the strong form of artin’s primitive root conjecture (1927
TRANSCRIPT
The strong form of Artin’s primitive root
conjecture (1927) for CM elliptic curves
Cristian Virdol
Department of Mathematics
Yonsei University
May 24, 2016
Abstract
Consider E a CM elliptic curve over Q. Assume that rankQE � g
where g � 0 is an integer, and let a1, . . . , ag 2 E(Q) be linearly inde-pendent points (so in particular a1, . . . , ag have infinite order, and wheng = 0 the set {a1, . . . , ag} is empty). For p a rational prime, we de-note by Fp the residue field at p. If E has good reduction at p, let E
be the reduction of E at p, let ai for i = 1, . . . , g, be the reduction ofai(modulo p), and let ha1, . . . , agi be the subgroup of E(Fp) generatedby a1, . . . , ag. Assume that Q(E[2], 2�1
a1, . . . , 2�1
ag) 6= Q. Then, in thispaper we obtain an asymptotic formula for the number of rational primes
p, with p x, for whichE(Fp)
ha1
,...,agi is cyclic, and we prove that the num-
ber of primes p, for whichE(Fp)
ha1
,...,agi is cyclic, is infinite. This result is
the right generalization of Hooley’s proof, under GRH, of the strong formof the classical Artin’s primitive root conjecture, in the context of ellip-tic curves, i.e. this result is an unconditional proof of the strong formof Artin’s primitive root conjecture for CM elliptic curves. Artin’s con-jecture (1927) states that, for any integer a 6= ±1 or a perfect square(or equivalently a 6= 1, and Q(±1,
pa) = Q(1[2], 2�1
a) 6= Q), there areinfinitely many primes p for which a is a primitive root (mod p), andan asymptotic formula for such primes is satisfied (this conjecture is notknown for any specific a; not even the infinity part of Artin’s conjecture isnot known for any specific a). So in this paper we prove the strong formof Artin’s primitive root conjecture (1927) for CM elliptic curves, i.e. weprove the infinity + the asymptotic formula for ALL a 6= ±1 or a perfectsquare, which for CM elliptic curves E is equivalent to the statement thatQ(E[2], 2�1
a1, . . . , 2�1
ag) = Q(0[2], 2�1a1, . . . , 2
�1ag) 6= Q.
We remark that unconditionally the infinity part of Artin’s primitiveroot conjecture (1927) for abelian varieties was proved in:
[V] C. Virdol, Artin’s conjecture for abelian varieties, Kyoto Journalof Mathematics, accepted.
1
2010 Mathematics subject classification: 11G10, 11G15.Keywords: Elliptic curves; Artin’s conjecture; primitive-cyclic points.
1 Introduction
Let E be an elliptic curve defined over Q, of conductor N . Let P
E
be the setof rational primes p of good reduction for E, (i.e. (p,N) = 1). For p 2 P
E
, wedenote by E the reduction of E at p.
We have that E(Fp
) ✓ E[m](Fp
) ✓ (Z/mZ)2 for any positive integer msatisfying |E(F
p
)||m. Hence
E(Fp
) ' Z/m1
Z⇥ Z/m2
Z, (1.1)
where mi
2 Z�1
, and m1
|m2
. Each Z/mi
Z is called a cyclic component ofE(F
p
). If m1
= 1 we say that E(Fp
) is cyclic.Assume that rankQE � g where g � 0 is an integer, and let a
1
, . . . , ag
2
E(Q) be linearly independent points (so in particular ai
, for i = 1, . . . , g, hasinfinite order). Let a
i
, for i = 1, . . . , g, be the reduction of ai
(modulo p), and letha
1
, . . . , ag
i be the subgroup of E(Fp
) generated by a1
, . . . , ag
. From above we
know that¯
E(Fp)
ha1
,...,agi has at most 2 cyclic components. We call a := (a1
, . . . , ag
)
a primitive-cyclic tuple for p if¯
E(Fp)
ha1
,...,agi is cyclic.
For x 2 R, define
fE,a
(x) = |{p 2 P
E
|p x,E(F
p
)
ha1
, . . . , ag
i
is cyclic}|.
Let Q(E[m]) be the field obtained by adjoining to Q the m-division points E[m]of E, and let Q(E[m],m�1a) := Q(E[m],m�1a
1
, . . . ,m�1ag
) be the field ob-tained by adjoining toQ them-division points E[m] of E, andm�1a
1
, . . . ,m�1ag
2
E(Q).In this paper we prove the following results (see also Remark 6.1 below):
Theorem 1.1. Let E be a CM elliptic curve defined over Q, of conductor N ,with CM by the full ring of integers O
K
of an imaginary quadratic filed K.Assume that rankQE � g where g � 0 is an integer, and let a
1
, . . . , ag
2 E(Q)be linearly independent points. Define a := (a
1
, . . . , ag
) and Q(E[m],m�1a) :=Q(E[m],m�1a
1
, . . . ,m�1ag
) for m positive integer. Then, we have
fE,a
(x) = cE,a
li x+O(x
(log x)(log log log x)),
where li x :=Rx
2
1
log t
dt, and
cE,a
=1X
m=1
µ(m)
[Q(E[m],m�1a) : Q],
where µ(·) is the Mobius function.
2
Theorem 1.2. Under the same assumptions as in Theorem 1.1, we have thatcE,a
6= 0 if and only if Q(E[2], 2�1a) 6= Q. Moreover if Q(E[2], 2�1a) = Q, thenthe function f
E,a
(x) is bounded.
Let a 2 Z, a 6= 0. Assume that a is linearly independent in the multiplicativegroup Q⇥, i.e. that a 6= ±1. For x 2 R, define
fa(x) := |{p| p x, hai = F⇥p
}|.
Note that Theorem 1.1 and 1.2 are precise analogues of the following two theo-rems, due to Hooley [H]:
Theorem 1.3. Let a 2 Z, a 6= 0. Assume that a is linearly independent inthe multiplicative group Q⇥, i.e. that a 6= ±1. Assume that the GeneralizedRiemann Hypothesis (GRH) holds for the Dedekind zeta functions of the fieldsQ( mp
1, mp
a) for all positive integers m. Then
fa(x) = cali x+O(x log log x
log2 x),
where
ca =1X
m=1
µ(m)
[Q( mp
1, mp
a) : Q].
Theorem 1.4. Under the same assumptions as in Theorem 1.3, we have thatca 6= 0 if and only if a 6= a perfect square, or equivalently if and only ifQ(1[2], 2�1
a) 6= Q. Moreover if Q(1[2], 2�1
a) = Q, then the function fa(x)is bounded.
We remark that if Q(E[2], 2�1a1
, . . . , 2�1ag
) = Q, then for all odd rational
primes p of good reduction for E, we have E[2](Fp
) ⇢¯
E(Fp)
ha1
,...,agi and E[2](Fp
) '
(Z/2Z)2, and thus in this case, as is stated in Theorem 1.2 above, fE,a
(x) isbounded. In a similar way one can prove the assertion from Theorem 1.4 above,that if Q(1[2], 2�1
a) = Q, then the function fa(x) is bounded.Here is a brief history of Artin’s primitive root conjecture (1927), and of
Artin’s primitive root conjecture (1927) for elliptic curves from this paper.It was Gauss who in the section of the Disquisiones Arithmeticae, consid-
ered the problem of determining the primes p for which a given number a isa primitive root, modulo p, for the particular case a = 10. On September 27,1927, according to Helmut Hasse’s diary, Emil Artin formulated his celebratedconjecture. This conjecture was proved in 1967, under GRH, by Hooley ([H]),in its strong form for all integers a 6= ±1 or a perfect square (see Theorems 1.3and 1.4 above). Unconditionally Artin’s conjecture is not known to be true forany specific a. Using the exact same decomposition as in Hooley’s paper [H] weprove UNCONDITIONALLY (see Theorems 1.1 and 1.2 above and also Remark6.1 below) the strong form of Artin’s primitive root conjecture (1927) for CMelliptic curves for ALL a 6= ±1 or a perfect square.
3
We remark also that this paper contains THE ONLY unconditional proofof strong form of Artin’s primitive root conjecture (1927) in characteristic 0 forALL a, and it is very probably that it will contain for a very long time: the reasonis that in 100+ years there has been no progress at all towards (quasi-)RiemannHypothesis (all we have unconditionally in 100+ years is something very similarto Lemma 4.2 below; not even Bombieri-Vinogradov theorem that was used forexample by Halberstam in [HA] to prove unconditionally Titchmarsh divisorproblem (1931) [TI] works in our case, as it did not work in Hooley’s paper [H]in his (conditional) proof of the strong form of Artin’s conjecture).
Similar questions were considered in [GM] and [GM1] which were based on ashort paper [LT] of Lang and Trotter from 1977, where the authors were able toprove, under GRH, in Theorem 1 of [GM] an asymptotic formula as in Theorems1.1 and 1.3 above for the very particular case of CM elliptic curves E definedover Q and ONLY for the ordinary primes of E. The methods of [GM] cannot beused to obtain an asymptotic form of Theorem 1 of [GM] for any non-CM ellipticcurve, or for any simple abelian variety of dimension at least 2; moreover, thecondition Q(E[2]) 6= Q imposed in Theorem 2 of [GM] (see also the subsequentremark) is necessary (in the classical Artin’s conjecture we have trivially thatQ(±1) = Q). These limitations arise from the fact that the statement consideredin [GM] is not the correct analogue of Artin’s conjecture, despite numerousclaims to the contrary (e.g., see page 63 of [MU1], the introduction of [C], the‘elliptic Artin’ section of [MO], the introduction of [AGM], etc.).
2 Elliptic curves
Let GQ := Gal(Q/Q). Let E be an elliptic curve over Q, of conductor N . LetP
E
be the set of rational primes p of good reduction for E (i.e. (p,N) = 1). Form � 1 an integer, let E[m] be the m-division points of E in Q, and let Q(E[m])be the field obtained by adjoining to Q the coordinates of the elements of E[m].We have
E[m] ' (Z/mZ)2.Assume that rankQE � g where g � 0 is an integer, and let a
1
, . . . , ag
2 E(Q) belinearly independent points. Throughout this paper we denote a := (a
1
, . . . , ag
),and Q(E[m],m�1a) := Q(E[m],m�1a
1
, . . . ,m�1ag
). Then we have naturalinjections
�m
: Gal(Q(E[m])/Q) ,! Aut(E[m]) ' GL2
(Z/mZ),
and
�a,m
: Gal(Q(E[m],m�1a)/Q) ,! Aut(E[m])n(E[m])g ' GL2
(Z/mZ)n((Z/mZ)2)g.
LetG
m
:= �m
(Gal(Q(E[m])/Q)),
andG
a,m
:= �a,m
(Gal(Q(E[m],m�1a)/Q)).
4
Definen(m) := |G
m
| = [Q(E[m]) : Q],
andna
(m) := |Ga,m
| = [Q(E[m],m�1a) : Q].
For a rational prime l, let
Tl
(E) = lim �
n
E[ln],
and Vl
(E) = Tl
(E)⌦Q. The Galois group GQ acts on
Tl
(E) ' Z2
l
,
where Zl
is the l-adic completion of Z at l, and also on Vl
(E) ' Q2
l
, and weobtain a representation
⇢E,l
:= lim �
n
�l
n : GQ ! Aut(Tl
(E)) ' GL2
(Zl
) ⇢ Aut(Vl
(E)) ' GL2
(Ql
),
which is unramified outside lN . If p 2 P
E
, let �p
be the Artin symbol ofp in GQ, and let l be a rational prime satisfying (l, p) = 1. We denote byPE,p
(X) = X2
� ap
X + p 2 Z[X] the characteristic polynomial of �p
on Tl
(E).Then P
E,p
(X) is independent of l. We have ap
= p + 1 � |E(Fp
)|. We know(Riemann Hypothesis) that P
E,p
(X) = (X � ⇡p
)(X � ⇡p
), where |⇡p
| =p
p.One can identify T
l
(E) with Tl
(E), where E is the reduction of E at p, andthe action of �
p
on Tl
(E) is the same as the action of the Frobenius ⇡p
of E onTl
(E) (throughout this paper we use the same symbol ⇡p
for a root of PE,p
(X)and also for the Frobenius ⇡
p
of E).Also for each rational prime l, one has a representation
⇢E,a,l
:= lim �
n
�a,l
n : GQ ! Aut(Tl
(E))n (Tl
E)g ' GL2
(Zl
)n (Z2
l
)g.
We know (see for example [SI] and [CW]):
Lemma 2.1. Let E be an elliptic curve defined over Q, of conductor N , andlet m be a positive integer. Assume that rankQE � g where g � 0 is an integer,and let a
1
, . . . , ag
2 E(Q) be linearly independent points. Then the extensionsQ(E[m])/Q and Q(E[m],m�1a)/Q are ramified only at places dividing mN .
We know (see [SI1]):
Lemma 2.2. Let E be an elliptic curve defined over Q. Then, with the samenotations as above, we have
|Gm
|� '(m)2,
where '(m) is the Euler function, and obviously
m4+2g
� |Ga,m
|.
5
Lemma 2.3. Let E be an elliptic curve over Q, of conductor N . Assumethat rankQE � g where g � 0 is an integer, let a
1
, . . . , ag
2 E(Q) be linearlyindependent points, and let a := (a
1
, . . . , ag
). Let p be the rational prime, and
let q 6= p be a distinct rational prime. Then¯
E(Fp)
ha1
,...,agi contains a (q, q)-type
subgroup, i.e. a subgroup isomorphic to Z/qZ ⇥ Z/qZ, if and only if p splitscompletely in Q(E[q], q�1a).
Proof : Since (p,Nq) = 1 from Lemma 2.1 we know that p is unramified inQ(E[q], q�1a). Then if
⇡p
: E(Fp
)! E(Fp
),
is the Frobenius endomorphism, we have that
Ker(⇡p
� 1) = E(Fp
).
But¯
E(Fp)
ha1
,...,agi contains a (q, q)-type subgroup if and only if E(Fp
)[q] ⇢ E(Fp
),
and there exists a bi
2 E(Fp
), for i = 1, . . . , g, such that q · bi
= ai
. Hence
we get that¯
E(Fp)
ha1
,...,agi contains a (q, q)-type subgroup if and only if E(Fp
)[q] ⇢
Ker(⇡p
� 1) and p has a first degree factor in Q(q�1a), which is equivalent tothe splitting of p in Q(E[q], q�1a). ⌅
Lemma 2.4. Let E be an elliptic curve over Q, of conductor N . Assumethat rankQE � g where g � 0 is an integer, let a
1
, . . . , ag
2 E(Q) be linearly
independent points, and let a := (a1
, . . . , ag
). Let p 2 P
E
. Then¯
E(Fp)
ha1
,...,agi is
cyclic if and only if p does not split completely in Q(E[q], q�1a) for any rationalprime q 6= p.
Proof : We know that E(Fp
) ✓ E(Fp
)[m] ✓ Z/mZ⇥Z/mZ for any positiveinteger m such that |E(F
p
)||m. But the p-primary part of E(Fp
)[m] is cyclic
(see for example II, §4 of [M]). Hence, we get that¯
E(Fp)
ha1
,...,agi is cyclic if and only
if it does not contain a (q, q)-type subgroup for any rational prime q 6= p. FromLemma 2.3, we deduce that this is equivalent to the fact that p does not splitcompletely in Q(E[q], q�1a) for any rational prime q 6= p. ⌅
3 CM elliptic curves
Let E be a CM elliptic curve over Q, of conductor N , with CM by the fullring of integers O
K
of an imaginary quadratic number field K = Q(p
�D),where D 2 Z�2
is square-free. Then it is known that K has class number 1 (seeExample 11.3.1 of Appendix C of [SI]). Let p 2 P
E
, and let X2
�ap
X+p 2 Z[X]be the characteristic polynomial at p. We say that p 2 P
E
is supersingular ifap
= 0 and ordinary otherwise. Let ⇡p
and ⇡p
be the roots of the polynomialX2
� ap
X + p 2 Z[X]. Then obviously p = ⇡p
⇡p
.From the theory of complex multiplication (see [SI]) we know:
6
Lemma 3.1. Let E be a CM elliptic curve over Q, of conductor N , with CMby the full ring of integers O
K
of an imaginary quadratic number field K =Q(p
�D). If p, with (p,N) = 1, is an ordinary prime, then p splits in OK
asp = (⇡
p
)(⇡p
), where ⇡p
and ⇡p
are the roots of the characteristic polynomialX2
�ap
X+p 2 Z[X] at p, and moreover one can assume that ⇡p
represents theFrobenius endomorphism of E.
We know (see Lemma 2.2 of [C]):
Lemma 3.2. Let E be a CM elliptic curve over Q, of conductor N , with CMby the full ring of integers O
K
of an imaginary quadratic number field K =Q(p
�D). If p, with (p,N) = 1, is an ordinary prime, then p splits completelyin Q(E[m]) if and only if ⇡p�1
m
2 OK
.
We know (Lemma 2.5 of [C]):
Lemma 3.3. Let x 2 R and let D, m be fixed positive integers with m <p
x�1.Then
S1
m
:= |{p| p x, p = (↵m+ 1)2 +D�2m2 for some ↵, � 2 Z}|
= O((
p
x
m+ 1)
p
x log log x
mp
D logpx�1
m
),
and
S2
m
:= |{p| p x, p = (↵
2m+ 1)2 +D
�2
4m2 for some ↵, � 2 Z}|
= O((
p
x
m+ 1)
p
x log log x
mp
D logpx�1
m
),
where the O-constants are absolute.
We know (Lemma 2.6 of [C]):
Lemma 3.4. With the same notations as in Lemma 3.3, for 1 i 2 and anym and x, we have
Si
m
⌧
p
x
mp
D(
p
x
m+ 1).
4 Chebotarev density theorem
Let L/F be a Galois extension of number fields, with Galois group G. Wedenote by n
L
and dL
the degree and the discriminant of L/Q, and by dF
thediscriminant of F/Q. Let ⌃
F
be the set of finite places of F . Let P(L/F ) bethe set of rational primes p which lie below places of F which ramify in L/F .
We know (see page 130 of [SE]):
7
Lemma 4.1. If L/F is Galois extension of number fields, then
log dL
|G| log dF
+ nL
(1�1
|G|
)X
p2P(L/F )
log p+ nL
log |G|.
Using the same assumptions as above, let C be a conjugacy class in G. Fora positive real number x, let
⇡C
(x, L/F ) := |{} 2 P
E
|NF/Q} x, } unramified in L/F, �
}
2 C}|,
where �}
is a Frobenius element at }. The Chebotarev density theorem saysthat
⇡C
(x, L/F ) ⇠|C|
|G|
li x ⇠|C|
|G|
x
log x,
and moreover we know (see for example Lemma 2 of [MU1]):
Lemma 4.2. (Lagarias and Odlyzko) If L/F is a Galois extension of numberfields, then there exists an e↵ective positive constant A, and there exists anabsolute constant c such that if
rlog x
nL
� cmax(log |dL
|, |dL
|
1/nL),
then
⇡C
(x, L/F ) =|C|
|G|
li x+O(|C|x exp(�A
slog x
|G|
)),
where C denotes the set of conjugacy classes contained in C, and the O-constantis absolute.
5 The proof of Theorem 1.1
We follow very closely [H] and [C] (we remark that we cannot apply the extensiondue to Huxley, i.e. Theorem 1 of [HU], of Bombieri-Vinogradov theorem as in[AM] in our proof of the strong form of Artin’s primitive root conjecture forCM elliptic curves; we have to apply the same classical decomposition, i.e. thedecomposition f
E,a
(x) = N(x, y)+O(M(x, y, 2p
x)) in (5.1) below, that appearsin Hooley’s paper [H] in which he proved, under GRH, the strong form of Artin’sconjecture (1927)).
From Lemma 2.4 we get
fE,a
(x) =1X
m=1
µ(m)⇡1
(x,Q(E[m],m�1a)/Q),
where the sum is over square-free positive integers, and ⇡1
(x,Q(E[m],m�1a)/Q)was defined in §4.
8
If p 2 ⇡1
(x,Q(E[m],m�1a)/Q), then p splits completely in Q(E[m]), andfrom Lemma 2.3 and from the fact that (Riemann hypothesis) |⇡
p
| =p
p p
x,for each root ⇡
p
of PE,p
(X), we get thatm2
||PE,p
(1)| < 22x. Hence it is su�cientto consider only positive square-free integers m satisfying m 2
p
x.For y 2
p
x a real number we have (throughout this paper q represents aprime number)
N(x, y)�M(x, y, 2p
x) fE,a
(x) N(x, y),
whereN(x, y) := |{p 2 P
E
| p x,
p does not split completely in any Q(E[q], q�1a)/Q, q y}|,
andM(x, y, 2
p
x) := |{p 2 P
E
| p x,
p splits completely in some Q(E[q], q�1a)/Q, y q 2p
x}|.
HencefE,a
(x) = N(x, y) + O(M(x, y, 2p
x)). (5.1)
We want to find a suitable y > 2 and to approximateN(x, y) andM(x, y, 2p
x).
5.1 A formula for N(x, y)
We haveN(x, y) =
X
m
0µ(m)⇡
1
(x,Q(E[m],m�1a)/Q),
where the sum is over all square-free positive integers m 2p
x whose primedivisors are y.
Let dm,a
the discriminant of Q(E[m],m�1a)/Q. Then from Lemmas 2.1 and4.1 we get that
na
(m)|dm,a
|
2
na(m)
⌧ m12+6gN2
andna
(m)(log |dm,a
|)2 ⌧ m12+6g(log(m4+2gN))2,
and so the two numbers above are ⌧ m13+6g. In order to apply Lemma 4.2we need m13+6g
⌧ log x. Since m is a product of distinct primes y, andthus m exp(2y) (this is because the number of primes y is < 2y
log y
, and
y2y
log y = exp (2y)), it is su�cient to choose
y =1
13 + 6glog log x. (5.2)
Then from Lemma 4.2 we obtain
N(x, y) = (X
m
0 µ(m)
na
(m)) li x+O(
X
m
0x exp(�A
slog x
na
(m)))
9
for some positive constant A. But na
(m) ⌧ m4+2g and there are at most
2y ⌧ (log x)1
13+6g square-free numbers composed of primes y, and thus weobtain
N(x, y) = (X
m
0 µ(m)
na
(m)) li x+O(
x
(log x)B) (5.3)
for any positive constant B.
5.2 A formula for M(x, y, 2
p
x)
We have that
M(x, y, 2p
x) X
y<q2
px
⇡1
(x,Q(E[q], q�1a)/Q) =
X
y<q2
px
⇡o
1
(x,Q(E[q], q�1a)/Q) +X
y<q2
px
⇡s
1
(x,Q(E[q], q�1a)/Q),
where⇡o
1
(x,Q(E[q], q�1a)/Q) := |{p| p x, ap
6= 0,
p splits completely in Q(E[q], q�1a)/Q}|,
and⇡s
1
(x,Q(E[q], q�1a)/Q) := |{p| p x, ap
= 0,
p splits completely in Q(E[q], q�1a)/Q}|.
From Lemmas 3.2 and 3.3 and from the fact that NK/Q⇡p
= p we get that
⇡o
1
(x,Q(E[q], q�1a)/Q) |{p 2 P
E
| p x,⇡p
� 1
q2 O
K
}| Sq
,
where Sq
is S1
q
if �D ⌘ 2, 3(mod 4), and S2
q
if �D ⌘ 1(mod 4).Let u <
p
x� 1 be a real number. From Lemma 3.4 we get thatX
u<q2
px
⇡o
1
(x,Q(E[q], q�1a)/Q) X
u<q2
px
Sq
⌧
X
u<q2
px
p
x
qp
D(
p
x
q+ 1)
=xp
D
X
u<q2
px
1
q2+
p
xp
D
X
u<q2
px
1
q
⌧
xp
D u log u+
p
x log log xp
D.
From Lemma 3.3, we get thatX
y<qu
⇡o
1
(x,Q(E[q], q�1a)/Q) X
y<qu
Sq
10
⌧
X
yqu
(x
q2p
D+
p
x
qp
D)log log x
logpx�1
q
⌧
x log log xp
D logpx�1
u
X
y<qu
1
q2+
p
x log log xp
D logpx�1
u
X
y<qu
1
q
⌧
x log log xp
D (logpx�1
u
)y log y+
p
x(log log x)(log log u)p
D logpx�1
u
.
If p 2 ⇡s
1
(x,Q(E[q], q�1a)/Q), then p splits completely in Q(E[q]) and hencefrom the proof of Lemma 2.3, we get that q2|p + 1 � a
p
= p + 1. Also becauseQ(⇣
q
) ⇢ Q(E[q]) (see [SI]) where ⇣q
is a primitive q-root of unity, we get that psplits completely in Q(⇣
q
) and hence q|p� 1. Thus q|2, and because 2 < y q,we deduce that ⇡s
1
(x,Q(E[q], q�1a)/Q) = 0.We choose
u = log x.
Because y = 1
13+6g
log log x, we obtain
M(x, y, 2p
x) = O(x
(log x)(log log log x)). (5.4)
5.3 The formula for fE,a(x)
From (5.1), (5.3) and (5.4) we get
fE,a
(x) = (X
m
0 µ(m)
na
(m)) li x+O(
x
(log x)B)
+O(x
(log x)(log log log x)), (5.5)
for any positive constant B.We have that
X
m
0 µ(m)
na
(m)=
X
m
µ(m)
na
(m)�
X
m
00 µ(m)
na
(m),
whereP
m
00 means that the sum is over the positive square-free integersm whichare divisible by a prime q > y. From Lemma 2.1 we obtain
X
m
00 µ(m)
na
(m)li x⌧
x
log x
X
q>y
1X
t=1
1
q2t5/4
⌧
x
(log x) y log y
= O(x
(log x)(log log x)(log log log x)). (5.6)
11
Hence, from (5.5) and (5.6) we get
fE,a
(x) = cE,a
li x+O(x
(log x)(log log log x)),
and we are done with the proof of Theorem 1.1. ⌅
6 The proof of Theorem 1.2
Obviously cE,a
> 0 implies Q(E[2], 2�1a) 6= Q, because otherwise¯
E(Fp)
ha1
,...,agicontains Z/2Z ⇥ Z/2Z for p 2 P
E
and p 6= 2, and thus it cannot be cyclic.Assume now that Q(E[2], 2�1a) 6= Q. Then [Q(E[2]) : Q] is 2, 3 or 6, or[Q(E[2]) : Q] = 1 and [Q(2�1a) : Q] > 1. If [Q(E[2]) : Q] > 1, then let Mbe the unique non-trivial abelian extension of Q contained in Q(E[2]), and if[Q(E[2]) : Q] = 1, then let M = Q(2�1a), which in this case is a non-trivialabelian extension of Q. We have K(E[q]) = Q(E[q]) for any prime q � 3 (seeLemma 6 of [MU]). Because [K : Q] = 2, we obtain that M \ K = Q orK ✓ M . If M \ K = Q, then using the fact that M ✓ Q(E[2], 2�1a) andK ✓ Q(E[q], q�1a) for any q � 3, we get that the density of the places p 2 P
E
that do not split completely in any of the fields Q(E[q], q�1a) is greater thanor equal to the density of the places p 2 P
E
that do not split completely in Mand K. Thus
cE,a
� (1�1
[M : Q])(1�
1
[K : Q]) > 0.
If K ✓M , then K ✓ Q(E[q], q�1a) for any prime q, and thus the density of theplaces p 2 P
E
that do not split completely in any of the fields Q(E[q], q�1a)is greater than or equal to the density of the places p 2 P
E
that do not splitcompletely in K. Hence
cE,a
� (1�1
[K : Q]) > 0,
and we are done with the proof of Theorem 1.2. ⌅
Remark 6.1. Theorems 1.1 and 1.2 actually hold true if one assumes that theelliptic curve E defined over Q has CM by an arbitrary order of the ring ofintegers O
K
of an imaginary quadratic field K, but throughout the paper, inorder to simplify the notations we have assumed that E has CM by the full ringof integer O
K
.
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