Some examples of quadratic fields with finite non-solvable maximal unramified extensions II

Abstract

Let K be a number field and $K_\mathrm{ur}$ be the maximal extension of K that is unramified at all places. In a previous article (Kim, J Number Theory 166:235–249, 2016), the first author found three real quadratic fields K such that $\mathrm {Gal}(K_\mathrm{ur}/K)$ is finite and non-abelian simple under the assumption of the generalized Riemann hypothesis (GRH). In this article, we extend the methods of Kim (2016) and identify more quadratic number fields K such that $\mathrm {Gal}(K_\mathrm{ur}/K)$ is a finite nonsolvable group and also explicitly calculate their Galois groups under the assumption of the GRH. In particular, we find the first imaginary quadratic field with this property.

1 Introduction

Let K be a number field and $K_\mathrm{ur}$ be the maximal extension of K that is unramified at all places. In [13], Yamamura showed that $K_\mathrm{ur}=K_\mathrm{l},$ where K denotes an imaginary quadratic field with absolute discriminant value $|d_K| \le 420,$ and $K_\mathrm{l}$ is the top of the class field tower of K and also computed $\mathrm {Gal}(K_\mathrm{ur}/K).$ Hence, we can find examples of abelian or solvable étale fundamental groups. It is then natural to wonder whether we can find examples with the property that $\mathrm {Gal}(K_\mathrm{ur}/K)$ is a finite nonsolvable group. In [3], we presented three explicit examples that provide an affirmative answer.

In this article, we will refine the previously used methods and identify two more quadratic number fields K such that $\mathrm {Gal}(K_\mathrm{ur}/K)$ is a finite nonsolvable group and also explicitly calculate their Galois groups under the generalized Riemann hypothesis (GRH). Under the assumption of GRH, we will show that $\mathrm {Gal}(K_\mathrm{ur}/K)$ is isomorphic to a finite nonsolvable group when $K=\mathbb {Q}(\sqrt{22268})$ (Theorem  4.1) and when $K=\mathbb {Q}(\sqrt{-1567})$ (Theorem 5.1).

In particular, to the best of the authors’ knowledge, $K=\mathbb {Q}(\sqrt{-1567})$ is the first example of an imaginary quadratic field which has a nonsolvable unramified extension and for which $\mathrm {Gal}(K_\mathrm{ur}/K)$ is explicitly calculated.

Tools used for the proof to identify certain unramified extensions with nonsolvable Galois groups, we use the database of number fields created by Klüners and Malle [4]. To exclude further unramified extensions, we use a wide variety of tools, including class field theory, Odlyzko’s discriminant bounds, results about low degree number fields with small discriminants, and various group-theoretical results. In particular, our examples demonstrate how to combine the methods of the previous paper [3] with more involved group-theoretical arguments to obtain conclusions for fields whose class numbers and discriminants do not yield immediate results via application of discriminant bounds.

2 Preliminaries

2.1 The action of Galois groups on class groups

If A is a finite abelian p-group, then $A \simeq \oplus \mathbb {Z}/p^{a_i}\mathbb {Z}$ for some integers $a_i.$ Let

$$\begin{aligned} n_a= & {} \text { number of }i\text { with }a_i=a,\\ r_a= & {} \text { number of }i\text { with }a_i\ge a. \end{aligned}$$

Then

$$\begin{aligned} r_1=p\text {-rank }A=\mathrm {dim}_{\mathbb {Z}/p\mathbb {Z}}\left( A/A^p\right) \end{aligned}$$

and, more generally,

$$\begin{aligned} r_a=\mathrm {dim}_{\mathbb {Z}/p\mathbb {Z}}\left( A^{p^{a-1}}/A^{p^a}\right) . \end{aligned}$$

The action of Galois groups on class groups can often be used to obtain useful information on the structure of class groups. We review the following lemma, often called p-rank theorem. By $\mathrm{cl}(K)$ we denote the class number of the number field K.

Lemma 2.1

(Theorem 10.8 of [11]) Let L / K be a cyclic extension of degree n. Let p be a prime, $p \not \mid n$ and assume that all fields E with $K \subseteqq E \varsubsetneqq L$ satisfy $p \not \mid \mathrm {Cl}(E).$ Let A be the p-Sylow subgroup of the ideal class group of L,  and let f be the order of p mod n. Then

$$\begin{aligned} r_a \equiv n_a \equiv 0\,\mathrm{mod}\,f \end{aligned}$$

for all a,  where $r_a$ and $n_a$ are as above. In particular, if $p | \mathrm {Cl}(L)$ then the p-rank of A is at least f and $p^f | \mathrm {Cl}(L).$

2.2 A remark on the class field tower

Lemma 2.2

(Theorem 1 of [10]) Let K be a number field and p any prime number. If the p-class group, i.e., the p-part of the class group of K is cyclic, then the p-class group of the Hilbert p-class field of K is trivial. Moreover, if $p=2$ and the 2-class group of K is isomorphic to $V_4,$ then the 2-class group of the Hilbert 2-class field of K is cyclic.

2.3 Root discriminant

Let K be a number field. We define the root discriminant of K to be $|d_K|^{1/n_K},$ where $n_K$ is $[K:\mathbb {Q}].$ Given a tower of number fields L / K / F,  we have the following equality for the ideals of F:

$$\begin{aligned} \begin{aligned} d_{L/F} = \left( d_{K/F}\right) ^{[L:K]} N_{K/F}\left( d_{L/K}\right) , \end{aligned} \end{aligned}$$

(2.1)

where $d_{L/F}$ denotes the relative discriminant (see [7, Corollary 2.10]). Set $F=\mathbb {Q}.$ It follows from (2.1) that, if L is an extension of K,  $|d_K|^{1/n_K} \le |d_L|^{1/n_L},$ with equality if and only if $d_{L/K}=1,$ i.e., L / K is unramified at all finite places.

2.4 Discriminant bounds

In this section, we describe how the discriminant bound is used to determine that a field has no nonsolvable unramified extensions.

2.4.1 Crucial proposition

Consider the following proposition, in which $K_\mathrm{ur}$ is the maximal extension of K that is unramified over all primes.

Proposition 2.3

(Proposition 1 of [13]) Let $B(n_K,\, r_1,\, r_2)$ be the lower bound for the root discriminant of K of degree $n_K$ with signature $(r_1,\, r_2).$ Suppose that K has an unramified normal extension L of degree m. If $\mathrm {Cl}(L) = 1,$ where $\mathrm {Cl}(L)$ is the class number of L,  and $|d_K|^{1/n_K} < B(60mn_K,\, 60mr_1,\, 60mr_2),$ then $K_\mathrm{ur} = L.$

If the GRH is assumed, much better bounds can be obtained. The lower bounds for number fields are stated in Martinet’s expository paper [6].

2.4.2 Description of [6, Table III]

Table III of [6] describes the following. If K is an algebraic number field with $r_1$ real and $2r_2$ complex conjugate embeddings, and $d_K$ denotes the absolute value of the discriminant of K,  then, for any b,  we have

$$\begin{aligned} \begin{aligned} d_K>A^{r_1}B^{2r_2}\mathrm{e}^{f-E}, \end{aligned} \end{aligned}$$

(2.2)

where $A,\,B,$ and E are given in the table, and

$$\begin{aligned} \begin{aligned} f=2\sum _{\mathfrak {p}}\sum ^{\infty }_{m=1}\frac{\log N(\mathfrak {p})}{N(\mathfrak {p})^{m/2}}F\left( \log N(\mathfrak {p})^m\right) , \end{aligned} \end{aligned}$$

(2.3)

where the outer sum is taken over all prime ideals of K,  N is the norm from K to $\mathbb {Q},$ and

$$\begin{aligned} F(x)=G(x/b) \end{aligned}$$

in the GRH case, where the even function G(x) is given by

$$\begin{aligned} \begin{aligned} G(x) = \Big (1- \frac{x}{2}\Big )\cos \frac{\pi }{2}x+\frac{1}{\pi }\sin \frac{\pi }{2}x \end{aligned} \end{aligned}$$

(2.4)

for $0 \le x \le 2$ and $G(x) = 0$ for $x>2.$

The values of A and B are lower estimates; the values of E have been rounded up from their true values, which are

$$\begin{aligned} \begin{aligned} 8\pi ^2b\Big (\frac{\mathrm{e}^{b/2}+\mathrm{e}^{-b/2}}{\pi ^2+b^2}\Big )^2 \end{aligned} \end{aligned}$$

(2.5)

in the GRH case.

3 Some group theory

In this section, we recall some facts from group theory.

3.1 Schur multipliers and central extensions

Definition 3.1

The Schur multiplier is the second homology group $H_2(G,\, \mathbb {Z})$ of a group G.

Definition 3.2

A stem extension of a group G is an extension

$$\begin{aligned} \begin{aligned} 1 \rightarrow H \rightarrow G_0 \rightarrow G \rightarrow 1, \end{aligned} \end{aligned}$$

(3.1)

where $H \subset Z(G_0) \cap G'_0$ is a subgroup of the intersection of the center of $G_0$ and the derived subgroup of $G_0.$

If the group G is finite and one considers only stem extensions, then there is a largest size for such a group $G_0,$ and for every $G_0$ of that size the subgroup H is isomorphic to the Schur multiplier of G. Moreover, if the finite group G is perfect as well, then $G_0$ is unique up to isomorphism and is itself perfect. Such $G_0$ are often called universal perfect central extensions of G,  or covering groups.

Proposition 3.3

Let H be a finite abelian group, and let $1 \rightarrow H \rightarrow G_0 \rightarrow G \rightarrow 1$ be a central extension of G by H. Then either this extension is a stem extension, or $G_0$ has a non-trivial abelian quotient.

Proof

By definition, if the extension is not a stem extension, then $H \not \subseteq G'_0,$ and thus $G_0/G'_0$ is a non-trivial abelian quotient. $\square$

Lemma 3.4

The Schur multiplier of $A_n$ is $C_2$ for $n=5$ or $n>7$ and it is $C_6$ for $n = 6$ or 7.

Proof

See [12, 2.7] $\square$

Lemma 3.5

The Schur multiplier of $\mathrm {PSL}_n(\mathbb {F}_{p^d})$ is a cyclic group of order $\mathrm {gcd}(n,\,p^d-1)$ except for $\mathrm {PSL}_2(\mathbb {F}_{4})$ (order 2), $\mathrm {PSL}_2(\mathbb {F}_{9})$ (order 6), $\mathrm {PSL}_3(\mathbb {F}_{2})$ (order 2), $\mathrm {PSL}_3(\mathbb {F}_{4})$ (order 48, product of cyclic groups of orders 3, 4, 4) and $\mathrm {PSL}_4(\mathbb {F}_{2})$ (order 2).

Proof

See [12, 3.3]. $\square$

3.2 Group extensions of groups with trivial centers

Let H and F be groups, with G a group extension of H by F:

$$\begin{aligned} 1 \rightarrow H \rightarrow G \rightarrow F \rightarrow 1. \end{aligned}$$

Then, it is well known that G acts on H by conjugation, and this action induces a group homomorphism $\psi _G : F \rightarrow \mathrm {Out}\text { }H,$ which depends only on G.

Lemma 3.6

((7.11) of [9]) Suppose that H has trivial center $(Z(H)=\{1\}).$ Then, the structure of G is uniquely determined by the homomorphism $\psi _G.$ For any group homomorphism $\psi$ from F to $\mathrm {Out}\text { }H,$ there exists an extension G of H by F such that $\psi _G =\psi .$ Moreover, the isomorphism class of G is uniquely determined by $\psi .$ [In particular, the class of $F \times H$ is determined by $\psi$ with $\psi (F)=1].$ All of the extensions are realized as a subgroup U of the direct product $F \times \mathrm {Aut}\text { }H$ satisfying the two conditions $U \cap \mathrm {Aut}\text { }H = \mathrm {Inn}\text { }H$ and $\pi (U) = F,$ where $\pi$ is the projection from $F \times \mathrm {Aut}\text { }H$ to F.

3.3 Prerequisites on $\mathrm {GL}_n(\mathbb {F}_q)$

3.3.1 General prerequisites

The following lemma is well known.

Lemma 3.7

Let $n\ge 2,$q be a prime power, and let $U\le \mathrm {GL}_n(\mathbb {F}_q)$ act irreducibly on $(\mathbb {F}_q)^n.$ Then the centralizer of U in $\mathrm {GL}_n(\mathbb {F}_q)$ is cyclic.

Proof

This follows immediately from Schur’s lemma. $\square$

Lemma 3.8

Let $n\ge 2,$q be a prime power and let $U\le \mathrm {GL}_n(\mathbb {F}_q)$ be cyclic, of order coprime to q. Assume that U acts irreducibly on $(\mathbb {F}_q)^n.$ Then the centralizer of U in $\mathrm {GL}_n(\mathbb {F}_q)$ is cyclic of order $q^n-1.$

Proof

This follows from [2, Hilfssatz II.3.11]. Namely, setting $G:=C_{\mathrm {GL}_n(\mathbb {F}_q)},$ the centralizer of U in $\mathrm {GL}_n(\mathbb {F}_q),$ that theorem states that G is isomorphic to $\mathrm {GL}_1(\mathbb {F}_{q^n}),$ and thus in particular cyclic of order $q^n-1.$$\square$

An important special case of the previous lemma is the following:

Lemma 3.9

Let $n\ge 2,$q be a prime power and let p be a primitive prime divisor of $q^n-1,$ that is p divides $q^n-1,$ but does not divide any of the numbers $q^k-1$ with $1\le k<n.$ Then the following hold:

• (i) There is a unique non-trivial linear action of $C_p$ on $(\mathbb {F}_q)^n,$ and this action is irreducible.

• (ii) The centralizer of a subgroup of order p in $\mathrm {GL}_n(\mathbb {F}_q)$ is cyclic, of order $q^n-1.$

Proof

Let $U<\mathrm {GL}_n(\mathbb {F}_q)$ be any subgroup isomorphic to $C_p.$ From Maschke’s theorem, it follows immediately that U acts irreducibly on $(\mathbb {F}_n)^q.$ From Lemma 3.8, the centralizer of U in $\mathrm {GL}_n(\mathbb {F}_q)$ is then cyclic, of order $q^n-1.$ Finally, every such U is the unique subgroup of order p of some p-Sylow subgroup of $\mathrm {GL}_n(\mathbb {F}_q)$ [note that, by assumption, the p-Sylow subgroups are of order dividing $q^n-1,$ and then in fact cyclic, since $\mathrm {GL}_1(\mathbb {F}_{q^n})\le \mathrm {GL}_n(\mathbb {F}_q)$ is cyclic]. Therefore, all such subgroups U are conjugate in $\mathrm {GL}_n(\mathbb {F}_q),$ proving the uniqueness in (i). $\square$

In the following sections, we collect some results about more specific linear groups.

3.3.2 Structure of $\mathrm {GL}_2(\mathbb {F}_p)$

Lemma 3.10

$\mathrm {GL}_2(\mathbb {F}_p)$ does not contain any non-abelian simple subgroups for any prime p.

Proof

Let S be non-abelian simple. Then it is known that S contains a non-cyclic abelian subgroup (see e.g., [5, Corollary 6.6]), and therefore even some subgroup $C_r\times C_r$ for some prime r. On the other hand, as a direct consequence of Schur’s lemma, any subgroup $C_r\times C_r$ of $\mathrm {GL}_2(\mathbb {F}_p)$ must intersect the center of $\mathrm {GL}_2(\mathbb {F}_p)$ non-trivially.¹ Since S has trivial center, it follows that S cannot be contained in $\mathrm {GL}_2(\mathbb {F}_p).$$\square$

3.3.3 Structure of $\mathrm {GL}_4(\mathbb {F}_2)$

This article uses the structure of $\mathrm {GL}_4(\mathbb {F}_2).$ Thus, we recall several structural properties of this group.

Proposition 3.11

$A_8$ is isomorphic to $\mathrm {PSL}_4(\mathbb {F}_2) = \mathrm {GL}_4(\mathbb {F}_2).$

Lemma 3.12

$A_8$ does not contain a subgroup isomorphic to $A_5 \times C_2$ or $\mathrm {SL}_2(\mathbb {F}_5).$

Proof

Both $A_5 \times C_2$ and $\mathrm {SL}_2(\mathbb {F}_5)$ contain an element of order 10,  but there is no element of order 10 in $A_8.$$\square$

Lemma 3.13

The class of (12345) is the unique conjugacy class of elements of order 5 in $A_8.$ In particular, there is a unique non-trivial linear $C_5$-action on $(\mathbb {F}_2)^4.$ This action is irreducible.

Proof

This is a special case of Lemma 3.9, with $q=2$ and $n=4.$$\square$

3.3.4 Structure of $\mathrm {GL}_4(\mathbb {F}_3)$

We also make use of the structure of $\mathrm {GL}_4(\mathbb {F}_3)$ in this article. So we recall several structural properties of this group. We proved the following lemmas, partially aided by the computer program Magma.

Lemma 3.14

$\mathrm {GL}_4(\mathbb {F}_3)$ contains a unique conjugacy class of subgroups isomorphic to $A_5 \times C_2.$

Proof

By computer calculation, we can check that $\mathrm {GL}_4(\mathbb {F}_3)$ has four conjugacy classes of subgroups of order 120. They are

$$\begin{aligned} \begin{aligned}&\Bigg \langle \left( {\begin{matrix}0&{}0&{}2&{}0\\ 0&{}2&{}0&{}1&{}\\ 1&{}0&{}0&{}0\\ 0&{}1&{}0&{}1 \end{matrix}}\right) ,\, \left( {\begin{matrix}1&{}1&{}1&{}2\\ 2&{}0&{}0&{}2\\ 2&{}1&{}0&{}0\\ 1&{}0&{}2&{}0\end{matrix}}\right) ,\,\left( {\begin{matrix}2&{}0&{}0&{}0\\ 0&{}2&{}0&{}0\\ 0&{}0&{}2&{}0\\ 0&{}0&{}0&{}2\end{matrix}}\right) \Bigg \rangle , \quad \Bigg \langle \left( {\begin{matrix}1&{}1&{}0&{}2\\ 0&{}2&{}0&{}0\\ 2&{}1&{}2&{}2\\ 0&{}0&{}0&{}2 \end{matrix}}\right) ,\, \left( {\begin{matrix}2&{}0&{}1&{}1\\ 1&{}2&{}0&{}1\\ 0&{}0&{}1&{}0\\ 0&{}0&{}1&{}2\end{matrix}}\right) \Bigg \rangle \\&\Bigg \langle \left( {\begin{matrix}2&{}1&{}2&{}2\\ 2&{}0&{}1&{}1\\ 1&{}1&{}2&{}0\\ 1&{}1&{}0&{}2\end{matrix}}\right) ,\,\left( {\begin{matrix}0&{}0&{}0&{}2\\ 2&{}1&{}0&{}2\\ 2&{}2&{}0&{}1\\ 0&{}2&{}2&{}0\end{matrix}}\right) ,\,\left( {\begin{matrix}2&{}0&{}0&{}0\\ 0&{}2&{}0&{}0\\ 0&{}0&{}2&{}0\\ 0&{}0&{}0&{}2\end{matrix}}\right) \Bigg \rangle \quad \text {and}\quad \Bigg \langle \left( {\begin{matrix}2&{}2&{}0&{}1\\ 0&{}1&{}0&{}0\\ 1&{}2&{}1&{}1\\ 0&{}0&{}0&{}1 \end{matrix}}\right) ,\, \left( {\begin{matrix}1&{}0&{}2&{}2\\ 2&{}1&{}0&{}2\\ 0&{}0&{}2&{}0\\ 0&{}0&{}2&{}1\end{matrix}}\right) \Bigg \rangle . \end{aligned} \end{aligned}$$

(3.2)

We use Magma to check that $\Big \langle \left( {\begin{matrix}2&{}1&{}2&{}2\\ 2&{}0&{}1&{}1\\ 1&{}1&{}2&{}0\\ 1&{}1&{}0&{}2\end{matrix}}\right) ,\,\left( {\begin{matrix}0&{}0&{}0&{}2\\ 2&{}1&{}0&{}2\\ 2&{}2&{}0&{}1\\ 0&{}2&{}2&{}0\end{matrix}}\right) ,\left( {\begin{matrix}2&{}0&{}0&{}0\\ 0&{}2&{}0&{}0\\ 0&{}0&{}2&{}0\\ 0&{}0&{}0&{}2\end{matrix}}\right) \Big \rangle$ is the only the conjugacy class of subgroup of order 120 which is isomorphic to $A_5 \times C_2.$$\square$

Lemma 3.15

$\mathrm {GL}_4(\mathbb {F}_3)$ does not contain a subgroup isomorphic to $A_5 \times V_4.$

Proof

$A_5\times V_4$ contains an abelian subgroup isomorphic to $C_{10}\times C_2.$ As a special case of Lemma 3.9 (with $q=3,$$n=4$), the centralizer of a cyclic group of order 5 in $\mathrm {GL}_4(\mathbb {F}_3)$ is cyclic, of order $3^4-1=80.$ Now of course, if $\mathrm {GL}_4(\mathbb {F}_3)$ contained a subgroup isomorphic to $C_{10}\times C_2,$ then the centralizer of a respective subgroup of order 5 would be non-cyclic. This ends the proof. $\square$

Lemma 3.16

There exist a unique conjugacy class of elements of order 5 in $\mathrm {GL}_4(\mathbb {F}_3).$ Furthermore, there is a unique non-trivial linear action of $C_5$ on $(\mathbb {F}_3)^4,$ and this action is irreducible.

Proof

This again follows directly from Lemma 3.9, with $q=3$ and $n=4.$$\square$

3.3.5 Structure of $\mathrm {GL}_3(\mathbb {F}_5)$

We will also use the structures of $\mathrm {GL}_3(\mathbb {F}_5).$

Lemma 3.17

$\mathrm {GL}_3(\mathbb {F}_5)$ contains a unique conjugacy class of subgroups isomorphic to $A_5 \times C_2.$

Proof

By computer calculation, we can check that $\mathrm {GL}_4(\mathbb {F}_3)$ has four conjugacy classes of subgroups of order 120. They are

$$\begin{aligned} \begin{aligned}&\Big \langle \left( {\begin{matrix}2&{}1&{}2\\ 3&{}0&{}0\\ 2&{}3&{}4 \end{matrix}}\right) ,\, \left( {\begin{matrix}0&{}1&{}1\\ 3&{}4&{}1\\ 4&{}2&{}1 \end{matrix}}\right) ,\,\left( {\begin{matrix}1&{}3&{}2\\ 1&{}3&{}1\\ 1&{}4&{}0 \end{matrix}}\right) \Big \rangle ,\quad \Big \langle \left( {\begin{matrix}4&{}0&{}1\\ 0&{}4&{}0\\ 4&{}1&{}0 \end{matrix}}\right) ,\, \left( {\begin{matrix}2&{}3&{}1\\ 3&{}0&{}3\\ 3&{}4&{}4 \end{matrix}}\right) \Big \rangle ,\\&\Big \langle \left( {\begin{matrix}1&{}0&{}0\\ 0&{}1&{}0\\ 1&{}4&{}4 \end{matrix}}\right) ,\, \left( {\begin{matrix}1&{}0&{}4\\ 2&{}1&{}1\\ 3&{}0&{}3 \end{matrix}}\right) ,\, \left( {\begin{matrix}4&{}0&{}0\\ 0&{}4&{}0\\ 0&{}0&{}4 \end{matrix}}\right) \Big \rangle ,\quad \text {and}\quad \Big \langle \left( {\begin{matrix}1&{}0&{}4\\ 0&{}1&{}0\\ 1&{}4&{}0 \end{matrix}}\right) ,\, \left( {\begin{matrix}3&{}2&{}4\\ 2&{}0&{}2\\ 2&{}1&{}1 \end{matrix}}\right) \Big \rangle . \end{aligned} \end{aligned}$$

(3.3)

We use Magma to check that $\Big \langle \left( {\begin{matrix}1&{}0&{}0\\ 0&{}1&{}0\\ 1&{}4&{}4 \end{matrix}}\right) ,\, \left( {\begin{matrix}1&{}0&{}4\\ 2&{}1&{}1\\ 3&{}0&{}3 \end{matrix}}\right) , \,\left( {\begin{matrix}4&{}0&{}0\\ 0&{}4&{}0\\ 0&{}0&{}4 \end{matrix}}\right) \Big \rangle$ is the only the conjugacy class of subgroup of order 120 which is isomorphic to $A_5 \times C_2.$

Lemma 3.18

$\mathrm {GL}_3(\mathbb {F}_5)$ does not contain a subgroup isomorphic to $A_5 \times V_4.$

Proof

By Lemma 3.10, any subgroup $A_5\le \mathrm {GL}_3(\mathbb {F}_5)$ has to act irreducibly. Since $A_5\times V_4$ has non-cyclic center, the claim now follows immediately from Lemma 3.7. $\square$

3.3.6 Structures of $\mathrm {GL}_5(\mathbb {F}_2)$ and $\mathrm {GL}_6(\mathbb {F}_2)$

Lemma 3.19

$\mathrm {GL}_5(\mathbb {F}_2)$ does not contain a subgroup isomorphic to $\mathrm {PSL}_2(8).$

Proof

The group $\mathrm {PSL}_2(\mathbb {F}_8)=\mathrm {SL}_2(\mathbb {F}_8)$ contains cyclic subgroups of order $\frac{8^2-1}{8-1}=9.$ However, $\mathrm {GL}_5(\mathbb {F}_2)$ does not contain any such subgroups. Indeed, since 9 is a prime power, Maschke’s theorem implies that the existence of such a cyclic subgroup would enforce the existence of an irreducible cyclic subgroup of order 9 in some $\mathrm {GL}_d(\mathbb {F}_2)$ with $d\le 5.$ Then $2^d-1$ would have to be divisible by 9,  which is not the case for any such d. This concludes the proof. $\square$

Lemma 3.20

$\mathrm {GL}_6(\mathbb {F}_2)$ contains a unique conjugacy class of subgroups isomorphic to $\mathrm {PSL}_2(\mathbb {F}_8).$

Proof

Since $\mathrm {PSL}_2(\mathbb {F}_8)=\mathrm {SL}_2(\mathbb {F}_8)\le \mathrm {GL}_2(\mathbb {F}_8),$ the existence follows immediately from the well-known fact that $\mathrm {GL}_{n\cdot d}(\mathbb {F}_q)$ contains subgroups isomorphic to $\mathrm {GL}_{n}(\mathbb {F}_{q^d}).$ The uniqueness can once again be verified with Magma. $\square$

Lemma 3.21

$\mathrm {GL}_6(\mathbb {F}_2)$ does not contain subgroups isomorphic to $\mathrm {PSL}_2(\mathbb {F}_8)\times C_2.$

Proof

By Maschke’s theorem (and using the proof of Lemma 3.19), any cyclic subgroup of order 9 in $\mathrm {GL}_6(\mathbb {F}_2)$ has to act irreducibly. By Lemma 3.8, the centralizer of such a subgroup is then cyclic of order $2^6-1=63.$ However, the centralizer of an order-9 subgroup in $\mathrm {PSL}_2(\mathbb {F}_8)\times C_2$ is of course of even order. This concludes the proof. $\square$

4 Example: $K=\mathbb {Q}(\sqrt{22268})$

Let K be the real quadratic number field $\mathbb {Q}(\sqrt{22268})$. We determine the Galois group of the maximal unramified extension of K.

Theorem 4.1

Let K be the real quadratic field $\mathbb {Q}(\sqrt{22268}).$ Then, under the assumption of GRH, $\mathrm {Gal}(K_\mathrm{ur}/K)$ is isomorphic to $A_5 \times C_2.$

The class number of K is 2,  i.e., $\mathrm {Cl}(K) \simeq C_2.$ Let $K_1$ be the Hilbert class field of K. Then $K_1$ can be written as $\mathbb {Q}(\sqrt{76},\,\sqrt{293}).$ By computer calculation, we know that the class group of $K_1$ is trivial, i.e., $K_1$ has no non-trivial solvable unramified extensions.

4.1 An unramified $A_5$-extension of $K_1$

Let $K=\mathbb {Q}(\sqrt{22268})$ and let L be the splitting field of

$$\begin{aligned} \begin{aligned} x^6 - 10x^4 - 7x^3 + 15x^2 + 14x + 3, \end{aligned} \end{aligned}$$

(4.1)

a totally real polynomial with discriminant $19^2 \cdot 293^2.$ We can also find the polynomial (4.1) from the database of [4] and check that the discriminant of a root field of the polynomial (4.1) is also $19^2 \cdot 293^2.$ Then, L is an $A_5$-extension over $\mathbb {Q}$ which is only ramified at 19 and 293. The factorizations of the above polynomial modulo 19 and 293 are

$$\begin{aligned} x^6 - 10x^4 - 7x^3 + 15x^2 + 14x + 3= & {} (x+12)^2 (x+15)^2 \left( x^2+3 x+12\right) \text { mod } 19,\nonumber \\ x^6 - 10x^4 - 7x^3 + 15x^2 + 14x + 3= & {} (x+66)^2 (x+103) (x+160) (x+242)^2 \text { mod } 293. \end{aligned}$$

Thus, 19 and 293 are the only primes ramified in this field with ramification index 2. By Abhyankar’s lemma, $LK_1/K_1$ is unramified at all primes, and 2,  19,  and 293 are the only primes ramified in $LK_1/\mathbb {Q}$ with ramification index 2 (note that $22268=4 \cdot 19 \cdot 293$). Since $A_5$ is a non-abelian simple group, $L \cap K_1 =\mathbb {Q}.$ Thus, $\mathrm {Gal}(LK_1/K_1) \simeq \mathrm {Gal}(L/\mathbb {Q}) \simeq A_5,$ i.e., $LK_1$ is an unramified $A_5$-extension of $K_1.$ We also know that $\mathrm {Gal}(LK_1/\mathbb {Q}) \simeq V_4 \times A_5.$ Define M as $LK_1.$

4.2 Determination of $\mathrm {Gal}(K_\mathrm{ur}/K)$

To prove Theorem 4.1, it suffices to show that M possesses no non-trivial unramified extensions. Since M / K is unramified, the root discriminant of M is $|d_M|^{1/n_M}=|d_K|^{1/n_K}=\sqrt{22268}= 149.2246\ldots$ If we assume GRH, then $|d_M|^{1/n_M}=|d_K|^{1/n_K}=\sqrt{22268}= 149.2246\cdots <153.252 \le B(31970,\,31970,\,0)$ (see [6, Table]). This implies that $[K_\mathrm{ur}:M]<\frac{31,970}{[M:\mathbb {Q}]}=133.2083\ldots$

We now first exclude the existence of non-trivial unramified abelian extensions of M. Suppose M possesses such an extension T / M. Without loss, T / M can be assumed cyclic of prime degree. Let $T'$ be its normal closure over $\mathbb {Q}.$ Then $T'$ is unramified and elementary-abelian over $K_1,$ and $\mathrm {Gal}(M/K_1)\simeq A_5$ acts on $\mathrm {Gal}(T'/M).$ The following intermediate result is useful.

Lemma 4.2

If T / M is an unramified cyclic $C_p$-extension, then the action of $A_5$ on $\mathrm {Gal}(T'/M)$ is faithful or $[T':M]=2.$

Proof

Since $A_5$ is simple, it suffices to exclude the case that the action of $A_5$ on $\mathrm {Gal}(T'/M)$ is trivial. In that case, the extension $1 \rightarrow \mathrm {Gal}(T'/M) \rightarrow \mathrm {Gal}(T'/K_1) \rightarrow A_5 \rightarrow 1$ would be a central extension. Assume that this extension is not a stem extension. In this case, $\mathrm {Gal}(T'/K_1)$ has a non-trivial abelian quotient by Proposition 3.3. Since $T'/K_1$ is unramified, this contradicts the fact that $K_1$ has class number 1. So the extension is a stem extension, whence Lemma 3.4 yields $\mathrm {Gal}(T'/M)\simeq C_2.$$\square$

Corollary 4.3

If T / M is an unramified cyclic $C_p$-extension, then $\mathrm {Gal}(T'/M)$ is one of $(C_2)^{k}$ with $k\in \{1,\,4,\,5,\,6,\,7\},$ or $(C_3)^4,$ or $(C_5)^3.$

Proof

Lemma 4.2 shows that either $[T':M]=2,$ or $A_5$ embeds into $\mathrm {Aut}(\mathrm {Gal}(T'/M)).$ Furthermore, we already know $[T':M]\le 133.$ Now it is easy to check that only the above possibilities for $\mathrm {Gal}(T'/M)$ remain (see in particular Lemma 3.10). $\square$

We now treat the remaining cases one by one.

4.2.1 2-Class group of M

With the above notation, suppose that $\mathrm {Gal}(T/M)\simeq C_2.$ Then, $T'/M$ is unramified and $\mathrm {Gal}(T'/M)$ is isomorphic to $(C_2)^m$ ($1 \le m \le 7$).

Let $E \subset L$ be a root field of the polynomial (4.1) and N be the compositum of E and $K_1,$ i.e., $N=EK_1.$ Then E can be defined by the composite of three polynomials: $x^2-19,$$x^2-293$ and the polynomial (4.1). By computer calculation, N is a root field of the following polynomial:

$$\begin{aligned} \begin{aligned}&x^{24} - 3784x^{22} - 28x^{21} + 6404076x^{20} + 53312x^{19} - 6401641814x^{18} \\&\quad -31411548x^{17} + 4204260566526x^{16} - 5837238288x^{15}\\ {}&\quad - 1908791963697448x^{14} +18501271313028x^{13}\\&\quad + 613640140988085895x^{12} - 11975084172112012x^{11} \\&\quad - 140616516271183965910x^{10} + 4264300576327196748x^9 \\&\quad +22779186389906647652933x^8 - 932994735936411884988x^7 \\&\quad -2542792801321996372912890x^6 \\&\quad + 124393633255686127917612x^5\\&\quad +185598619641359536180924174x^4\\&\quad - 9237397310199896463461164x^3\\&\quad -7951324489796939270027088092x^2\\&\quad + 291464252731787840722883096x\\&\quad +151174316045577424616769218057. \end{aligned} \end{aligned}$$

(4.2)

We also know that $\mathrm {Gal}(M/N)$ is isomorphic to $D_5.$

By computer calculation, we know that the class group of N is isomorphic to $C_2$ under GRH. Let $N'$ be the Hilbert class field of N. (Note that, $N'$ is a subfield of M,  since M / N is unramified.)

By Lemma 2.2, the 2-class group of $N'$ is trivial. Thus the rank m of the 2-class group of M is a multiple of 4 by Lemma 2.1, i.e., m is equal to 0 or 4.

Suppose that $m=4.$ Then, $\mathrm {Gal}(T'/K_1)$ is an extension of $A_5$ by $(C_2)^4.$ By Lemma 4.2, $\mathrm {Gal}(M/K_1)$ acts faithfully on $\mathrm {Gal}(T'/M).$ Consider $\mathrm {Gal}(T'/K).$ This group is an extension of $\mathrm {Gal}(M/K) (\simeq A_5 \times C_2)$ by $\mathrm {Gal}(T'/M) (\simeq (C_2)^4)$ and an extension of $\mathrm {Gal}(K_1/K) (\simeq C_2)$ by $\mathrm {Gal}(T'/K_1)$ simultaneously. Therefore, it is natural to examine how $\mathrm {Gal}(K_1/K)$ acts on $\mathrm {Gal}(T'/M) (\simeq (C_2)^4).$ By Lemma 3.12, $\mathrm {Gal}(M/K) (\simeq A_5 \times C_2)$ does not act faithfully on $\mathrm {Gal}(T'/M) (\simeq (C_2)^4).$ Since $\mathrm {Gal}(M/K_1) (\simeq A_5)$ acts non-trivially on $\mathrm {Gal}(T'/M),$ we obtain that $\mathrm {Gal}(K_1/K) (\simeq \mathrm {Gal}(M/LK))$ acts trivially on $\mathrm {Gal}(T'/M) (\simeq (C_2)^4).$

$\mathrm {Gal}(T'/LK) \simeq (C_2)^5$ Since $\mathrm {Gal}(M/LK)$ acts trivially on $\mathrm {Gal}(T'/M),$$\mathrm {Gal}(T'/LK)$ is $(C_2)^3 \times C_4$ or $(C_2)^5.$ Let $\mathrm {Gal}(T'/LK)$ be $(C_2)^3 \times C_4.$ Then, $\mathrm {Gal}(T''/LK)$ is isomorphic to $(C_2)^4,$ where $T''/LK$ is the maximal elementary abelian 2-subextension of $T'/LK.$ By the maximality of $T'',$$T''$ is also Galois over $\mathbb {Q}$ and $\mathrm {Gal}(T''/K)$ is an extension of $A_5$ by $(C_2)^4.$ By restriction, this $A_5$-actions on $(C_2)^4$ comes from the $\mathrm {Gal}(M/K)$-actions on $\mathrm {Gal}(T'/M)$ mentioned above. Since $\mathrm {Gal}(T'/K_1)$ does not have any abelian quotient, $\mathrm {Gal}(T''/K)$ also has no abelian quotients, i.e., $T'' \cap K_1 =K.$ Thus, $\mathrm {Gal}(T'/K)$ is a direct product of $\mathrm {Gal}(T''/K)$ and $\mathrm {Gal}(K_1/K),$ i.e., $\mathrm {Gal}(T'/LK)$ is a direct product of $\mathrm {Gal}(T''/LK) \simeq (C_2)^4$ and $\mathrm {Gal}(K_1/K)\simeq C_2.$ This contradicts the fact that $\mathrm {Gal}(T'/LK)$ is $(C_2)^3 \times C_4.$ Thus, $\mathrm {Gal}(T'/LK)$ is isomorphic to $(C_2)^5,$ and there exists some S / LK / K such that $SK_1=T'$ and $\mathrm {Gal}(S/K) \simeq (C_2)^4 \rtimes A_5.$

In a similar manner, we can prove that there exists some $S'/L/\mathbb {Q}$ such that $S'K_1=T'$ and $\mathrm {Gal}(S'/\mathbb {Q}) \simeq (C_2)^4 \rtimes A_5.$

Since $S'K$ is contained in $T',$$S'K/K$ is an unramified extension. Therefore, the only ramified primes in $S'/L/\mathbb {Q}$ are 2,  19, and 293 with ramification index 2. Since 19 and 293 are already ramified in $L/\mathbb {Q},$ the only ramified prime in $S'/L$ is 2.

Unramifiedness of$S'/L$ Suppose that 2 is ramified in $S'/L.$ The ramification index of 2 should then be 2. Let $\bar{\mathfrak {p}}$ (resp. $\mathfrak {p}$) be a prime ideal in $S'$ (resp. L) satisfying $\bar{\mathfrak {p}} | 2$ (resp. $\mathfrak {p}|2$). The factorization of the polynomial (4.1) modulo 2 is

$$\begin{aligned} \begin{aligned} x^6 - 10x^4 - 7x^3 + 15x^2 + 14x + 3 \equiv (x+1) \big (x^5+x^4+x^3+x+1\big ) \text { mod }2. \end{aligned} \end{aligned}$$

(4.3)

Thus, we know that $\mathrm {Gal}(L_{\mathfrak {p}}/\mathbb {Q}_2)$ is isomorphic to $C_5 \simeq \langle (12345) \rangle ,$ where $L_{\mathfrak {p}}$ is the $\mathfrak {p}$-completion of L. Consider $\mathrm {Gal}(S'_{\bar{\mathfrak {p}}}/L_{\mathfrak {p}}).$ Since the ramification index of $\mathfrak {p}$ is 2,  $\mathrm {Gal}(S'_{\bar{\mathfrak {p}}}/L_{\mathfrak {p}})$ is $C_2$ or $(C_2)^2,$ i.e., the proper subgroup of $(C_2)^4.$ Hence, $\mathrm {Gal}(S'_{\bar{\mathfrak {p}}}/\mathbb {Q}_3) = \mathrm {Gal}(S'_{\bar{\mathfrak {p}}}/L_{\mathfrak {p}}) \rtimes \langle (12345) \rangle \subsetneq (C_2)^4 \rtimes \langle (12345) \rangle .$ This contradicts the statement that there is no proper subgroup of $(C_2)^4$ that is invariant under the action of $\langle (12345) \rangle$ (see Lemma 3.13). Thus, $S'/L$ should be unramified at all places. In conclusion, $S'/\mathbb {Q}$ is a $(C_2)^4 \rtimes A_5$-extension of $\mathbb {Q}$ that has ramification index 2 at only 19 and 293. Let us now consider the root discriminant of $S'.$ Since $S'/L$ is unramified at all places,

$$\begin{aligned} \left| d_{S'}\right| ^{1/n_{S'}}=\left| d_{L}\right| ^{1/n_{L}}=\left( 19^{30} \cdot 293^{30}\right) ^{1/60}= \sqrt{19\cdot 293} =74.6123\ldots \end{aligned}$$

This implies that $|d_{S'}|^{1/n_{S'}} < 106.815\cdots \le B(960,\,960,\,0)$ under the GRH (see [6, Table]). This contradicts the definition of the lower bound for the root discriminant. Thus, the 2-class group of M is trivial.

4.2.2 3-Class group of M

Suppose that T / M is an unramified $C_3$-extension. Then, as seen above, $T'$ is unramified over M and $\mathrm {Gal}(T'/M)$ is isomorphic to $(C_3)^4.$ Then, $\mathrm {Gal}(T'/\mathbb {Q})$ is an extension of $\mathrm {Gal}(M/\mathbb {Q}) \simeq A_5 \times V_4$ by $(C_3)^4.$ Therefore, it is natural to examine how $\mathrm {Gal}(M/\mathbb {Q})$ acts on $\mathrm {Gal}(T'/M)\simeq (C_3)^4.$ By Lemmas 3.14 and 3.15, we know that there are three possibilities of the actions of $\mathrm {Gal}(M/\mathbb {Q})$ on $\mathrm {Gal}(T'/M).$ [Note that $\mathrm {Aut}((C_3)^4) \simeq \mathrm {GL}_4(\mathbb {F}_3)].$ Each action is induced by the following three group homomorphisms $\psi : A_5 \times V_4 \rightarrow \mathrm {GL}_4(\mathbb {F}_3)$:

• $\psi$ is trivial.

• $\psi (A_5 \times V_4)\simeq A_5.$

• $\psi (A_5 \times V_4)\simeq A_5 \times C_2.$

By Lemma 4.2, $\mathrm {Gal}(M/K_1)$ acts faithfully on $\mathrm {Gal}(T'/M).$ Therefore, $\psi$ cannot be trivial.

$\psi (A_5 \times V_4)\simeq A_5$ This means that $\mathrm {Gal}(M/K_1)(\simeq A_5)$ acts non-trivially on $\mathrm {Gal}(T'/M)$ and $\mathrm {Gal}(M/L) \simeq V_4$ acts trivially on $\mathrm {Gal}(T'/M).$ Since $|\mathrm {Gal}(T'/M)|$ and $|\mathrm {Gal}(M/L)|$ are coprime, $\mathrm {Gal}(T'/L)$ is isomorphic to $V_4 \times (C_3)^4.$ Let S be the subfield of $T'$ fixed by $V_4.$ Then $\mathrm {Gal}(S/\mathbb {Q})$ is a group extension of $A_5$ by $(C_3)^4.$

Since 19 and 293 are already ramified in $L/\mathbb {Q},$ the only ramified prime in S / L is 2. If 2 is ramified in S / L,  its ramification index should be 2. But it is impossible, because the degree of [S : L] is odd. Thus S / L is unramified over all places. By a similar argument as in Sect. 4.2.2.1, we can check that this contradicts the definition of the lower bound for the root discriminant.

$\psi (A_5 \times V_4)\simeq A_5 \times C_2$ First of all, let us see the intermediate fields in M / L. Since $\mathrm {Gal}(M/L)$ is isomorphic to $V_4,$ there are three proper intermediate fields in M / L.

Suppose that $\mathrm {Gal}(M/L(\sqrt{76}))$ acts trivially on $\mathrm {Gal}(T'/M).$ This means that $\mathrm {Gal}(T'/L(\sqrt{76}))$ is isomorphic to $C_2 \times (C_3)^4,$ i.e., there exists a subfield S in $T'/L(\sqrt{76})$ such that $\mathrm {Gal}(S/L(\sqrt{76}))$ is isomorphic to $(C_3)^4.$

We easily check that $S/L(\sqrt{76})$ is unramified over all places. Let $\bar{\mathfrak {p}}$ (resp. $\mathfrak {p}',$$\mathfrak {p}$) be a prime ideal in S [resp. $L(\sqrt{76}),$$\mathbb {Q}(\sqrt{76})]$ satisfying $\bar{\mathfrak {p}} | 2$ (resp. $\mathfrak {p}'|2,$$\mathfrak {p}|2$). We had already show that the factorization of the polynomial (4.1) modulo 2 is

$$\begin{aligned} \begin{aligned} x^6 - 10x^4 - 7x^3 + 15x^2 + 14x + 3 \equiv (x+1) \big (x^5+x^4+x^3+x+1\big ) \text { mod }2. \end{aligned} \end{aligned}$$

(4.4)

Thus, we know that $\mathrm {Gal}(L(\sqrt{76})_{\mathfrak {p}'}/\mathbb {Q}(\sqrt{76})_{\mathfrak {p}})$ is isomorphic to $C_5 \simeq \langle (12345) \rangle ,$ where $L(\sqrt{76})_{\mathfrak {p}'}$ [resp. $\mathbb {Q}(\sqrt{76})_{\mathfrak {p}}]$ is the $\mathfrak {p}'$-completion of $L(\sqrt{76})$ [resp. the $\mathfrak {p}$-completion of $\mathbb {Q}(\sqrt{76})_{\mathfrak {p}}].$

Let us consider $\mathrm {Gal}(S_{\bar{\mathfrak {p}}}/L(\sqrt{76})_{\mathfrak {p}'}).$ We know that $S/L(\sqrt{76})$ is unramified. Thus, $S_{\bar{\mathfrak {p}}}/L(\sqrt{76})_{\mathfrak {p}'}$ is a cyclic extension, i.e., $\mathrm {Gal}(S_{\bar{\mathfrak {p}}}/L(\sqrt{76})_{\mathfrak {p}'})$ is isomorphic to $C_3$ or a trivial group.

Suppose that $\mathrm {Gal}(S_{\bar{\mathfrak {p}}}/L(\sqrt{76})_{\mathfrak {p}'})$ is isomorphic to $C_3.$ Then $\mathrm {Gal}(S_{\bar{\mathfrak {p}}}/\mathbb {Q}(\sqrt{76})_{\mathfrak {p}})=\mathrm {Gal}(S_{\bar{\mathfrak {p}}}/L(\sqrt{76})_{\mathfrak {p}'}) \rtimes \langle (12345) \rangle \subsetneq (C_3)^4 \rtimes \langle (12345) \rangle .$ This contradicts the statement that there is no proper subgroup of $(C_3)^4$ that is invariant under the action of $\langle (12345) \rangle$ (see Lemma 3.16). In conclusion, $\mathrm {Gal}(S_{\bar{\mathfrak {p}}}/L(\sqrt{76})_{\mathfrak {p}'})$ is trivial.

Thus, for a number field $S/\mathbb {Q},$$e_2=2$ and $f_2=5$ where $e_2$ is the ramification index of 2 and $f_2$ is the inertia degree for 2. Let us recall the function (2.3)

$$\begin{aligned} f=2\sum _{\mathfrak {p}}\sum ^{\infty }_{m=1}\frac{\log N(\mathfrak {p})}{N(\mathfrak {p})^{m/2}}F\left( \log N(\mathfrak {p})^m\right) . \end{aligned}$$

Since every term of f is greater than or equal to 0,  the following holds for the number field S.

$$\begin{aligned} \begin{aligned} f \ge 2\sum ^{972}_{j=1}\sum ^{100}_{i=1}\frac{\log N(\bar{\mathfrak {q}}_j)}{N(\bar{\mathfrak {q}}_j)^{i/2}}F\left( \log N\left( \bar{\mathfrak {q}}_j\right) ^i\right) , \end{aligned} \end{aligned}$$

(4.5)

where the $\bar{\mathfrak {q}}_j$ denote the prime ideals of S satisfying $\bar{\mathfrak {q}}_j | 2.$ Since $f_2=5,$$N(\bar{\mathfrak {q}}_j)=2^5$ for all j. Set $b=8.8.$ By a numerical calculation, we have

$$\begin{aligned} \begin{aligned} f \ge 2\cdot 972 \sum ^{100}_{i=1}\frac{\log 2^5}{2^{5i/2}}F\left( \log 2^{5i}\right) =1111.46\ldots \end{aligned} \end{aligned}$$

(4.6)

Let us recall (2.2). For $b=8.8,$ we have

$$\begin{aligned} \begin{aligned} \left| d_{S}\right| ^{1/n_{S}}&>149.272\cdot \mathrm{e}^{(f-604.89)/9720}\\&\ge 149.272\cdot \mathrm{e}^{(1111.46-604.89)/9720}=157.258\ldots \end{aligned} \end{aligned}$$

(4.7)

$|d_{S}|^{1/n_{S}}=|d_K|^{1/n_K}=\sqrt{22268}$ contradicts the fact that $|d_{S}|^{1/n_{S}}= 149.2246\ldots$

Next, suppose that $\mathrm {Gal}(M/LK)$ acts trivially on $\mathrm {Gal}(T'/M).$ This means that $\mathrm {Gal}(T'/LK)$ is isomorphic to $C_2 \times (C_3)^4,$ i.e., there exists a subfield $S'$ in $T'/LK$ such that $\mathrm {Gal}(S'/LK)$ is isomorphic to $(C_3)^4.$

By the same argument as in the above, we can get

$$\begin{aligned} \begin{aligned} \left| d_{S'}\right| ^{1/n_{S}}&>157.258\ldots \end{aligned} \end{aligned}$$

(4.8)

and this contradicts the fact that $|d_{S}|^{1/n_{S}}= 149.2246\ldots$

Finally, suppose that $\mathrm {Gal}(M/L(\sqrt{293}))$ acts trivially on $\mathrm {Gal}(T'/M).$ This means that $\mathrm {Gal}(T'/L(\sqrt{293}))$ is isomorphic to $C_2 \times (C_3)^4,$ i.e., there exists a subfield $S''$ in $T'/L(\sqrt{293})$ such that $\mathrm {Gal}(S''/L(\sqrt{293}))$ is isomorphic to $(C_3)^4.$

We easily know that 19 and 293 are the only ramified primes in $S''/\mathbb {Q}.$ By a similar argument as in Sect.  4.2.1.2, we can check that this contradicts the definition of the lower bound for the root discriminant.

In conclusion, the 3-class group of M is trivial.

4.2.3 5-Class group of M

Suppose that T / M is an unramified $C_5$-extension. Then, $T'$ is unramified over M and $\mathrm {Gal}(T'/M)$ is isomorphic to $(C_5)^3.$ Thus, $\mathrm {Gal}(T'/\mathbb {Q})$ is an extension of $\mathrm {Gal}(M/\mathbb {Q}) \simeq A_5 \times V_4$ by $(C_5)^3.$ Therefore, it is natural to examine how $\mathrm {Gal}(M/\mathbb {Q})$ acts on $\mathrm {Gal}(T'/M)\simeq (C_5)^3.$ By Lemmas 3.17 and 3.18, we know that there are three possibilities of the actions of $\mathrm {Gal}(M/\mathbb {Q})$ on $\mathrm {Gal}(T'/M).$ Each action is induced by the following three group homomorphisms $\psi : A_5 \times V_4 \rightarrow \mathrm {GL}_3(\mathbb {F}_5)$:

• $\psi$ is trivial.

• $\psi (A_5 \times V_4)\simeq A_5.$

• $\psi (A_5 \times V_4)\simeq A_5 \times C_2.$

By a similar argument as in Sect. 4.2.2, we just need to think about the case $\psi (A_5 \times V_4)\simeq A_5 \times C_2.$

$\psi (A_5 \times V_4)\simeq A_5 \times C_2$ Consider again the intermediate fields of M / L as in Sect. 4.2.2.2. Suppose that $\mathrm {Gal}(M/L(\sqrt{76}))$ acts trivially on $\mathrm {Gal}(T'/M).$ This means that $\mathrm {Gal}(T'/L(\sqrt{76}))$ is isomorphic to $C_2 \times (C_5)^3,$ i.e., there exists a subfield S in $T'/L(\sqrt{76})$ such that $\mathrm {Gal}(S/L(\sqrt{76}))$ is isomorphic to $(C_5)^3.$

From [4], we know that L can also be defined as the splitting field of following polynomial, corresponding to an imprimitive degree-12 action of $A_5$:

$$\begin{aligned} \begin{aligned}&x^{12} + 11x^{11} - 59x^{10} - 647x^9 - 295x^8 + 5446x^7 + 4294x^6 \\&\quad - 14727x^5 - 4960x^4 + 16477x^3 - 4028x^2 - 1813x + 324. \end{aligned} \end{aligned}$$

(4.9)

Let $E\subset L$ be a root field of the polynomial (4.9). We know that the discriminant $d_E$ of E is $19^6 \cdot 293^6.$ Since $|d_E|^{1/n_E}=|d_L|^{1/n_L},$L / E is unramified.

Define N as the compositum of E and $\mathbb {Q}(\sqrt{76}).$ Then N is a subfield of $L(\sqrt{76})$ and $\mathrm {Gal}(L(\sqrt{76})/N)$ is isomorphic to $C_5.$

By Abhyankar’s lemma, we easily know that $L(\sqrt{76})/N$ is unramified. Using a computer calculation, we can check that N is a root field of the following polynomial:

$$\begin{aligned}&x^{24} - 111x^{22} + 4394x^{20} - 83286x^{18} + 818659x^{16} - 4122356x^{14} \nonumber \\&\quad +\;9878557x^{12} -10688099x^{10} + 5561624x^8 - 1360039x^6\nonumber \\&\quad +\;130854x^4 - 2499x^2 + 1. \end{aligned}$$

(4.10)

By the calculation of Sage, we can check that the class group of N is equal to $C_{10},$ i.e., 5-class group of N is $C_5$ and Hilbert 5-class field of N is $L(\sqrt{76}).$ We know that $\mathrm {Gal}(T'/L(\sqrt{76}))$ is isomorphic to $C_2 \times (C_3)^5,$ i.e., 5-class group of $L(\sqrt{76})$ is not trivial. This contradicts Lemma 2.2.

Suppose that $\mathrm {Gal}(M/LK)$ acts trivially on $\mathrm {Gal}(T'/M).$ Define $N'$ as the compositum of E and K. Then N can be defined by the following polynomial:

$$\begin{aligned} \begin{aligned}&x^{24} - 98x^{22} + 4073x^{20} - 94,476x^{18} + 1354898x^{16} - 12553566x^{14} \\&\quad +76075696x^{12} - 297782263x^{10} + 723063287x^8 - 1000608193x^6 \\&\quad +654400814x^4 - 110097135x^2 + 3818116. \end{aligned} \end{aligned}$$

(4.11)

By a computer calculation with Magma, we can check, assuming GRH, that the class group of N is equal to $C_{10},$ i.e., the 5-class group of N is $C_5$ and the Hilbert 5-class field of N is LK. By the same argument as above, we obtain a contradiction.

Suppose that $\mathrm {Gal}(M/L(\sqrt{293}))$ acts trivially on $\mathrm {Gal}(T'/M).$ This means that $\mathrm {Gal}(T'/L(\sqrt{293}))$ is isomorphic to $C_2 \times (C_5)^3,$ i.e., there exists a subfield $S''$ in $T'/L(\sqrt{293})$ such that $\mathrm {Gal}(S''/L(\sqrt{293}))$ is isomorphic to $(C_5)^3,$ and such that 19 and 293 are the only ramified primes in $S''/\mathbb {Q}.$ By a similar argument as in Sect.  4.2.2.1, we can check that this contradicts the lower bound for the root discriminant.

In conclusion, 5-class group of M is also trivial under the assumption of the GRH. We have therefore obtained:

Proposition 4.4

The class number of M is 1,  under the assumption of the GRH.

4.2.4 $A_5$-unramified extension of M

Since the class number of M is one, there is no solvable unramified extension over M. The last thing we have to do is to show that there is no nonsolvable unramified extension over M. Since $[K_\mathrm{ur}:M]<133.2083\ldots ,$ our task is to show that K does not admit an unramified $A_5$-extension.

Suppose that M admits an unramified $A_5$-extension F. Because $[K_\mathrm{ur}:M]<134,$F is the unique unramified $A_5$-extension of M,  i.e., F is Galois over $\mathbb {Q}.$ It is well known that $A_5$ is isomorphic to $\mathrm {PSL}_2(\mathbb {F}_5)$ and $S_5$ is isomorphic to $\mathrm {PGL}_2(\mathbb {F}_5).$ By Lemma 3.6, $\mathrm {Gal}(F/K_1) \simeq A_5 \times A_5,$ i.e., $K_1$ admits another $A_5$-unramified extension $F_1.$

(Note that, $F_1$ is also Galois over $\mathbb {Q},$ or otherwise $K_1$ would have further unramified $A_5$-extensions, contradicting Odlyzko’s bound.) Then, by Lemma 3.6, there are only two possibilities for $\mathrm {Gal}(F_1/K)$: $A_5 \times C_2$ or $S_5.$

Case 1: $\mathrm {Gal}(F_1/K) \simeq A_5 \times C_2$ By a similar argument in the above, K admits an $A_5$-unramified extension $F_2.$ Then, $\mathrm {Gal}(F_2/\mathbb {Q})$ is also isomorphic to $A_5 \times C_2$ or $S_5.$

Case 1.1: $\mathrm {Gal}(F_2/\mathbb {Q}) \simeq A_5 \times C_2$ This implies that there exists an $A_5$-extension $F_3/\mathbb {Q}$ with all ramification indices $\le 2$ and unramified outside of $\{2,\,19,\,293\}.$ Assume first that 19 is unramified in $F_3/\mathbb {Q}.$ Let E be a quintic subfield of $F_3/\mathbb {Q}.$ Then, by a well known result of Dedekind, we get the upper bound $|d_E|\le 2^6\cdot 293^2 < 5.5\cdot 10^{6}$ for the discriminant of E. However, from [8, Table 2 in Sect. 4.1] no extension with this discriminant bound and ramification restrictions exists. We may therefore assume that 19 is ramified in $F_3/\mathbb {Q}.$ Since its inertia group is generated by a double transposition in $A_5,$ the inertia degree of 19 in the extension $F_2/\mathbb {Q}$ (with Galois group $A_5 \times C_2$) is at most 2. The same holds for the inertia degree of 19 in the extension $L/\mathbb {Q},$ and therefore eventually also in the compositum $LF_2/\mathbb {Q}.$

Let us recall the function (2.3)

$$\begin{aligned} f=2\sum _{\mathfrak {p}}\sum ^{\infty }_{m=1}\frac{\log N(\mathfrak {p})}{N(\mathfrak {p})^{m/2}}F\left( \log N(\mathfrak {p})^m\right) . \end{aligned}$$

Since every term of f is greater than or equal to 0,  the following holds for the number field $LF_2.$

$$\begin{aligned} \begin{aligned} f \ge 2\sum ^{1800}_{j=1}\sum ^{100}_{i=1}\frac{\log N(\bar{\mathfrak {q}}_j)}{N(\bar{\mathfrak {q}}_j)^{i/2}}F\left( \log N(\bar{\mathfrak {q}}_j)^i\right) , \end{aligned} \end{aligned}$$

(4.12)

where the $\bar{\mathfrak {q}}_j$ denote the prime ideals of $LF_2$ satisfying $\bar{\mathfrak {q}}_j | 19.$ Since $f_{19}=2,$$N(\bar{\mathfrak {q}}_j)=19^2$ for all j. Set $b=8.8.$ By a numerical calculation, we have

$$\begin{aligned} \begin{aligned} f \ge 2\cdot 1800 \sum ^{100}_{i=1}\frac{\log 19^2}{19^{i}}F\big (\log 19^{2i}\big )=683.225\ldots \end{aligned} \end{aligned}$$

(4.13)

Let us recall (2.2). For $b=8.8,$ we have

$$\begin{aligned} \begin{aligned} \left| d_{LF_2}\right| ^{1/n_{LF_2}}&>149.272\cdot \mathrm{e}^{(f-604.89)/7200}\\&\ge 149.272\cdot \mathrm{e}^{(683.225-604.89)/7200}=150.905\ldots \end{aligned} \end{aligned}$$

(4.14)

$|d_{LF_2}|^{1/n_{LF_2}}=|d_K|^{1/n_K}=\sqrt{22268}$ contradicts the fact that $|d_{LF_2}|^{1/n_{LF_2}}= 149.2246\ldots$

Case 1.2: $\mathrm {Gal}(F_2/\mathbb {Q}) \simeq S_5$ By the unramifiedness of $F_2/K,$ and since the only involutions of $S_5$ not contained in $A_5$ are the transpositions, a quintic subfield E of $F_2$ must have the discriminant 22268. However, such a quintic number field does not exist, from [8]. This is a contradiction.

Case 2: $\mathrm {Gal}(F_1/K) \simeq S_5$ By Lemma 3.6, $\mathrm {Gal}(F_1/\mathbb {Q}) \simeq S_5 \times C_2.$ Consequently, $F_1$ is the compositum of K and an $S_5$-extension $F_2$ of $\mathbb {Q}.$ Furthermore, $F_2/\mathbb {Q}$ has a quadratic subextension contained in $K_1,$ but linearly disjoint from K. Therefore, it is either $\mathbb {Q}(\sqrt{293})$ or $\mathbb {Q}(\sqrt{76}).$ Consider now a quintic subfield E of $F_2/\mathbb {Q}.$ Of course, $E/\mathbb {Q}$ is unramified outside $\{2,\,19,\,293\}.$ Furthermore, all non-trivial inertia subgroups are generated either by transpositions or by double transpositions. Finally, the inertia subgroups at those primes which ramify in the quadratic subfield of $F_2/\mathbb {Q}$ are generated by transpositions. By a similar argument as in Sect. 4.2.4.2, we then get one of the following two upper bounds for the discriminant of E: either $|d_E| \le 2^3\cdot 19 \cdot 293^2$ [namely, if the quadratic subfield is $\mathbb {Q}(\sqrt{76})],$ or $|d_E| \le 2^6\cdot 19^2 \cdot 293.$ Such a quintic number field does not exist, from [8, Sect. 4.1]. This is a contradiction.

In conclusion, M admits no unramified $A_5$-extensions, i.e., we have that $\mathrm {Gal}(K_{ur}/K_1) \cong A_5$ under the assumption that the GRH holds. This concludes the proof of Theorem 4.1.

5 Example: $K=\mathbb {Q}(\sqrt{-1567})$

Until now, we dealt with real quadratic fields. In this section, we will give the first case of an imaginary quadratic field.

Let K be the imaginary quadratic number field $\mathbb {Q}(\sqrt{-1567}).$ We show the following:

Theorem 5.1

Let K be the imaginary quadratic field $\mathbb {Q}(\sqrt{-1567})$ and $K_\mathrm{ur}$ be its maximal unramified extension. Then $\mathrm {Gal}(K_\mathrm{ur}/K)$ is isomorphic to $\mathrm {PSL}_2(\mathbb {F}_8) \times C_{15}$ under the assumption of the GRH.

The class number of K is 15,  i.e., $\mathrm {Cl}(K) \simeq C_{15}.$ Let $K_1$ be the Hilbert class field of K.

5.1 Class number of $K_1$

The first thing we have to do is show that the class number of $K_1$ is one. It can be computed that $K_1$ is the splitting field of the polynomial

$$\begin{aligned} \begin{aligned}&x^{15} + 14x^{14} + 56x^{13} + 105x^{12} + 497x^{11} + 832x^{10} + 1157x^9 + 1274x^8 \\&\quad +644x^7 - 971x^6 - 2582x^5 - 177x^4 + 7x^3 + 1187x^2 - 20x + 1. \end{aligned} \end{aligned}$$

(5.1)

We can then check with Magma that the class number of $K_1$ is 1,  under GRH.

5.2 An unramified $\mathrm {PSL}_2(\mathbb {F}_8)$-extension of $K_1$

Let $K=\mathbb {Q}(\sqrt{-1567})$ and let L be the splitting field of

$$\begin{aligned} \begin{aligned} x^9 - 2x^8 + 10x^7 - 25x^6 + 34x^5 - 40x^4 + 52x^3 - 45x^2 + 20x - 4, \end{aligned} \end{aligned}$$

(5.2)

a polynomial with complex roots. Then L is a $\mathrm {PSL}_2(\mathbb {F}_8)$-extension of $\mathbb {Q}$ and 1567 is the only prime ramified in this field with ramification index two. By Abhyankar’s lemma, LK / K is unramified at all primes. Since $\mathrm {PSL}_2(\mathbb {F}_8)$ is a non-abelian simple group, $L \cap K_1 =\mathbb {Q}.$ So $\mathrm {Gal}(LK_1/K_1) \simeq \mathrm {Gal}(L/\mathbb {Q}) \simeq \mathrm {PSL}_2(\mathbb {F}_8),$ i.e., $LK_1$ is a $\mathrm {PSL}_2(\mathbb {F}_8)$-extension of $K_1$ which is unramified over all places. It follows that $\mathrm {Gal}(LK_1/\mathbb {Q})$ is isomorphic to $\mathrm {PSL}_2(\mathbb {F}_8) \times D_{15}.$

5.3 The determination of $\mathrm {Gal}(K_\mathrm{ur}/K)$

Define M as $LK_1.$ Since M / K is unramified at all places, the root discriminant of M is $|d_{M}|^{1/|M|}=|d_{K}|^{1/{|K|}}=\sqrt{1567}= 39.5853\ldots$ If we assume GRH, then $|d_{M}|^{1/|M|}=|d_{K}|^{1/{|K|}}=\sqrt{1567}= 39.5853<39.895\cdots = B(1000000,\,0,\,500000)$ (see [6, Table]). This implies that $[K_\mathrm{ur}:M]<\frac{1,000,000}{[M:\mathbb {Q}]}=66.1375\ldots$ We now proceed similarly as in Sect. 4. Let T be a non-trivial unramified $C_p$-extension of M,  and let $T'$ be its Galois closure over $\mathbb {Q}.$ First, we obtain the following analog of Lemma 4.2.

Lemma 5.2

If T / M is a non-trivial unramified cyclic $C_p$-extension, then the action of $\mathrm {PSL}_2(\mathbb {F}_8)$ on $\mathrm {Gal}(T'/M)$ is faithful.

Proof

As in Lemma 4.2, and using additionally that $\mathrm {PSL}_2(\mathbb {F}_8)$ has trivial Schur multiplier (see Lemma 3.5). $\square$

Corollary 5.3

If T / M is a non-trivial unramified cyclic $C_p$-extension, then $p=2$ and $\mathrm {Gal}(T'/M)\simeq (C_2)^6.$

Proof

Use Lemma 5.2, the bound $[T':M]\le 66,$ and Lemmas 3.10 and 3.19 in order to obtain that $(C_2)^6$ is the only elementary-abelian group in the relevant range which allows a non-trivial $\mathrm {PSL}_2(\mathbb {F}_8)$-action. $\square$

We deal with the remaining case below.

5.3.1 2-Class group of M

Suppose that M has an unramified $C_2$-extension T and let $T'$ be its normal closure over $\mathbb {Q}.$ As shown above, $T'$ is unramified over M and $\mathrm {Gal}(T'/M)$ is isomorphic to $(C_2)^6.$

Let $\bar{\mathfrak {p}}$ (resp. $\mathfrak {p}$) be a prime ideal in $L'$ (resp. L) satisfying $\bar{\mathfrak {p}} | 2$ (resp. $\mathfrak {p}|2$). The factorization of the polynomial (5.2) modulo 2 is

$$\begin{aligned} \begin{aligned} x^2 \big (x^7 + x^4 + 1\big ) \text { mod }2. \end{aligned} \end{aligned}$$

(5.3)

Since $\mathrm {PSL}_2(\mathbb {F}_8)$ contains no elements of order 14,  we thus know that $\mathrm {Gal}(L_{\mathfrak {p}}/\mathbb {Q}_2)$ is isomorphic to $C_7,$ where $L_{\mathfrak {p}}$ is the $\mathfrak {p}$-completion of L. Consider $\mathrm {Gal}(L'_{\bar{\mathfrak {p}}}/L_{\mathfrak {p}}).$ Because $L'/L$ is unramified, $\mathrm {Gal}(L'_{\bar{\mathfrak {p}}}/L_{\mathfrak {p}})$ is either trivial or $C_2.$

By Lemma 3.20, there is a unique class of subgroups $\mathrm {PSL}_2(\mathbb {F}_8)$ inside $\mathrm {GL}_6(\mathbb {F}_2).$ The cyclic subgroups of order 7 in these subgroups act fixed-point-freely on $(C_2)^6$ (in fact, the vector space decomposes into a direct sum of two irreducible modules of dimension 3 under their action). Therefore, the corresponding group extension of $C_7$ by $(C_2)^6$ has trivial center, and in particular contains no element of order 14. Thus, $\mathrm {Gal}(L'_{\bar{\mathfrak {p}}}/L_{\mathfrak {p}})$ is trivial, i.e., $\mathfrak {p}$ splits completely in $L'.$

Define S to be the compositum $L'K.$ Since $-1567 \equiv 1$ modulo 8,  2 splits completely in K. Then, for the number field $S/\mathbb {Q},$ we have that $f_2=7,$ where $f_2$ is the inertia degree of 2. Let us recall the function (2.3) again.

$$\begin{aligned} f=2\sum _{\mathfrak {p}}\sum ^{\infty }_{m=1}\frac{\log N(\mathfrak {p})}{N(\mathfrak {p})^{m/2}}F\left( \log N(\mathfrak {p})^m\right) . \end{aligned}$$

Since every term of f is greater than or equal to 0,  the following holds for the number field S.

$$\begin{aligned} \begin{aligned} f \ge 2\sum ^{9216}_{j=1}\sum ^{100}_{i=1}\frac{\log N(\bar{\mathfrak {q}}_j)}{N(\bar{\mathfrak {q}}_j)^{i/2}}F\left( \log N\left( \bar{\mathfrak {q}}_j\right) ^i\right) , \end{aligned} \end{aligned}$$

(5.4)

where the $\bar{\mathfrak {q}}_j$ denote the prime ideals of S satisfying $\bar{\mathfrak {q}}_j | 2.$ Since $f_2=7,$$N(\bar{\mathfrak {q}}_j)=2^7$ for all j. Set $b=11.6.$ By a numerical calculation, we have

$$\begin{aligned} \begin{aligned} f \ge 2\cdot 9216 \sum ^{100}_{i=1}\frac{\log 2^7}{2^{7i/2}}F\left( \log 2^{7i}\right) =6814.41\ldots \end{aligned} \end{aligned}$$

(5.5)

Let us recall (2.2). For $b=11.6,$ we have

$$\begin{aligned} \begin{aligned} \left| d_{S}\right| ^{1/n_{S}}&>39.619 \cdot \mathrm{e}^{(f-4790.3)/64,512}\\&\ge 39.619 \cdot \mathrm{e}^{(6814.41-4790.3)/64,512}=40.8818\ldots \end{aligned} \end{aligned}$$

(5.6)

Since S / K is unramified, $|d_{S}|^{1/n_{S}}=|d_K|^{1/n_K}=\sqrt{1567}=39.5853\ldots$ This is a contradiction. Therefore, the 2-class group of M is trivial. In conclusion, the class number of M is one.

5.3.2 $A_5$-unramified extension of M

Since $[K_\mathrm{ur}:M]<66.1375\ldots ,$ our final task is to show that M does not admit an unramified $A_5$-extension. By an analogous argument as in Sect. 4.2.4, K admits an $A_5$-extension F and $\mathrm {Gal}(F/\mathbb {Q})$ is also isomorphic to $A_5 \times C_2$ or $S_5.$

Case 1: $\mathrm {Gal}(F/\mathbb {Q}) \simeq A_5 \times C_2$ This implies that there exists an $A_5$-extension $F_1/\mathbb {Q}$ with ramification index 2 at 1567,  and unramified at all other finite primes. However, from [1, Tables] no such extensions exists. This is a contradiction.

Case 2: $\mathrm {Gal}(F/\mathbb {Q}) \simeq S_5$ By the unramifiedness of F / K,  a quintic subfield E of F must have the discriminant $-1567.$ However, the minimal negative discriminant of quintic fields with Galois group $S_5$ is $-4511$ [8, Table 3]. This is a contradiction.

Therefore, we know that $K_\mathrm{ur}=M$ under the assumption of the GRH. This concludes the proof of Theorem 5.1.