Bob's Talk: October 28th
Theorem 1.2: Let K be a number field, E/K an elliptic curve with E(K)[2] = {0}. Let r be a non-negative integer, and
suppose that E has a quadratic twist E^{F_0} for which d_2(E^{F_0}/K) = r. Then:
a) If Gal(K(E[2])/K)) \cong S_3, then N_r(E,X) \gg X/(log X)^{2/3}.
b) If Gal(K(E[2])/K)) \dont Z/3Z, then N_(E,X) \gg X/(log X)^{1/3}.
(Recall d_2(E/K) = dim Sel_2(E/K).)
N_r(E,X) = # {quadratic F/K: d_2(E^F/K) = r, N_{K/Q} f(F/K) \leq X }.
Key Input from Section 3:
Corollary 3.4(ii) : Let K be a number field, E/K an elliptic curve with E(K)[2] = 0, F/K a quadratic extension. Suppose:
A)(1) all places v of additive reduction split in F
(2) all places v of multiplicative reduction and ord_v(Delta_E) is even, split in F.
(3) all v|2 split in F.
(4) All Archimedean v for which ord_v(Delta_E) is odd split in F.
B) All places v of multiplicative reduction where ord_v (Delta_E) is odd are unramified in F.
C) For each finite v whrere F/K is ramified, E(K_v)[2] = 0.
then d_2(E^F/K) = d_2(E/K).
Bob says: the argument may not be so clear unless the extensions M and N are linearly disjoint over K.
We want to produce a lot of quadratic F/K for which d_2(E^F/K) = r.
Construction: M/K finite Galois
N/K finite abelian
MN/K extension
In applications, M = K(E[2]).
N = ray class field to K for the conductor 8*Delta_E*oo. (The 8 esnures that 2 splis in F/K.)
Let S be a normal subset of Gal(M/K).
In the application, we take S = {s in Gal(M/K) such that |s| = 3}.
Let PP_S be the set of all primes p of K such that p is unramified in MN/K and for which the
Artin map [p,M/K] is contained in S.
Let NN = {prod_{i=1}^r p_i: p_1,...,p_r in PP_S are distinct}
NN_1 = {a in NN: [a,N/K] = 1}.
The consequence of this is that: if a is in NN_1, then since N is the (8oo \Delta_E)-ray class field of K, there
is an alpha in K such that a = (alpha) with alpha \equiv 1 \pmod 8 oo \Delta_E: in particular alpha is totally
positive and congruent to 1 mod 8. Take F = K(\sqrt{\alpha}). Then the conditions A) and B) of Corollary 3.4(ii)
are satisfied, so remains to look at C). Regarding C), the only places of K which are ramified in K(\alpha)/K
are the ones corresponding to primes p | (alpha) = p_1*...*p_r. (Note alpha \equiv 1 mod 8oo Delta_E, so no p_i
divides 2 \Delta_E.) For p = p_i, we required that the Frobenius class [p_i,M/K] \subset S, i.e., the Frobenius
has order 3. So the completion of K(E[2])/K is nonsplit over such p_i's.
Thus, for F = K(\sqrt{alpha}) with alpha the above generator of A in NN_1, we have d_2(E^F/K) = d_2(E/K).
Analytic claim: # {a in NN_1: N_{K/Q}(a) \leq X} > X/(log X)^a}.
where a = 2/3 in the S_3 case, 1/3 in the Z/3Z case.
Comment: Easy to reduce to the case where d_2(E/K) = r: If not, simply replace E by E^{F_0} and observe that the
conductor bounds change only by a constant. For the analytic bounds, observe that
the conductor of K(\sqrt{\alpha \alpha_0}) divides the product of the conductors of K(\sqrt{\alpha}) and
the conductors of K(\sqrt{\alpha_0}).
Now to the proof of the analytic claim:
Lemma 4.1: There exists a positive real constant C such that
|a in NN_1 : N_{K/Q} a < X| = C(1 + o(1)) X/(log X)^{1-|S|/[M:K]}.
Bob says: This is true when M and N are linearly disjoint over K. It is in general false!
So let's first suppose the linear disjointness condition. Put
A = Gal(N/K). If chi: A -> C^{\times} is a character, put
f_{chi}(s) = \sum_{a in NN} \chi(a)/(Na)^s = \prod_{p in PP_S} (1+\chi(p) (Np)^{-s}).
Absolutely convergent for Re(s) > 1.
Note: different from a usual Euler product which has a minus sign and an inverse, due to the restriction to
squarefree ideals.
log f_{chi}(s) = sum_{p in P_S} ==(Taylor expansion)== sum_{p in P_S} (\chi(p)/Np^s - 1/2 \chi(p)^2/Np^{2s} + 1/3 ...)
The sum of -1/2 \chi(p)^2/Np^{2s} + ... is holomorphic for Res) > 1/2.
sum_p sum_{m=2}^{oo} 1/p^{m\sigma} = (1 + 1/p^{sigma})(sum_{k=1}^{oo} 1/p^{2k sigma}) <= 2 \sum_n 1/n^{2\sigma}.
Therefore, up to a function which is analytic at s = 1, this is equal to
sum_{p in PP_S}) \chi(p)/Np^s.
Mazur and Rubin Claim: This is equivalent to \delta_X log(1/s-1), where
d_{\chi} = {0, \chi nontrivial
{|S|/[M:K], if \chi = \chi_0 is the trivial character
Suppose for simplicity that S = [sigma] is a single conjugacy class in Gal(M/K).
Write S_M = S, S_{MN} = { \tilde{sigma} in Gal(MN/K): \tilde{sigma}|M is in S}.
If \tilde{sigma} = (\sigma,eta_1), \tilde{\tau} = (\tau,\eta_2),
then \tilde(\tau} \tilde{\sigma} \tilde{tau}^{-1} = (\tau \sigma \tau^{-1}, \eta_1).
So conjugacy classes upstairs are conjugacy classes downstairs crossed with single elements of the abelian group A.
Then S_{MN} is a union of [N:K] classes in Gal(MN/K).
By Chebotarev for MN/K, {p of K: [MN/K,p] = class of (sigma,n)} has density |S|/([M:K]*[N:K]).
Bob says that some version of Chebotarev implies:
sum_{[MN/K,p] = Class(\sigma x \eta)} 1/Np^s \sim |S|/([M:K][N:K]) log(1/s-1)
So
sum_{[M/K,p] = [sigma] \chi(p)/Np^s = sum_{\tau in Gal(N/K)} \sum){[MN/K/p] = Class(\sigma x \tau) \chi(tau)/Np^s
\sim
(\sum_{\tau} \chi(\tau) |S|/([M:K][N:K]) log(1/s-1)
\sim {|S|/[M:K] log(1/s-1), chi = chi_0
{0, otherwise
Now consider a case of non-disjointness: M/K is abelian and M is contained in N.
N
| H
M } A
| G
K
Then, for sigma in Gal(M/K), Class(sigma) = sigma.
If sigma_1 in Gal(N/K) is a class lifting sigma, then the other elements are in sigma_1 + H.
sum_{[M/K,p] = [sigma]} \chi(p)/Np^s = \sum_{t \in H} \summ_{[N/K,p] =
\sigma_1 + \tau} \chi(\sigma_1 + \tau)/Np^s
chi: A -> C^{\times} is a homomorphism.
If chi is trivial on H, then all the terms are equal, and the sum becomes
\sum_{\tau in H} \chi(s) 1/[N:K] log(1/s-1) =
\chi(sigma) [N:M]/[N:K] log(1/s-1) = \chi(sigma)/[M:K] log(1/s-1).
Whereas if chi is not induced by a character of Gal(M/K),
then
\sum_{[M/K,p]} \chi(p)/Np^s \sim \chi(sigma) 1/[M:K] log(1/s-1)
whereas if \chi is not induced by a character of Gal(M/K), a similar computation shows that the sum is
equivalent to 0.
There exists a possibly different \delta'_{chi} such that
log f_{\chi}(s) = \delta'_{chi} log(1/s-1) + g_{chi}(s)
for g_{\chi}(s) holomorphic on Re(s) = 1.
Hence
f_{chi}(s) = (s-1)^{- delta'_{chi}} + exp(g_{chi}(s))
note the exponent is in general a complex number.
sum_{chi} f_{chi}(s) = \sum_{a in N} \sum_{chi} chi(a)/Na^s
=
{[N:K], if [a,N/K] = 1
{0 , if [a,N/K =/= 1
=
sum_{chi} (s-1)^{-\delta_{chi}} exp(g_{chi}(s))
So we need to know something about how many exponents -\delta_{chi} are nonzero.
Can work around this by looking carefully at all the cases.