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18.782 Arithmetic Geometry Lecture Note 26

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18.782 Arithmetic Geometry Lecture Note 26 18.782 Introduction to Arithmetic Geometry Fall 2013 Lecture #26 12/10/2013 26.1 Genus 1 curves with no rational points Let C=k be a (smooth, projective, geometrically irreducible) curve of genus 1 over a perfect field k. Let n be the least positive integer for which Divk C contains an effective divisor D of degree n (such divisors exist; take the pole divisor of any non-constant function in k(C), for example). If C has a k-rational point, then n = 1 and C is an elliptic curve. We now consider the case where C does not have a rational point, so n > 1. We have deg(D) = n > 2g−2 = 0, so the Riemann-Roch theorem implies `(D) = deg(D) + 1 − g = n; and for any positive integer m we have `(mD) = deg(mD) + 1 − g = mn: We now analyze the situation for some specific values of n. 26.1.1 The case n = 2 We have `(D) = 2, so let f1; xg be a basis for L(D). Then `(2D) = 4, so in addition to f1; x; x2g, the Riemann-Roch space L(2D) contains a fourth linearly independent function y. We then have f1; x; x2; y; xy; x3g as a basis for L(3D), but L(4D) is an 8-dimensional vector space containing the 9 functions f1; x; x2; y; xy; x3; x2y; x4; y2g, so there is a linear relation among them, and this linear relation must have nonzero coefficient on both y2 and x4. Assuming we are not in characteristic 2, we can complete the square in y to obtain an equation of the form y2 = f(x) where f is a quartic polynomial over k. The polynomial f must be squarefree, and it cannot have any k-rational roots (otherwise we would have a rational point). Note that the homegenization of this equation is singular at (0 : 1 : 0), but its desingularization is a curve 3 in P . Using the same argument as used on the problem set for hyperelliptic curves, one can show that every curve defined by an equation of this form has genus 1. 26.1.2 The case n = 3 We have `(D) = 3, so let f1; x; yg be a basis for L(D). The 10 functions f1; x; y; x2; xy; y2; x3; x2y; xy2; y3g all lie in the 9-dimensional Riemann-Roch space L(3D), hence there is a linear relation among them that defines a plane cubic curve without any rational points. Conversely, every plane cubic curve has genus 1, since over a finite extension of k we can put the curve in Weierstrass form, which we have already proved has genus 1 (recall that genus is preserved under base extension of a perfect field). An example of a plane cubic curve with no rational points was given on the problem set, and here is another one: 3x3 + 4y3 + 5z3 = 0: 1 Andrew V. Sutherland Unlike the example on the problem set, this curve has a rational point locally everywhere, that is, over every completion of Q. As noted back in Lecture 3, every geometrically irreducible plane curve has rational points modulo p for all sufficiently large primes p, and in this example the only primes that we need to check are 2, 3, and 5; it is easy to check that there are rational solutions modulo each of these primes, and modulo 33. Using Hensel's 3 lemma, solutions modulo p (or p , for p = 3) can be lifted to Qp, and there are clearly solutions over R = Q1 26.1.3 The case n = 4 We have `(D) = 4, so let f1; x; y; zg be a basis for L(D). The 10 functions f1; x; y; z; x2; y2; z2; xy; xz; yzg all lie in the 8-dimensional Riemann-Roch space L(2D), hence there are two independent 3 linear relations among them, each corresponding to a quadratic form in P , and C is the intersection of two quadric hypersurfaces (its clear that C is contained in the intersection, and one can show that it is equal to the intersection by comparing degrees). 26.2 The case n > 4 One can continue in a similar fashion for n > 4; indeed, by a theorem of Lang and Tate, over Q there are genus 1 curves that exhibit every possible value of n. But the situation becomes quite complicated already for n = 5: we have f1; w; x; y; zg as a basis for L(D) and in L(2D) we get 15 functions in a Riemann-Roch space of dimension 10.1 26.3 Twists of elliptic curves A genus one curve C=k with no k-rational points is not an elliptic curve, but for some finite extension L=k the set C(L) will be nonempty; thus if base-extend C to L, we obtain an elliptic curve over L. We will show, this elliptic curve can be defined by a Weierstrass equation whose coefficients actually lie in k, so it is also the base-extension of an elliptic curve E=k. The curves E and C are clearly not isomorphic over k, since E has a k-rational point and C does not, but they become isomorphic when we base-extend to L. In other words, the isomorphism ': C ! E is defined over L, but not over k, so the distinguished k-rational point O on E is the image of an L-rational point on C that is not defined over k. Definition 26.1. Two varieties defined over a field k that are related by an isomorphism defined over k¯ are said to be twists of each other. In order to characterize the curves that are twists of a given elliptic curve E=k, we introduce the j-invariant. For simplicity, we will assume henceforth that char(k)6 = 2; 3, so that we can put our elliptic curves in short Weierstrass form. But the j-invariant can also be defined in terms of a general Weierstrass equation and except where we explicitly note otherwise, all the theorems we will prove are true in any characteristic. 1 3 Note that while every curve can be smoothly embedded in P , this embedding will not necessarily be n−1 defined over k. Over k, P is the best we can do. 2 2 3 Definition 26.2. Let E=k be an elliptic curve with Weierstrass equation y = x +a4x+a6. The j-invariant of E is 3 4a4 j(E) := 1728 3 2 : 4a4 + 27a6 3 2 Note that the denominator is always nonzero, since ∆(E) = −16(4a + 27a6)6 = 0. Theorem 26.3. For every j 2 k there exists an elliptic curve E=k with j(E) = j. 2 3 Proof. We define such an E=k via an equation y = x + a4x + a6 as follows. If j = 0, let a4 = 0 and a6 = 1, and if j = 1728, let a4 = 1 and a6 = 0. Otherwise, let a4 = 3j(1728 − j) 2 and a6 = 2j(1728 − j) . One can check that ∆(E)6 = 0 and j(E) = j in each case. Theorem 26.4. Two elliptic curves defined over k have the same j-invariant if and only if they are isomorphic over k¯. 2 3 2 3 0 0 Proof. For the forward implication, let y = x + a4x + a6 and y = x + a4x + a6 be Weierstrass equations for elliptic curve E=k and E0=k, respectively, with j(E) = j(E0) = j. 0 0 If j = 0 then a4 = a4 = 0, and we can make a6 = a6 by a linear change of variables 0 0 defined over a suitable extension of k, hence E 'k¯ E . If j = 1728 then a6 = a6 = 0, 0 and we can similarly make a4 = a4 via a change of variables over a suitable extension of k. 0 Otherwise, over a suitable extension of k we can make a4 and a4 both equal to 1, and then 0 0 0 0 j(E) = j(E ) ) a6 = a6. Thus in every case, j(E) = j(E ) ) E 'k¯ E . 3 For the reverse implication, we note that the cubic x + a4x + a6 is uniquely determined by its roots, which are precisely the x-coordinates fx1; x2; x3g of the three points of order 2 ¯ 0 2 0 in E(k). If E 'k¯ E , then both curves can be embedded in P so that E[2] = E [2], and they will then have the same Weierstrass equation, hence the same j-invariant. Corollary 26.5. Let C=k be a genus one curve and let O and O0 be any two points in C(k¯). Then the elliptic curves (C; O) and (C; O0) over k¯ have the same j-invariant. Proof. The translation-by-O0 map on (C; O) is an isomorphism from (C; O) to (C; O0). It follows from the corollary that the j-invariant of an elliptic curve (E; O) is independent of the choice of O, it depends only on the curve E. Definition 26.6. Let C=k be a curve of genus one. The j-invariant j(C) of C is the j-invariant of the elliptic curve (C; O) over k¯, for any O 2 C(k¯). Theorem 26.7. Let C=k be a curve of genus one. Then j(C) 2 k. Proof. Let us pick O 2 C(L), where L is some finite Galois extension L=k, and let E=L be the elliptic curve (C; O).
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