Arithmetically equivalent fields in a Galois extension with Frobenius Galois group of 2-power degree

Abstract Let 
$F_{2^n}$
 be the Frobenius group of degree 
$2^n$
 and of order 
$2^n ( 2^n-1)$
 with 
$n \ge 4$
 . We show that if 
$K/\mathbb {Q} $
 is a Galois extension whose Galois group is isomorphic to 
$F_{2^n}$
 , then there are 
$\dfrac {2^{n-1} +(-1)^n }{3}$
 intermediate fields of 
$K/\mathbb {Q} $
 of degree 
$4 (2^n-1)$
 such that they are not conjugate over 
$\mathbb {Q}$
 but arithmetically equivalent over 
$\mathbb {Q}$
 . We also give an explicit method to construct these arithmetically equivalent fields.


Introduction
The following theorem concerning coincidence of Hecke L-functions is proved in [6]. A natural question arises from this theorem.
Question Are there arbitrarily large number of number fields whose Dedekind zeta functions coincide?
Two number fields K and K ′ are called arithmetically equivalent (over Q) if the Dedekind zeta functions of K and K ′ coincide. Conjugate number fields obviously have the same Dedekind zeta functions; thus, we are interested in nonconjugate arithmetically equivalent fields. Many examples of such fields are known until now (see [8, Examples in III.1.b]), but examples of three or more arithmetically equivalent fields seem not to be known. The aim of this paper is to give such examples systematically.
To state our result more precisely, we introduce some notation. Let F 2 n be a finite field of 2 n elements. We consider the Frobenius group F 2 n defined by where F × 2 n acts faithfully on F 2 n . The group F 2 n can be described also as an affine linear group over F 2 n : The Frobenius kernel N is isomorphic to and a Frobenius complement H is isomorphic to Let K/Q be a Galois extension with Galois group isomorphic to the Frobenius group F 2 n . Such an extension K/Q is called an F 2 n -extension. The fixed field L of K by the Frobenius kernel N is a cyclic extension of degree 2 n − 1 over Q and Gal(K/L) is isomorphic to an elementary abelian 2-group of rank n. Our main theorem is the following. As a matter of fact, there are several nonconjugate arithmetically equivalent fields of degree 2 s (s = 2, . . . , n − 2) over L inside K. We concentrate the smallest degree fields for simplicity both in the proof and in the construction. Our proof and construction are explicit and specific throughout, and this enables us to find families of a large number of nonconjugate arithmetically equivalent fields explicitly.
The outline of the paper is as follows. In the next section, we prove Theorem 1.2 in a refined form (Theorem 2.4) by using mainly the representation theory of finite groups. In Section 3, we discuss how to construct F 2 n -extensions. We show that if a cyclic extension L is constructed, then we can construct infinitely many F 2 n -extensions containing L by using Kummer theory (Theorem 3.3). Our discussion here is explicit and constructive for the argument in the following section. In Section 4, we explain how to find nonconjugate arithmetically equivalent fields in an F 2 n -extension and give an explicit description of these fields (Proposition 4.1).
Throughout this paper, we use the following notation. We fix an integer n greater than 3. The Frobenius group of degree 2 n and of order 2 n (2 n − 1) is denoted by F 2 n = H ⋉ N with H and N defined in the above.

The proof of the main theorem
In this section, we shall prove Theorem 1.2. Let K/Q be an F 2 n -extension. We fix an isomorphism between Gal(K/Q) and F 2 n = H ⋉ N and identify them by the isomorphism. Let L be the fixed field K N . The extension L/Q is a cyclic extension of degree 2 n − 1. The Galois group of K/L is isomorphic to an elementary abelian 2-group N of rank n, since the additive group of the field F 2 n is isomorphic to F n 2 . In this section, we use the following notation from the representation theory of finite groups. For a finite group G, we denote by Irr(G) the set of the irreducible character of G and by 1 G the principal character of G. For a character ψ of a subgroup E of G, we denote the induced character from ψ to G by ψ G and for a character χ of G, the restriction of χ to E by χ E .
We Note that the group E in Proposition 2.1 is core-free, that is, Core G (E) = ⋂ g∈G (gEg −1 ) = 1, and thus the character 1 G E is a faithful permutation character. This also implies that the Galois closure of K E coincides with K.
To prove the proposition, we use the following fact on the representation of Frobenius groups, which is a special case of [4, Satz V.16.13].

Lemma 2.2
The irreducible characters of F 2 n = H ⋉ N consist of linear characters μ i (i ∈ {0, . . . , 2 n − 2}) which are extensions of η i ∈ Irr(H) with ker μ i ⊃ N and a character ψ of degree 2 n − 1 induced from a nontrivial character φ of N such that ψ N = ∑ h∈H φ h .
Proof of Proposition 2. 1 We compute the inner product of 1 G E and χ ∈ Irr(G) by Frobenius reciprocity: Let μ i and ψ be the characters as in Lemma 2.2. Since E ⊂ N, we have If χ = ψ, then we can write ψ = φ G with φ(≠ 1 N ) ∈ Irr(N). It is clear that

383
Here, the second equality holds since the action of H on N is transitive and faithful, and therefore the set {hxh −1 | h ∈ H} coincides with N − {1}. Moreover, the third equality follows from the fact that φ is nontrivial. We conclude Consequently, we obtain the decomposition of 1 G E : We now enumerate such quartic fields up to conjugacy. We shall prove a more precise version of Theorem 1.2.   We have already showed their arithmetic equivalence in Corollary 2.3. By Galois theory, we only have to prove the following group-theoretic version of Theorem 2.4.

Theorem 2.5
Let E be the set of the subgroups of order 2 n−2 of N. The group G acts on E through F × 2 n . Let D be a subgroup of N of order 2 n−1 . We have |E ∩ D| = 2 n−1 − 1 with obvious abuse of notation. (ii) If n is even, then the set E is divided into conjugacy classes C satisfy |C ∩ D| = 3, and the rest of the classes C ′ satisfies |C ′ ∩ D| = 1.
Proof It is well known that the number of t-dimensional subspaces in an s-dimensional vector space over F 2 is given by the q-binomial coefficient with q = 2, which we denote by Using this formula, we can compute and Now, we identify N with the additive group of F 2 n . If g is a generator of F × 2 n and E ∈ E , then we can write E = {0, g i1 , . . . , g i 2 n−2 −1 } with {i 1 , . . . , i 2 n−2 −1 } ⊂ {1, 2, . . . , 2 n − 1}. For notational convenience, we write it as E = (i 1 , . . . , i 2 n−2 −1 ). If we represent τ ∈ G by a product g ν (g ∈ H, ν ∈ N), then it is easy to see that the conjugate E τ is given by E τ = E g = (i 1 + , . . . , i 2 n−2 −1 + ). We compute the normalizer N G (E). We have τ = g ν ∈ N G (E) if and only if there exists a permutation γ ∈ S 2 n−2 −1 such that Summing up both the sides for j, we obtain and this yields (2 n−2 − 1) ≡ 0(mod 2 n − 1). In this connection, we see Therefore, if n is odd, then we conclude that = 0 and N G (E) = N. Hence, the orbit length of every E ∈ E is 2 n − 1, and the set E is divided into If n is even, then we obtain = 0 or 2 n − 1 3 . In the latter case, the element g in N G (E) is of order 3. Accordingly, the orbit length of E is either 2 n − 1 or 2 n − 1 3 . Let u Arithmetically equivalent fields 385 (resp. v) be the number of orbits of length 2 n − 1 (resp. 2 n − 1 3 ). It obviously yields We compute the total number of orbits u + v by using the lemma of Burnside- As is seen in the above, we have Fix( Moreover, if the order of g i is neither 1 nor 3, then Fix(x) = ∅. Obviously, if the order of g i is equal to 1, then we have i = 0 and Fix(1) = E . We now suppose that the order of g i is 3, and thus i = (2 n − 1)/3. Since the minimal ]. Therefore, we conclude |Fix(g i )| = 2 n − 1 3 .
Let O be an orbit in E . Since G acts on the set of D's transitively by Singer's theorem [3, Theorem 11.3.1], the number |O ∩ D| is independent of the choice of D.
We first consider the case where n is odd. Let O i (i = 1, . . . , (2 n−1 − 1)/3) be the conjugacy classes. Since Hence, we conclude that |O i ∩ D| = 3, namely each D contains three conjugate fields. Next, we consider the case where n is even. Let O i (i = 1, . . . , 2(2 n−2 − 1)/3) be the conjugacy classes of length 2 n − 1, and let P be the conjugacy class of length (2 n − 1)/3. If E ∈ P, then it is invariant by an element of order 3 in F × 2 n . Therefore, such E is contained in three different D's. Since there are 2 n − 1 nonconjugate D's, we conclude that |P ∩ D| = 1. This also yields an equation like (2.3):

Construction of F 2 n -extensions
In this section, we construct F 2 n -extensions for every n. The method is an extension of those used in [7,9], where only metacyclic extensions are constructed. The method fully works for a general base field k whose characteristic is not 2. Thus, we assume that G = Gal(K/k) = F 2 n = H ⋉ N and L = K N , and that a cyclic extension L/k has been constructed.
In the case k = Q, if we take a prime number satisfying ≡ 1(mod 2 n − 1), there is a unique cyclic field L of degree 2 n − 1 inside the th cyclotomic field. Furthermore, there exist infinitely many such prime numbers for each n by Dirichlet's theorem on arithmetic progression.
Let us return to the general case. We now have to construct an elementary abelian 2-extension of degree 2 n over L which is an F 2 n -extension over k. We fix a generator g of F × 2 n and consider the F 2 n -valued characters Here, we consider the cyclic group C 2 n −1 as a Galois group of L/k, and σ is a fixed generator of C 2 n −1 . In this situation, it is necessary to distinguish Gal(L/k) and F × 2 n . We define

387
They are the primitive orthogonal idempotents, and we have a direct sum decomposition of the group ring into one-dimensional irreducible modules by Maschke's theorem.
We now further define where F 2 (χ i ) is the field generated by the character values of χ i . There are as many different ε i as the Galois conjugacy class of the characters {χ i } (see [5,Lemma 9.17]), and they are nonzero by [5,Corollary 9.22]. If we factor the polynomial X 2 n −1 − 1 = ∏ t ϕ t (X) into irreducibles in the polynomial ring F 2 [X], then we have a direct sum decomposition over F 2 : If we choose the index t so that ϕ t is a minimal polynomial of χ t (σ), then Hence, we obtain Lemma 3.1 Let the notation be as above. If (i, 2 n − 1) = 1, then the module Proof If we assume that (i, 2 n − 1) = 1, then the order of χ i is exactly 2 n − 1 and the value of χ i is not contained in any proper subfields of F 2 n . Thus, we observe that ε i F 2 [C 2 n −1 ] is an n-dimensional subspace over F 2 .
Since ε i 's are orthogonal idempotents, V i is apparently an F 2 [C 2 n −1 ]-module.
To show its irreducibility, suppose to the contrary that V i is not irreducible. There is a proper submodule W of V i . Since V i splits over F 2 n , the module W also splits over F 2 n . Therefore, the character of W is a proper subsum of ε i . However, such a subsum does not have its values in F 2 ; therefore, W cannot be defined over F 2 . This is a contradiction. ∎ It is readily seen that there are φ(2 n − 1)/n ε i 's with (i, 2 n − 1) = 1, where φ is the Euler's totient function.

Lemma 3.2 Let the notation be as in Lemma
as abstract groups.

M. Kida
Proof By (1.1), it suffices to show that C 2 n −1 ⋉ V i is isomorphic to AGL 1 (F 2 n ).
By the isomorphism (3.2), we identify V i with F 2 [X]/(ϕ i (X)), where ϕ i (X) is the minimal polynomial of g i = χ i (σ) and hence is of degree n. We define a map κ from . By noting that σ acts on V i by the multiplication of g, the map κ sends On the other hand, we compute Therefore, κ is a homomorphism. We see that (σ j , U(x)) ∈ ker κ if and only if g i j = 1 and U(g i ) = 0. The condition g i j = 1 is equivalent to j ≡ 0(mod 2 n − 1) since (i, 2 n − 1) = 1. The condition U(g i ) = 1 is equivalent to the minimal polynomial ϕ i (X) of g i divides U(X). Therefore, the kernel consists of the trivial element only. Since the orders of both the groups are the same, the map κ is an isomorphism. ∎ We can now state our method of construction of F 2 n -extensions.

Theorem 3.3
Recall that L is a cyclic extension of k of degree 2 n − 1 and that ε i is an idempotent defined by (3.1). Assume that (i, 2 n − 1) = 1. If θ ∈ ε i (L × /(L × ) 2 ) is nontrivial, then the Galois closure of L ( √ θ) over k is an F 2 n -extension over k.
We have an exact sequence induced from the restriction map. The Galois group Gal(L/k) acts on Gal(K/L): for γ ∈ Gal(K/L) and σ ∈ Gal(L/k), we choose an extensionσ in Gal(K/k) and we define σ ⋅ γ =σγσ −1 . This action is well defined because Gal(K/L) is abelian. For σ ∈ Gal(L/k), we define s(σ) ∈ Gal(K/k) by It is easy to verify that this map s gives a splitting homomorphism and the above exact sequence splits. By Kummer theory, there exists a bilinear nondegenerate pairing defined as where the map μ 2 → F 2 is an isomorphism whose inverse map is F 2 ∋ x ↦ (−1) x . This yields an isomorphism Gal(K/L) ≅ Hom(M, F 2 ), γ ↦ ⟨γ, θ⟩. (3.5) Both the sides of (3.5) are Gal(L/k)-modules. The action on the right-hand side is given by σ(θ ↦ ⟨γ, θ⟩) = (θ ↦ ⟨γ, σ θ⟩).
We shall show that Gal(K/L) is an irreducible F 2 [Gal(L/k)]-module isomorphic to ε j F 2 [Gal(L/k)] for some integer j prime to 2 n − 1. Then, from Lemma 3.2, Gal(K/k) ≅ F 2 n follows. To do this end, we compute the action of σ ∈ Gal(L/k) on Gal(K/L) in terms of (3.3). In the above, we have shown thatσ( √ θ i ) = ± √ σ θ i , and thus we can define e i ∈ F 2 byσ

Using (3.3), we can compute further
The relation of e i 's and f i 's is derived by computingσσ −1 ( √ θ i ) = √ θ i . In fact, the left-hand side is equal to Hence, we obtain Now, let (g 1 , . . . , g n ) be the dual basis of Gal(K/L) with respect to the paring ⟨⋅, ⋅⟩. We compute the action σ ⋅ g i on Combining with (3.6), we have This means that σ acts on Gal(K/L) by A −1 . Therefore, Gal(K/L) is isomorphic to an irreducible module ε −i F 2 [Gal(L/k)]. Since (−i, 2 n − 1) = 1, we conclude Gal(K/k) ≅ F 2 n . This completes the proof of Theorem 3.3. ∎

Remark 3.4
Our proof shows that if σ acts on M by A as (3.3), then it acts on Gal(K/L) by A −1 . This argument does not depend on the assumption that (i, 2 n − 1) = 1. If we drop this assumption, then we obtain Galois extensions whose Galois groups are various semidirect products of Gal(L/k) and Gal(K/L) including the direct product. See [7, Section 6] for example.
The following corollary follows from the proof of Theorem 3.3.
The following corollary guarantees that there are infinitely many F 2 n -extensions containing L. Proof This follows from the Kummer duality (3.4). ∎

Identifying arithmetically equivalent fields
In the previous section, we have constructed F 2 n -extensions. In this section, we explain how to identify arithmetically equivalent fields inside the F 2 n -extensions. We continue to use the notation used in the proof of Theorem 3.3. For convenience, we recall some of them. Let L/k be a cyclic extension of degree 2 n − 1. We assume that i is an integer prime to 2 n − 1 and consider an irreducible where ε i is the idempotent defined by (3.1). The module M is generated by θ and has a basis (θ = θ 1 , . . . , θ n ) over F 2 . We now fix a generator σ of Gal(L/k) and assume that σ acts on the above basis by (3.3). The Galois group Gal(K/L) is isomorphic to the dual group Hom(M, F 2 ) of M (see (3.5)). We want to find quadratic extensions of L( √ θ) which are arithmetically equivalent but not conjugate. In Proposition 2.1, we have shown that all such quadratic extensions are arithmetically equivalent, and hence we only have to identify the conjugacy classes of these fields.

Definition 4.1
We denote byF n 2 the quotient space of F n 2 by the subspace generated by e 1 = t (1, 0, . . . , 0). Namely, column vectors e and f ∈ F n 2 are equal inF n 2 if they coincide except for the first coordinate. Ifṽ = t (e 1 , . . . , e n ) ∈F n 2 , then a quadratic extension is well defined and there is a one-to-one correspondence between such quadratic extensions and the setF n 2 − {0}. The conjugate field of Q(ṽ) by σ is then given by is equivalent to that θ 1 coincides with either ∏ n k=1 θ e2 a k2 +⋯+en a kn k or the product ∏ n k=1 θ a k1 +e2 a k2 +⋯+en a kn k since ∏ n k=1 θ a k1 k does not coincide with θ 1 . It is easy to observe that this condition holds if and only if Aṽ = e 1 . Ifṽ satisfies this condition, then Q(ṽ) = Q(A −1 e 1 ) is conjugate to Q(Ae 1 ). Note that since Gal(L/k) acts transitively onF n 2 , for every element v ∈F n 2 , there exists an integer j such thatṽ = A j e 1 . In accordance with this observation, we define the following equivalence relation onF n 2 .

Definition 4.2
Let us fix σ as a generator of Gal(L/k), and let A be the matrix defined by (3.3). The elementsṽ = A i e 1 andf = A j e 1 inF n 2 are said to be equivalent if i + j ≡ 0(mod 2 n − 1).
Using these definitions, we obtain the following proposition. For explicit computation, it remains to give a basis of the irreducible module M = ⟨θ⟩. We use the isomorphism

M. Kida
in (3.2) for that purpose. Recall that ϕ i (X) is the minimal polynomial of g i over F 2 , where g is a fixed generator of F × 2 n , and that σ ∈ Gal(L/k) acts on the righthand side by the multiplication by X. Thus, if we take (1, X, . . . , X n−1 ) as a basis of F 2 [X]/(ϕ i (X)), then σ acts by the companion matrix of ϕ i (X) = a 0 + a 1 X + ⋯ + a n−1 X n−1 + X n ∶ This matrix action is compatible if we take a basis (θ, σ θ, . . . , σ n−1 θ) for M.
To illustrate how the above method works, we give an explicit description for the case n = 4.
are not conjugate but arithmetically equivalent.