Vol. 5 No 2 - Pi Mu Epsilon

-. n(P 2 m and n(P ) > m, we have n(P3) 2 m, so P3 is in rm.1 2 -rm istherefore a subgroup of T.Theorem 3.3 I' has finite index in r, for integers m > 0.mLet Pl = (xl,yl) and P = (x2,y2) be two points in r.need to consider conditions under which (P2-Pl) E rm,We may assume the P,P2 f! r , say n(Pl) = n(P2) = n < m. Put-2n -3n) -2n -3n)P + ( CP ,6^p and P2 = (C2p ,i2p with612 = C13 t A5p4" t Bp6" and 6; = 6; + AC2p4n 6nt BP , where El,C2, 6. and i2 are units.From the addition formula, n(P-P) 2 m if and only ifWeIn particular, when n = m - 1 this shows that rm-l/l'm is finite form > 1. [The lifting procedure described in 2. shows that the indexof rm-l in rm is exactly 0. Also, it should be noted that these argumentsrequire that singularities be avoided.] This shows that I' is of finiteindex in rl.We still must show that r/rl is finite. Let n(P) = n(P) = 0.We claim that n(P 2- P 1 )> m when C2 2 El and 6- E 61 modulo a sufficientlyhigh power of p. Say C2 2 El, 6 2 5 6 1 (mod for sufficientlylarge r.2If not, n(P 2 -P 1 ) < m implies ~ (6~t6~) i. r and ~ ( c ~ ~ +A)> ~ C r ~ C ~by the above argument. Then, ~(26~) = ~((6,+6~)-(6,-6~)).> r andBut62+61 2^1so n P ) m if and only if v(-) = n+v&) < n-m.2 1 62-61 2-"1Thus, v(C -6 ) .> m-n and ~ (6~-6~) 2 m-n is clearly a necessary2 - 1 -condition for n(P2-P) i m.If P # 2, S t 6 is a unit, so v(6,+6)is a sufficient condition.If p = 2, write(62+61)(62-61) = 6^-6 1 = 6; -£4n+ A(E2-E1)p .= 0. Then, (E2-C1) i m-n2 2 2 4nSince P # 3, E2 +62~ltC12 is a unit; and then so is 5 tC2CltCl tAp .From the above, we knowHowever, we have the identity2-@A3 - 27B2 = (x3tAx+B)~(x) + (3x tA)Q(x) where P(x) = 1BAx - 27B.2and Q(x) = -6Ax + 9Bx - @A2.&'