Elliptic Curves Dr T. Fisher Lent 2008 LATEXed by Sebastian Pancratz ii These notes are based on a course of lectures given by Dr T. Fisher in Part III of the Mathematical Tripos at the University of Cambridge in the academic year 2007(cid:21)2008. ThesenoteshavenotbeencheckedbyDrT.Fisherandshouldnotberegardedaso(cid:30)cial notes for the course. In particular, the responsibility for any errors is mine (cid:22) please email Sebastian Pancratz (sfp25) with any comments or corrections. Contents 1 Fermat’s Method of Descent 1 2 Some Remarks on Plane Cubics 5 3 Weierstrass Equations 9 4 The Group Law 13 5 Isogenies 17 6 The Invariant Di(cid:27)erential 21 7 Formal Groups 25 8 Elliptic Curves over Local Fields 29 9 Elliptic Curves over Number Fields: The Torsion Subgroup 35 10 Kummer Theory 39 11 Elliptic Curves over Number Fields: The Mordell(cid:21)Weil Theorem 41 12 Heights 43 13 Dual Isogenies 47 14 Galois Cohomology 49 15 Weil Pairing 53 16 Decent by Cyclic Isogeny 57 Chapter 1 Fermat’s Method of Descent Consider a right-angled triangle ABC A ooooooocoooooo b B a C with a2+b2 = c2 and area (ABC) = ab/2. Lemma 1.1. Every primitive triangle is of the form oooouo2o+oovo2oooo 2uv u2−v2 for some u,v ∈ Z with u > v > 0. Proof. Withoutlossofgenerality,aisodd,bisevenandcisodd. Nowrewritea2+b2 = c2 as(cid:0)b(cid:1)2 = c+ac−a where c±a arecoprimepositiveintegerswithproductasquare. Unique 2 2 2 2 factorisation in Z gives c+a = u2, c−a = v2 for some u,v ∈ Z so a = u2 −v2, b = 2uv, 2 2 c = u2+v2. De(cid:28)nition. D ∈ Q is congruent if there is a rational triangle with area D. >0 Remark. It su(cid:30)ces to consider D a square-free integer. Example. 5 and 6 are congruent numbers. Remark. D is congruent if and only if Dw2 = uv(u2 −v2) for some u,v,w ∈ Z with w 6= 0, which is equivalent to Dy2 = x3 −x by setting x = u/v, y = w/v2 for some x,y ∈ Q with y 6= 0. Theorem 1.2 (Fermat). 1 is not congruent, that is, there is no solution to w2 = uv(u+v)(u−v) (∗) with u,v,w ∈ Z and w 6= 0. 2 Fermat’s Method of Descent Proof. Without loss of generality, u and v are coprime with u > 0. Moreover, if v < 0 then we replace (u,v) with (−v,u), and if u ≡ v (mod 2) then we replace (u,v,w) by (cid:0)u+v, u−v, w(cid:1). 2 2 2 Now u, v, u+v and u−v are positive coprime integers with product a square. Unique factorisation in Z gives u = a2, v = b2, u+v = c2, u−v = d2 for some a,b,c,d ∈ N. As u 6≡ v (mod 2), c and d are odd. Consider the triangle ooooooaooooooo c−2d c+d 2 This is a primitive triangle with area 1(cid:16)c+d(cid:17)(cid:16)c−d(cid:17) c2−d2 v (cid:16)b(cid:17)2 = = = . 2 2 2 8 4 2 Put w = b/2 and note that this is an integer. By Lemma 1.1, 1 w2 = u v (u2−v2) 1 1 1 1 1 for some u ,v ∈ Z. So (u ,v ,w ) is another solution to (∗). But 4w2 = b2 = v | w2 1 1 1 1 1 1 so |w | < |w|. Thus, by Fermat’s method if in(cid:28)nite descent, there are no solutions to 1 (∗). Remark. For a right-angled triangle with sides a, b and c the three numbers (cid:0)a−b(cid:1)2, 2 (cid:0)c(cid:1)2 and (cid:0)a+b(cid:1)2 form an arithmetic progression of length 3 with common di(cid:27)erence 2 2 D = ab/2. 1.1 A Variant for Polynomials Let K be a (cid:28)eld with char(K) 6= 2. Theorem 1.3. Let u,v ∈ K[t] be coprime. If αu+βv is a square for distinct ratios (α : β) ∈ P1 then u,v ∈ K. Proof. We may assume that K = K¯. Changing coordinates on P1 via M(cid:246)bius maps, we may assume that the four ratios are (1 : 0), (0 : 1, (1 : −1) and (1 : −λ) for some λ ∈ K with λ 6= 0,1. u = a2, u−v = c2 = (a−b)(a+b) v = b2, u−λv = d2 = (a−µb)(a+µb) √ where µ = λ. u and v are coprime so a and b are coprime, hence a−b and a+b are coprime. Unique factorisation in K[t] gives that a−b and a+b are squares. Similarly, a−µb and a+µb are squares. Now max{deg(a),deg(b)} < max{deg(u),deg(v)} and so we are done by Fermat’s method of in(cid:28)nite descent. 3 De(cid:28)nition. (i) E/K is the projective closure of an a(cid:30)ne curve y2 = f(x) where f(X) ∈ K[X] is a monic degree 3 polynomial with distinct roots over K¯. (ii) For any (cid:28)eld extension L/K, E(L) = {(x,y) ∈ L2 : y2 = f(x)}∪{O } E where O is the point at in(cid:28)nity. E Fact. E(L) is naturally an abelian group with identity O . E Example. E: y2 = x3−x, E(Q) = {O ,(0,0),(±1,0)}. E Corollary 1.4. If E/K is an elliptic curve then E(K(t)) = E(K). Proof. Without loss of generality, K = K¯. By substitutions in x, we may assume E is of the form E: y2 = x(x−1)(x−λ) for some λ ∈ K−{0,1}. If (x,y) ∈ E(K(t)), write x = u/v with u,v ∈ K[t] coprime so that w2 = uv(u−v)(u−λv) for some w ∈ K[t]. Unique factorisation in K[t] implies that u, v, u − v and u − λv are all squares. By Theorem 1.3, u,v ∈ K so x,y ∈ K. Chapter 2 Some Remarks on Plane Cubics For now, consider a (cid:28)eld K = K¯ with char(K) 6= 2. De(cid:28)nition. An algebraic curve C is rational if it is birational to P1. Example. A plane a(cid:30)ne curve C = {f(x,y) = 0} ⊂ A2 is rational if there exists φ,ψ ∈ K(t) such that (i) A1 → A2,t 7→ (φ(t),ψ(t)) is injective on A1−{(cid:28)nite set}, (ii) f(φ(t),ψ(t)) = 0. Example. • Any non-singular conic is rational. (−1,0l)l•llllllllllllllll•ll y=t(x+1) and (x,y) = (cid:0)1−t2, 2t (cid:1). 1+t2 1+t2 • Any singular cubic is rational. y2=x3 y2=x2(x+1) •y=tx (x,y)=(t2,t3) • • • By Corollary 1.4, elliptic curves are not rational. 6 Some Remarks on Plane Cubics Theorem 2.1. Let C ⊂ P2 be a smooth projection cubic. Then we can change coordi- nates on P2 such that C: Y2Z = Z3f(X/Z) where f(X) ∈ K[X] is a monic cubic polynomial with distinct roots over K¯, i.e., C is an elliptic curve. Theorem 2.2 (Bezout’s Theorem). Plane curves C,D ⊂ P2 of degrees m,n with no common components meet in exactly mn points, counted with multiplicity: X (C.D) = mn. P P∈C∩D Here are some properties of (C.D) : P (i) (C.D) ≥ 1 if and only if P ∈ C ∩D. P (ii) (C.D) = 1 if and only of P is a smooth point of C and D, and T C 6= T D. P P P (iii) If C = {F = 0} ⊂ P2 and D is the line through P and P then (C.D) = 0 1 P0 ord F(P +tP ). t=0 0 1 De(cid:28)nition. P ∈ C is a point of in(cid:29)ection, or (cid:29)ex, if (C.T C) ≥ 3. For a smooth P P plane curve C = {F = 0} ⊂ P2 of degree d, the Hessian is detH(X ,X ,X ) = 1 2 3 det(∂2F/∂X ∂X ) and has degree 3(d−2). i j i,j=1,2,3 Lemma 2.3. Assume char(K) - 2(d−1). Then P ∈ C is a (cid:29)ex if and only if H(P) = 0. Proof. Taylor expand to obtain X ∂F 1 X ∂2F F(P +X) = F(P)+ (P)X + (P)X X +··· . i i j ∂X 2 ∂X ∂X i i j i i,j The tangent line is nX ∂F o T C = (P)X = 0 P i ∂X i i and a conic Q de(cid:28)ned by nX ∂2F o Q = (P)X X = 0 . i j ∂X ∂X i j i,j Then P is a (cid:29)ex if and only if T C ⊂ Q =⇒ Q is singular if and only if H(P) = 0. P We claim that P is a smooth point on Q and T Q = T C. Granted the claim, if Q is P P singular then T C ⊂ Q. To show the claim, note that F is homogeneous of degree d so P F(λX) = λdF(X). Taking ∂/∂λ and set λ = 1, X ∂F X = dF i ∂X i i where ∂F/∂X is homogeneous of degree d−1. Repeating this gives i X ∂F X X = d(d−1)F i j ∂X ∂X i j i,j