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).

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 ∈ {−1, 1}, 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 extension of this for abelian varieties when the ground field is finitely generated over its prime field, and an even further generalization called the Lang-Nérontheorem.

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 ∈ A. There is a constant C1, depending on A and Q, so that for all P ∈ 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 ∈ A,

2 h(mP ) ≥ m h(P ) − C2

(c) For every constant C3, P ∈ 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 ∈ 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 ∈ A is a linear combination of points in the following ﬁnite set

0 {Q1,...,Qr} ∪ {Q ∈ A : h(Q) ≤ 1 + (C1 + C2)/2}

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 (|p|, |q|). 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.

The Weak Mordell-Weil Theorem Lemma. Let L/K be a finite Galois extension. E(L)/mE(L) finite implies E(K)/mE(K) is also finite. Sketch. Assume E(L)/mE(L) is finite. There is a natural map E(K)/mE(K) → E(L)/mE(L) and let Φ be the kernel of this map. Φ = E(K) ∩ mE(L)/mE(K). We will be done if we show Φ is finite. This is done by showing there is an injective map Φ → Map (GL/K ,E[m]), where E[m] is the set of m torsion points. 3 Since both GL/K and E[m] are finite , so is the set of maps between them, and this shows Φ is also finite. The map is defined as follows: Let P ∈ Φ, pick a QP ∈ E(L) such that mQP = P , and the map λP corresponding to P is that one that sends σ ∈ GL/K to σ(QP ) − QP . Then it is easy to check that this gives the desired injective map. 4 So without loss of generality, we may assume E[m] ⊂ E(K) where E[m] is the set of m-torsion points. The next step is to translate the finiteness of E(K)/mE(K) to a statement about the finiteness of a certain field extension of K. This is done via the Kummer pairing. Definition. The Kummer pairing

κ : E(K) × GK/K → E[m] is deﬁned as follows. For P ∈ E(K), let Q ∈ E(K) such that [m]Q = P . Then

κ(P, σ) = Qσ − Q

If you are wondering where all this comes from, all of these pop out quite naturally from the Galois cohomology groups associated to the exact sequence

0 → E[m] → E → E → 0

Proposition. (a) The Kummer pairing is well-deﬁned.

(b) The Kummer pairing is bilinear. (c) The kernel of the Kummer pairing on the left is mE(K).

−1 (d) The kernel of the Kummer pairing on the right is GK/L, where L = K(m E(K)) is the compositum of the ﬁelds K(Q) as Q ranges over the points of E(K) satisfying [m]Q ∈ E(K). Hence the Kummer pairing induces a perfect bilinear pairing

E(K)/mE(K) × GL/K → E[m]

3It can be shown that the set of m torsion points on an elliptic curve is a finite set (order m2 over characteristic zero) - easy to see this in the case of an elliptic curve over C, and in fact this suffices over any field of characteristic zero by the Lefschetz principle. 4 In group cohomological terms, the image of Φ in is actually a 1-cocyle on GL/K with values in the abelian group E[m]. This lemma can easily be seen to follow from the inflation-restriction sequence.

4 It is quite straightforward to check all of these. We will need E[m] ⊂ E(K) to check linearity in GK/K . So it now suffices to show L/K is finite. This is done by first showing the field L/K has certain nice properties, and then one shows that any field with these properties has to be finite (this step is done using some Kummer theory, and some theorems from a first course in Algebraic Number Theory - the finiteness of class group, and Dirichlet’s Unit theorem).

Proposition. Let L = K([m]−1E(K)) be the ﬁeld deﬁned earlier.

(a) L/K is an abelian extension of exponent m (i.e. GL/K is abelian and every element has order dividing m). (b) Let 0 0 ∞ S = {ν ∈ MK : E has bad reduction at ν ∪ {ν ∈ MK : ν(m) 6= 0} ∪ MK }

Then L/K is unramiﬁed outside S (i.e. ν ∈ MK and ν∈ / S, then L/K is unramiﬁed at ν.)