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Rational Points on Curves Rational Points on Curves Padmavathi Srinivasan 16th Octber 2012 References: The Arithmetic of Elliptic Curves by Silverman, Hartshorne Chapter 4, Wikipedia Notation By a curve, we will mean a smooth, projective, geometrically integral variety over a field k (which may not be algebraically closed). A natural invariant associated to the curve is the genus - there are many equivalent ways of defining this (which you will see see at some point). Here is one definition, geometric genus is 1 g := dimk Γ(X; Ω ), where Ω1 is the sheaf of differentials. If you are over the complex numbers, the curve is just a compact Riemann surface, and g is just the number of holes (the complex dimension of the space of global holomorphic differentials) . By a rational point, we will just mean a closed point with residue field k (your ground field). Or more n concretely, if you are thinking of the curve as being embedded in projective space Pk , a rational point on the curve is a point will all its coordinates in k. For example, on the curve defined by x2 + y2 − z2 = 0 in 2 , [1 : 0 : 1] is a rational point. On the other PR hand, the curve defined by x2 + y2 + z2 = 0 in 2 has no rational points. PR In this talk, we will mostly focus on results over Q or a finite extension of Q (a number field). Genus 0 and Genus ≥ 2 Genus 0 Over an algebraically closed field, there is only one genus zero curve and that is P1. All isn't lost if your field isn't algebraically closed. Any genus zero curve can be embedded as a smooth conic in P2 (you will see this when you study divisors). In this case, if there is a single rational point P , then there are infinitely many!1 To see this, you note that projection from the point P gives an isomorphism onto P1 (a quadratic equation over the field k, where one of the roots is in k will also have its second root in k). For example, over R, every smooth conic in P2 is of the form ax2 + by2 + cz2, where all three coefficients are non-zero. We can rescale our coordinate axes so that a; b; c 2 {−1; 1g, and now at least two of them will have to have same sign - so up to isomorphism, we have x2 + y2 − z2 which is just the Riemann sphere 1 PR and x2 + y2 + z2 which has zero rational points. g ≥ 2 It's difficult to give a general answer for all fields, but there are results if we restrict our attention to a smaller class of fields. Theorem (Mordell Conjecture/Faltings theorem). Any curve of genus > 1 over Q (or a finite extension of Q) has only finitely many rational points. This theorem is hard - Faltings won a Fields medal for it! Its proof involves relating this to other conjectures of Shafarevich and Tate. (Show graph) 1 1 It is infinite if the ground field is infinite. In any case, you get the isomorphism to P 1 Genus 1 curves A genus one curve may not have any rational points at all. If it has a rational point, it is then called an elliptic curve (the name comes from elliptic integrals). In this case, E(K) can be given the structure of an abelian group! Over a characteristic zero field (like Q or a number field), the elliptic curve can be embedded as a plane cubic curve cut out by a single homogeneous equation y2z = x3 +Axz2 +Bz3 and the distinguished rational point is taken to be the point at infinity [0 : 1 : 0]. The group law in this case can be described geometrically. We will not be giving a proof of why this gives a well defined group law now - you will see this when you study divisors on curves. Over the complex numbers, E(C) is just the torus C=Γ for some rank two lattice Γ in C and the addition in the group is just the usual addition in C. Theorems and Conjectures Some natural questions to ask after we know that the set of rational points form an abelian group: (a) Is E(K) is a finitely generated abelian group? More geometrically, can we get all rational points by starting with a finite number of them using the construction of secants and tangents? (b) If yes, what are the possibilities for the rank and torsion subgroup? (c) Given an elliptic curve, is there an effective way to compute its rank and torsion? It's difficult to give a general answer when we ask this question in this generality, , but as before there are answers if we look at Q or number fields. Theorem (The Mordell-Weil Theorem). E(Q) is a finitely generated abelian group. This holds for number fields as well. It is not always true. For example E(C) is nowhere close to being finitely generated.2 Theorem (Mazur's theorem). There are only finitely many possibilities for E(Q)tor. There is a generalization of this to number fields due to Merel. There is a folklore conjecture on the possibilities for the rank of E(Q). People believe that it is unbounded. The highest possible rank that has been found so far is 18 (by Elkies, in 2006). y2+xy = x326175960092705884096311701787701203903556438969515x+51069381476131486489742177100373772089779103253890567848326 Only relatively recently (2010) has it even been proved that the average rank of E(Q) is bounded - this result is due to Bhargava and Shankar. There is no known effective algorithm to compute the rank of a given elliptic curve, though computing the torsion is relatively easy (by looking at reductions modulo various primes). Sketch of proof of Mordell-Weil over Q The proof of the Mordell-Weil can be broken down into two parts: the Weak Mordell-Weil theorem (which says E(Q)=mE(Q) is finitely generated for every m ≥ 2) and the Infinite descent argument using height functions. 2http://mathoverflow.net/questions/108543/over-which-fields-does-the-mordell-weil-theorem-hold says that there is an ex- tension of this for abelian varieties when the ground field is finitely generated over its prime field, and an even further general- ization called the Lang-N´erontheorem. 2 Infinite Descent Even if E(Q)=mE(Q) is finite, it is not enough to guarantee E(Q) is finitely generated. For instance R=mR = 0. Similarly, E(Qp) has a finite index subgroup isomorphic to the additive group Zp, and therefore the quotient E(Qp)=mE(Qp) is finite even though E(Qp) is not. The problem in both these cases is because there is a large number of elements divisible by m. So the rough idea is to show multiplication by m increases the 'size' of a point and therefore there are only finitely many elements of bounded height that are divisible by m. Theorem (Descent theorem, Proposition VIII:3:1, Silverman). Let A be an abelian group. Suppose there is a 'height' function h : A ! R with the following properties: (a) Let Q 2 A. There is a constant C1, depending on A and Q, so that for all P 2 A, h(P + Q) ≤ 2h(P ) + C1 (b) There is an integer m ≥ 2 and a constant C2, depending on A, so that for all P 2 A, 2 h(mP ) ≥ m h(P ) − C2 (c) For every constant C3, P 2 A : h(P ) ≤ C3 is a finite set. Suppose further for the integer m in (b), the quotient group A=mA is finite. Then A is finitely generated. Proof. (a) Choose representatives Q1;Q2;:::;Qr for A=mA and let P 2 A be an arbitrary element. Idea: Show the difference between P and an appropriate linear combination of Q1;:::;Qr is a multiple of a point whose height is less than a constant, independent of P . Then, Q1;:::;Qr and the finitely many points with height less than this constant will generate A. (b) Inductively define Pi. P0 = P , P = mP1 + Qi1 ;Pn = mPn+1 + Qin . (c) 1 h(P ) ≤ [h(mP ) + C ] j m2 j 2 1 = [h(P − Q ) + C ] m2 j−1 ij 2 1 ≤ [2h(P ) + C0 + C ] m2 j−1 1 2 0 0 where C1 is the maximum of the constants from (a) for Q = −Qij , 1 ≤ j ≤ r. C1 and C2 do not depend on P . Use this repeatedly, to see −n 0 h(Pn) < 2 h(P ) + (C1 + C2)=2 (d) For n sufficiently large, it follows 0 h(Pn) ≤ 1 + (C1 + C2)=2 and n n X j−1 P = m Pn + m Qij j=1 3 (e) Any P 2 A is a linear combination of points in the following finite set 0 fQ1;:::;Qrg [ fQ 2 A : h(Q) ≤ 1 + (C1 + C2)=2g All the constants here are effectively computable once we have a set of generators for A=mA. The height function that works on E(Q) is defined as follows: For a rational number t = p=q in lowest terms, the height of t, H(t) := max (jpj; jqj). The height on E(Q) (relative to a given Weierstrass equation) is the function hx(P ) := log H(x(P )) if P 6= O (the origin) and is defined to be zero if P = O. There are explicit formulas for doubling and translation on an elliptic curve that help us check that this height function satisfies the conditions that we want it to satisfy with m = 2. Details can be found in Silverman, Chapter VIII.
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