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contain all the points of order n on E(Qal). The next lemma shows that, in order to prove that S(n)(E/Q) is finite, it suffices to prove that S(n)(E/L) is finite. Lemma 14.8. For any finite Galois extension L of Q and any n, the kernel of S(n)(E/Q) ’! S(n)(E/L) is finite. Proof. Since S(n)(E/Q) and S(n)(E/L) are subgroups of H1(Q, En) and H1(L, En) respec- tively, it suffices to prove that the kernel of H1(Q, En) ’! H1(L, En) is finite. But (cf. 12.6), this kernel is H1(Gal(L/Q), En(L)), which is finite because both Gal(L/Q) and En(L) are finite. It remains to consider (c). The proof of its analogue for L requires the three fundamental theorems in any course on algebraic number theory. We review their statements. Review of algebraic number theory. In the following, L is a finite extension of Q and R is the ring of all algebraic integers in L (see p53). Every element of R is a product of irreducible (i.e., unfactorable ) elements, but this " factorization may not be unique. For example, in Z[ -5] we have " " 6 = 2 · 3 = (1 + -5)(1 - -5) " " and 2, 3, 1 + -5, 1- -5 are irreducible with no two associates. The idea of Kummer and Dedekind to remedy this problem was to enlarge the set of numbers with ideal numbers , now called ideals, to recover unique factorization. For ideals a and b, define ab = { aibi | ai " a, bi " b}. It is again is an ideal. Theorem 14.9 (Dedekind). Every ideal in R can be written uniquely as a product of prime ideals. " For example, in Z[ -5], " " " " (6) = (2, 1 + -5)(2, 1 - -5)(3, 1 + -5)(3, 1 - -5). 64 J.S. MILNE For an element a " R and a prime ideal p in R, let ord (a) be the exponent of p in the unique factorization of the ideal (a), so that (a) = pord (a). a For x = " L, define ord (x) = ord (a) - ord (b). The ideal class group C of R is defined b to be the cokernel of the homomorphism L× ’! Z ’! C ’! 0 ‚"R, prime x ’! (ord (x)). It is 0 if and only if R is a principal ideal domain, and so C can be regarded as giving a measure of the failure of unique factorization of elements in R. Theorem 14.10 (Finiteness of the class number). The ideal class group C is finite. We next need to "understand the group U"of units in R. For R = Z, U = {±1}, but " already for R = Z[ 2], U is infinite because 2 + 1 is a unit in Z[ 2]. One can show that " " Z[ 2]× = {±(1 + 2)n | n " Z} H" Z/2Z •" Z. Theorem 14.11 (Dedekind unit theorem). The group U of units of R is finitely gener- ated. In fact, the full theorem gives a formula for the rank of U. As in any commutative ring, a is a unit in R if and only if (a) = R. In our case, this is equivalent to saying that ord (a) = 0 for all prime ideals p, and so we have an exact sequence 0 ’! U ’! L× ’! •" Z ’! C ’! 0 with U finitely generated and C finite. The fundamental theorems of algebraic number theory show, more generally, that, when T is a finite set of prime ideals in L, the groups UT and CT defined by the exactness of a ’!(ord (a)) 0 ’! UT ’! L× - ’! •" "T Z ’! CT ’! 0 ------ / are, respectively, finitely generated and finite. Completion of the proof of the finiteness of the Selmer group. Lemma 14.12. Let N be the kernel of a ’! (ord (a) mod n) : Ker(L×/L×n) ’! •" "T Z/nZ). / Then there is an exact sequence n 0 ’! UT /UT ’! N ’! (CT )n Proof. Let ± " N. Then n|ord (±) for all p " T , and so we can map ± to the class c of / (ord (±)) in CT . Clearly nc = 0, and any element of CT killed by n arises in this way. If n c = 0, then there exists a ² " L× such that ord (²) = ord (±)/n for all p. Now ±/²n lies in n UT , and is well-defined up to an element of UT . Now the argument used in the special case shows that S(n)(E/L) is finite. ELLIPTIC CURVES 65 Remark 14.13. The above proof of the finiteness of the Selmer group is taken from my book, Etale Cohomology, p133. It is simpler than the standard proof (see [S1] p190 196) which unnecessarily translate[s] the putative finiteness of E(K)/mE(K) into a statement about certain field extensions of K. 15. Heights Let P = (a0 : . . . : an) " Pn(Q). We shall say that (a0 : . . . : an) is a primitive representative for P if ai " Z, gcd(a0, . . . , an) = 1. The height H(P ) of P is then defined to be H(P ) = max |ai|. j Here | " | is the usual absolute value. The logarithmic height h(P ) of P is defined to be log H(P ). Heights on P1. Let F (X, Y ) and G(X, Y ) be homogeneous polynomials of degree m in Q[X, Y ], and let V (Q) be the set of their common zeros. Then F and G define a map Õ : P1(Q) \ V (Q) ’! P1(Q), (x : y) ’! (F (x, y) : G(x, y)). Proposition 15.1. If F (X, Y ) and G(X, Y ) have no common zero in P1(Qal), then there exists a constant B such that |h(Õ(P )) - mh(P )| d" B, all P " P1(Q). Proof. We may suppose that F and G have integer coefficients. Let (a : b) be a primitive rep- m-i resentative for P . Then, for a monomial H(X, Y ) = cXiY , |H(a, b)| d" |c| max(|a|m, |b|m), and so |F (a, b)|, |G(a, b)| d" C (max(|a|, |b|)m with C = (m + 1) max(|coeff. of F or G|). Now H(Õ(P )) d" max(|F (a, b)|, |G(a, b)|) d" C(max(|a|, |b|)m = C · H(P )m. On taking logs, we obtain the inequality h(Õ(P )) d" mh(P ) + log C. The problem with proving a reverse inequality is that F (a, b) and G(a, b) may have a large common factor, and so the first inequality in the second last equation may be strict. We use the hypothesis that F and G have no common zero in Qal to limit this problem. Let R be the resultant of F and G the hypothesis says that R = 0. Consider -m -m Y F (X, Y ) = F (X , 1) and Y G(X, Y ) = G(X , 1). When regarded as polynomials in the Y Y X single variable , F (X , 1) and G(X , 1) have the same resultant as F (X, Y ) and G(X, Y ), Y Y Y and so (see p55), there are polynomials U(X ), V (X ) " Z[X ] of degree m - 1 such that Y Y Y X X X X U( )F ( , 1) + V ( )G( , 1) = R. Y Y Y Y 66 J.S. MILNE 2m-1 m-1 m-1 On multiplying through by Y and renaming Y U(X ) as U(X, Y ) and Y V (X ) as Y Y V (X, Y ), we obtain the equation 2m-1 U(X, Y )F (X, Y ) + V (X, Y )G(X, Y ) = RY . Similarly, there are homogenous polynomials U (X, Y ) and V (X, Y ) of degree m - 1 such that U (X, Y )F (X, Y ) + V (X, Y )G(X, Y ) = RX2m-1. Substitute (a, b) for (X, Y ) to obtain the equations U(a, b)F (a, b) + V (a, b)G(a, b) = Rb2m-1, U (a, b)F (a, b) + V (a, b)G(a, b) = Ra2m-1. From these equations we see that gcd(F (a, b), G(a, b)) divides gcd(Ra2m-1, Rb2m-1) = R. Moreover, as in the first part of the proof, there is a C > 0 such that U(a, b), U (a, b), V (a, b), V (a, b) d" C (max |a|, |b|)m-1 . Therefore 2C (max |a|, |b|)m-1 (max |F (a, b)|, |G(a, b)|) e" |R||a|2m-1, |R||b|2m-1. Together with gcd(F (a, b), G(a, b))|R, these inequalities imply that 1 1 H(Õ(P )) e" max(|F (a, b)|, |G(a, b)|) e" H(P )m. |R| 2C On taking logs, we obtain the inequality h(Õ(P )) e" mh(P ) - log 2C. There is a well-defined map (special case of the Veronese map) (a : b), (c : d) ’! (ac : ad + bc : bd) : P1 × P1 ’! P2. Let R be the image of (P, Q). Lemma 15.2. 1 H(R) d" d" 2. 2 H(P )H(Q) Proof. Choose (a : b) and (c : d) to be primitive representatives of P and Q. Then H(R) d" max(|ac|, |ad + bc|, |bd|) d" 2 max(|a|, |b|) max(|c|, |d|) = 2H(P )H(Q). [ Pobierz caÅ‚ość w formacie PDF ] |