a; : : : ; zn+ag contains at least 2f(n) + n unit distances, since there are f(n) among each of the sets fz1; : : : ; zng and fz1 + a; : : : ; zn + ag and jzi  (zi + a)j = 1 for each i. This new set may have three points on a line, but we show that we can choose a appropriately so that this is not the case. Consider a pair of points zi and zj . For each zk, the set fzk+a : jaj = 1g intersects the line through zi and zj in at most two points. So there are at most two values of a that will give three points on a line. There are  n 2  pairs of points and n choices for zk so there are at most 2n  n 2  = n2(n 1) values of a that make a point zk + a collinear with two points zi and zj . Similarly we have n2(n 1) values of a that make a point zk collinear with zi + a and zj + a. Thus there are only  nitely many values of a that give three points on a line, but there are in nitely many choices for a, so we are done. This shows that f(2n)  2f(n) + n for n > 2 and clearly f(2) = 1. From this we get that f(2k)  2k 1(k  1) = 2k 1 log2(2 k 1). Taking 2k  n < 2k+1 we get that f(n)  cn log n for all n. This construction gives a lower bound for Theorems 4.2.1 and 4.2.2. 404.5. Lower bounds The bound in Theorem 4.2.3 is not far from optimal. In fact we can get a lower bound of cn1+ . Consider n1  lines parallel to the x-axis, and choose n rational points on each line such that no three points on di erent lines are collinear (this can always be done since there are in nitely many rational points to choose from). There are cn2 rational distances on each horizontal line and n1  such lines giving at least cn1+ rational distances with rational angles (all the angles are zero). The bound in Theorem 4.2.4 is tight up to a constant factor as can be seen by considering an n1   n square grid. Then there are at least cn2 rational distances on each of the n1  horizontal lines in the grid containing n points. This gives at least cn1+ rational distances with rational angles. 41Chapter 5 Simultaneous arithmetic progressions 5.1 Introduction There are interesting problems in number theory related to arithmetic pro- gressions on elliptic curves (see Section 5.4 for the de nition). An example of such an open problem is: what is the maximum number (if it exists) of ratio- nal points on an elliptic curve such that their x-coordinates are in arithmetic progression? In [8], Bremner found elliptic curves in Weierstrass form with arithmetic progressions of length 8 on them, and Campbell [10] found ellip- tic curves of the form y2 = f(x), with f a quartic, that contain arithmetic progressions of length 12. In [7], Bremner described how these arithmetic progressions are related to 3 3 magic squares with square entries. Bremner, Silverman, and Tzanakis noted in [9] that points in arithmetic progression on elliptic curves are often independent with respect to the group structure, which suggests a relation with the much-researched rank of the curve. In [30], Garcia-Selfa and Tornero looked instead for \simultaneous" arith- metic progressions on elliptic curves, which are de ned as follows. De nition 5.1.1. A simultaneous arithmetic progression (SAP) of length k consists of points (xi; y (i)) in R2, where xi = a1 + i  d1 and yi = a2 + i  d2 for i = 0; 1; : : : ; k 1 are arithmetic progressions, and  is a permutation on the numbers 0; 1; : : : ; k  1. Note that the appearance of this permutation is quite natural, since points with both coordinates in arithmetic progression would all lie on a line. Garcia-Selfa and Tornero gave examples of elliptic curves over Q that con- tain an SAP of length 6. They also showed that there are only  nitely many such curves, and there are none with an SAP of length 7. Extending their methods to SAP’s of length 8 did not seem computationally feasible, and they were not able to  nd an elliptic curve with an SAP of length 8, or prove that none exists. The  nal open problem they suggested is  nding a universal bound for the length of SAP’s on elliptic curves over Q. 425.2. First proof for algebraic curves In Section 4 we prove that 4319 is an upper bound for the length of an SAP on an elliptic curve over R in Weierstrass form, using a combinatorial approach. This solves the open problem above. We  rst approach the more general problem of bounding the k for which an algebraic curve (as de ned in section 1.3.2) can contain k points from a k k grid. We do this in two di erent ways: in Section 2 we give a short proof based on a theorem of Jarn  k, which will give the bound k  cd9, where d is the degree of the curve. Then in Section 3 we give a di erent proof using the well-known crossing inequality from graph theory, in combination with B ezout’s Theorem. This will result in the improved bound k  cd7. We also extend this result to complex plane algebraic curves. Finally in Section 4 we specialize the second proof to the case of an elliptic curve in Weierstrass form, with some adjustments, resulting in our upper bound of 4319. 5.2 First proof for algebraic curves Our  rst theorem shows that the length of an SAP on a curve is bounded by a function of the degree d of the curve. The proof is an application of a theorem of Jarn  k. In the next section the dependence of the bound on d will be improved with a di erent approach. By a k  k grid we will mean the cartesian product of two arithmetic progressions of length k; so an SAP consists of k elements from a k k grid with exactly one element on each row and on each column. In the theorems below, we will not speci cally deal with SAP’s, but more generally with any collection of k points from a k  k grid. Theorem 5.2.1. If f is a curve of degree d  2 with no linear factor, and f has at least k points from a k  k grid, then there is an absolute constant c such that k  cd9. If f is also irreducible, the bound improves to k  c0d6. The dependence on d of the bound in the theorem cannot be removed. Consider any k points from a k k grid, and take d such that k  d(d+3)=2. Then there is a curve of degree d passing through all k of the points (see Section 5.2 in [29]). Our proof uses the following result of Jarn  k [35]. Theorem 5.2.2 (Jarn  k). If f is a strictly convex di erentiable curve of length N , then the number of integer points on f is less than  N2=3 for some constant  . 435.2. First proof for algebraic curves For algebraic curves, this bound is by no means optimal. In fact, Bombieri and Pila proved in [5] that we can get the bound c(d; ")N1=d+" for any " > 0 if the curve is irreducible. This clearly gives a better bound for large degree. To be able to apply Jarn  k’s bound, we need to break up our alge- braic curve into convex or concave pieces (of course Jarn  k’s Theorem works equally well for concave curves). This can be achieved by cutting the curve at all in ection points and singularities (points on f where the  rst deriva- tives fx and fy both vanish). We will show that the resulting number of pieces is bounded by a function of the degree of f , by reducing to a bounded number of irreducible curves, and then using the following bounds. Lemma 5.2.3. Suppose f is an irreducible curve of degree d. Then f has at most 3d(d 2) in ection points, and at most (d 1)(d 2)=2 singularities. For a proof of the bound on in ection points see Proposition 3.33 in [37], and for the bound on singularities see Section 5.4 in [29]. Note that the bound on singularities also follows from the genus formula in Chapter 3, since singularities have  P  1 and genus cannot be negative. It follows that we need at most 3d(d 2) + (d 1)(d 2)=2 < 4d2 cuts to break an irreducible f into convex parts. Proof of Theorem 5.2.1. First assume that f is irreducible and contains k points from a k  k grid. We scale and translate f so that the gap in the k k grid is 1 in both the x- and y-directions, and the points of the grid are integral. Now we can separate f into convex parts using at most 4d2 cuts. One of these parts has at least the average number k=4d2 of points from the grid on f . Since the grid has gap 1 and length k we can bound the length of this part of the curve by 2k. Thus we get, by Theorem 5.2.2, that k 4d2 <  (2k)2=3: This gives k < Cd6 for an irreducible curve f . For a reducible curve f with k points from a k  k grid, we have a factorization f = f 11 f  2 2 : : : f  r r where each fi is irreducible of degree di. Since f has no linear factor, we have di  2 for all i, as well as r  d=2. We take the factor fj with the most points from the grid on it, which is at least k r  2k d . Repeating the argument above with 2k=d points from a k k grid, we get the inequality 2k=d4d2 <  (2k) 2=3, which leads to k < cd9. 445.3. Second proof for algebraic curves 5.3 Second proof for algebraic curves In this section we obtain the following improvement of Theorem 5.2.1, using graph theory. Theorem 5.3.1. If f is a curve of degree d  2 with no linear factor, such that f has at least k points from a k  k grid, then there is an absolute constant c such that k  cd7. If f is also irreducible, the bound improves to k  c0d4. The idea behind the proof is to construct a graph G (actually, a multigraph) out of translates of the curve in a grid, with the grid points as vertices, and with edges between points which occur consecutively on a translate. To get the stated upper bound for k, we obtain a lower bound on the number of intersections of G from the Crossing Inequality (Theorem 5.3.2), and com- pare this with an upper bound that we get from B ezout’s Theorem (5.3.4). Construction of the graph G. For convenience suppose that the grid is [1; k]  [1; k], and extend it to [1; 2k]  [1; 2k]; this will be the vertex set of our graph. Consider the k2 translates of f obtained by shifting i in the x-direction and j in the y direction, for all pairs i; j 2 [1; k]. We will basically draw an edge along the curve between two grid points if they occur as consecutive points on some translate of f , but there are several things we have to watch out for. First note that some parts of the curve do not occur among the edges, namely the parts that go o to in nity, and components of f that have no grid point or a single grid point on it. Especially this last case will have to be accounted for in the proof. Second, we may have more than one edge connecting two vertices. We show that the maximum edge multiplicity for a pair of vertices is d2. If we have more than one edge connecting two vertices then these two vertices appear as consecutive points on di erent translates. These points are given as (l;m) and (l0;m0) for some l; l0;m;m0 2 Z. If these points appear on r translates then the vector (l  l0;m  m0) occurs as a di erence vector between r pairs of points on the original elliptic curve. But this is equivalent to having r points on the original curve intersecting r points on its translate by (l  l0;m m0). Hence r  d2 by B ezout’s Theorem. Third, since we are considering translates of a curve in a grid, a vertex may occur on a number of translates. Suppose v1 and v2 are consecutive points on a translate. We may have a point v3 on another translate which is actually between v1 and v2 on the  rst translate. In this case the edge 455.3. Second proof for algebraic curves from v1 to v2 passes through the vertex v3. This is not allowed in a graph so we have to alter our graph slightly. In this case we remove the edge in consideration from v1 to v2 and add an edge from v1 to v3. Performing this change where necessary we end up with a graph with the same number of vertices and edges but without the problem of an edge passing through a vertex to which it is not adjacent. We call this graph G. We now introduce the results from graph theory that we will use to get a lower bound on the crossing number of our graph. Given a simple graph G, the crossing number cr(G) is the minimum number of pairs of crossing edges in a planar drawing of G. Theorem 5.3.2 (Crossing Inequality). Suppose G is a simple graph with n vertices and e edges. If e > 7:5n then cr(G)  e3 33:75n2 : The crossing inequality was  rst proved independently by Ajtai, Chv atal, Newborn and Szemer edi [2] and by Leighton [40]. The version with the best bound to date, presented above, was given by Pach and T oth [44]. Pach and T oth also gave the following crossing inequality for multi- graphs, which is the result we will use here. Theorem 5.3.3 (Crossing inequality for multigraphs). Suppose G is a multigraph with n vertices and e edges (counting multiplicity). Suppose there are at most m edges between any pair of vertices in G. If e > 7:5mn then cr(G)  e3 33:75mn2 : To get an upper bound on the number of intersections in our graph, we use B ezout’s Theorem. For details see [29]. Theorem 5.3.4 (B ezout). Suppose F and G are curves of degree m and n. If F and G do not have a common factor, then they intersect in at most mn points. The main consequence that we will use is that if f(x; y) = 0 is an irreducible curve of degree d, then f and a translate of f intersect in at most d2 points. Finally, we will need the following result. For details, see [28], p. 218. 465.3. Second proof for algebraic curves Lemma 5.3.5 (Harnack). The number of connected components in R2 of an irreducible curve is at most (d 1)(d 2)2 + 1. Proof of Theorem 5.3.1. We  rst assume that f is irreducible. Suppose that f contains k points from a k  k grid. Typically, a component of f with m vertices on it will give m edges in our graph, except that we get m 1 edges if the component does not form a closed loop (i.e. contains a point at in nity), and 0 edges if the component contains only one vertex. Fortunately, by Harnack’s result there are at most (d  1)(d  2)=2 + 1 components, and by B ezout the line at in nity and our curve can have at most d intersections, so the number of such "bad" components is  d2=2. If we assume k  d2, then at least k=2 of the vertices are on "good" components, hence give us k=2 edges. All translates together then give k3=2 edges in our graph. The crossing number of the graph is the number of intersections between translates plus the number of self-intersections of translates. By B ezout’s Theorem, for any pair of translates there are at most d2 intersections. A self-intersection is a singularity, hence the number of these is bounded above by (d 1)(d 2)=2 by Lemma 5.2.3. There are k2 translates and  k2 2  pairs of translates so we get cr(G)  d2  k2 2  + k2 (d 1)(d 2) 2 = 1 2 k2(d2k2  3d+ 2)  1 2 d2k4: By the crossing inequality for multigraphs we get the lower bound: (k3=2)3 33:75d2(4k2)2  e3 33:75d2v2  cr(G): Combining these we get k  C1d4, with C1 = 26  33:75, for an irreducible curve. For a reducible curve f with k points from a k k grid we proceed as in Section 5.2. We can  nd an irreducible factor of f with at least 2k=d grid points on it. That gives us k3=d edges in our graph, so that the inequality becomes (k3=d)3 33:75d2(4k2)2  1 2 d2k4; which gives us k  C2d7, with C2 = 33:75  23. 475.4. SAPs on elliptic curves Theorem 5.3.1 can be extended to any complex plane algebraic curve, i.e. where f 2 C[x; y] and we consider the zero set Cf (C) = f(x; y) 2 C2 : f(x; y) = 0g: We use the result for the reals to prove the result for this complex case. By a k k grid in C2 we mean a cartesian product of two arithmetic progressions in C; by an arithmetic progression in C we mean points  + i with  ;  2 C and i = 0; 1; : : : ; k  1. Theorem 5.3.6. If f is a complex plane algebraic curve of degree d  2 with no linear factor such that Cf (C) contains at least k points from a k k grid, then there is an absolute constant c such that k  cd7. Proof. Suppose f(w; z) is our complex curve with k points on a k  k grid, given by  + j in one direction and  + j in the other direction, where  ;  ;  ;  2 C and j = 0; 1; : : : ; k 1. Now consider the polynomial g(x; y) = f( +x ;  +y ). This is a polynomial with complex coe cients in two real variables, and the curve g(x; y) = 0 has k points on the k k grid consisting of the points (i; j) with i; j = 0; 1; : : : ; k  1. The real and imaginary parts of g are real algebraic curves of degree  d, each having k points on the grid. Thus Theorem 5.3.1 gives the stated bound. 5.4 SAPs on elliptic curves In this section we use the method from Section 5.3, with some modi cations, to give a universal bound on the size of an SAP on a real elliptic curve, answering the question of Garcia-Selfa and Tornero. By elliptic curve, we will mean a nonsingular irreducible cubic curve, with Weierstrass equation y2 + axy+ by = x3 + cx2 + dx+ e; however, we will not make any use of the arithmetic theory of elliptic curves. The upper bound that we would get using Jarn  k’s Theorem as in Section 5.2 is roughly 4  105. With the approach from Section 5.3 we would get an upper bound of roughly 2  105, using k  C1  d4 from the proof. For the speci c case of elliptic curves, we can make several improvements. First of all, we know that the curve is irreducible, and that there are no singularities. Second, we know exactly what the components can look like. Finally, the following lemma shows that we can do better than the intersection bound d2 = 9 B ezout’s Theorem. 485.4. SAPs on elliptic curves Lemma 5.4.1. An elliptic curve and a translate of that curve can intersect in at most 4 points (excluding points at in nity.) Proof. Suppose the curve is given by y2 + axy + by = x3 + cx2 + dx + e. A translate is given by (y + v)2 + a(x + u)(y + v) + b(y + v) = (x + u)3 + c(x+ u)2 + d(x+ u) + e where at least one of u; v does not equal 0. If (x; y) is an intersection point of these curves then subtracting one equation from the other we get 2vy + v2 + avx+ auy + auv + bv = 3ux2 + 3u2x+ u3 + 2cux+ cu2 + du: If 2v + au = 0 then all terms involving y disappear. In this case we have a quadratic in x which can have at most 2 real roots. Putting these values into the original equation we get at most 4 intersection points. If 2v + au 6= 0 then we can solve for y to get y = 3ux2 + (3u2 + 2cu av)x+ (u3 + cu2 + du auv  bv  v2) 2v + au : Substituting this into the original equation we get f(x) = 0 where f is a quartic polynomial in x. This polynomial has at most 4 roots. Thus we cannot have more than 4 intersection points of our elliptic curve and its translate. The main result is: Theorem 5.4.2. Suppose we have an elliptic curve given by y2+axy+by = x3 + cx2 + dx+ e, containing an SAP of length k. Then k  4319. Modi ed construction of G. The graph we will use is essentially the same, but we will make a small adjustment based on a closer analysis of the possible components. A cubic curve of the type above will either consist of a single component containing a point at in nity, or it will consist of two components, one a closed loop and the other containing a point at in nity. We will treat these two cases simultaneously. We will add the point at in nity to our graph, along with the two edges that approach it; this will improve our bounds below. Suppose our curve has t points from the SAP on the loop component (if it exists; otherwise t = 0) and k  t points on the in nite component. If there is more than one point on the loop, so t > 1, then it gives a t-cycle with t edges in the graph. If t = 1 or t = 0, then we get no edges from the loop component. 495.4. SAPs on elliptic curves Consider the k t points on the component containing a point at in nity. They give us k t vertices and k t 1 edges. We add the curve’s point at in nity to the graph, then we connect the rightmost point on the top part of the curve to the point at in nity, and we do the same for the rightmost point on the bottom part of the curve. This component now contains k  t + 1 vertices and k t+ 1 edges. Together with the other component we end up with k + 1 vertices in one translate, and we have k + 1 edges if t 6= 1 and k edges if t = 1. By the same argument as in the construction of G in the previous section and by Lemma 5.4.1, edges have multiplicity at most 4. The new edges (to the point at in nity) will not increase the multiplicity. The only way we can have more than one edge going from a point in the grid to the point at in nity is if that point is the rightmost point on the top half of one translate and the rightmost point on the bottom half of another translate. Thus these edges have multiplicity at most 2. Proof of Theorem 5.4.2. Our graph now has 4k2+1 vertices. Let t be de ned as above. If t 6= 1, then the number of edges, counting multiplicity, in G is k2(k + 1), while if t = 1 then the number of edges is k3. We need only consider the case with fewer edges, so we assume we have k3 edges. The crossing inequality now gives (k3)3 4(33:75)(4k2 + 1)2  cr(G): By Lemma 5.4.1, any pair of translates intersects in at most 4 points in the grid, and there are  k2 2  such pairs. Thus the crossing number is bounded above by cr(G)  4  k2 2  : Putting these two inequalities together we get (k3)3 4(33:75)(4k2 + 1)2  4  k2 2  : Solving for k in this inequality and noting that k is a positive integer, we get k  4319. Note that in Theorem 5.3.3 we require e > 4(7:5)n = 30n. This certainly holds since otherwise e  30n, i.e. k2(k + 1)  30(4k2 + 1). A quick calculation shows that this gives k  120. 50Bibliography [1] S.S. Abhyankar, Algebraic geometry for scientists and engineers, Math- ematical Surveys and Monographs 38, AMS, 2000. [2] M. Ajtai, V. Chv atal, M.M. Newborn, and E. Szemer edi, Crossing-free subgraphs, Theory and practice of combinatorics 60 (1982), 9{12. [3] T.M. Apostol, Introduction to analytic number theory, Springer, 1976. [4] N.H. Anning and P. Erd}os, Integral distances, Bull. Amer. Math. Soc. 51 (1945), 598{600. [5] E. Bombieri and J. Pila, The number of integral points on arcs and ovals., Duke Mathematical Journal 59 (1989), 337{357. [6] P. Brass, W. Moser, and J. Pach, Research problems in discrete geom- etry, Springer, 2006. [7] A. Bremner, On squares of squares., Acta Arithmetica 88 (1999), 289{ 297. [8] A. Bremner, On arithmetic progressions on elliptic curves., Experimen- tal Mathematics 8 (1999), 409{413. [9] A. Bremner, J. Silverman, and N. Tzanakis, Integral points in arith- metic progression on y2 = x(x2  n2), J. Number Theory 80 (2000), 187{208. [10] G. Campbell, Points on y = x2 at rational distance, Math. Comp. 73 (2004), 2093{2108. [11] L. Caporaso, J. Harris, and B. Mazur, Uniformity of rational points, J. Amer. Math. Soc. 10 (1997), 1{35. [12] A. Choudhry, Points at rational distances on a parabola, Rocky Moun- tain J. Math. 36 (2006), 413{424. 51Bibliography [13] K. Conrad, Arithmetic progressions of three squares, Online expository paper, available at www.math.uconn.edu/~kconrad/blurbs/ugradnumthy/3squarearithprog.pdf [14] K. Conrad, Arithmetic progressions of four squares, Online expository paper, available at www.math.uconn.edu/~kconrad/blurbs/ugradnumthy/4squarearithprog.pdf [15] L.E. Dickson, History of the theory of numbers. Vol. II: Diophantine analysis, Chelsea Publishing Co., New York, 1966. [16] H. Darmon and L. Merel, Winding quotients and some variants of Fer- mat’s last theorem, J. reine angew. Math. 490 (1997), 81{100. [17] P. Erd}os, Integral distances., Bull. Amer. Math. Soc. 51 (1945), 996. [18] P. Erd}os, On sets of distances of n points, Am. Math. Mon. 53 (1946), 248{250. [19] P. Erd}os, Verchu niakoy geometritchesky zadatchy,(On some geometric problems, in Bulgarian), Fiz.-Mat. Spis. B ulgar. Akad. Nauk. 5 (1962), 205{212. [20] P. Erd}os, On some problems of elementary and combinatorial geometry, Annali di Matematica pura ed applicata 4 (1975), 99{108. [21] P. Erd}os, N eh any elemi geometriai probl em ar ol (On some problems in elementary geometry, in Hungarian), K oz. Mat. Lapok 61 (1980), 49{ 54. [22] P. Erd}os, Combinatorial problems in geometry, Math. Chronicle 12 (1983), 35{54. [23] P. Erd}os, Some combinatorial and metric problems in geometry, Collo- quia Mathematica Societatis J anos Bolyai 48 (1985), 167{177. [24] P. Erd}os, Ulam, the man and the mathematician, J. Graph Theory 9 (1985) no. 4, 445{449. [25] P. Erd}os and G.B. Purdy, Extremal problems in combinatorial geometry, Handbook of Combinatorics, Elsevier Science (1995), 809{874. [26] J.-H. Evertse, H.P. Schlickewei, and W.M. Schmidt, Linear equations in variables which lie in a multiplicative group, Ann. Math. 155 (2002), 807{836. 52Bibliography [27] G. Faltings, Endlichkeitss atze f ur abelsche Variet aten  uber Zahlk orpern, Invent. Math. 73 (3) (1983), 349{366. [28] G. Fischer, Plane Algebraic Curves, Student Mathematical Library 15, AMS, 2001. [29] W. Fulton, Algebraic curves: an introduction to algebraic geometry, 1969. [30] I. Garcia-Selfa and J.M. Tornero, On simultaneous arithmetic progres- sions on elliptic curves, Experimental Mathematics 15 (2006), 471{478. [31] R. Guy, Unsolved problems in number theory, Problem Books in Math- ematics Subseries: Unsolved Problems in Intuitive Mathematics 1, Springer, 3rd ed., 2004. [32] R. Hartshorne, Algebraic geometry, Graduate Texts in Mathematics 52, Springer, 1977. [33] H. Harborth, A. Kemnitz, and M. M oller, An upper bound for the min- imum diameter of integral point sets, Discrete & Comput. Geom. 9 (1993), 427{432. [34] G.B. Hu , Diophantine problems in geometry and elliptic ternary forms, Duke Math. J. 15 (1948), 443{453. [35] V. Jarn  k,  Uber die Gitterpunkte auf konvexen Kurven, Mathematische Zeitschrift 24 (1926), 500{518. [36] A. Kemnitz, Punktmengen mit ganzzahligen Abst anden, Habilitationss- chrift, TU Braunschweig, 1988. [37] F.C. Kirwan, Complex algebraic curves, Cambridge University Press, 1992. [38] T. Kreisel and S. Kurz, There are integral heptagons, no three points on a line, no four on a circle, Discrete & Computational Geometry 39 (2008), 786{790. [39] S. Lang, Algebra, Addison-Wesley, 3rd ed., 1994. [40] F.T. Leighton, New lower bound techniques for VLSI, Math. Systems Theory 17 (1984), 47{70. 53Bibliography [41] H.B. Mann, On linear relations between roots of unity, Mathematika 12 (1965), 107{117. [42] J. Matou sek, The number of unit distances is almost linear for most norms, Adv. Math. 226 (2011), 2618{2628. [43] W. D. Peeples Jr., Elliptic curves and rational distance sets, Proc. Am. Math. Soc. 5 (1954), 29{33. [44] J. Pach and G. T oth, Graphs drawn with few crossings per edge, Com- binatorica 17 (1997), 427{439. [45] W.M. Schmidt, Diophantine approximation, LNM 785, Springer Verlag, 1980. [46] R. Schwartz, J. Solymosi, and F. de Zeeuw, Simultaneous arithmetic progressions on algebraic curves, International Journal of Number The- ory 7 (2011), 921{931. [47] R. Schwartz, J. Solymosi, and F. de Zeeuw, Rational distances with rational angles, accepted for publication in Mathematika; arXiv:1008.3671v2, [math.CO], 2011. [48] J. Silverman, The arithmetic of elliptic curves, Graduate Texts in Math- ematics 106, Springer, 1994. [49] J. Solymosi, Note on integral distances, Discrete & Comput. Geom. 30 (2003), 337{342. [50] J. Solymosi and F. de Zeeuw, On a Question of Erd}os and Ulam, Dis- crete and Computational Geometry 43 (2010), 393{401. [51] J. Spencer, E. Szemer edi, and W. Trotter, Unit distances in the Eu- clidean plane, Graph Theory and Combinatorics: Proceedings of the Cambridge Combinatorial Conference, in Honour of Paul Erd}os, Aca- demic Press, 1984, 293{303. [52] L. Sz ekely, Crossing numbers and hard Erd}os problems in discrete ge- ometry, Combinatorics, Probability and Computing 6 (1997), 353{358. [53] S.M. Ulam, A collection of mathematical problems, Interscience Tracts in Pure and Applied Mathematics 8, Interscience Publishers, 1960. 54

Cite

Citation Scheme:

    

Usage Statistics

Country Views Downloads
China 33 0
United States 12 7
Republic of Korea 8 0
Spain 2 0
Switzerland 1 4
Russia 1 0
Turkey 1 0
Canada 1 0
United Kingdom 1 0
Poland 1 0
City Views Downloads
Beijing 25 0
Shenzhen 7 0
Seoul 6 0
Ashburn 5 0
Unknown 4 5
San Mateo 3 0
Daejeon 2 0
Redmond 2 2
Cartagena 2 0
Richmond 1 0
Guelph 1 0
Bethesda 1 0
Saint Petersburg 1 0

{[{ mDataHeader[type] }]} {[{ month[type] }]} {[{ tData[type] }]}
Download Stats

Share

Embed

Customize your widget with the following options, then copy and paste the code below into the HTML of your page to embed this item in your website.
                        
                            <div id="ubcOpenCollectionsWidgetDisplay">
                            <script id="ubcOpenCollectionsWidget"
                            src="{[{embed.src}]}"
                            data-item="{[{embed.item}]}"
                            data-collection="{[{embed.collection}]}"
                            data-metadata="{[{embed.showMetadata}]}"
                            data-width="{[{embed.width}]}"
                            async >
                            </script>
                            </div>
                        
                    
IIIF logo Our image viewer uses the IIIF 2.0 standard. To load this item in other compatible viewers, use this url:
http://iiif.library.ubc.ca/presentation/dsp.24.1-0072373/manifest

Comment

Related Items