Iterated Galois Groups over Quadratic Number Field

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(1)國立臺灣師範大學理學院數學系碩士論文 Department of Mathematics, College of Science. National Taiwan Normal University Master's Thesis. Iterated Galois Groups over Quadratic Number Field. 高智強 Kao, Chih-Chiang. 指導教授：夏良忠博士 Advisor: Hsia, Liang-Chung, Ph.D. 中華民國 109 年 7 月 July 2020.

(2) Abstract Consider the base field K is a real quadratic number field and a polynomial X 2 + c where c lies in the ring of integer OK . We will give some criteria on the iterated polynomial f n (X) of X 2 + c to determine whether the Galois group of f n (X) over K is isomorphic to the wreath product of cyclic group of order 2. Next, we will focus on the following three cases: √ 1. K = Q( 2); √ 2. K = Q( 2p) where p is a prime and p ≡ 3 (mod 4); √ 3. K = Q( p) where p is a prime and p ≡ 1 (mod 4). √ The class number of Q( 2) is one, for the other two cases, we need to assume hK = 1. We will give sufficient conditions on c such that the Galois group of the iterated polynomial over K is isomorphic to the iterated wreath product. In the last part, we will prove some 2-independent property of an integer set over a quadratic number field. Key word: iterated polynomial, arboreal Galois group, iterated wreath product, 2independent.. i.

(3) Contents Abstract. i. 1 Introduction. 1. 2 Preliminaries. 3. 3 Criteria for Ωn ∼ = [C2 ]n. 10. 4 Iteration sequences associated to even integer polynomials. 17. 2 5 Some values of c with √ |bi | 6∈ K for all i ≥ 2 5.1 Case 1: K = Q(√2) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2 Case 2: K = Q( 2p) and p ≡ 3 (mod 4) . . . . . . . . . . . . . . . . . . . . √ 5.3 Case 3: K = Q( p) and p ≡ 1 (mod 4) . . . . . . . . . . . . . . . . . . . .. 18 18 21 23. √ 6 Ωn ∼ = [C2 ]n for K = Q( 2). 27. 7 2-independent property of integers over quadratic number field. 31. References. 39. ii.

(4) 1. Introduction. Let K be a field of characteristic zero, we consider a sequence of polynomials f1 (X), f2 (X), · · · ∈ K[X] where deg fi ≥ 2 for all i ∈ N. Define F1 (X) = f1 (X) and Fn+1 (X) = Fn (fn+1 (X)) for n ≥ 2, such Fn (X) is called the iterated polynomial of fn (X). Let Kn be the field obtained by adjoining the roots of Fn (X) to K and denoted the Galois group of Kn over K by Ωn , called the iterated Galois group. We want to know how large the Galois group can be? This type of problem is first considered by R. W. K. Odoni [1], he focuses on a function field K(X, T ) where X and T are algebraically independent over K. Moreover, we regard X as a variable and T as a parameter. The polynomials which we care about being fn (X, T ) = X kn + T where kn are positive integer greater than 1. The main result is the following: ¯ be an algebraic closure of K. Then Theorem (R. W. K. Odoni. [1, Theorem 1]). Let K ¯ the Galois group of Fn (X, T ) over K(T ) is isomorphic to the (natural-)wreath product Ωn ∼ = Ckn oUn−1 (· · · (Ck3 oU2 (Ck2 oU1 Ck1 )) · · · ). where Cki is the cyclic group of order ki , for each i = 1, · · · , n, and Ui is the set of roots of Fi (X, T ) for i = 1, · · · , n − 1. In this case, we called such kind of wreath product the iterated wreath product. If all ki are equal, namely Cki = Cm for some m then we denote it by [Cm ]n where n means n times wreath product and called it the n-th iterated wreath product of Cm . There is more detail about iterated wreath product in [8, Section 4]. Note that T can be regarded as a parameter, we can restrict T to any given set. In particular, the first case is to consider ki = 2 for all i, and T varies among integers. The question has been investigated by M. Stoll [2], and in his paper, the basic field is the field of the rational number. The polynomial which we consider is f (X) = X 2 + c where c varies in the set of Z. In this case, we denote the n-th iterated polynomial of f (X) by f n (X). By Lemma 1.1 in Odoni’s paper [1], we can show that the iterated Galois group of f n (X) over Q can be embedded into [C2 ]n . The key point is that for n ≥ 2, Ωn ,→ C2 oUn−1 Ωn−1 ,→ [C2 ]n , by induction. Moreover, he gives some sufficient conditions on c to determine whether the iterated Galois group is isomorphic to the n-th iterated wreath product of C2 . Theorem (M. Stoll. [2, Theorem]). If c ∈ Z has one of the following properties: 1. c > 0, and c ≡ 1 (mod 4); 1.

(5) 2. c > 0, and c ≡ 2 (mod 4); 3. c < 0, c ≡ 0 (mod 4) and −c is not square in Z. then Ωn ∼ = [C2 ]n for all n ≥ 1. In the conclusion of Odoni and Stoll, they both prove that the iterated Galois group is isomorphic to an iterated wreath product under some condition. On the other hand, there is a relation between tree automorphism group and iterated wreath product of some symmetric groups. Hence, in the next section, we will give an exposition to connect among the three of them. Stoll’s results give us some sufficient conditions to guarantee that the iterated Galois group in our question is the largest possible group, namely the n-th iterated Galois group of a cyclic group C2 . Naturally, we can restrict the parameter T on a set other than integers. Hence, throughout this paper, we assume our basic field to be a real quadratic number field K, and still focus on X 2 + c but c here is a quadratic algebraic integer. We want to know that the largest Galois group of f n (X) over K and find a sufficient condition on c such that the Galois group as large as possible. In Odoni’s [1] and Stoll’s paper [2], a sufficient and necessary condition to determine whether Ωn is isomorphic to [C2 ]n or not is to consider the constant term of f ( X), called cn . They prove {c1 , · · · , cn } is linearly independent over F2 in Q∗ /(Q∗ )2 if and only if Ωn ∼ = [C2 ]n . However, it is hard to decide whether {c1 , · · · , cn } is linearly independent over F2 in Q∗ /(Q∗ )2 or not, so we use the Möbius inversion formula on cn to get bn . Then all bn are pairwise coprime and {b1 , · · · , bn } is linearly independent over F2 in Q∗ /(Q∗ )2 if and only if Ωn ∼ = [C2 ]n . Hence it is enough to show that, for every bi , there exists a prime element that divides bi with odd power. In the three cases of Stoll’s paper, |bi | is not a square in Q for i ≥ 2 which means bi appears a new prime element with odd power for i ≥ 2 and 2 . So {b1 , · · · , bn } is linearly independent over F2 in Q/Q2 for all n. b1 = −c 6∈ OK Nevertheless, in the case of a real quadratic number field, the unit group of the ring of integer is infinite, so the condition |bi | is not square in K for i ≥ 2 which is not enough to guarantee {b1 , · · · , bn } is linearly independent over F2 in K ∗ /(K ∗ )2 where K is the base field. If we assume |NK/Q (bi )| is not square in Q for i ≥ 2 then we can deduce that Ωn ∼ = [C2 ]n for n ≥ √ 1. But it is hard to prove |NK/Q (bi )| is not square in Q for i ≥ 2. However, for K = Q( 2), we can use another method to prove that bi can be divided by some prime element with odd power for i ≥ 3, and we can show that {b1 , b2 } is linearly independent √ over n ∗ ∗ 2 ∼ F2 in K /(K ) by direct computation. Hence Ωn = [C2 ] for all n ≥ 1 as K = Q( 2). In my thesis, we will follow Odoni and Stoll’s method. Due to technical reasons, we assume that the ring of integer of K is a unique factorization domain. First, we will show that the iterated Galois group of f n (X) also can be embedded into [C2 ]n so the largest possible Galois group is isomorphic to [C2 ]n . Next, we will focus on some real quadratic 2 number fields that |bi | 6∈ OK for i ≥ 2. Moreover, √ and give sufficient conditions on c such n ∼ for K = Q( 2), such kind of c will imply Ωn = [C2 ] . The main theorem in my thesis is the following:. 2.

(6) Theorem. Assume that K is a real quadratic number field of class number one. Consider f (X) = X 2 + c ∈ OK [X] with NK/Q (c) > 0 where OK is the ring of integer of K, for the following three cases: √ √ 1. K = Q( 2), and c = u + v 2 where u, v are positive odd integer. √ √ 2. K = Q( 2p) where p is a prime and p ≡ 3 (mod 4), and c = u + v 2p where u, v are positive integer and u ≡ −1 (mod 2p), v ≡ 1 (mod 2). √ √ 1+ p where u, v 3. K = Q( p) where p is a prime and p ≡ 1 (mod 4), and c = u + v 2 are positive integer and u ≡ −1 (mod 4p), v ≡ 2p (mod 4p). |bi | is not a square in OK for i ≥ 2. Moreover, if c satisfies case 1, then Ωn ∼ = [C2 ]n . In the next section, we will give some preliminary knowledge that contains wreath product, tree automorphism, and some results in algebraic number theory. In the third section, we will give some criteria to determine an iterated Galois group whether is isomorphic to the n-th iterated wreath product of C2 or not. In the fourth section, we extend the condition of Lemma 2.1 in Stoll’s paper [2] from integer into quadratic algebraic integer. The fifth section, let c satisfies the conditions in Section 5, we give some sufficient conditions on c such that |b√i | is not square in OK for i ≥ 2. In the sixth section, we will prove Ωn ∼ = [C2 ]n as K = Q( 2). In the last section, we get back to consider c is an integer, and determine whether the set {b1 , b2 , · · · } is 2-independent over quadratic number field or not. Moreover, √ we will √ give some sufficient conditions on c to prove Gal(f n (X)/K) ∼ = [C2 ]n for K = Q( 2) and Q( −2) as n ≥ 1.. 2. Preliminaries. First, we will introduce the wreath product and tree automorphism. The main reference is from J. Rotman’s book [4, Chapter 7, p.167 − p.177]. Definition. A group G is a semidirect product of K by Q, denoted by G = K ⋊ Q, if K ⊴ G and there exists a subgroup Q′ of G where Q′ ∼ = Q such that K ∩ Q′ = {e} and KQ′ = G. Q Definition. Let D and Q be groups, Λ be a finite Q-set, and K = λ∈Λ Dλ , where Dλ ∼ =D for all λ ∈ Λ. Then the wreath product of D by Q, denoted by D oΛ Q, is the Q semidirect product of K by Q, where Q acts on K by q · (dλ ) = (dqλ ) for q ∈ Q and (dλ ) ∈ λ∈Λ Dλ . Definition. A graph Γ is a nonempty set V , called vertices, together with an adjacency relation on V , denoted by v ∼ u, that is symmetric (v ∼ u implies u ∼ v for all u, v ∈ V ) and irreflexive (v 6∼ v for all v ∈ V ). Definition. A tree is a graph Γ which is connected and there does not exist a loop in Γ, which means there does not exist {v1 , · · · , vk } ⊂ V such vi ∼ vi+1 for i = 1, · · · , k − 1 and vk ∼ v1 . 3.

(7) We say a graph is a rooted tree if Γ is a tree in which a special vertex is singled out and called root which is always denoted by 0. We define a partial order on Γ by considering the root which is a minimal element and it is the only vertex of level zero. If v ∼ 0 then v ∈ V is level one, all level one vertices are denoted by L1 . The set of level n vertices denoted by Ln . For v ∈ V \Ln−2 , if v ∼ v ′ for some v ′ ∈ Ln−1 then v ∈ Ln . Here if v ′ ∈ Ln−1 , v ∈ Ln and v ′ ∼ v then we say v ′ is a parent and v is a child. If there exists the largest level k in Γ then we call the level of Γ is k. In the figure of Γ, the root will be drawn in the bottom. A rooted binary tree is a rooted tree in which every vertex has at most two children. A rooted binary tree of level k which has exactly 2k − 1 vertices is called a rooted full binary tree of level k. Definition. A tree automorphism for a rooted tree T with vertices V is a bijection ϕ : V → V such that u, v ∈ V are adjacent if and only if ϕ(u) and ϕ(v) are adjacent, thus ϕ will preserve the order of vertices. Here, the root 0 is the fixed point of any automorphism. It is plain that the set of all tree automorphism of a tree T detonated by Aut0 (T ), is a group under composition. Here, we show an example first to understand the operation of tree automatism. Give a rooted tree, T as follows: a4. b4. 4. a3. b3. a2. b2. 3. 2. a1. b1. 1. 0 If ϕ ∈ Aut0 (T ), then ϕ fixes 0 since it is the only vertex adjacent to 4 vertices, ϕ e permutes on Γ = {1, 2, 3, 4}, and for each i ∈ Γ, ϕ induces a bijective map from {ai , bi } to {aφ(i) e , bφ(i) e }. Clearly, S4 can act on Γ and S2 can act on Λ = {a, b}. We can identify {ai , bi : i ∈ S} with Γ × Λ by writing ai as (a, i) and bi as (b, i). Hence, we have Aut0 (T ) ∼ = S2 oΓ S4 . Now, we will explain the relationSbetween the iterated Galois group of X 2 + c and the tree automorphism group. Let Vn = ni=0 {f −i (0)}, we consider a tree Tn with vertices, Vn . If v, v ′ ∈ Vn and v ′ = f (v) then v ∼ v ′ where v ′ is parent and v is child. Note that since our basic field is characteristic zero if f n (X) has multiple roots then it must be 0. Assume none of f −n (0) is equal, Vn has 2n+1 − 1 elements and so Tn is a rooted fully binary tree.. 4.

(8) Here, we consider an example for the Galois group of f 2 (X) over Q where f (X) = X 2 +1. √ K2 = K1 ( −1 ± i). √. √ √ √ −1 + i − −1 + i −1 − i − −1 − i. K1 = Q(i). −i. i. Q. 0. The left-hand side is the tower of field extension of f 2 (X) where f (X) = X 2 + 1. On the other hand, on the right-hand side, we have a 2-level rooted full binary tree, T2 . The root is zero, the first level is roots of f (X) and the second level is roots of f 2 (X). For any ϕ ∈ Gal(K2 /Q), we can verify it as a tree automorphism on T2 by permuting the vertices. Hence we can embed Gal(K2 /Q) into Aut0 (T2 ) by Galois theory. Moreover, we have a fact that if Tn is a rooted full binary tree then Aut0 (Tn ) ∼ = [C2 ]n . Therefore, Stoll tells us under some conditions of c, the iterated Galois group is isomorphic to Aut0 (Tn ). According to this relation, we know that the largest Galois group is isomorphic to n-th iterated wreath product of C2 . The results of Stoll shows that for c satisfy those conditions, then Gal(Kn /Q) is as large as possible. Next, we will give some preliminary about algebraic number theory, the following topic can be found in most books about algebraic number theory. The main textbooks I use are M. Rosen [3, Chapter 6,12,13] and J. Neukirch [5, Chapter 4]. There is also some knowledge about commutative algebra which refers to M. F. Atiyah [7, Chapter 5]. Definition. A Kummer extension is an extension of a field k of characteristic p ≥ 0, of the type 1/n 1/n K = k(a1 , · · · , at ) where a1 , . . . , at ∈ k, n is some natural number, and it is assumed that k contains the n-th root of unities (in particular, if p 6= 0 then n is prime to p) Definition. The exponent of a group G is the least natural number n such that for all g ∈ G, g n = e. If L/K is an abelian extension and the exponent of Gal(L/K) is n then we say that the abelian extensions L over K of exponent n. Theorem. Let n be a natural number which is relatively prime to characteristic of the field k, and assume that k contains all n-th roots of unity. Then the abelian extensions K over k of exponent n correspond one-to-one to the finite subgroups 4 of k ∗ /(k ∗ )n , via the rules p √ 1 n 4 7→ K = k( n 4) where ∆ = {a n : a ∈ k ∗ , a · (k ∗ )n ∈ ∆} 5.

(9) and. K 7→ (k ∗ ∩ (K ∗ )n )/(k ∗ )n .. Moreover, we have. Gal(K/k) ∼ = Hom(4, µn ). where µn is the group of n-th roots of unity. Remark. By Kummer theory, we define K/k to be a 2-Kummer extension if char k 6= 2 and √ √ √ K = k( a1 , a2 , · · · , an ) for some a1 , a2 , · · · , an ∈ k, n ∈ N. Definition. Let L be a finite field extension of degree n over K. Suppose α1 , · · · , αn is a basis for L/K and α ∈ L. Then ααi =. n X. aij αj for i = 1, · · · n where aij ∈ K.. j=1. The norm of α for L to K is defined to be NL/K (α) = det(aij ), and the trace of α is TL/K (α) = a11 + · · · + ann . Definition. If α1 , · · · , αn is an n-tuple of elements of L, the discriminant ∆(α1 , · · · , αn ) is defined to be det(TL/K (αi αj )). Definition. A subfield K of the complex numbers is called an (algebraic) number field if [K : Q] < ∞. We call an algebraic number field K a quadratic number field if [K : Q] = 2. Definition. Let B be a ring and A be a subring of B. An element α ∈ B is said to be integral over A if there exists a nonzero monic polynomial f (x) ∈ A[x] such that α is a root of f (x). Here, α is called an algebraic integer if α is integral over Z. If K is an algebraic number field, the subset of K consisting of algebraic integers called the ring of (algebraic) integers of K, denoted by OK . Theorem. The ring of integer, OK , of K forms a ring. Proof. It is sufficient to show that if α, β ∈ OK then α + β, αβ ∈ OK . First, we claim that Z[α] is a finitely-generated Z-module if and only if α is integral over Z. Suppose α is an algebraic integer, then there exists a nonzero monic polynomial f (x) ∈ Z[x] such that f (α) = αn + an−1 αn−1 + · · · + a1 α + a0 where ai ∈ Z, n ∈ N. For any αn+r ∈ Z[α] where r ≥ 0, we have αn+r = −(an1 αn−1+r + · · · + a1 αr+1 + a0 αr ), 6.

(10) and by induction, all positive powers of α lie in Z[α] which are generated by 1, α, · · · , αn−1 . Hence Z[α] is finitely-generated. Conversely, suppose Z[α] is a finitely-generated Z-module and the generators of Z[α] are α1 , · · · , αn , we consider that ααi =. n X. aij αj where aij ∈ Z.. i=1. Set a n × n matrix A = (aij ), then α is an eigenvalue of A. Hence α is a root of the characteristic polynomial of A which is a monic integer-valued polynomial and so α is an algebraic integer. Suppose α, β are algebraic integers, then Z[α], Z[β] are both finitely-generated and denoted their generators by {1, α, · · · , αm } and {1, β, · · · , β n }, respectively. Then {αi β j : 0 ≤ i ≤ m, 0 ≤ j ≤ n} are generators of Z[α, β]. By Hilbert’s basis theorem, Z[α, β] is Noetherian Z-module. Hence Z[α + β] and Z[αβ] are finitely-generated submodules of Z[α, β]. Hence α + β and αβ are algebraic integers. Definition. An integral domain R, that is not a field, is called a Dedekind domain if R is Noetherian, integrally closed and every nontrivial prime ideal is a maximal ideal of R. Proposition. The ring of integers of a number field, OK is a Dedekind domain. A basic fact is that every nonzero proper ideal in a Dedekind domain, R, can be uniquely factored into a product of prime ideals up to the order of the product. Moreover, if A, B are two ideals in R and A ⊂ B then there exists C ⊴ R such that B = AC. Definition. A set {α1 , · · · , αn } ⊂ OK is called an integral basis of OK if OK = Zα1 + · · · + Zαn . Definition. Let K be a number field, a fractional ideal of K is a finitely generated nonzero OK -submodule of K. Note that the fractional ideals form an abelian group, called the ideal group JK of K. The fractional principal ideal (a) = aOK , a ∈ K ∗ , form a subgroup of the group of ideals JK , which will be denoted PK . The quotient group ClK = JK /Pk is called the ideal class group. Note that ClK is finite, and its order hK = (JK : PK ) is called the class number of K. Proposition. Let K be a number field, then hK = 1 if and only if OK is a unique factorization domain. 7.

(11) Proof. If hK = 1 then every ideal is principal ideal, thus OK is a principal ideal domain and so a unique factorization domain. Conversely, we want to show that if OK is a unique factorization domain then hK = 1. Let p be a nonzero proper prime ideal of R and choose 0 6= x ∈ p. Since p is a proper ideal so x is not a unit. Since OK is UFD so we can write x as pa11 · · · pakk where pi are distinct prime in OK , ai ∈ N and k ≥ 1. Since x ∈ p and p is a prime ideal so one of pi ∈ p, say p1 . Then (p1 ) ⊂ p. Note that since p1 is prime so (p1 ) is a prime ideal and OK is a Dedekind domain so (p1 ) is maximal. However p is a proper ideal of OK which contains (p1 ), thus p = (p1 ) and so every prime ideal in OK is principal. Since every ideal in OK can be uniquely factored into a product of prime ideals. Hence every ideal is principle. Let K be a number field, if p is a prime ideal of OK , then p ∩ Z is a nonzero prime ideal in Z. Hence, it is natural to consider for any nonzero prime ideal in Z, how can it be decomposed in OK . Definition. Let (p) be a prime ideal of Z and pi be prime ideals of OK contains pOK , then (p) = pe11 · · · pegg where ei ≥ 1. The number ei is called the ramification index of pi . Since pi is maximal ideal for all i so OK /pi is a finite field which contains Z/pZ. Thus the degree of OK /pi over Z/pZ is called the degree of pi . e. Theorem. For any prime ideal (p) in Z, if (p) = pe11 · · · pgg with degree of OK /pi over Z/pZ, fi . Then g X ei fi = n. i=1. Proof. First, we claim that if p is a prime ideal in OK , e is the ramification index of p and f is the degree of OK /p over Z/pZ then the number of elements in OK /pe is pef . Since its degree is f , so the number elements in OK /p is pf . This statement is true for e = 1. Since pe−1 /pe is a subgroup of OK /pe , by the second law of isomorphism, (OK /pe )/(pe−1 /pe ) ∼ = Ok /pe−1 . It is enough to show that pe−1 /pe has pf elements, then we get the result by induction. Since pe ⊊ pe−1 so there exists α ∈ pe−1 \pe . We want to show that (α) + pe = pe−1 . Note that, we have pe ⊊ (α) + pe ⊆ pe−1 , so there exists A, B ⊴ OK where A is nontrivial such that ((α) + pe )A = pe and Bpe−1 = (α) + pe . By the unique factorization property in Dedekind domain, AB = p, thus A = p and so (α) + pe = pe−1 . Consider a map from OK to pe−1 /pe by γ 7→ γα + pe . It is easy to show that it is a homomorphism and surjective. γ is in the kernel if and only if γα ∈ pe if and only if vp (γα) ≥ e. Since vp (γα) = vp (γ) + vp (α) = vp (γ) + e − 1, 8.

(12) so γ is in the kernel if and only if vp (γ) ≥ 1 which equivalent to saying γ ∈ p. Hence OK /p ∼ = pe−1 /pe and so the latter group has pf elements. Theorem. Suppose K/Q is a Galois extension. Let (p) be a prime ideal in Z and (p) = pe11 · · · pegg . Then e1 = · · · = eg and f1 = · · · = fg . If e and f denote these common values, then ef g = n. Proof. Note that if ϕ ∈ Gal(K/Q) and A ⊴ OK then ϕA also is an ideal in OK . Also ϕOK = OK . Thus OK /ϕA = ϕOK /ϕA ∼ = OK /A. In particular, if P is a prime ideal in OK e then ϕP also is a prime ideal in OK . Moreover, we have a fact that if (p) = pe11 · · · pgg then there exists ϕ ∈ Gal(K/Q) such that ϕpi = pj for any i = 6 j. Hence, for a given index i there exists ϕ ∈ Gal(K/Q) such that ϕp1 = pi . Since Ok /p1 ∼ = Ok /ϕp1 = Ok /pi so f1 = fi for all i. Thus, all fi are equal. e Apply ϕ to both sides of (p) = pe11 · · · pgg . Since p ∈ Z it is clear that ϕ(p) = p. Thus (p) = (ϕp1 )e1 · · · (ϕpg )eg . In this product, we see the exponent of pi = ϕp1 is e1 . However, in the original expression, the exponent of pi is ei . By the uniqueness of prime factorization, we must have e1 = ei for all i. P Finally, since gi=1 ei fi = n we see that ef g = n. Now, we focus on the quadratic number field. The following useful results are well-known in algebraic number theory. We skip the proof and it can be found in M. Rosen’s book [3, Chapter 13, p.188 − p.191]. √ Proposition. If K is a quadratic number field, then K = Q( d) where d ∈ Z and d is square-free. Moreover, OK = Z + ωd Z (√. where. if d ≡ 2, 3 (mod 4) 1+ d if d ≡ 1 (mod 4) 2 √ denote the discriminant of K = Q( d). Then ( 4d if d ≡ 2, 3 (mod 4), δK = d if d ≡ 1 (mod 4). ωd =. Proposition. Let δK. d√. Since n = 2 so ef g = n = 2 by previous Theorem, and we have three cases; e = 2, f = g = 1, or g = 2, e = f = 1, or f = 2, e = g = 1. We say, respectively, that p ramifies, splits, or is inertial. Moreover, we have a method to determine which kind of a prime ideal in Z can be factored in OK . 9.

(13) √ Proposition. Consider K = Q( d), suppose p is an odd prime. 1. If p ∤ δK and ( dp ) = 1 then (p) is split. 2. If p ∤ δK and ( dp ) = −1 then (p) is inertial. 3. If p | δK then (p) is ramified. √ Proposition. Consider K = Q( d). 1. If 2 ∤ δK and d ≡ 1 (mod 8) then (2) is split. 2. If 2 ∤ δK and d ≡ 5 (mod 8) then (2) is inertial. 3. If 2 | δK then (2) is ramified.. 3. Criteria for Ωn ∼ = [C2]n. We follow the idea from M. Stoll [2] to study a similar question for c in real quadratic number field of class number one. Here we denote the real quadratic number field by K and the ring of integer of K is OK which we assume to be a unique factorization domain. The following theorems mainly follow from M. Stoll’s paper [2]. As we consider parameter c ranges over a real quadratic number field which is different from Stoll’s case, for the reader’s convenience, we will give a detailed proof if necessary. Throughout this paper, c is a quadratic integer such that −c is not a square in OK . We take f (X) := X 2 +c ∈ OK [X], f 0 (X) := X, and f n+1 (X) := f (f n (X)) = (f n (X))2 +c for all n ≥ 0, which is the iterate polynomial of f (X) as the composition f ◦ f ◦ · · · ◦ f of polynomials, {z } | n times. also we have f n+1 (X) = f n (f (X)) = f n (X 2 + c). Let c1 := −c and cn+1 := f (cn ) = c2n + c = f n+1 (0)(= f n (−c) = f n (c)) for n ≥ 1. Let Kn be the splitting field of f n (X) over K and denote by Ωn := Gal(f n (X)/K) = Gal(Kn /K) its Galois group over K. First, we have the following proposition: Proposition (c.f. [2, Fact 1.0]). 1. f n (X) has degree 2n . √ 2. For n ≥ 0, Kn+1 = Kn ( α − c | α are roots of f n ) where K0 = K. 10.

(14) 3. [Kn+1 : Kn ] ≤ 22 . n. Proof. These facts are stated in M. Stoll’s paper, for the sake of completeness we give a brief proof below. 1. This can be proved by induction. 2. Note that if γ is a root of f n+1 (X) then 0 = f n+1 (γ) = f n (γ 2 + c). √ Hence γ 2 + c = α for some root α of f n , and so γ = ± α − c. Since Kn is the splitting field of f n over K so √ Kn+1 = K(γ | γ are roots of f n+1 ) = K( α − c | α are roots of f n ). On the other hand, since Kn = K(α | α are roots of f n ) so and. √ Kn ⊂ K( α − c | α are roots of f n ), √ √ Kn ( α − c | α are roots of f n ) ⊂ K( α − c | α are roots of f n ).. Because K ⊂ Kn , √ √ Kn ( α − c | α are roots of f n ) = K( α − c | α are roots of f n ) = Kn+1 . Moreover, by Kummer theory, Kn+1 /Kn is a 2-Kummer extension. √ 3. Here α − c is a fixed root of x2 = α − c, then √ #{ α − c | α are roots of f n } = 2n . Let hα1 − c, . . . , α2n − ci be the subgroup of Kn∗ /(Kn∗ )2 generated by α1 −c. · · · , α2n −c. Since Kn+1 /Kn is a 2-Kummer extension, by Kummer theory, Gal(Kn+1 /Kn ) ∼ = hα1 − c, . . . , α2n − ci which is isomorphic to a subgroup of (C2 )n .. Lemma 3.1 (c.f. [1, Lemma 4.4] and [2, Lemma 1.1]).. √ 1. If c 6∈ (−2, 0) then cn 6= 0 for all n ∈ N. Moreover, if K = Q( d) where d ≡ 2, 3 (mod 4) and c ∈ OK then cn 6= 0 for all n ∈ N. 11.

(15) 2. There exists a sequence {bn }(n ≥ 1) in OK such that : Q For all n ≥ 1, cn = bd , and bn are pairwise coprime. d|n. Proof. 1. First, we assume that c 6∈ (−2, 0). We want to show that cn ≥ |c| for n ≥ 2. For n = 2, c2 = c2 + c = c(c + 1) ≥ |c| ≥ 0. Suppose for n = k − 1, ck−1 ≥ |c| then consider n = k, we have ck = c2k−1 + c ≥ c2 + c ≥ |c| > 0. By induction, cn ≥ |c| > 0 for all n ≥ 2 and c 6= 0 so cn 6= 0 for all n ∈ N. √ √ Next, we deal with c ∈ OK where K = Q( d) and d ≡ 2,√3 (mod 4) then c = u + v d where u, v ∈ Z for all c ∈ OK . We denoted cn = un + vn d, since cn = c2n−1 + c, so √ √ 2 + u) + (2un−1 vn−1 + v) d. cn = un + vn d = (u2n−1 + dvn−1 We claim that |vn | ≥ |v| for all n. We only need to consider c ∈ Ok \Z so v 6= 0 and clearly v1 = −v. Moreover, |v2 | = |2uv + v| = |v| · |2u + 1| ≥ |v| since u ∈ Z. Now, we assume that for n = k − 1, |vk−1 | ≥ |v| then consider n = k, we have |vk | = |2uk−1 vk−1 + v| ≥ ||2uk−1 vk−1 | − |v||. If uk−1 = 0 then |vk | ≥ |v|. If not, then |vk | ≥ ||2uk−1 vk−1 | − |v|| ≥ ||2vk−1 | − |v|| ≥ ||2v| − |v|| = |v|. Hence |vn | ≥ |v| > 0 for all n ∈ N and so cn 6= 0 for all n ∈ N. 2. Since cn 6= 0, we can define bn :=. Y. µ(n/d). cd. ∈K. d|n. where µ is the Möbius function, and have to show that bn ∈ OK . Let π be a prime element in OK which divides at least one of cn . Let m = min{n ≥ 1 : π | cn } and e = vπ (cm ). First, we claim that if m | n then vπ (cn ) = e and vπ (cn ) = 0 if m ∤ n.. 12.

(16) Note that π ∤ ci for i = 1, · · · , m − 1 and vπ (cm ) = e. Since cm = f m (0) so we denote f m (x) by g(x), thus f m (x) = g(x) = g(0) + g ′ (0)x +. g ′′ (0) 2 x + ··· . 2. Moreover, since g ′ (x) = f ′ (f n−1 (x)) · · · · · f ′ (f (x)) · f ′ (x) so g ′ (0) = 0. First, we prove that if m | n then vπ (cn ) = e, by induction. Let n = km and suppose vπ (c(k−1)m ) = vπ (g k−1 (0)) = e. Then cn = g n (0) = g(0) + (g k−1 (0))2 h(0) where h(X) ∈ OK [X]. Hence vπ (cn ) = vπ (g n (0)) = min{vπ (g(0)), vπ ((g k−1 (0))2 h(0))} = e, since vπ ((g k−1 (0))2 h(0)) ≥ 2e. For any n ≥ 2, let n = km + r where r = 1, · · · , m − 1. Then cn = f r (f km (0)) = f r (ckm ) ≡ f r (0). (mod π),. and f (0) = −c1 , f r (0) = cr for r = 2, · · · , n − 1. Hence vπ (cn ) = 0 if m ∤ n. So, vπ (cn ) = e if m | n, and 0 otherwise. Thus X vπ (bn ) = vπ (cd ) · µ(n/d) = e, if n = m. d|n. For otherwise case, if m ∤ n then vπ (cd ) = 0 for all d. Thus, we only need to consider m | n as following: X X X n/m µ = 0. vπ (bn ) = vπ (cd ) · µ(n/d) = e · µ(n/lm) = e · l n d|n. lm|n. l| m. Hence, for any prime element π, vπ (bn ) ≥ 0 for all n ∈ N, and so bn ∈ OK . Moreover, for fixed π, vπ (bm ) = e and vπ (bn ) = 0 for n 6= m. That Q means bn are pairwise coprime. Clearly, use Möbius inversion formula, we get cn = d|n bd .. Remark. In the previous proposition and lemma, we don’t need to assume OK is a unique factorization domain. However, to prove the following lemma, we need the hypothesis of class number one in K. Lemma 3.2 (c.f. [2, Lemma 1.2]). If none of c1 , c2 , · · · , cn is a square in OK , then f n (X) is irreducible in OK [X].. 13.

(17) Proof. Assume f n (X) is reducible, let m := min{k : f k (X) is reducible}, then 1 < m ≤ n. We have f m−1 (X) is irreducible, and f m (X) = f m−1 (X 2 + c) is reducible. We claim that if h(X) is an even polynomial and h(X) | f m (X) then h(X) is a constant. By our hypothesis, f m (X) = h(X)g(X) for some g(X) ∈ OK [X]. Since both f m (X) and h(X) are even, we can deduce that g(X) is an even polynomial via h(X)g(X) = f m (X) = f m (−X) = h(−X)g(−X) = h(X)g(−X). ˆ g (y) . Hence f m−1 (X) Thus we can change the variable X 2 into y, then f m−1 (y + a) = h(y)ˆ is reducible which contradicts the hypothesis. Since f m (X) is an even polynomial, if h(X) is a nontrivial divisor of f m (X), then so is h(−X). Suppose g(X) = gcd(h(X), h(−X)) ∈ K[X] then we can find q(X), r(X) ∈ OK [X] such that ( ( h(X) = g(X)q(X) h(−X) = g(−X)q(−X) ⇒ h(−X) = g(X)r(X) h(X) = g(−X)r(−X) where we use the hypothesis of OK is a unique factorization domain. Since g(X) is the greatest common divisor of h(X), h(−X) and g(−X) is a common divisor of h(X), h(−X), thus g(−X) | g(X). Hence g(X) = ±g(−X) by comparing their leading coefficients. If g(X) = −g(−X), that means the degree of g(X) is odd and so X | g(X). However g(X) | h(X) and h(X) | f m (X) so X | f m (X) but f m (0) = cm 6= 0 for all m ≥ 1 which is absurd. Hence g(X) = g(−X), that means g(X) is an even polynomial. But we have shown that m f (X) has no nontrivial even degree divisor, so g(X) must be a constant. Since the leading coefficient of f m (X) is 1, and g(X) | f m (X) so g(X) | 1. Hence g(X) is a unit and h(X) and h(−X) are coprime. Thus, f m (X) being reducible, and there exists a polynomial k(X) with f m (X) = k(X) · k(−X). In particular, cm = f m (0) = k(0)2 is a square in OK . Lemma 3.3 ([2, Lemma 1.4]). For all n ∈ N, we have Ωn ,→ [C2 ]n . In particular, if n [Kn+1 : Kn ] = 22 then Ωn+1 ∼ = [C2 ]n+1 ⇔ Ωn ∼ = [C2 ]n . Proof. See Lemma 1.4 in Stoll’s paper [2]. Definition. For nonzero numbers a1 , · · · , an in a field K are called 2-independent over K if their residue classes in the F2 -vector space K ∗ /(K ∗ )2 are linearly independent. Lemma 3.4 (c.f. [2, Lemma 1.5]). If Ωn ∼ = [C2 ]n and c1 , c2 , · · · , cn are 2-independent over K, then for all γ ∈ K ∗ : γ 6∈ (Kn∗ )2 ⇔ c1 , c2 , · · · , cn , γ are 2-independent over K. Proof. Note that we have the following fact which is proved by induction : for a transitive permutation group G, one has (H o G)ab ∼ = Gab × H ab 14.

(18) where Gab is the largest abelian factor group of G which means G/G′ where G′ is the commutator subgroup of G. n ab ∼ n According √ to this fact, we have ([Cn2 ] ) = (C2 ) . Thus the largest√2-Kummer √ extension of Q( d) within Kn has degree 2 . Consider the natural map Q( d)∗ /(Q( d)∗ )2 → Kn∗ /(Kn∗ )2 , the kernel of this map is √ √ √ (Q( d)∗ ∩ (Kn∗ )2 )/(Q( d)∗ )2 ∼ = Hom(Gal(Kn /Q( d), {±1}) ∼ = Hom((C2 )n , {±1}). Since |Hom((C2 )n , {±1})| = 2n so it has F2 -dimension n. Denoting the set of zeros of f m by Um , since f m is even so α ∈ Um implies −α ∈ Um . We have Y Y cm = f m (0) = f m−1 (−c) = (−c − α) = (−c + α) for m > 1 α∈Um−1. α∈Um−1. which is a square in Km because for α ∈ Um−1 , α − c is a square in Km . Thus, all of c1 , · · · , cn are squares in Kn . Since c1 , · · · , cn are 2-independent over K, the above kernel must be generated by the classes of c1 , · · · , cn . Hence c ∈ (Kn∗ )2 if and only if the √ √ residue residue class of c in (Q( d)∗ ∩ (Kn∗ )2 )/(Q( d)∗ )2 can be generated by c1 , · · · , cn if and only if c1 , · · · , cn , c are not 2-independent over K. Lemma 3.5 ([2, Lemma 1.6]). For all n: [Kn+1 : Kn ] = 22 ⇔ cn+1 is not a square in Kn . n. Proof. The proof follows from M. Stoll’s paper [2, Lemma 1.6]. In the following theorem, we denote the norm of K over Q by N (·) if no confusion will arise. Theorem 1 (c.f. [2, Theorem]). 1. For all n, the following statements are equivalent: (a) Ωn ∼ = [C2 ]n (b) c1 , c2 , · · · , cn are 2-independent over K; (c) b1 , b2 , · · · , bn are 2-independent over K. 2. If none of |N (b2 )|, |N (b3 )|, · · · , |N (bn )| is a square in Q, then Ωn ∼ = [C2 ]n . Proof. 1. We use induction to prove this statement. For k = 1 is trivial, so suppose these statements hold for k = n − 1 ”(a) ⇐⇒ (b)”: By Lemma 3.3, we have n−1 Ωn ∼ = [C2 ]n if and only if Ωn−1 ∼ = [C2 ]n−1 and [Kn : Kn−1 ] = 22 .. 15.

(19) Use the induction hypothesis, we apply Lemma 3.5, then ∗ )2 . Ωn ∼ = [C2 ]n if and only if cn 6∈ (Kn−1 ∗ By our induction hypothesis and Lemma 3.4, cn 6∈ (Kn−1 )2 if and only if c1 , · · · , cn−1 , cn n are 2-independent over K if and only if Ωn ∼ = [C2 ] . Q ”(b) ⇐⇒ (c)”: Since cn = d|n bd , suppose c1 , · · · , cn are not 2-independent over K then there exists {ci1 , · · · , cim } ⊂ {c1 , · · · , cn } where im is the largest index such that. ci1 · · · cim = κ2 where κ ∈ K. We can write the left-hand side as a product of {b1 , · · · , bn }. Because bim | κ2 and b2im ∤ ci1 · · · cim , thus such subset {b1 , · · · , bn } which contains bim is not 2-independent over K. On the other hand, suppose c1 , · · · , cn are 2-independent K. Note that the residue Q overµ(m/d) −1 ∗ ∗ 2 class of ci is equal to ci in K /(K ) . Since bm = d|m cd so the residue class Q Q ∗ 2 of bm is equal to d|m cd . But d|m cd 6∈ (K ) by the hypothesis and b1 , · · · , bn are pairwise coprime. Hence {b1 , · · · , bn } is 2-independent over K. 2. Since b1 , · · · , bn are pairwise coprime by Lemma 3.1, thus 1.(c) is equivalent to the following conditions: ( 1. None of bi is square in K. 2. There is at most one of bi multiply ± u is square in K. where u is a fundamental unit in OK . Note that |N (bi )| = |N (−bi )| = |N (ubi )| = |N (−ubi )|. Since none of |N (bi )| is a square in Q for i = 2, · · · , n, so ±bi and ±ubi are not square in K for i ≥ 2. Hence there must appear a new prime with odd power in bi for i ≥ 2. Moreover, we assume b1 = −c is not a square, thus {b1 , · · · , bn } is 2-independent over K.. Remark. The second part of Theorem 3.1 is different from M. Stoll’s Theorem [2, Theorem]. That’s because the unit group in Z is {±1} but there are infinitely many units in a quadratic ring of integer. Hence, bi and bj might be both forms of square times a unit with odd power. Then bi and bj are coprime but bi bj ∈ K 2 . Therefore, |bi | 6∈ K 2 is not enough to show that {b1 , · · · , bn } is 2-independent over K.. 16.

(20) 4. Iteration sequences associated to even integer polynomials. Let g(X) ∈ OK [X 2 ] be an even polynomial whose constant term is ±1. Put γ1 := g(0) = ±1, Q µ(n/d) and γn+1 := g(γn ) for n ≥ 1. Assume that all γn 6= 0 then set δn := γd for n ≥ 1. We d|n. have Lemma 4.1 (c.f [2, Lemma 2.1]). Suppose that for all n ≥ 1, there is a m ∈ OK , such that m | γn + γ2n , m is prime to γn , and −1 is not a square mod m. Then for all i ≥ 2, δi is not a square in K. Proof. For n ≥ 2, n = pe11 pe22 · · · perr where r ≥ 1 and all ej ≥ 1. Set n′ := p1 p2 · · · pr > 1 and k := n/n′ . Put α := γ2k and let m ∈ OK such that m | γk + γ2k , m is prime to γk and −1 is not a square mod m. We have m is prime to γ2k , if not, then there exists a prime p such that p | m and p | γ2k . Since γk + γ2k = mx for some x ∈ OK so p | γk , which is a contradiction. Thus, α = γ2k ≡ −γk (mod m) implies γ3k = g k (γ2k ) ≡ g k (−γk ) = g k (γk ) = γ2k = α. (mod m),. and by induction, γlk ≡ α (mod m) for all l ≥ 2. Note that µ(kn′ /d) 6= 0 if and only if d is a multiple of k. Hence Y µ(n/d) Y µ(kn′ /kt) Y µ(n′ /t) δn = γd = γkt = γkt kt|kn′. d|n µ(n′ ). ≡ (−1). ·. Y. α. µ(n′ /t). t|n′ µ(n′ ). ≡ (−1). P. α. t|n′. µ(n′ /t). ≡ −1. (mod m).. t|n′. Since −1 is not a square mod m, δn cannot be a square in K. Lemma 4.2. Take m = γk + γk+1 in Lemma 4.1, then m | γk + γ2k and m is prime to γk for k ≥ 1. Proof. Since −γk ≡ γk+1 (mod m) so γk+2 = g(γk+1 ) ≡ g(−γk ) ≡ g(γk ) = γk+1. (mod m).. Use induction on k, γ2k ≡ γk+1 ≡ −γk (mod m). Hence γk + γ2k ≡ γk − γk ≡ 0 (mod m). Since γk+1 = g(γk ) ≡ g(0) = ±1 (mod γk ) so γk is prime to γk+1 . Hence m is prime to γk for all k ≥ 1.. 17.

(21) 5. Some values of c with |bi| 6∈ K 2 for all i ≥ 2. In this section, we deal with the following 3 cases, √ 1. K = Q( 2); √ 2. K = Q( 2p) where p is a prime and p ≡ 3 (mod 4); √ 3. K = Q( p) where p is a prime and p ≡ 1 (mod 4). √ We have a fact that the class number of Q( 2) is one. For the other two cases, we also need to assume hK = 1. Let g(X) = |c|X 2 + sgn(c) then γ1 = g(0) = ±1. In addition, since cn = |c| · γn for n ≥ 2 Q and cn 6= 0 so γn 6= 0 for all n. Hence we can define δn = d|n γ µ(n/d) , moreover, δn = |bn | for n ≥ 2. √ Remark. In H. Hasse’s paper [9], he proved that if K = Q( d) is real and hK = 1 then d = p, 2q or qr, where p, q, r are primes and q ≡ r ≡ 3 (mod 4). However, for the technical √ reason, we cannot apply a similar method on K = Q( p) where p is a prime and p ≡ 3 (mod 4). The case which d = qr where q, r are primes and q ≡ r ≡ 3 (mod 4) is harder than the other two cases. Hence, we only deal with the above three cases.. 5.1. √ Case 1: K = Q( 2). √ √ In this subsection, since O = Z[ 2] so we write c ∈ O by u + v 2 and γn ∈ OK √ by K n K n n √ αn + βn 2. Because K is fixed, we denote the norm of K over Q by N (·). Let c = u + v 2 with N (c) > 0 where u, v are odd positive integers. We have the following lemmas. Lemma 5.1. None of c1 , · · · , cn is square in OK √ Proof. If x + y 2 is a square in OK , then there exists α, β ∈ Z such that √ √ √ x + y 2 = (α + β 2)2 = (α2 + 2β 2 ) + 2αβ 2, thus y must be even. √ Since c1 = −c = −u − v 2 < 0 so c1 is not a square. For n ≥ 2, we have cn = c2n−1 + c. √ √ = (un−1 + vn−1 2)2 + (u + v 2). √ 2 = (u2n−1 + 2vn−1 + u) + (2un−1 vn−1 + v) 2. Since v is odd, so 2un−1 vn−1 + v is odd, it follows that cn is not a square in OK . Corollary. All f n (X) are irreducible over OK . Proof. Apply Lemma 3.2 and 5.1, we get the result. 18.

(22) √ √ Lemma 5.2. For any γ = α + β 2 ∈ Z[ 2], N (γ) = α2 − 2β 2 ≡ 7. (mod 8) ⇔ α, β are odd.. Proof. Suppose α, β are odd then α2 ≡ β 2 ≡ 1 (mod 8) and so α2 − 2β 2 ≡ 7 (mod 8). Conversely, for any integer n ∈ Z, n2 ≡ 0, 1, 4 (mod 8). Because we assume that α2 −2β 2 ≡ 7 (mod 8), the only possible of α, β are both odd. Lemma 5.3. N (γn + γn+1 ) > 0 for all n ∈ N. Proof. Note that for n ≥ 1, we have γn+1 = c · γn2 + 1 √ √ = (u + v 2)(αn + βn 2)2 + 1 √ √ = (u + v 2)(Σn + Λn 2) + 1 where Σn = αn2 + 2βn2 √ = (uΣn + 2vΛn + 1) + (uΛn + vΣn ) 2. Hence. Λn = 2αn βn. ( αn+1 = uΣn + 2vΛn + 1, βn+1 = uΛn + vΣn .. Since we assume u, v are positive, so it can deduce that αn , βn are both positive for all n ≥ 1. First, we will prove N (γn ) > 0 for all n ∈ N by induction. Clearly, N (γ1 ) = 1 > 0. Suppose N (γn ) > 0, then 2 2 N (γn+1 ) = αn+1 − 2βn+1 = (uΣn + 2vΛn + 1)2 − 2(uΛn + vΣn )2 = 2uΣn + 4vΛn + N (c)(Σ2n − 2Λ2n ) + 1 = 2uΣn + 4vΛn + N (c)(αn2 − 2βn2 )2 + 1 = 2uΣn + 4vΛn + N (c)N (γn )2 + 1 > 0,. since u, v > 0, so αn , βn > 0 and Σn , Λn ≥ 0 for all n ∈ N. Now, for any n ∈ N, N (γn + γn+1 ) = (αn + αn+1 )2 − 2(βn + βn+1 )2 2 2 = (αn2 − 2βn2 ) + (αn+1 − 2βn+1 ) + (2αn αn+1 − 4βn βn+1 ) = N (γn ) + N (γn+1 ) + (2αn αn+1 − 4βn βn+1 ) > 0 The last term is nonnegative because we have proved that N (γn ) = αn2 − 2βn2 > 0 and √ αn , βn > 0. Thus αn > 2βn for all n ∈ N. √ Lemma 5.4. If c = u + v 2 where u, v are odd, then N (γn + γn+1 ) ≡ 7 (mod 8) for all n ∈ N. 19.

(23) Proof. For n = 1, we have N (γ1 + γ2 ) = N (c + 2) = (u + 2)2 − 2v 2 ≡ 7 (mod 8). If n ≥ 1, then ( αn+1 = u(αn2 + 2βn2 ) + 4αn βn v + 1 ≡ αn + 1 (mod 2), βn+1 = 2αn βn u + v(αn2 + 2βn2 ) ≡ αn (mod 2). Therefore, for n ≥ 2, we have ( αn + αn+1 ≡ αn + (αn + 1) ≡ 1 (mod 2), βn + βn+1 ≡ αn−1 + αn ≡ 1 (mod 2). √ Since γn + γn+1 = (αn + αn+1 ) + (βn + βn+1 ) 2, we have N (γn + γn+1 ) ≡ 7 (mod 8), by Lemma 5.2. √ Theorem 2. Let m ∈ √ Z[ 2] satisfy N (m) ≡ 7 (mod 8), and N (m) > 0, then −1 is not a square modulo m in Z[ 2]. √ Proof. Note that the only ramified prime in Z[ 2] is 2. If p is an odd prime √ in Z and p ≡ 3, 5 2 (mod 8) then p is inertial, so N (p) = p ≡ 1 (mod √ 8). Hence, if m ∈ Z[ 2] with N (m) ≡ 7 (mod 8) then there exists a split prime π ∈ Z[ 2] such that π | m, N (π) is a prime in Z and N (π) ≡ 7 (mod 8). √ Suppose −1 is a quadratic residue modulo m in Z[ 2] then −1 also is a square in √ √ Z[ 2]⧸ where (π) is the ideal in Z[ 2] generated by π. Consider the natural map ϕ : √ (π) √ √ Z[ 2] → Z[ 2]/(π) which induces a map ϕ˜ : Z → Z[ 2]/(π). Thus, the kernel of ϕ˜ is Z ∩ (π) = (N (π)) which is the ideal in Z generated by N (π). Hence we have √ Z[ 2]⧸ ∼ Z⧸ ∼ (π) = (N (π)) = FN (π) . However, −1 is not a square in FN (π) √ since N (π) ≡ 3 (mod 4), which is absurd. Therefore −1 is not a square modulo m in Z[ 2]. √ Corollary. If u, v are odd positive integers and N (c) > 0 then δn is not a square in Q( 2) for n ≥ 2. Proof. We can take m = γn + γn+1 in Lemma 4.1, then −1 is not a square modulo m. Hence √ by Lemma 4.1, δn is not a square in Q( 2) for n ≥ 2.. 20.

(24) 5.2. √ Case 2: K = Q( 2p) and p ≡ 3 (mod 4). √ √ In this √ subsection, since OK = Z[ 2p] so we write cn ∈ OK by un + vn 2p and γn ∈ OK√by αn +βn 2p. Because K is fixed, we denote the norm of K over Q by N (·). Let c = u+v 2p with N (c) > 0 where u, v are positive integers, u ≡ −1 (mod 2p) and v ≡ 1 (mod 2). We have the following lemmas. Lemma 5.5. None of c1 , · · · , cn is a square in OK √ Proof. If x + y 2p is a square in OK , then there exists α, β ∈ Z such that p p p x + y 2p = (α + β 2p)2 = (α2 + 2pβ 2 ) + 2αβ 2p, thus y must be even. √ Since c1 = −c = −u − v 2p and v is odd so c1 is not a square. For n ≥ 2, we have cn = c2n−1 + c. p 2 + u) + (2un−1 vn−1 + v) 2p = (u2n−1 + 2pvn−1. since v is odd so 2un−1 vn−1 + v is odd and cn is not a square in OK . Corollary. All f n (X) are irreducible over OK . Proof. By Lemma 3.2 and 5.5, we get the result. √ √ Lemma 5.6. For any γ = α + β 2p ∈ Z[ 2p], If α ≡ ±1 (mod 2p) and β ≡ 1 (mod 2) then N (γ) ≡ 6p + 1 (mod 8p). Proof. Suppose α = 2pk ± 1 and β = 2l + 1 where k, l ∈ Z then N (γ) = α2 − 2pβ 2 = (4p2 k 2 ± 4pk + 1) − 2p(4l2 + 4l + 1) ≡ 4pk(pk ± 1) + 1 − 2p (mod 8p) ≡ 1 − 2p (mod 8p) since one of pk, pk ± 1 is even. ≡ 6p + 1 (mod 8p). Lemma 5.7. N (γn + γn+1 ) > 0 for all n ∈ N. √ Proof. Note that for n ≥ 1, if γn = αn + βn 2p then γn+1 = c · γn2 + 1 p p = (u + v 2p)(αn + βn 2p)2 + 1 p p = (u + v 2p)(Σn + Λn 2p) + 1 where Σn = αn2 + 2pβn2 p = (uΣn + 2pvΛn + 1) + (uΛn + vΣn ) 2p. 21. Λn = 2αn βn.

(25) ( αn+1 = uΣn + 2pvΛn + 1, βn+1 = uΛn + vΣn .. Hence. Clearly, since u, v > 0, so αn , βn > 0 and Σn , Λn ≥ 0 for all n ∈ N. First, we will prove N (γn ) > 0 for all n ∈ N by induction. Clearly, N (γ1 ) = 1 > 0. Suppose N (γn ) > 0, then 2 2 N (γn+1 ) = αn+1 − 2pβn+1 = (uΣn + 2pvΛn + 1)2 − 2p(uΛn + vΣn )2 = 2uΣn + 4pvΛn + N (c)(Σ2n − 2pΛ2n ) + 1 = 2uΣn + 4pvΛn + N (c)(αn2 − 2pβn2 )2 + 1 = 2uΣn + 4pvΛn + N (c)N (γn )2 + 1 > 0. Now, for any n ∈ N, N (γn + γn+1 ) = (αn + αn+1 )2 − 2p(βn + βn+1 )2 2 2 = (αn2 − 2pβn2 ) + (αn+1 − 2pβn+1 ) + (2αn αn+1 − 4pβn βn+1 ) = N (γn ) + N (γn+1 ) + (2αn αn+1 − 4pβn βn+1 ) > 0 √ The last term is nonnegative because αn ≥ 2pβn for all n ∈ N. √ Lemma 5.8. If c = u + v 2p where u ≡ −1 (mod 2p), and v ≡ 1 (mod 2) then N (γn + γn+1 ) ≡ 6p + 1 (mod 8p). Proof. Note that, for n ≥ 1 ( αn+1 = u(αn2 + 2pβn2 ) + 4vpαn βn + 1 ≡ −αn2 + 1 (mod 2p) βn+1 = 2uαn βn + v(αn2 + 2pβn2 ) ≡ αn (mod 2). Since γ1 = g(0) = 1 so α1 ≡ 1 (mod 2p) and β1 ≡ 0. (mod 2).. For n = 2, we have α2 ≡ −α12 + 1 ≡ 0 (mod 2p) and β2 ≡ α1 ≡ 1 (mod 2). As n = 3, α3 ≡ −α22 + 1 ≡ 1 ≡ α1 Thus. ( αn ≡ 1 (mod 2p) βn ≡ 0 (mod 2). (mod 2p) and β3 ≡ α2 ≡ 0 ≡ α1. ( αn ≡ 0 (mod 2p) for n is odd, and βn ≡ 1 (mod 2). (mod 2).. for n is even.. Hence αn + αn+1 ≡ 1 (mod 2p) and βn + βn+1 ≡ 1 (mod 2) for any n ∈ N. Therefore, N (γn + γn+1 ) ≡ 6p + 1 (mod 8p) by Lemma 5.6. 22.

(26) √ Theorem 3. Let m ∈ Z[ 2p] satisfy √ N (m) ≡ 6p + 1 (mod 8p), and N (m) ≥ 0, then −1 is not quadratic residue modulo m in Z[ 2p]. √ Proof. Note that the discriminant of Z[ 2p] is 8p so the only ramified primes are 2, p. But N (m) ≡ 6p + 1 (mod 8p) thus √ p ∤ N (m) and 2 ∤ N (m). Thus, all prime factors of m√are split or inertial. Since our Z[ 2p] is a unique factorization domain, so for any m ∈ Z[ 2p] we can factor m in some primes pa11 · · · pakk q1b1 · · · qbl l where pi are inertial and qj are split. Note that N (pi ) is a square of prime, so N (pi ) ≡ 1 (mod 4). Hence, if all prime factors of m are inertial then N (m) ≡ 1 (mod 4) which is a contradiction. So l ≥ 2, that means there exists a split prime q1 divides m such that N (q1 ) ≡ 3 (mod 4). √ 2p]⧸ Z[ Suppose −1 is a quadratic residue modulo m in OK then −1 also is a square in (qj ) √ 2p] which generated by q . Consider the natural map ϕ : where (q ) is the ideal in Z[ j √ √ √j Z[ 2p] → Z[ 2p]/(qj ) which induces a map ϕ˜ : Z → Z[ 2p]/(qj ). Thus, the kernel of ϕ˜ is Z ∩ (qj ) = (N (qj )) which is the ideal in Z generated by N (qj ). Hence we have p Z[ 2p]⧸ ∼Z ∼ (q ) = ⧸(N (q )) = FN (qj ) . j. j. However, −1 is not a square in FN (q√j ) since N (qj ) ≡ 3 (mod 4), which is absurd. Therefore −1 is not a square modulo m in Z[ 2p]. Corollary. If u ≡ 1 (mod 2p), and v ≡ 1 (mod 2) and N (c) > 0 then δn is not a square in √ Q( 2p) for n ≥ 2. Proof. We can take m = γk + γk+1 in Lemma 4.1, then −1 is not a square modulo m. Hence √ by Lemma 4.1, δn is not a square in Q( 2p) for n ≥ 2.. 5.3. √ Case 3: K = Q( p) and p ≡ 1 (mod 4). h √ i √ 1+ p 1+ p In this subsection, since OK = Z 2 so we write cn ∈ OK by un + vn and 2 √ 1+ p . Because K is fixed, we denote the norm function of K over γn ∈ OK by αn + βn 2 √ 1+ p Q by N (·). Let c = u + v with N (c) > 0 where u, v are positive integers, u ≡ −1 2 (mod 4p) and v ≡ 2p (mod 4p). Note that since p ≡ 1 (mod 4) so p = 4k + 1 for some k ∈ Z. We have the following lemmas. Lemma 5.9. None of c1 , · · · , cn is square in OK √ 1+ p Proof. If x + y is a square in OK , then there exists α, β ∈ Z such that 2 x+y. √ √ 2 √ 1+ p 1+ p 1+ p 2 2 2 = α+β = (α + kβ ) + (2αβ + β ) . 2 2 2. If β is even then 2αβ + β 2 ≡ 0 (mod 4), if β is odd then 2αβ + β 2 ≡ 1 or 3 (mod 4). 23.

(27) √ 1+ p Since c1 = −c = −u − v and v ≡ 2p (mod 4p) so v ≡ 2 (mod 4) and so c1 is not 2 a square. For n ≥ 2, suppose vn−1 ≡ 2 (mod 4) and use induction, we have cn = c2n−1 + c √ 2 √ 1+ p 1+ p = un−1 + vn−1 + u+v 2 2 √ 1+ p 2 2 2 = (un−1 + kvn−1 + u) + (2un−1 vn−1 + vn−1 + v) 2 2 and 2un−1 vn−1 + vn−1 + v ≡ 2 (mod 4) so cn is not a square in OK .. Corollary. All f n (X) are irreducible over OK . Proof. By Lemma 3.2 and Lemma 5.9, we get the result. √ 1+ p Lemma 5.10. Let c = u + v where u, v ∈ Z. If u ≡ ±1 (mod 4p) and v ≡ 2p 2 (mod 4p) then N (c) ≡ 2p + 1 (mod 4p). Proof. Note that . . √ √ 1+ p 1− p N (c) = u + v u+v 2 2 2 v 2 v p = u+ − 2 4 2 v v2p = u2 + uv + − 4 4 2 2 = u + uv − kv Suppose u = 4pm ± 1 and v = 4pn + 2p where m, n ∈ Z then N (c) = u2 + uv − kv 2 = (4pm ± 1)2 + (4pm ± 1)(4pn + 2p) − k(4pn + 2p)2 ≡ 1 + 2p (mod 4p) . √ 1+ p 2. where u, v ∈ Z. If u ≡ −1 (mod 4p) and v ≡ 2p Lemma 5.11. Let c = u + v (mod 4p) then N (γn + γn+1 ) ≡ 2p + 1 (mod 4p).. 24.

(28) Proof. Note that for n ≥ 1, if γn = αn + βn. . √ 1+ p 2. then. γn+1 = c · γn2 + 1 √ 2 √ 1+ p 1+ p αn + βn +1 = u+v 2 2 √ √ 1+ p 1+ p = u+v Σn + Λn +1 2 2 √ 1+ p = (uΣn + vkΛn + 1) + (uΛn + vΣn + vΛn ) , 2 where Σn = αn2 + kβn2 and Λn = 2αn βn + βn2 . Thus ( αn+1 = u(αn2 + kβn2 ) + vk(2αn βn + βn2 ) + 1 βn+1 = (u + v)(2αn βn + βn2 ) + v(αn2 + kβn2 ). Since γ1 = g(0) = 1, so α1 ≡ 1. (mod 4p) and β1 ≡ 0 (mod 4p). √ √ 1+ p 1+ p = α2 + β2 , so For n = 2, we have γ2 = (u + 1) + v 2 2 ( α2 = u(α12 + kβ12 ) + vk(2α1 β1 + β12 ) + 1 ≡ u + 1 ≡ 0 (mod 4p) β2 = (u + v)(2α1 β1 + β12 ) + v(α12 + kβ12 ) ≡ v ≡ 2p (mod 4p). For n = 3, we have ( α3 = u(α22 + kβ22 ) + vk(2α2 β2 + β22 ) + 1 ≡ 1 ≡ α1 (mod 4p) β3 = (u + v)(2α2 β2 + β22 ) + v(α22 + kβ22 ) ≡ 0 ≡ β1 (mod 4p). Thus. ( αn ≡ 1 (mod 4p) βn ≡ 0 (mod 4p). ( αn ≡ 0 (mod 4p) for n is odd, and βn ≡ 2p (mod 4p). for n is even.. Hence αn + αn+1 ≡ 1 (mod 4p) and βn + βn+1 ≡ 2p (mod 4p) for any n ∈ N. Therefore, N (γn + γn+1 ) ≡ 2p + 1 (mod 4p) by Lemma 5.10. Lemma 5.12. If u, v > 0 then αn , βn ≥ 0 for all n ≥ 1. Proof. Since γ1 = 1 so α1 ≥ 0 and β1 ≥ 0. Assume it is hold for n = m, then consider n = m + 1, we have ( 2 2 2 αm+1 = u(αm + kβm ) + vk(2αm βm + βm )+1 2 2 2 βm+1 = (u + v)(2αm βm + βm ) + v(αm + kβm ). Since k = (p − 1)/4 so k is bigger than 0 and αm , βm ≥ 0. By induction, we have αn , βn ≥ 0 for all n ∈ N. 25.

(29) Lemma 5.13. If N (c) > 0, u, v > 0, then N (γn + γn+1 ) > 0 for all n ∈ N. Proof. We will prove N (γn ) > 0 for all n ∈ Z by induction. Note that since u, v ≥ 0 so αn , βn ≥ 0 for all n ∈ N, thus Σn , Λn ≥ 0 for all n ≥ 1. Σ2n + Σn Λn − kΛ2n = (αn2 + kβn2 )2 + (αn2 + kβn2 )(2αn βn + βn2 ) − k(2αn βn + βn2 )2 = αn4 − 2kαn2 βn2 + k 2 βn4 + 2αn3 βn + αn2 βn2 − 2kαn βn3 = (αn2 + αn βn − kβn2 )2 = N (γn )2 . N (γ1 ) = N (1) > 0 and Suppose N (γn ) > 0, then 2 2 N (γn+1 ) = αn+1 + αn+1 βn+1 − kβn+1 = (uΣn + vkΛn + 1)2 + (uΣn + vkΛn + 1)(uΛn + vΣn + vΛn ) − k(uΛn + vΣn + vΛn )2 = (Σ2n + Σn Λn − kΛ2n )N (c) + 2uΣn + vΣn + uΛn + vΛn + 2vkΛn + 1 = N (γn )2 N (c) + 2uΣn + vΣn + uΛn + vΛn + 2vkΛn + 1 > 0.. By induction, N (γn+1 ) > 0 for all n ∈ N. Because N (γn + γn+1 ) = (αn + αn+1 )2 + (αn + αn+1 )(βn + βn+1 ) − k(βn + βn+1 )2 = N (γn ) + N (γn+1 ) + αn βn+1 + αn+1 βn + 2αn αn+1 − 2kβn βn+1 , if we can prove αn βn+1 + αn+1 βn + 2αn αn+1 − 2kβn βn+1 ≥ 0 then N (γn + γn+1 ) > 0. Note that N (γn ) > 0 so we have N (γn ) = αn2 + αn βn − kβn2 > 0, thus αn (αn + βn ) > kβn2 for all n ∈ N. Since αn , βn ≥ 0 for all n so αn βn+1 + αn+1 βn + 2αn αn+1 = αn (αn+1 + βn+1 ) + αn+1 (αn + βn ) p ≥ 2 αn αn+1 (αn + βn )(αn+1 + βn+1 ) q 2 > 2 k 2 βn2 βn+1 = 2kβn βn+1 Hence N (γn + γn+1 ) ≥ 0 for all n > 1. h √ i 1+ p Theorem 4. Let m ∈ Z 2 = OK satisfy N (m) ≡ 2p + 1 (mod 4p), and N (m) > 0, then −1 is not quadratic residue modulo m in OK . Proof. Note that the discriminant of OK is p so the only ramified primes are p. But N (m) ≡ 2p + 1 (mod 4p) thus p ∤ N (m), moreover, N (m) ≡ 3 (mod 4). Hence all prime factors of m is split or inertial. Moreover, if p is inertial then N (p) is a square of a rational prime and N (p) ≡ 1 (mod 4). Hence if all prime factor of m are inertial then N (m) ≡ 1 (mod 4) which is a contradiction. so there exists a split prime divides m. 26.

(30) Since our OK is a unique factorization domain so for any m ∈ OK we can factor m in some primes pa11 · · · pakk qb11 · · · qbl l where pi are inertial and qj are split. Note that N (pi ) is a square of prime, so N (pi ) ≡ 1 (mod 4). Hence there exists at least one qj such that N (qj ) ≡ 3 (mod 4) where N (qj ) is a rational prime. Suppose −1 is a quadratic residue modulo m in OK then −1 also is a square in OK⧸(q ) j where (qj ) is the ideal in OK which generated by qj . Consider the natural map ϕ : OK → OK /(qj ) which induces a map ϕ˜ : Z → OK /(qj ). Thus, the kernel of ϕ˜ is Z ∩ (qj ) = (N (qj )) which is the ideal in Z generated by N (qj ). Hence we have OK⧸ ∼Z ∼ (qj ) = ⧸(N (qj )) = FN (qj ) . However, −1 is not a square in FN (qj ) since N (qj ) ≡ 3 (mod 4), which is absurd. Therefore −1 is not a square modulo m in OK . Corollary. If u ≡ −1 (mod 4p), and v ≡ 2p (mod 4p) and N (c) > 0 then δn is not a square √ in Q( p) for n ≥ 2. Proof. We can take m = γk + γk+1 in Lemma 4.1, then −1 is not a square modulo m. Hence √ by Lemma 4.1, δn is not a square in Q( p) for n ≥ 2.. 6. √ n ∼ Ωn = [C2] for K = Q( 2). √ √ Note that√if c = u + v 2 is a square in Z[ 2] then v must be even. However, we consider c = u + v 2 where u, v are odd positive integers, so none of b1 = −c,. b2 = −(c + 1),. b 1 b2 = c 2 + c. is a square. In this section, we will prove that there must appear √ a new prime with odd power in bi √ for i > 2, thus {b1 , · · · , bn } is 2-independent over Q( 2). Since Z[ 2] is a unique factorization domain, bi can be unique factored into products of units and primes as following: bi = ±ue pr11 · · · prkk √ √ where u = 1 + 2 is a fundamental unit in Z[ 2]. By the corollary to Theorem 5.1, at least one of e, r1 , · · · , rk must be odd. We will prove, for i ≥ 3, there does not exist √ the case which e is odd and r1 , · · · , rk are even, that means ′ ′ 2 bi = ±u(bi ) where bi ∈ Z[ 2]. Hence, there appears √ a new prime with odd power in bi for i ≥ 3 and so {b1 , · · · , bn } is 2-independent over Q( 2) for √ n ≥ 1.√ 2 ∈ Z[√ 2], let [γ]2 be the residue We define some notations as follows: for γ = α + β √ class of γ in Z[ 2]/(2) and [γ]4 be the residue class of γ in Z[ 2]/(4). Moreover, we also denote [γ]2 := [α mod 2, β mod 2]2 and [γ]4 := [α mod 4, β mod 4]4 . √ For example, if γ = 10 + 23 2 then [γ]2 = [0, 1]2 and [γ]4 = [2, 3]. 27.

(31) √ √ √ For any γ = α + β 2, γ ′ = α′ + β ′ 2 ∈ Z[ 2], we define [γ]4 ∗ [γ ′ ]4 := [γγ ′ ]4 = [αα′ + 2ββ ′ mod 4, αβ ′ + α′ β mod 4]4 and. [γ]2 ∗ [γ ′ ]2 := [γγ ′ ]2 = [αα′ mod 2, αβ ′ + α′ β mod 2]2 .. First, we observe the following result: √ Lemma 6.1. If γ = ±u(γ ′ )2 where γ ′ ∈ Z[ 2] then [γ]4 ∈ {[0, 0]4 , [2, 2]4 , ±[1, 1]4 , ±[3, 1]4 }. √ Proof. Suppose γ ′ = x + y 2 then √ √ γ = ±(1 + 2)(x + y 2)2 √ √ = ±(1 + 2)(x2 + 2y 2 + 2xy 2) √ = ±((x2 + 2y 2 + 4xy) + (x2 + 2y 2 + 2xy) 2). Hence.  [0, 0]4    ±[3, 1] 4 [γ]4 = [2, 2]4    ±[1, 1]4. if if if if. x, y are even, x, y are odd, x is even and y is odd, x is odd and y is even.. √ Next, since c = u+v 2 where u, v are odd positive integers, so [c]4 ∈ {[1.1]4 , [1, 3]4 , [3, 1]4 , [3, 3]4 }, and we want to compute [cn ]4 for all n ≥ 1. Proposition. If [c]4 = [a, b]4 ∈ {[1.1]4 , [1, 3]4 , [3, 1]4 , [3, 3]4 } then [c1 ]4 = [−a, −b]4 . Moreover, ( [a + 3, b + 2]4 if n is even, [cn ]4 = [a + 2, b]4 if n is odd and n ≥ 3. Proof. Obviously, since c1 = −c so [c1 ]4 = [−a, −b]4 . Note that a, b are odd so a2 ≡ b2 ≡ 1 (mod 4) and 2ab ≡ 2. (mod 4).. Since [c1 ]4 = [−a, −b]4 and cn = c2n−1 + c for n ≥ 2 so [c2 ]4 = [−a, −b]4 ∗ [−a, −b]4 + [a, b]4 = [a2 + 2b2 , 2ab]4 + [a, b]4 = [a2 + 2b2 + a, 2ab + b]4 = [a + 3, b + 2]4 ,. 28.

(32) and [c3 ]4 = [a + 3, b + 2]4 ∗ [a + 3, b + 2]4 + [a, b]4 = [(a + 3)2 + 2(b + 2)2 + a, 2(a + 3)(b + 2) + b]4 = [2(b + 2)2 + a, b]4 = [a + 2, b]4 because a + 3 is even and b + 2 is odd. Moreover, since [c4 ]4 = [a + 2, b]4 ∗ [a + 2, b]4 + [a, b]4 = [(a + 2)2 + 2b2 + a, 2(a + 2)b + b]4 = [a + 3, b + 2]4 = [c2 ]4 , thus [cn ]4 = [a + 2, b]4 for any odd integer n ≥ 3, and [cn ]4 = [a + 3, b + 2]4 for all even n. By Lemma 6.1, if we can prove that [bi ]4 6∈ {[0, 0]4 , [2, 2]4 , ±[1, 1]4 , ±[3, 1]4 } for i ≥ 3, then there must appear a new prime with odd power in bi . Since b1 = −c and b2 = −(c + 1) so [b1 ]2 = [1, 1]2 and [b2 ]2 = [0, 1]2 for any case of c. Moreover, we have Lemma 6.2. [bi ]2 6= [0, 0]2 , [0, 1]2 for i ≥ 3. Proof. Note that, for any [a, b]2 where a, b ∈ {0, 1}, [a, b]2 ∗[0, 0]2 = [0, 0]2 and [a, b]2 ∗[0, 1]2 = [0, 0]2 or [0, 1]2 . Suppose [bn ]2 = [0, 0]2 or [0, 1]2 for some n ≥ 3, note that Y cn = bd . d|n. If n ≥ 3 is odd, then [cn ]2 = [b1 ]2 ∗ · · · ∗ [bn ]2 = [0, 0]2 or [0, 1]2 . However, [cn ]2 = [1, 1]2 , by previous Proposition, which is absurd. On the other hand, if n ≥ 3 is even then [0, 1]2 = [cn ]2 = [b1 ]2 ∗ [b2 ]2 ∗ · · · ∗ [bn ]2 . However, [b1 ]2 ∗[b2 ]2 ∗[bn ]2 = [1, 1]2 ∗[0, 1]2 ∗[bn ]2 = [0, 1]2 ∗[bn ]2 = [0, 0]2 whenever [bn ]2 = [0, 0]2 or [0, 1]2 . Hence the right-hand side of above equation is [0, 0]2 which is absurd. Theorem 5. For n ≥ 3, [bn ]4 6∈ {[1, 1]4 , [1, 3]4 , [3, 1]4 , [3, 3]4 }. Proof. Note that, we have the following multiplication table: * [1, 0]4 [3, 0]4 [1, 2]4 [3, 2]4. [1, 0]4 [1, 0]4 [3, 0]4 [1, 2]4 [3, 2]4. [3, 0]4 [3, 0]4 [1, 0]4 [3, 2]4 [1, 2]4 29. [1, 2]4 [1, 2]4 [3, 2]4 [1, 0]4 [3, 0]4. [3, 2]4 [3, 2]4 [1, 2]4 [3, 0]4 [1, 0]4.

(33) Hence {[1, 0]4 , [3, 0]4 , [1, 2]4 , [3, 2]4 } is closed under multiplication. By Lemma 6.2, we know that [bn ]4 ∈ {[1, 1]4 , [1, 3]4 , [3, 1]4 , [3, 3]4 , [1, 0]4 , [3, 0]4 , [1, 2]4 , [3, 2]4 } for n ≥ 3. Now, suppose there exists n such that [bn ]4 ∈ {[1, 1]4 , [1, 3]4 , [3, 1]4 , [3, 3]4 } and such n is minimal. Since n is the least number greater than 2 such that [bn ]4 6∈ {[1, 0]4 , [3, 0]4 , [1, 2]4 , [3, 2]4 } so [bi ]2 = [1, 0]2 for all 3 ≤ i < n. If n is odd then [1, 1]2 = [cn ]2 = [b1 ]2 ∗ · · · ∗ [bn ]2 = [1, 1]2 ∗ · · · ∗ [1, 1]2 = [1, 0]2 which is absurd. On the other hand, we consider n is even. Since n is the first number such that [bn ]4 6∈ {[1, 0]4 , [3, 0]4 , [1, 2]4 , [3, 2]4 } so Y [bd ]4 ∈ {[1, 0]4 , [3, 0]4 , [1, 2]4 , [3, 2]4 }, d|n 2<d<n. since they are closed under multiplication. Hence we need to consider the following cases: 1. For [c]4 = [1, 1]4 , we have [cn ]4 = [0, 3]4 for n is even. Note that [b1 ]4 ∗ [b2 ]4 = [3, 3]4 ∗ [2, 3]4 = [0, 3]4 so [b1 ]4 ∗ [b2 ]4 ∗ [bn ]4 = [2, 1]4 or [2, 3]4 . Moreover, [cn ]4 = [b1 ]4 ∗ [b2 ]4 ∗ · · · ∗ [bn ]4 = [2, 1]4 or [2, 3]4 , which is a contradiction since [cn ]4 = [0, 3]4 . 2. For [c]4 = [1, 3]4 , we have [cn ]4 = [0, 1]4 for n is even. Note that [b1 ]4 ∗ [b2 ]4 = [3, 1]4 ∗ [2, 1]4 = [0, 1]4 so [b1 ]4 ∗ [b2 ]4 ∗ [bn ]4 = [2, 1]4 or [2, 3]4 . Moreover, [cn ]4 = [b1 ]4 ∗ [b2 ]4 ∗ · · · ∗ [bn ]4 = [2, 1]4 or [2, 3]4 , which is a contradiction since [cn ]4 = [0, 1]4 . 3. For [c]4 = [3, 1]4 , we have [cn ]4 = [2, 3]4 for n is even. Note that [b1 ]4 ∗ [b2 ]4 = [1, 3]4 ∗ [0, 3]4 = [2, 3]4 so [b1 ]4 ∗ [b2 ]4 ∗ [bn ]4 = [0, 1]4 or [0, 3]4 . Moreover, [cn ]4 = [b1 ]4 ∗ [b2 ]4 ∗ · · · ∗ [bn ]4 = [0, 1]4 or [0, 3]4 , which is a contradiction since [cn ]4 = [2, 3]4 . 30.

(34) 4. For [c]4 = [3, 3]4 , we have [cn ]4 = [2, 1]4 for n is even. Note that [b1 ]4 ∗ [b2 ]4 = [1, 1]4 ∗ [0, 1]4 = [2, 1]4 so [b1 ]4 ∗ [b2 ]4 ∗ [bn ]4 = [0, 1]4 or [0, 3]4 . Moreover, [cn ]4 = [b1 ]4 ∗ [b2 ]4 ∗ · · · ∗ [bn ]4 = [0, 1]4 or [0, 3]4 , which is a contradiction since [cn ]4 = [2, 1]4 . √ Corollary. If c = u + v 2 where u, v are odd positive integers, then Ωn ∼ = [C2 ]n for any n ≥ 1. Proof. By Lemma 6.1 and Theorem 5, we show that bi will appear √ a new prime with odd power for i ≥ 3. Hence {b1 , · · · , bn } is 2-independent over Q( 2) for any n ∈ Z and so Ωn ∼ = [C2 ]n by Theorem 1.. 7. 2-independent property of integers over quadratic number field. We split this section into two parts, first, for f (X) = X 2 + c ∈√Z[X], we √ will find some suffin cient conditions such that the Galois group of f (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. µ(n/d). cd. ,. d|n. 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); 31.

(35) 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 2independent over Q. Moreover, Ωn ∼ = [C2 ]n for all n ∈ N. Proof. Since (bi , bj ) = 1 for i 6= 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, ( √ nao −1 if S is 2-independent over Q( a), = √ S 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 #{b ∈ B : b ≤ n} n→∞ #{a ∈ A : a ≤ n}. dA (B) = lim 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 32.