We split this section into two parts, first, for f (X) = X2+ c∈ Z[X], we will find some suffi-cient conditions such that the Galois group of fn(X) over Q(√
2) or Q(√
−2) is isomorphic to [C2]n. Next, we will prove some properties of 2-independent over a quadratic number field.
we denote d∈ Z where d is square-free. Give c ∈ Z, we define c1 =−c and cn= c2n−1+ c for n≥ 2.
Let
bn =Y
d|n
cµ(n/d)d ,
then bn ∈ Z for all n and bn are pairwise coprime (by [2, Lemma 1.1]). Note that b1 =−c, and we define that
Bc ={bn: n∈ N and b1 =−c}.
Definition. A subset S in a field K is called 2-independent over K if their residue classes in the F2-vector space K∗/(K∗)2 are linearly independent. If S is not 2-independent over K, then we say that S is 2-dependent over K.
M. Stoll [2, Theorem] shows that:
Theorem. If c∈ Z has one of the following properties:
1. c > 0, and c≡ 1 (mod 4);
2. c > 0, and c≡ 2 (mod 4);
3. c < 0, c≡ 0 (mod 4) and −c is not a square in Z.
then none of |b2|, · · · , |bn| is a square in Q.
Corollary. If c satisfies one of the properties in the previous theorem, then Bc is 2-independent over Q. Moreover, Ωn ∼= [C2]n for all n∈ N.
Proof. Since (bi, bj) = 1 for i6= j, so Bc is 2-independent over Q which is equivalent to for any subset {bi1,· · · , bik} ⊂ Bc, none of bij is square and at most one of −bij is a square in Q. Moreover, since b1 =−c is not a square in Q for all cases. Hence none of |b2|, · · · , |bn| is a square in Q implies Bc is 2-independent over Q.
By Theorem in [2], since {b1,· · · , bn} is 2-independent over Q for all n, so Ωn ∼= [C2]n for all n∈ N.
It is natural to consider whether Bc is 2-independent over another field or not. The easiest case is to consider the basic field is a quadratic number field. For convenience, if S ⊂ K = Q(√
a) then we denoted, na
S o
=
(−1 if S is 2-independent over Q(√ a), 1 if S is 2-dependent over Q(√
a).
Moreover, we take a = 1 if K = Q and only consider the case {B1c} = −1 in the following statements. To study this problem, we need to investigate all prime numbers in which divides some elements in Bc.
Definition. We denote the set of all prime numbers by P. Consider S ⊂ Z, we define the set of prime divisors of S by
P (S) ={p ∈ P : p divides s for some s ∈ S with s 6= 0}.
Moreover, if we have B ⊂ A ⊂ Z then we can define the density of B in A by dA(B) = lim
n→∞
#{b ∈ B : b ≤ n}
#{a ∈ A : a ≤ n}
By the property of bn, we have:
Proposition 1. If c satisfies one of the properties in Stoll’s theorem. Then, for any bi where i≥ 2, there exists a prime pi such that vpi(bi) is odd where vpi(·) is the valuation.
Proof. If vp(b) is even for all prime p, then |b| is a square. Since none of |bi| is square for i≥ 2 so there exists prime number pi such that vpi(bi) is odd.
Moreover, since bi are pairwise coprime so pi 6= pj if i 6= j in previous proof. Hence P (Bc) is an infinite set. However, in R. Jones’ paper [10, Theorem 1.2], he shows that
Theorem. dP(P (Bc)) = 0 for c∈ Z\{−1}.
Hence, there also exist infinitely many prime numbers that do not divide any one of bn. Actually, it is far more than the number of primes which can divide some bn. To deal with this problem, for arbitrary a, we have Q(√
a) =Q(√
d) where d is the radical of a. Hence it is enough to consider Q(√
d) where d is square-free. Here, we prove a useful lemma first.
Lemma 7.1. {Bdc} = 1 for some square-free integer d if and only if there exists {bi1,· · · , bik} ⊂ Bc such that
bi1· · · bik = dn2 where n∈ Z.
Proof. Suppose Bc is not 2-independent over Q(√
d) then there exists {bi1,· · · , bik} ⊂ Bc
and s, t∈ Q such that
bi1· · · bik = (s + t√
d)2 = (s2+ dt2) + 2st√ d.
Since the left-hand side is an integer so s = 0 or t = 0.
If t = 0 then bi1· · · bik = s2. However, Bc is 2-independent over Q which makes a contradiction. Hence s = 0 and we have
bi1· · · bik = dt2.
If t is not an integer then so is dt2. But the left-hand side is an integer which makes a contradiction, thus t∈ Z.
On the other hand, if there exist some elements in Bc such that bi1· · · bik = dn2 = (0 + n√
d)2 where n∈ Z.
Then Bc is not 2-independent overQ(√ d).
The easiest case is to consider d is a prime. Here we give a sufficient and necessary condition to determine the value of {Bpc} where p is a prime.
Theorem 6. Let p be a prime number, {Bpc} = 1 if and only if bn = pt2 or bn =−pt2 and c∈ Q2 where t∈ Z and n ∈ N.
Proof. If bn= pt2 then bn = (0 + t√
p)2. For the other case, if bn=−pt2 and c =−b1 = s2 where s∈ Q then b1bn= (0 + st√
p)2. Hence, for above two cases,{Bpc} = 1 . Conversely, if {Bpc} = 1, by Lemma 1, there exists {bi1,· · · , bik} ⊂ Bc such that
bi1· · · bik = pt2 where t ∈ Z.
Without loss of generality, let p| bn, then for any i > 1 and i6= n, there exists a prime pi 6= p such that vpi(bi) is odd by Proposition 1. Hence bi 6∈ {bi1,· · · , bik}, otherwise, vpi(bi1· · · bik) is odd but vpi(pt2) is even which is absurd. Hence {bi1,· · · , bik} = {bn} or {b1, bn}.
If {bi1,· · · , bik} = {bn} then bn = pt2. For the second case, we can assume n 6= 1, otherwise, it can be reduced to the first case. Thus, we have b1bn = pt2. Since vp(b1) = 0 so b1 must be −s2 for some s∈ Z and s | t. Hence c = −b1 ∈ Q2 and bn= p· (t/s)2 where t/s ∈ Z.
Because P\P (Bc) is an infinitely set, by Theorem 6, it is obviously that{±pBc} = −1 for any p 6∈ P (Bc). Hence, there exists infinitely many quadratic number field such that Bc is 2-independent over it. In fact, we can describe more precisely about primes, not in P (Bc).
Proposition 2. If (−cp ) = −1 then p 6∈ P (Bc) where (··) is the Legendre symbol, thus {±pBc} = −1.
Proof. Note that (−cp ) =−1 so p ∤ c1. If p| cn for some n then cn = c2n−1+ c≡ 0 (mod p).
Thus (−cp ) = 1 which is a contradiction. Hence p∤ cn and so p∤ bn for all n∈ N.
Hence, if b1 is not a quadratic residue modulo p then Bcis 2-independent over Q(ñp).
However, all such prime numbers do not lie in P (Bc). In Theorem 6, we have a sufficient and necessary condition to determine whether {Bpc} = 1 or not. Later, we will show that there is infinitely many such primes exist under some conditions.
Obviously, since one of b1 or b2 is even, so we can give a sufficient and necessary condition to determine whether or not Bn is 2-independent over Q(√
2) or Q(√
−2). In the following two results, since we assume {B1c} = −1 so we only consider c ∈ Z which satisfies one of the following properties:
1. c > 0, c≡ 1 (mod 4);
2. c > 0, c≡ 2 (mod 4);
3. c < 0, c≡ 0 (mod 4) and −c is not a square in Z.
Definition. An NSW(Newman-Shanks-Williams) number is an integer m such that the Diophantine equation
2x2 = m2+ 1 can be solved in integers.
Theorem 7. {B2c} = 1 if and only if √
c is an NSW number in case 1 or v2(c) is odd and vp(c) is even for all odd prime p in case 3.
Proof. Note that, by Theorem 6.3,{B2c} = 1 if and only if bn= 2t2 or bn=−2t2 and c∈ Q2. Moreover, since one of b1 or b2 is even so n = 1 or 2.
Suppose √
c is an NSW number in case 1 then c is a square and 2x2 = c + 1 has an integer solution, called t. Hence b2 =−(c + 1) = −2t2 and so {B2c} = 1.
For another case, since c < 0, v2(c) is odd and vp(c) is even for all odd prime p so c =−2t2 where t ∈ Z. Again, by Theorem 6.3, b1 =−c = 2t2 implies {B2c} = 1.
Conversely, suppose {B2c} = 1 then Bc must satisfy one of the following case:
1. b1 = 2t2;
2. b2 = 2t2;
3. −b1 = c∈ Q2 and b2 =−2t2 where t∈ Z.
In the first case, b1 =−c = 2t2 then c < 0, v2(c) is odd and vp(c) is even for all prime p. The second case is absurd since b2 =−(c + 1) = 2t2 imply c < 0. However, we assume that if c < 0 then 4| c = −1 − 2t2. For the last case, since c∈ Q2 so we can write c as (c′)2 where c′ ∈ Q. Then b2 =−(c + 1) = −2t2 which means 2x2 = c + 1 has an integer solution and so √
c = c′ is an NSW number.
Corollary. If c ∈ Z has one of the following properties:
1. c > 0, and c≡ 1 (mod 4) and √
c is not an NSW number;
2. c > 0, and c≡ 2 (mod 4);
3. c < 0, c ≡ 0 (mod 4) and −c is not square in Z. Moreover, v2(c) is even or vp(c) is odd for some odd prime p.
then Gal(fn(X)/Q(√
2)) ∼= [C2]n for all n≥ 1 where f(X) = X2+ c.
Proof. By Theorem 7, if c satisfies one of properties in hypothesis then {B2c} = −1. Hence {b1,· · · , bn} is 2-independent over Q(√
2) for all n which is equivalent to Gal(fn(X)/Q(√ 2)) ∼= [C2]n, by Theorem 1.
Theorem 8. {−2Bc} = 1 if and only if √
c is an NSW number in case 1 or c/2 is a square in case 2.
Proof. For case 1, Suppose Bcis not 2-independent overQ(√
−2) then there exists {bi1,· · · , bik} ⊂ Bc such that
bi1· · · bik =−2t2 where t∈ Z.
Since c is odd and b2 =−(c + 1) so b2 is the only even integer in Bc then b2 ∈ {bi1,· · · , bik}.
If there exists bi ∈ {bi1,· · · , bik} where i ≥ 3 then there exists an odd prime pi such that vpi(bi1· · · bik) is odd, but vpi(−2t2) is even for any odd prime pi, which is a contradiction.
Hence {bi1,· · · , bik} = {b1, b2} or {b2}.
If {bi1,· · · , bik} = {b2} then
b2 =−(c + 1) = −2t2.
That means t is a solution of 2x2 = c + 1 which is a contradiction since we assume √
c is not an NSW number.
If {bi1,· · · , bik} = {b1, b2} then
−2t2 = b1b2 = c(c + 1)≥ 0 which is a contradiction. Hence Bc is 2-independent over Q(√
−2).
For case 2, Suppose Bcis not 2-independent overQ(√
−2) then there exists {bi1,· · · , bik} ⊂ Bn such that
bi1· · · bik =−2t2 where t∈ Z.
Since b1 = −c is the only even number in Bn so b1 ∈ {bi1,· · · , bik}. If there exists bi ∈ {bi1,· · · , bik} where i ≥ 2 then there exists an odd prime pi such that vpi(bi1· · · bik) is odd, but vpi(−2t2) is even for any odd prime pi, which is a contradiction. So{bi1,· · · , bik} = {b1}.
Hence b1 =−c = −2t2,i.e., c/2 = t2, but we assume that c/2∈ Q2 which is a contradiction.
Hence Bc is 2-independent overQ(√
−2).
For case 3, Suppose Bcis not 2-independent overQ(√
−2) then there exists {bi1,· · · , bik} ⊂ Bc such that
bi1· · · bik =−2t2 where t∈ Z.
Since b1 = −c is the only even number in Bc so b1 ∈ {bi1,· · · , bik}. If there exists bi ∈ {bi1,· · · , bik} where i ≥ 2 then there exists an odd prime pisuch that vpi(bi1· · · bik) is odd, but vpi(2n2) is even for any odd prime pi, which is a contradiction. Hence{bi1,· · · , bik} = {b1} and we have
b1 =−2t2.
But b1 =−c > 0, hence such {bi1,· · · , bik} does not exist. Hence Bc is 2-independent over Q(√
−2).
Conversely, consider c > 0, c≡ 1 (mod 4) and √
c is an NSW number then 2x2 = c + 1 has solution. Hence
b2 =−(c + 1) = −2x2, which means Bc is not 2-independent overQ(√
−2).
If c > 0, c≡ 2 (mod 4) and c/2 ∈ Q2 that means c = 2t2 where t is an odd integer, then Bc is not 2-independent overQ(√
−2) since
b1 =−c = −2t2.
Corollary. If c ∈ Z has one of the following properties:
1. c > 0, c≡ 1 (mod 4) and c is not an NSW number;
2. c > 0, c≡ 2 (mod 4) and c/2 6∈ Q2;
3. c < 0, c≡ 0 (mod 4) and −c is not a square in Z.
then Gal(fn(X)/Q(√
−2)) ∼= [C2]n for all n≥ 1 where f(X) = X2+ c.
Proof. By Theorem 8, if c satisfies one of properties in hypothesis then {−2Bc} = −1. Hence {b1,· · · , bn} is 2-independent over Q(√
−2) for all n which is equivalent to Gal(fn(X)/Q(√
−2)) ∼= [C2]n, by Theorem 1.
Now, we want to know whether there exists infinitely many primes, p∈ P (Bc) such that Bn is not 2-independent overQ(√p) or not.
Theorem 9. there exists infinitely many d such Bc is 2-independent over Q(√
d) if c has one of the following property:
1. c > 0, c≡ 1 (mod 4) and c is not a square;
2. c > 0, c≡ 2 (mod 4);
3. c < 0, c≡ 0 (mod 4) and −c is not a square in Z.
Proof. We have shown that for all i ≥ 3 there is an odd prime pi such that vpi(bi) is odd.
We claim that if di =− sgn(bi)pi then Bc is 2-independent overQ(√
di) for all i.
Suppose Bc is not 2-independent over Q(√
di) for some i then there exists a subset {bi1,· · · , bik} ⊂ Bc such that
bi1· · · bik = dit2 for some t∈ Q.
Since vpi(dit2) is odd so bi ∈ {bi1,· · · , bik} and vp(dit2) is even for another prime so bk 6∈
{bi1,· · · , bik} for k ≥ 2 and k 6= i.
Hence {bi1,· · · , bik} = {bi} or {b1, bi}. If {bi1,· · · , bik} = {bi} then bi = dit2 for some t∈ Z,
thus
sgn(bi) = sgn(dit2) = sgn(di) =− sgn(bi) which is a contradiction.
If bi ∈ {bi1,· · · , bik} = {b1, bi} then
b1bi = dit2 for some t∈ Z,
and so |b1| must be a square. However, we have assumed that |b1| = |c| is not a square for all case.
Hence Bc is 2-independent overQ(√
di) for all i.
Remark. By Theorem 9, we know that for the above three cases, there exists infinite many quadratic number field such that Gal(fn(X)/Q(√
d)) ∼= [C2]n. However, it is hard to find all elements in P (Bc) for given c∈ Z. Hence, we cannot give a criterion for p, c to determine whether Gal(fn(X)/Q(√p)) ∼= [C2]n or not.
In the second part, we have some property about 2-independent over quadratic number field.
Proposition. {−1Bc} = 1 if and only if c ∈ Q2. Proof. If c ∈ Q2, let c = (c′)2 where c′ ∈ Q, then
b1 =−c = −(c′)2.
Hence Bc is 2-dependent over Q(√
−1), by Lemma 7.1.
Conversely, if Bcis 2-dependent overQ(√
−1), then there exists {bi1,· · · , bik} ⊂ Bcsuch that
bi1· · · bik =−t2 where t∈ Z.
Note that vp(−t2) is even for any prime p and none of |bi| is a square for i ≥ 2. Hence {bi1,· · · , bik} = {b1} and so b1 =−c = −t2. Therefore, c∈ Q2.
Proposition. If {B1c} = 1 then {Bac} = 1 for any a ∈ Z.
Proof. Suppose Bn is not 2-independent over Q then there exists {bi1,· · · , bik} ⊂ Bn such that
bi1· · · bik = s2 = (s + 0·√
a)2 where s∈ Q.
So Bn is not 2-independent over Q(√ a).
Proposition. If P ({a}) 6⊂ P (Bc) then {Bac} = {−aBc} = −1.
Proof. If P ({a}) 6⊂ P (Bc) then there exists a prime p such that p| a but p 6∈ P (Bc). Suppose {Bac} or {−aBc} = 1, then there exists {bi1,· · · , bik} ⊂ Bc such that
bi1· · · bik = at2 or − at2 where t∈ Z.
However, take vp(·) on both sides, the left-hand side is zero. On the other hand, the right-hand side is nonzero, which is a contradiction.
Proposition. Suppose {Bdc} = 1 then {−dBc} = 1 if and only if {−1Bc} = 1.
Proof. It is clear when d = 1. Hence we only need to consider d > 1. Since {Bdc} = 1 so there exists {bi1,· · · , bik} ⊂ Bc such that
bi1· · · bik = dt2 where t∈ Z.
Suppose {−dBc} = 1 then there exists {bj1,· · · , bjl} ⊂ Bc such that bj1· · · bjl =−ds2 where s∈ Z.
Hence we have
bi1· · · bikbj1· · · bjl =−(dst)2, by Lemma 7.1, {−1Bc} = 1.
Conversely, suppose {−1Bc} = 1 then c ∈ Q2 by Lemma 7.3. Let c = s2 where s∈ Q so b1 =−s2. Now, we discuss the following two cases:
First, if b1 ∈ {bi1,· · · , bik} then without loss of generality, let bik = b1. Since d > 1 so {b1} ⊊ {bi1,· · · , bik}. Hence
bi1· · · bik−1 = dt2
b1 = dt2
−s2 =−d · (t/s)2
which means {−dBc} = 1.
For the second case, we consider b1 6∈ {bi1,· · · , bik} then b1bi1· · · bik =−s2· dt2 =−d(st)2. Hence {−dBc} = 1, by Lemma 1.
Proposition. For any a, b ∈ Z, if {Bac} = {Bbc} = 1 then {Babc} = 1.
Proof. By Lemma, there exists {bi1,· · · , bik} and {bj1,· · · , bjl} ⊂ Bc such that bi1· · · bik = at2 and bj1· · · bjl = bs2 where s, t∈ Z.
So
bi1· · · bik · bj1· · · bjl = ab(st)2. Hence Bc is not 2-independent overQ(√
ab).
Corollary. Suppose bn= pt2 or bn =−pt2 and c∈ Q2 where t∈ Z and n ∈ N then {−pBc} = 1 if and only if {−1Bc} = 1.
Proof. By Theorem 6.3, {Bpc} = 1. Hence by Lemma 6.3, {−pBc} = 1 if and only if {−1Bc} = 1.
Proposition. For any a, b ∈ Z, if {Bac}{Bbc} = −1 then {Babc} = −1.
Proof. Without loss of generality, Let {Bac} = 1 and {Bbc} = −1, and suppose {Babc} = 1. By Lemma, there exists {bi1,· · · , bik} and {bj1,· · · , bjl} ⊂ Bc such that
bi1· · · bik = at2 and bj1· · · bjl = abs2 where s, t∈ Z.
So
bi1· · · bik· bj1· · · bjl = b(ast)2
which implies {Bbc} = 1. However, we assume {Bbc} = −1 which is absurd. Hence Bc is 2-independent over Q(√
ab).
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