[ Pobierz całość w formacie PDF ]
way round. In either case, it doesn t divide ad + bc, and so (ac : ad + bc : bd) is a primitive representative for R. It remains to show that 1 max(|ac|, |ad + bc|, |bd|) e" (max(|a|, |b|) (max |c||d|) , 2 but this is an elementary exercise. ELLIPTIC CURVES 67 Heights on E. Let E be the elliptic curve 2 E : Y Z = X3 + aXZ2 + bZ3, a, b " Q, " = 4a3 + 27b2 = 0. For P " E(Q), define H((x(P ) : z(P ))) if z(P ) = 0 H(P ) = 0 if P = (0 : 1 : 0). and h(P ) = log H(P ). Other definitions of h are possible, but they differ by bounded amounts, and therefore lead to the same canonical height (see below). Lemma 15.3. For any constant B, the set of P " E(Q) such that h(P ) Proof. Certainly, for any constant B, {P " P1(Q) | H(P ) d" B} is finite. But for every point (x0 : z0) " P1(Q), there are at most two points (x0 : y : z0) " E(Q), and so {P " E(Q) | H(P ) d" B} is finite. Proposition 15.4. There exists a constant A such that |h(2P ) - 4h(P )| d" A. Proof. Let P = (x : y : z) and 2P = (x2 : y2 : z2). According to the duplication formula (p37), (x2 : z2) = (F (x, z) : G(x, z)) where F (X, Z) and G(X, Z) are polynomials of degree 4 such that F (X, 1) = (3X2 + a)2 - 8X(X3 + ax + b) G(X, 1) = 4(X3 + aX + b). Since X3 + aX + b and its derivative 3X2 + b have no common root, neither do F (X, 1) and G(X, 1), and so Proposition 15.1 shows that |h(2P ) - 4h(P )| d" A for some constant A. Theorem 15.5. There exists a unique function h : E(Q) ’! R satisfying the conditions (a) and (b): (a) h(P ) - h(P ) is bounded; (b) h(2P ) = 4h(P ). In fact, h(2nP ) h(P ) = lim n’! " 4n and it has the following additional properties: (c) for any C e" 0, the set {P " E(Q) | h(P ) d" C} is finite; (d) h(P ) e" 0, with equality if and only if P has finite order. 68 J.S. MILNE Proof. We first prove uniqueness. If h satisfies (a) with bound B, then |h (2nP ) - h(2nP )| d" B. If in addition it satisfies (b), then h(2nP ) B d" , h (P ) - 4n 4n and so h(2nP )/4n converges to h (P ). To prove the existence, we first verify that h(2nP )/4n is a Cauchy sequence. From Propo- sition 15.4, we know that there exists a constant A such that |h(2P ) - 4h(P )| d" A for all P . For N e" M e" 0 and P0 " E(Q), h(2NP0) h(2MP0) N-1 h(2n+1P0) h(2nP0) - = - 4N 4M n=M 4n+1 4n N-1 1 d" |h(2n+1P0) - 4h(2nP0)| 4n+1 n=M N-1 1 d" A 4n+1 n=M A 1 1 d" 1 + + + · · · 4M+1 4 42 A = . 3 · 4M This shows that the sequence h(2nP )/4n is Cauchy, and we define h(P ) to be its limit. Because H(P ) is an integer e" 1, h(P ) e" 0 and h(P ) e" 0. When M = 0 the displayed equation becomes h(2NP ) A - h(P ) d" , 4N 3 and on letting N ’! " we obtain (a). For (b), note that h(2n+1P ) h(2n+1P ) h(2P ) = lim = 4 · lim = 4 · h(P ). n’! " n’! " 4n 4n+1 The set of P for which h(P ) d" C is finite, because h has this property and the difference h(P ) - h(P ) is bounded. If P is torsion, then {2nP | n e" 0} is finite, so h is bounded on it, by D say, and h(P ) = h(2nP )/4n d" D/4n for all n. On the other hand, if P has infinite order, then {2nP | n e" 0} is infinite and h is unbounded on it. Hence h(2nP ) > 1 for some n, and so h(P ) > 4-n > 0. ELLIPTIC CURVES 69 The function h is called18 the canonical, or Néron-Tate, height. If was defined indepen- dently by Tate using the above method, and by Néron using a much more elaborate method which, however, gives more information about h. Let f : M ’! K be a function from an abelian group M into a field K of characteristic = 2. Such an f is called a quadratic form if f(2x) = 4f(x) and B(x, y) =df f(x + y) - f(x) - f(y) is bi-additive. Then B is symmetric, and it is the only symmetric bi-additive form B : 1 M × M ’! K such that f(x) = B(x, x). We shall need the following criterion: 2 Lemma 15.6. A function f : M ’! K from an abelian group into a field K of characteristic = 2 is a quadratic form if and only if it satisfies the parallelogram19 law: f(x + y) + f(x - y) = 2f(x) + 2f(y) all x, y " M. Proof. On taking x = y = 0 in the parallelogram law, we find that f(0) = 0, on taking x = y we find that f(2x) = 4f(x), and on taking x = 0 we find that f(-y) = f(y). By symmetry, it remains to show that B(x + y, z) = B(x, z) + B(y, z), i.e., that f(x + y + z) - f(x + y) - f(x + z) - f(y + z) + f(x) + f(y) + f(z) = 0. Now four applications of the parallelogram law show that: f(x + y + z) + f(x + y - z) - 2f(x + y) - 2f(z) = 0 f(x - y + z) + f(x + y - z) - 2f(x) - 2f(y - z) = 0 f(x - y + z) + f(x + y + z) - 2f(x + z) - 2f(y) = 0 2f(y + z) + 2f(y - z) - 4f(y) - 4f(z) = 0. The alternating sum of these equations is the required equation. Proposition 15.7. The height function h : E(Q) ’! R is a quadratic form. We have to prove the parallelogram law. Lemma 15.8. There exists a constant C such that H(P1 + P2)H(P1 - P2) d" C · H(P1)2H(P2)2 for all P1, P2 " E(Q). Proof. Let P1 + P2 = P3 and P1 - P2 = P4, and let Pi = (xi : yi : zi). Then (x3x4 : x3z4 + x4z3 : z3z4) = (W0 : W1 : W2) where (see p37) W0 = (X2Z1 - X1Z2)2 4 4 W1 = 2(X1X2 + aZ1Z2)(X1Z2 + X2Z1) + 4bZ1Z2 2 2 2 2 2 2 W2 = X1 X2 - 2aX1X2Z1Z2 - 4b(X1Z1Z2 + X2Z1Z2) + a2Z1Z2. It follows that H(W0 : W1 : W2) d" CH(P1)2H(P2)2. 18 Unfortunately, there are different definitions of the canonical height, which differ by a constant factor. 19 2 In elementary linear algebra, the parallelogram law says that, for vectors u and v in Rn, u + v + 2 2 2 u - v = 2 u + 2 v . 70 J.S. MILNE According to Lemma 15.2, 1 H(W0 : W1 : W2) e" H(P3)H(P4). 2 Lemma 15.9. The canonical height function h : E(Q) ’! R satisfies the parallelogram law: h(P + Q) + h(P - Q) = 2h(P ) + 2h(Q). Proof. On taking logs in the previous lemma, we find that h(P + Q) + h(P - Q) d" 2h(P ) + 2h(Q) + B. On replacing P and Q with 2nP and 2nQ, dividing through by 4n, and letting n ’! ", we obtain the inequality h(P + Q) + h(P - Q) d" 2h(P ) + 2h(Q). Putting P = P + Q and Q = P - Q in this gives the reverse inequality: P + Q P - Q 1 1 h(P ) + h(Q ) d" 2h + 2h = h(P + Q ) + h(P - Q ). 2 2 2 2 Aside 15.10 (For the experts). Let K be a number field. For each prime v of K, let | · |v be the normalized valuation, for which the product formula holds: |a|v = 1, a " K×. v Define the height of a point P = (a0 : a1 : . . . : an) " Pn(K) to be H(P ) = max(|ai|v). i v Because of the product formula, H(P ) doesn t depend on the choice of (a0 : . . . : an) representing P . When K = Q, we can choose the ai to be integers with no common factor, which makes maxi |ai|p = 1 for all p, and leaves H(P ) = maxi |ai| ". With this definition, all the above results extend to elliptic curves over number fields. 16. Completion of the Proof of the Mordell-Weil Theorem, and Further Remarks Let P1, . . . , Ps be a set of representatives for E(Q)/2E(Q). For any Q " E(Q) there exists an i such that Q ± Pi " 2E(Q) for both choices of signs. According to the parallelogram law, h(Q ± Pi) d" h(Q) + h(Pi) for (at least) one choice of signs. For that choice, let Q ± Pi = 2Q . Then 4h(Q ) = h(Q ± Pi) d" h(Q) + h(Pi) d" h(Q) + C 1 where C = max h(Pi). Hence h(Q ) C. Now the argument 2 sketched in Section 11 shows that E(Q) is generated by P1, . . . , Ps and the Q with h(Q) d" C. ELLIPTIC CURVES 71 The Problem of Computing the Rank of E(Q). According to André Weil, one of the [ Pobierz caÅ‚ość w formacie PDF ] |