Table Of ContentPRIMITIVE ROOTS IN QUADRATIC FIELDS II
5
JOSEPH COHEN
0
0
2
n
a Abstract. This paper is continuation of the paper ”Primitive
J
rootsinquadraticfield”. WeconsiderananalogueofArtin’sprim-
9
itive root conjecture for algebraic numbers which is not a unit in
real quadratic fields. Given such an algebraic number, for a ratio-
]
T nalprimepwhichisinertinthefieldthemaximalorderoftheunit
N modulopisp2 1. AnextensionofArtin’sconjectureisthatthere
−
. are infinitely many such inert primes for which this order is maxi-
h
mal. weshowthatforanychoiceof85algebraicnumberssatisfying
t
a a certain simple restriction, there is at least one of the algebraic
m
numbers which satisfies the above version of Artin’s conjecture.
[
1 1. Introduction
v
0 This paper is continuation of the paper ”Primitive rootsin quadratic
2
field” ([8]). In this paper we considered an analogue of Artin’s prim-
1
1 itive root conjecture for units in real quadratic fields. Given such a
0 nontrivial unit, for any rational prime p which is inert, the maximal
5
order of the unit modulo p is p + 1. An extension of Artin’s conjec-
0
/ ture is that there are infinitely many such inert primes for which this
h
t order is maximal. This is known at present only under the General-
a
ized Riemann Hypothesis. Unconditionally, we showed that for any
m
choice of 7 units in different real quadratic fields satisfying a certain
:
v
simple restriction, there is at least one of the units which satisfies the
i
X above version of Artin’s conjecture. In this paper we want to extend
r this result for any algebraic integer modulo inert prime p. we will prove
a
Theorem 1.1. Let K = Q(√∆) be a quadratic field and α 85 be a
{ i}i=1
set of 85 integers of K such that
(1) The norms N(α ) = α σ(α ), of the α′s, are multiplicatively in-
i i i i
dependent.
Date: February 1, 2008.
In partial fulfillment for the Ph.D. degree. Supported by grants from the
Technion-IsraelInstitute of Technology.
1
2 JOSEPH COHEN
(2) 5N(α )∆, N(α ) are not perfect squares.
i i
(3) M(α ) = σ(α )/α are multiplicatively independent.
i i i
Then at least one of the 85 integers has at least order p2−1 mod p
24
for infinitely many inert primes p in K.
Notethatinthecaseofsplit primes, Narkiewicz ([12])provedamuch
stronger result.
Since in our case the order is p2 1, which is not “linear”, some of
−
their divisors are too big and we cannot use the method of [6]. But
since p2 1 = (p 1)(p + 1) can be factored into two linear factors,
− −
with the following remark we can still use the method of [6]
Remark 1.2. Consider an algebraic number α in K = Q(√∆). Let
p ∤ α be an inert prime in K. Since
(1) M(α) αp−1 (mod (p)),
≡
(2) N(α) αp+1 (mod (p)),
≡
we have:
ord(M(α)) ord(α) (mod (p)) and ord(N(α)) ord(α) (mod (p))
| |
In addition:
ord(M(α)) p+1 and ord(N(α)) p 1
| | −
But,
(p−1,p+1) = 1 or (p 1, p+1) = 1.
2 − 2
So,
ord(M(α))ord(N(α)) 2ord(α) (mod (p))
|
Let e and e be some integers. If we prove that M(α) and N(α)
1 2
have simultaneously at least orders p+1 and p−1, respectively, then we
e1 e2
will obtain that α has at least order p2−1. This way we reduce the
2e1e2
problem to a “linear” problem.
PRIMITIVE ROOTS IN QUADRATIC FIELDS II 3
1.1. Notationand Preliminaries. Letπ(y;m,s)thenumberofprimes
p y such that p s (mod m) where m and s are some integers, and
≤ ≡
Li(y)
E(y;m,s) := π(y;m,s)
− ϕ(m)
where Li(y) = y dt . Also set
R
2 logt
E(x;m) := max max E(y;m,s) .
1≤y≤x (s,m)=1| |
Denote = p2 1 p x,p u (mod v) where u, v are some
A { − | ≤ ≡ }
given integers such that (u,v) = 1, and take X = Li(x).
ϕ(v)
For a square-free integer d, (d,v) = 1, denote
:= a : a 0 mod d
d
|A | |{ ∈ A ≡ }|
= p2 1 : p x, p umodv, p2 1 0 mod d
|{ − ≤ ≡ − ≡ }|
d
= X p p x,p umodv, p m mod d
|{ | ≤ ≡ ≡ }|
m=1
m2−1≡0modd
d
= X p p x,p u mod v, p m mod d
|{ | ≤ ≡ ≡ }|
m=1
m2−1≡0 mod d
(m,d)=1
By the Chinese Remainder Theorem, for each m there exists an
integer l such that
m
d
d = X p p x, p lm (mod dv) ;
|A | |{ | ≤ ≡ }|
m=1
m2−1≡0 mod d
(m,d)=1
Since p p x, p l (mod dv) is asymptotically independent
m
|{ | ≤ ≡ }|
of m, there exists some integer l such that
d
d = π(x;dv,l) X 1 = π(x;dv,l)ρ(d)
|A |
m=1
m2−1≡0 mod d
(m,d)=1
4 JOSEPH COHEN
d
where ρ(d) = 1.
P
m=1
m2−1≡0 mod d
(m,d)=1
We note that ρ(q) = 2 for any prime q, and hence for any square-free
d, ρ(d) = 2ν(d) where ν(d) denotes the number of prime divisors of d.
By the definition of E(x;dv,l),
ρ(d) Lix 2ν(d)
= +ρ(d)E(x;dv,l)) = X +2ν(d)E(x;dv,l).
d
|A | ϕ(d)ϕ(v) ϕ(d)
For any prime q define ω(q) := 2q , ω(d) = ω(q) = 2ν(d)d and
ϕ(q) Q ϕ(d)
q|d
ω(d)
R := X = 2ν(d)E(x;dv,l)
d d
|A |− d
Finally, we define the Mo¨bius function, µ(1) = 1 and for a square-free
d = p p , µ(d) = ( 1)k.
1 k
··· −
Now we want to prove two lemmas.
Lemma 1.3. For any prime q > 3 which is relatively prime to v we
have:
2 1
(1.1) 0 .
≤ q 1 ≤ 2
−
2 z
(1.2) X logq 2log = O(1) (2 w z)
q 1 − w ≤ ≤
w≤q<z −
where O does not depend on z or w.
2 1
(1.3) Y (1 ) .
− q 1 ≫ log2z
2<q<z −
q∤v
Proof. Since q > 3, it is clear that (1.1) holds.
Asforthesecondequality, 2 logq = 2 logq q = 2 logq(1+
P q−1 P q q−1 P q
w≤q<z w≤q<z w≤q<z
1 ) = 2 logq + 2 logq = 2log z + O(1) ( logp =
q−1 P q P q(q−1) w P p
w≤q<z w≤q<z p<x
logx+O(1)).
PRIMITIVE ROOTS IN QUADRATIC FIELDS II 5
Hence we get (1.2). Finally,
2 2
Y (1 ) Y (1 )
− q 1 ≫ − q 1
2<q<z − 2<q<z −
q∤v
2
= exp(log Y (1 ))
− q 1
2<q<z −
2
= exp( X log(1 ))
− q 1
2<q<z −
2 4
exp( X ( ))
≫ −q 1 − (q 1)2
2<q<z − −
Since
2 2 2 2 2
= + +
q 1 q q(q 1) ≤ q (q 1)2
− − −
and 2 converges, we get
P (q−1)2
2<q<z
2 2
Y (1 ) exp( X )
− q 1 ≫ − q
2<q<z − 2<q<z
Since
2
X 2loglogz
q ∼
2<q<z
we have
2 1
exp( X ) exp( 2loglogz) =
− q ≫ − log2z
2<q<z
(cid:3)
Lemma 1.4. For any square-free natural number d, (d,v) = 1, and a
real number A > 0, there exist constants c ( 1) and c ( 1) such that
2 3
≥ ≥
X
(1.4) X µ2(d)3ν(d)|Rd| ≤ c3logAX, (X ≥ 2)
1
d<(loXgx2)c2
Proof. Denote by S the term which we need to estimate.
R
SR = X µ2(d)3ν(d) Rd
| |
1
d<(loXgx2)c2
6 JOSEPH COHEN
By the definitions of R and E(x;dv)
d
SR X µ2(d)6ν(d) E(x;dv) .
≤ | |
1
d<(loXgx2)c2
Since E(x;dv) x if d x, we get that
≪ dv ≤ v
µ2(d)6ν(d)
1 1
SR ≪ x2 X d1 |E(x;dv)|2.
2
1
d<(loXgx2)c2
By Cauchy’s Inequality,
µ2(d)62ν(d)
1 1 1
SR x2( X )2( X E(x;dv) )2.
≪ d | |
1 1
d<X2 dv<(lovgXx2)c2
For sufficiently large x we obtain
µ2(d)62ν(d)
1 1 1
SR x2(X )2( X E(x;dv) )2.
≪ d | |
1 1
d<x2 dv<(logxx2)c2
With Bombieri-Vinogradov Theorem [2] (given any positive con-
stant e , there exists a positive constant e such that E(x;d) =
1 2 P
1
d<logxe22x
O( x )) for the last sum and the inequality µ2(d)36ν(d) (logw+
loge1x P d ≤
d<w
1)36, [5], p.115, equation (6.7)) we find that for given constant B there
exists c such that
2
x
S .
R ≪ logBx
So, for given A there exists c such that
2
X
S
R ≪ logAX
where depends on v and c . (cid:3)
2
≪
1.2. Proof of Theorem 1.1 – the sieve part. In this section we use
the Selberg lower bound sieve and show that there is some small real
number δ and some constant c(δ ) > 0 (which depends on δ ) such
1 1 1
that for at least c(δ ) x primes p x, p u (mod v), if q p2 1
1 log3x ≤ ≡ | −
then either q > x1/8+δ1 or q v.
|
PRIMITIVE ROOTS IN QUADRATIC FIELDS II 7
Now, define S( ,z,v) = a a ,(a, p) = 1 and define a
Q
A |{ | ∈ A }|
p<z
p∤v
function g by g(t ) = 1,t = 4.42 and g(t) < 1 for t > t . Then (see [5,
0 0 0
Theorem 7.4, page 219]):
Lemma 1.5. We have
ω(q) logX (loglog3X)8
(1.5) S( ,z,v) XY(1 ) 1 g( )+O( )
A ≥ − q { − 2logz logX }
q<z
q∤v
where the O-term does not depend on X or on z.
By Lemmas 1.3 and 1.4, (1.1), (1.2) and (1.4) hold. Hence we can
use Lemma 1.3 with z = X18+δ0
2 1 logX (loglog3X)8
S( ,X81+δ0,v) X Y (1 ) 1 g( )+O( ) .
A ≥ −q 1 { − 2logX18+δ0 logX }
q<X81+δ0 −
q∤v
By Lemma 1.3 (1.3) we have for δ sufficiently small
0
X x
S( ,X81+δ0,v)
A ≫ log2X ≫ log3x
Thus for such δ we obtain that there is a constant c(δ ) > 0 (which
0 0
depends on δ ) such that for at least c(δ ) x primes p x, p
0 0 log3x ≤ ≡
u (mod v) if q p2 1 then either q > x1/8+δ0 or q v. Hence we obtain
| − |
that for all 0 < δ < δ there is a constant c(δ ) > 0 (which depends
1 0 1
on δ ) such that for at least c(δ ) x primes p x, p u (mod v),
1 1 log3x ≤ ≡
if q p2 1 then either q > x1/8+δ1 or q v.
| − |
1.3. Proof of Theorem 1.1 – The algebraic part.
1.3.1. Construction of the arithmetic sequence. Let K = Q(√∆) be
any quadratic field, the integers ring of K, α any algebraic
O ∈ O
integer and a = N(α). In this section we want to construct integers u
and v, (u,v) = 1 such that for all primes p such that p u (mod v),
≡
the discriminant ∆ of Q(√∆) and a satisfy
∆ a
( ) = ( ) = 1.
p p −
This means that p is inert and a is not a quadratic residue (mod p).
In addition we want to obtain by the construction that (p2−1,v) = 1
24
(since after sieving the small factors of p2−1 we may be left with small
24
factors which divide v, see previous section).
8 JOSEPH COHEN
Inordertofulfillthesedemands, wewillfirstshowthatthereexist in-
finitely many primes p satisfying the following simultaneous conditions
(1.6) (−1) = (5) = (a) = (∆) = 1.
p p p p −
This condition is equivalent to the condition:
1 5 a ∆
B(p) = (1 (− ))(1 ( ))(1 ( ))(1 ( )) = 0
− p − p − p − p 6
Since the Legendre symbol is a multiplicative function, we obtain,
1 a ∆ a∆ 5 5a 5∆ 5a∆
(1 (− ))(1 ( ) ( )+( ) ( )+( )+( ) ( )).
− p − p − p p − p p p − p
Let S be the set of all integers of the form n = ( 1)b05b1ab2∆b3, b
i
− ∈
0,1 . Then
{ }
(1.7) B(p) = ( 1)b0+b1+b2+b3 (n), b 0,1 .
P P − P p i ∈ { }
p≤Z n∈S p≤Z
By the assumption in the theorem each n S is not a perfect square
∈
3
when b is odd.
P i
i=0
This assumption, together with the fact that for n not a perfect
square (by the reciprocity law for Legendre symbol)
(n) = o(π(Z)) as Z
P p → ∞
p≤Z
implies that B(p) is asymptotic to at least π(Z) (since all the neg-
P
p≤Z
ative summands contribute o(π(Z)) and at least the natural number 1
contributes π(Z)). This shows that the simultaneous conditions have
infinitely many solutions p.
We fix some particular p satisfying the condition (1.6) and for each
0
odd prime l = 3, such that l 24a∆, we define u = p if l ∤ p2 1, and
6 | l 0 0 −
u = 9p otherwise.
l 0
Claim 1.6. l ∤ u2 1.
l −
Proof. If u = p then by the assumption l ∤ p2 1, so l ∤ u2 1.
l 0 0 − l −
If u = 9p , assume, by reductio ad absurdum, that l u2 1. Hence
l 0 | l −
l 81p2 1. Since, l p2 1, we obtain that l 80p2. On the other
| 0 − | 0 − | 0
hand, by our condition, ( 5 ) = 1 so (p0) = 1 (p 1 (mod 4)).
p0 − 5 − 0 ≡
Hence p 2 or 3 (mod 5). Since l p2 1 and p 2 or 3 (mod 5)
0 ≡ | 0 − 0 ≡
we conclude that l ∤ 5. Using the assumption that l p2 1 we deduce
| 0−
that l = p (if l = p then l ∤ p2 1). Hence (l = 2,3) l ∤ 80p2, which
6 0 0 0 − 6 0 (cid:3)
is a contradiction.
PRIMITIVE ROOTS IN QUADRATIC FIELDS II 9
In addition let u = p if 8 p 2 1 and u = p 8 if 16 p 2 1.
2 0 0 2 0 0
| − − | −
Likewise we take u = p if 3 p 2 1 and u = p 3 if 9 p 2 1.
3 0 0 3 0 0
| − − | −
Let v = 24a∆ and u be the common solution of u u (mod 16),
2
≡
u u (mod 9) and all the congruences u u (mod l). Such a solu-
3 l
≡ ≡
tion exists, by the Chinese Remainder Theorem.
Since l ∤ u2 1 for every odd prime l = 2,3, l v, and by the con-
− 6 |
struction (u2−1,6) = 1 we conclude that (u2−1,v) = 1.
24 24
Finally, if p u (mod v), then p p (mod 24) and p p or
0 0
≡ ≡ ≡
4p (mod l) for all odd primes l v. So, (∆) = (∆) = 1, and similarly
0 | p p0 −
for a. This completes the construction of u and v.
Note that by the construction of the integers u and v we have that
(u,v) = 1. (take l an odd prime number, l v = 24a∆ and assume
|
that l u. Since u u (mod l), l u . Hence l p or 9p (in this
l l 0 0
| ≡ | |
case l = 2,3). In other words l = p . But p ∤ 24a∆ (p fulfills the
0 0 0
6
simultaneous condition (1.6)) and l 24a∆).
|
1.3.2. The last step of the proof. As we saw at the previous subsec-
tions, for at least c(δ ) x primes p x, p u (mod v), if q p2 1
1 log3x ≤ ≡ | −
then q > x1/8+δ1 or q v. Since (p2−1,v) = 1, if q p2−1 then q > x1/8+δ1.
| 24 | 24
Since by the construction of u and v, p 1 (mod 4) and (p2−1,v) =
≡ 24
1, we have that p−1 and p+1 are odd. In addition if p 1 (mod 3)
4 2 ≡
then (p−1,v) = 1 and if p 1 (mod 3) then (p+1,v) = 1
12 ≡ − 6
If we conclude the result about p 1 and p+1 we have for at least
−
c (δ ) x primes p x, p u (mod v), if q p−1 or q p+1 then
1 1 log3x ≤ ≡ | d− | d+
q > x1/8+δ1, where d = 4 or 12 and d = 6 or 2, respectively.
− +
For the last step of the proof we need to use a version of Lemma 4
from Narkiewicz [12], which generalized Lemma 2 in [3].
Lemma 1.7. Ifa ,...a are multiplicativelyindependentalgebraicnum-
1 k
bers of an algebraic number-field K, G the subgroup of K⋆ generated by
a ,...a , and for any prime ideal P not dividing a , a we denote
1 k 1 k
···
by GP the reduction of G (modP), then for all positive y one can have
GP < y for at most O(y1+k1) prime ideals P, with the implied constant
| |
being dependent on the a ’s and K.
i
10 JOSEPH COHEN
Proof. According to [12],
For any real number T denote by M = M(T) the set of all k-element
sequences (r ,...,r ) of non-negative integers satisfying
1 k
r + r +...+ r T
1 2 k
| | | | | | ≤
It is easy to see that for T tending to infinity M(T) = (c+o(1))Tk
| |
with suitable positive constant c = c . If now P is a prime ideal for
k
which G < y, then select T to be the smallest rational integer with
P
| |
cTk > 2y and let a = bi for i = 1,...,k. There exist two distinct
i ci
sequences Z = (z ), W = (w ) in M(T) for which
i i
bz1 bzk bw1 bwk
P 1 ··· k 1 ··· k
|cz1 czk − bw1 cwk
1 ··· k 1 ··· k
Hence for sufficient large P (since the a′s are multiplicatively inde-
i
pendent).
bz1 bzk bw1 bwk
P 1 ··· k − 1 ··· k = D = 0
| c[z1,w1] c[zk,wk] 6
1 ··· k
Thus for sufficient large P
ν (Πazi−wi 1) 0
P i − ≥
where ν denotes the P-adic valuation and it follows that for fixed
P
z w ,...,z w we obtain log(max a¯ 2T) T possibilities for
1 1 k k j j
− − ≪ | | ≪
P. Finally we obtain
P GP < y T1+k y1+k1
|{ | | | }| ≪ ≪ (cid:3)
Look at the p 1 case (the case of p + 1 is similar). We have
−
for at least c (δ ) x primes p x, p u (mod v) such that
1 1 log3x ≤ ≡
if q p−1 then q > x1/8+δ1 where d = 4 or 12. Let p 1 =
| d− − −
d q (p)q (p) q (p), q (p) > q (p) > ... > q (p), m 7, and
− 1 2 m m m−1 1
· · · ≤
let a be some integer where a¯ its image in F∗.
p
Denote by S the set S = a ,...,a where a ,...,a aremultiplica-
n n 1 n 1 n
{ }
tively independent integers and take seven integers a ,...,a from S
i1 i7 n
and assume that at least one prime, say q (p), which is greater than
1
x1/8+δ1, divides [F∗ : a¯ ] for k = 1,...,7.
p h iki
