Extended Genus Fields of Abelian Extensions of Rational Function Fields

Abstract

In this paper, we obtain the extended genus field of a finite abelian extension of a global rational function field. We first study the case of a cyclic extension of prime power degree. For the general case, we use the fact that the extended genus fields of a composition of two cyclotomic extensions of a global rational function field is the same as the composition of their corresponding extended genus fields. In the main result of the paper, we give the extended genus field of finite abelian extensions of a global rational function field explicitly in terms of the field and extended genus field of its “cyclotomic projection”.

1. Introduction

The concepts of genus field and extended (or narrow) genus field depend on the respective concepts of Hilbert class field (HCF) and extended (or narrow) Hilbert class field. The theory of the genus goes back to Gauss. The HCF concept is much more recent. The first to translate Gauss’ genus theory into genus to “modern terms” was Hilbert. Currently, it may be used to study the “easy” part of the HCF of a finite extension of the field of rational numbers.

The notion of genus field for number fields was first introduced by H. Hasse, who defined the genus field of a quadratic extension of $\mathbb{Q}$. Since the genus field is related to the HCF, one natural way to study genus fields is by means of class field theory. However, we may study genus fields of abelian extensions of the rational field by using Dirichlet characters.

For number fields, the definition of Hilbert and extended Hilbert class field are canonically given as the maximal abelian unramified and the maximal abelian unramified at the finite primes of the field, respectively. The definition of the genus field is not absolute, as is the HCF, but depends on an extension of the base field. A. Fröhlich gave a general definition of genus fields for any number field K as follows: $K_{\mathfrak{ge}}:=KF$, where F is the maximal abelian extension of the field of the rational numbers $\mathbb{Q}$ contained in the Hilbert class field of K. Fröhlich’s definition is also canonical.

We are interested in global function fields. In this context, there are several different definitions of the HCF of a global field K depending on which aspect we are interested in. In this paper, we study the extended genus field of a finite abelian extension $K/k$, where $k={\mathbb{F}}_q(T)$ is a global rational function field. Let ${\mathfrak{p}}_{\infty }$ be the infinite prime of k. Then we define the HCF of K as the maximal unramified abelian extension $K_H$ of K such that the infinite primes of K (those above ${\mathfrak{p}}_{\infty }$) decompose fully in $K_H$. B. Anglès and J.-F. Jaulent [1] give the same concept by means of the idèle norm subgroup corresponding to $K_H$. They also define the extended HCF of any global field K by means of the norm subgroup of $K_{H^{+}}$ in the idèle group $J_K$ of K. We use the definitions of Anglès and Jaulent of $K_H$ and $K_{H^{+}}$ to define the genus field $K_{\mathfrak{ge}}$ and the extended genus field $K_{\mathfrak{gex}}$ of K with respect to the extension $K/k$, by means of their corresponding idéle subgroups. We compare our findings with the concepts of extended genus fields given by Ramírez–Rzedowski–Villa and by R. Clement in [2].

The definition of the extended genus field $K^{ext}$ of a finite abelian extension $K/k$ given by Ramírez–Rzedowski–Villa differs from the one defined by Anglès and Jaulent. In the general case, we have that $K_{\mathfrak{gex}}\subseteq K^{\mathrm{ext}}$.

In this paper, we explicitly describe the extended genus field $K_{\mathfrak{gex}}$ via Anglès and Jaulent’s approach. This result reinforces the connection between these two approaches and highlights the robustness of genus field theory, since we obtain that $K_{\mathfrak{gex}}=K^{ext}$. We explore the implications of this equality in the context of algebraic number theory and its applications to the study of class field theory.

We first study the case of a cyclic extension of k of prime power degree $l^n$. We consider four possible types of primes l: (1) $l=p$, the characteristic of k (Artin–Schreier–Witt case); (2) $l\nmid q-1$, $l\ne p$; (3) $l^n|q-1$ (Kummer case); and (4) $l^{\rho }|q-1$, $1\le \rho <n$ and $l^n\nmid q-1$ (“semi-Kummer” case).

Our main results are Theorems 11 and 12. We obtain that $K_{\mathfrak{gex}}=E_{\mathfrak{gex}}K$, where E is the cyclotomic projection of K (see (1) below) and $E_{\mathfrak{gex}}$ is the maximal abelian extension of E contained in a cyclotomic extension field unramified at the finite primes.

The main tools used in this paper are the Carlitz theory of cyclotomic function fields and class field theory, particularly the concepts of HCF and genus fields developed by Anglès and Jaulent.

2. Notations and General Results

For the general Carlitz–Hayes theory of cyclotomic function fields, we refer to [3]. For local and global class field theory, we refer to [4,5], respectively.

We will be using the following notation. Let $k={\mathbb{F}}_q(T)$ be a global rational function field, where ${\mathbb{F}}_q$ is the finite field of q elements. Let $R_T={\mathbb{F}}_q\left[T\right]$ and let $R_T^{+}$ denote the set of monic irreducible elements of $R_T$. For $N\in R_T$, $k({\Lambda }_N)$ denotes the Nth cyclotomic function field where ${\Lambda }_N$ is the Nth torsion of the Carlitz module. For a $D\in R_T$, we define $D^{*}:={(-1)}^{\mathrm{deg}D}D$. The relevance of $D^{*}$ is that $k(\sqrt[n]{D^{*}})\subseteq k({\Lambda }_D)$ in the case $n|q-1$.

We will call a field F a cyclotomic function field if there exists $N\in R_T$ such that $F\subseteq k({\Lambda }_N)$. In this way, in the case $n|q-1$ and $D\in R_T$, we have that $k(\sqrt[n]{D^{*}})$ is a cyclotomic function field.

Let $N\in R_T$. The Dirichlet characters $\chi modN$ are the group homomorphisms $\chi :{(R_T/\langle N\rangle )}^{*}⟶{\mathbb{C}}^{*}$. Given a group X of Dirichlet characters modulo N, the field associated to X is the fixed field $F=k{({\Lambda }_N)}^H$, where $H={\bigcap }_{\chi \in X}ker\chi$. We say that F corresponds to the group X and that X corresponds to F. We have that $X\cong Hom(Gal(F/k),{\mathbb{C}}^{*})$. When X is a cyclic group generated by $\chi$, we have that the field associated to X is equal to $F=k{({\Lambda }_N)}^{ker\chi }$, and we say that F corresponds to $\chi$.

Let $N\in R_T\backslash \{0\}$, and let $N={\prod }_{i=1}^r{P}_i^{{\alpha }_i}$ be its decomposition as a product of irreducible polynomials. Then ${(R_T/\langle N\rangle )}^{*}\cong {\prod }_{i=1}^r{(R_T/\langle P_i^{{\alpha }_i}\rangle )}^{*}$. Then, if $\chi$ is a Dirichlet character $modN$, we have that $\chi ={\prod }_{i=1}^r{\chi }_{P_i}$, where ${\chi }_{P_i}$ is a character $mod{P}_i^{{\alpha }_i}$.

Given a cyclotomic function field F with Dirichlet group characters X, we have that the ramification index of $P\in R_T^{+}$ in $F/k$ equals $|X_P|$, where $X_P=\{{\chi }_P\mid \chi \in X\}$ and ${\chi }_P$ is the Pth component of $\chi$. The maximum cyclotomic extension of F, unramified at the finite prime divisors, is the field that corresponds to $Y:={\prod }_{P\in R_T^{+}}{X}_P$ This field is denoted as $F_{\mathfrak{gex}}$.

We denote the infinite prime of k by ${\mathfrak{p}}_{\infty }$. That is, ${\mathfrak{p}}_{\infty }$ is the pole divisor of T, and $1/T$ is a uniformizer for ${\mathfrak{p}}_{\infty }$.

Given a finite extension $K/k$, a definition of HCF $K_H$, and a definition of extended HCF $K_{H^{+}}$ of K, the respective genus field and extended genus field of K with respect to k are the extensions $KL$ such that L is the maximal abelian extension of k contained in $K_H$ and in $K_{H^{+}}$, respectively.

We will use both notations $e_P(F|k)$ or $e_{F/k}(P)$ to denote the ramification index of the prime P of k in F. For the place ${\mathfrak{p}}_{\infty }$ we use the notation $e_{\infty }(F|k)$.

When $K/k$ is a finite abelian extension, it follows from the Kronecker–Weber–Hayes Theorem that there exist $N\in R_T$, $n\in \mathbb{N}\cup \{0\}$, and $m\in \mathbb{N}$ such that $K\subseteq {}_{n}k{({\Lambda }_N)}_m$, where, for any F, $F_m:=F{\mathbb{F}}_{q^m}$, for any $N\in R_T$, ${}_{n}k({\Lambda }_N):=L_{n}k({\Lambda }_N)$, and $L_n$ is the maximum subfield of $k({\Lambda }_{1/T^n})$ where ${\mathfrak{p}}_{\infty }$ is totally and wildly ramified. Then we define

$$\begin{array}{c}E:=\mathcal{M}K\cap k({\Lambda }_N)\end{array}$$

(1)

where $\mathcal{M}=L_n{k}_m$. We call E the “cyclotomic projection of K”, since E is a cyclotomic function field with similar arithmetic properties as K, except for wild ramification of the infinite prime.

When $K=k(\sqrt[\ell ]{\gamma D})$ is a Kummer extension, where $\ell |q-1$, $\gamma \in {\mathbb{F}}_q^{*}$ and $D\in R_T$ is free of ℓ- powers, then we have that $K\subseteq k({\Lambda }_D)$ if and only if $\gamma \equiv {(-1)}^{\mathrm{deg}D}mod{({\mathbb{F}}_q)}^{\ell }$. If $\gamma \neg \equiv {(-1)}^{\mathrm{deg}D}mod{({\mathbb{F}}_q)}^{\ell }$, we have that $E=k(\sqrt[\ell ]{D^{*}})$ where $D^{*}{(-1)}^{\mathrm{deg}D}D$ is the cyclotomic projection of K.

Let H be the decomposition group of the infinite primes in $K{E}_{\mathfrak{ge}}/K$, and ${\mathcal{H}=H|}_{E_{\mathfrak{ge}}}$. Since $E_{\mathfrak{ge}}\cap K=E\cap K$, from the Galois correspondence, we have that $H\cong \mathcal{H}$. The group ${\mathcal{H}}_1{=\mathcal{H}|}_E$ is also the decomposition group of the infinite primes of K in $KE/K$. Furthermore, $\mathcal{H}\subseteq I_{\infty }(k({\Lambda }_N)/k)\cong C_{q-1}$, where $I_{\infty }$ denotes the inertia group of ${\mathfrak{p}}_{\infty }$. Therefore, $\mathcal{H}$ is a cyclic group and $H\cong \mathcal{H}\cong {\mathcal{H}}_1$. We have that $K_{\mathfrak{ge}}={(E_{\mathfrak{ge}}K)}^H=E_{\mathfrak{ge}}^{\mathcal{H}}K$.

For any global function field L, ${\mathbb{P}}_L$ denotes the set of all places of L.

For $x\in \mathbb{Z}$, $v_l(x)$ denotes the valuation of x at l. That is, $v_l(x)=\gamma$ if $l^{\gamma }|x$ and $l^{\gamma +1}\nmid x$. We write $v_l(0)=\infty$.

3. Basic Results

One result of ramification of tamely ramified extensions, used frequently, is the following theorem.

Theorem 1

(Abhyankar’s Lemma). Let $L/K$ be a separable extension of global function fields. Assume that $L=K_1{K}_2$ with $K\subseteq K_i\subseteq L$, $1\le i\le 2$. Let $\mathfrak{p}$ be a prime divisor of K and $\mathfrak{P}$ a prime divisor in L above $\mathfrak{p}$. Let ${\mathfrak{P}}_i:=\mathfrak{P}\cap K_i$, $i=1,2$. If at least one of the extensions $K_i/K$ is tamely ramified at $\mathfrak{p}$, then

$$e_{L/K}(\mathfrak{P}|\mathfrak{p})=lcm[e_{K_1/K}({\mathfrak{P}}_1|\mathfrak{p}),e_{K_2/K}({\mathfrak{P}}_2|\mathfrak{p})],$$

where $e_{L/K}(\mathfrak{P}|\mathfrak{p})$ denotes the ramification index.

Next, we present some basic facts on finite cyclic groups and we apply them to the case of a finite field.

Let G be a cyclic group of order n, say $G=\langle a\rangle$. Let ${\Lambda }_m$ be the unique subgroup of G of order m where $m|n$. We have $\Lambda =\langle a^{n/m}\rangle$. Let $t\in \mathbb{N}$ and let $G^t:=\{x\in G\mid x=y^t\mathrm{for}\mathrm{some}y\in G\}$. We have $G^t=im{\phi }_t$, where ${\phi }_t:G\to G$ is given by ${\phi }_t(x)=x^t$.

Note that if $t\in \mathbb{N}$ and $d=gcd(t,n)$, then $G^d=G^t$, namely, if $\alpha ,\beta \in \mathbb{Z}$ are such that $\alpha t+\beta n=d$, we have

$$G^d=G^{\alpha t+\beta n}={(G^{\alpha })}^t{(G^n)}^{\beta }\subseteq G^t·1=G^t.$$

Conversely, let $t=\kappa d$ with $\kappa \in \mathbb{N}$. Then $G^t={(G^{\kappa })}^d\subseteq G^d$.

We also have that ${\Lambda }_d={\Lambda }_t$ since, if $x\in {\Lambda }_t$, then $x^t=1={(x^t)}^{\alpha }·{(x^n)}^{\beta }=x^{\alpha t+\beta n}=x^d$ so that ${\Lambda }_t\subseteq {\Lambda }_d$. Conversely, if $t=\kappa d$ and if $x\in {\Lambda }_d$, then $1=x^t={(x^d)}^{\kappa }=1^{\kappa }=1$.

Now, if $d|n$, then $G^d={\Lambda }_{n/d}$ because we have the exact sequence

$$1⟶{\Lambda }_d⟶G\stackrel{{\phi }_d}{⟶}{G}^d⟶1,$$

obtaining $|G^d|={\displaystyle \frac{|G|}{|{\Lambda }_d|}}={\displaystyle \frac{n}{d}}=|{\Lambda }_{n/d}|$ and, if $x\in G^d$, there exists $y\in G$ such that $x=y^d$ that implies $x^{n/d}={(y^d)}^{n/d}=y^n=1$ so that $G^d\subseteq {\Lambda }_{n/d}$. Thus $G^d={\Lambda }_{n/d}$.

We apply the previous basic results to the multiplicative group of the finite field ${\mathbb{F}}_q^{*}$, which is a cyclic group of $q-1$ elements.

Lemma 1.

Let l be a prime number and $n\in \mathbb{N}$, with $l^n|q-1$. Let $F:={\mathbb{F}}_q\left(\sqrt[l^n]{\beta }\right)$ with $\beta \in {\mathbb{F}}_q^{*}$. Then $F={\mathbb{F}}_{q^{l^s}}$ for some $0\le s\le n$.

Proof. 

Let $\mu =\sqrt[l^n]{\beta }$. Then ${\mu }^{l^n}=\beta \in {\mathbb{F}}_q^{*}$. Set s, $0\le s\le n$, to be the minimal non-negative integer such that ${\mu }^{l^s}=\theta \in {\mathbb{F}}_q^{*}$. If $s=0$, then $\mu =\theta \in {\mathbb{F}}_q^{*}$ and $f(X)=X^{l^s}-\theta =X-\theta$ is irreducible.

For any s, we will see that $f(X)=X^{l^s}-\theta$ is an irreducible polynomial. We have $f(X)=X^{l^s}-\theta ={\prod }_{j=1}^{l^s}(X-{\zeta }_{l^s}^j\mu )$, where ${\zeta }_m$ denotes a primitive m-th root of unity. Let $\mathcal{G}:=Gal({\mathbb{F}}_q(\mu )/{\mathbb{F}}_q)$. Let $\sigma \in \mathcal{G}$, $\sigma \ne Id$ and let $\sigma (\mu )={\zeta }_{l^s}^j\mu$, where $j=j_0{l}^b$, with $gcd(j_0,l)=1$. We choose an element $\sigma \in \mathcal{G}$ such that b is minimal.

We have $\sigma (\mu )={\zeta }_{l^s}^{j_0{l}^b}\mu ={\zeta }_{l^{s-b}}^{j_0}\mu$. Let $i_0\in \mathbb{Z}$ be such that $j_0{i}_0\equiv 1mod{l}^s$. Then ${\sigma }^{i_0}(\mu )={\zeta }_{l^{s-b}}^{j_0{i}_0}(\mu )={\zeta }_{l^{s-b}}\mu$. This $\mathcal{G}$ is a cyclic group of order $l^{s-b}$ and

$$\prod _{\epsilon =1}^{l^{s-b}}(X-{\zeta }_{l^{s-b}}^{\epsilon }\mu )=X^{l^{s-b}}-{\mu }^{l^{s-b}}\in {\mathbb{F}}_q\left[X\right].$$

Hence ${\mu }^{l^{s-b}}\in {\mathbb{F}}_q^{*}$. Therefore $b=0$, $|\mathcal{G}|=l^s$, and $X^{l^s}-\theta =Irr(\mu ,X,{\mathbb{F}}_q)$ is irreducible.

It follows that $[{\mathbb{F}}_q(\mu ):{\mathbb{F}}_q]=|\mathcal{G}|=l^s$ and $F={\mathbb{F}}_q(\mu )={\mathbb{F}}_q(\sqrt[l^n]{\beta })={\mathbb{F}}_q(\sqrt[l^s]{\theta })={\mathbb{F}}_{l^s}$. □

Remark 1.

We have ${\mathbb{F}}_q(\sqrt[l^n]{\beta })={\mathbb{F}}_{q^{l^s}}$, where s is the minimal non-negative integer such that ${\mu }^{l^s}\in {\mathbb{F}}_q^{*}$, with $\mu =\sqrt[l^n]{\beta }$.

Corollary 1.

With the above notations, if $m\in \mathbb{N}$, we have $[{\mathbb{F}}_q(\sqrt[l^{n+m}]{\beta }):{\mathbb{F}}_q]=l^{s+m}$.

Proof. 

Set $\delta =\sqrt[l^{n+m}]{\beta }$, then ${\delta }^{l^m}=\mu =\sqrt[l^n]{\beta }$, and ${\mu }^{l^s}={\delta }^{l^{m+s}}=\theta \in {\mathbb{F}}_q^{*}$. Clearly $m+s$ is minimal. □

Next, we consider $\mu =\sqrt[l^n]{\beta }$, with $\beta \in {\mathbb{F}}_q^{*}$ and $[{\mathbb{F}}_q(\sqrt[l^n]{\beta }):{\mathbb{F}}_q]=l^s$ for some $0\le s\le n$. Set ${\mu }^{l^s}=\theta \in {\mathbb{F}}_q^{*}$, $\beta ={\mu }^{l^n}={({\mu }^{l^s})}^{l^{n-s}}={\theta }^{l^{n-s}}$. Hence $\beta \in G^{l^{n-s}}$, where $G={\mathbb{F}}_q^{*}$.

In case that there would exist $\epsilon \in {\mathbb{F}}_q^{*}$ such that $\beta ={\epsilon }^{l^{n-s+1}}$, it would imply that ${\mu }^{l^n}={\epsilon }^{l^{n-s+1}}$ and thus $\mu ={({\epsilon }^{l^{n-s+1}})}^{1/l^n}={\epsilon }^{l^{n-s+1-n}}={\epsilon }^{l^{-s+1}}=\sqrt[l^{s-1}]{\epsilon }$. Therefore, $X^{l^{s-1}}-\epsilon$ would have $\mu$ as a root. It would follow that $[{\mathbb{F}}_q(\mu ):{\mathbb{F}}_q]\le l^{s-1}<l^s$, contrary to our hypothesis. Therefore, $\beta \in G^{l^{n-s}}\backslash G^{l^{n-s+1}}$.

Conversely, if $\beta \in G^{l^{n-s}}\backslash G^{l^{n-s+1}}$, then $\beta ={\kappa }^{l^{n-s}}$ where $\kappa$ is not an l-power. Therefore

$${\mu }^{1/l^n}={(\kappa )}^{l^{n-s}}{)}^{1/l^n}={\kappa }^{l^{-s}}=\sqrt[l^s]{\kappa },$$

hence ${\mathbb{F}}_q(\sqrt[l^n]{\mu })\subseteq {\mathbb{F}}_{q^{l^s}}$. In the case that ${\mathbb{F}}_q(\sqrt[l^n]{\mu })\subseteq {\mathbb{F}}_{q^{l^{s-1}}}$, it would follow that ${\mu }^{l^{s-1}}\in {\mathbb{F}}_q^{*}$ contrary to our hypothesis.

We have proved:

Theorem 2.

We have that $[{\mathbb{F}}_q(\sqrt[l^n]{\beta }):{\mathbb{F}}_q]=l^s$ if and only if $\beta \in {({\mathbb{F}}_q^{*})}^{l^{n-s}}\backslash {({\mathbb{F}}_q^{*})}^{l^{n-s+1}}$.

A basic result on cyclic groups of prime power degree that we need is the following.

Proposition 1.

Let G be a cyclic group of order $l^{\tau }$, with l a prime number. Given $H_1,H_2<G$, then $H_1\subseteq H_2$ or $H_2\subseteq H_1$. In particular, $H_1\cap H_2=H_j$ with $j=1$ or $j=2$ and if $H_1\ne \{Id\}$ and $H_2\ne \{Id\}$, then $H_1\cap H_2\ne \{Id\}$.

Proof. 

By cyclicity, G has a unique subgroup of each of the divisors of $|G|=l^{\tau }$. These subgroups are ${\Lambda }_0\subseteq {\Lambda }_1\subseteq \dots \subseteq {\Lambda }_{\tau }$ with $|{\Lambda }_i|=l^i$. The result follows. □

4. Extended Genus Fields and Class Field Theory

First, we establish the definition of extended genus fields according to Anglès and Jaulent [1].

We have that $k_{\infty }\cong {\mathbb{F}}_q\left(\left({\displaystyle \frac{1}{T}}\right)\right)$ is the completion of k at ${\mathfrak{p}}_{\infty }$. Let $x\in k_{\infty }^{*}$. Then x is written uniquely as

$$x={\left({\displaystyle \frac{1}{T}}\right)}^{n_x}{\lambda }_x{\epsilon }_x\mathrm{with}{n}_x\in \mathbb{Z},{\lambda }_x\in {\mathbb{F}}_q^{*}\mathrm{and}{\epsilon }_x\in U_{\infty }^{(1)},$$

where $U_{\infty }^{(1)}=U_{{\mathfrak{p}}_{\infty }}^{(1)}$ is the group of one units of $k_{\infty }$. We write ${\pi }_{\infty }:=1/T$, which is a uniformizer at ${\mathfrak{p}}_{\infty }$.

The sign function is defined as ${\varphi }_{\infty }:k_{\infty }^{*}⟶{\mathbb{F}}_q^{*}$ given by ${\varphi }_{\infty }(x)={\lambda }_x$ for $x\in k_{\infty }^{*}$.

We have that ${\varphi }_{\infty }$ is an epimorphism and $ker{\varphi }_{\infty }=\langle {\pi }_{\infty }\rangle \times U_{\infty }^{(1)}$.

For a finite separable extension L of $k_{\infty }$, we define the sign of $L^{*}$ by the morphism ${\varphi }_L:={\varphi }_{\infty }\circ N_{L/k_{\infty }}:L^{*}⟶{\mathbb{F}}_q^{*}$. We have ${\displaystyle \frac{L^{*}}{ker{\varphi }_L}}\cong \mathcal{A}\subseteq {\mathbb{F}}_q^{*}$.

For a global function field L, let $\mathcal{P}$ be the set of places of L dividing ${\mathfrak{p}}_{\infty }$. We define the following subgroups of the group of idèles $J_L$ as

$$\begin{array}{c}{U}_L:=\prod _{v|\infty }{L}_v^{*}\times \prod _{v\nmid \infty }{U}_{L_v}\mathrm{and}{U}_L^{+}:=\prod _{v|\infty }ker{\varphi }_{L_v}\times \prod _{v\nmid \infty }{U}_{L_v},\end{array}$$

(2)

where we denote $v\nmid \infty$ if $v\notin \mathcal{P}$ and $v|\infty$ if $v\in \mathcal{P}$. The groups $U_L{L}^{*}$ and $U_L^{+}{L}^{*}$ are open subgroups of $J_L$, the idèle group of L.

Definition 1.

Let $K/k$ be a finite abelian extension. Then the Hilbert class field (HCF)$$ and the extended HCF $K_{H^{+}}$ of K are the fields corresponding to the idèle subgroups $U_K{K}^{*}$ and $U_K^{+}{K}^{*}$ of $J_K$, respectively. By class field theory, the respective genus $K_{\mathfrak{ge}}$ and extended genus fields $K_{\mathfrak{gex}}$ with respect to the extension $K/k$ correspond to the idèle subgroups $(N_{K/k}{U}_K)k^{*}$ and $(N_{K/k}{U}_K^{+})k^{*}$ of $J_k$, respectively.

By class field theory, we have

$$Gal(K_H/K)\cong J_K/U_K{K}^{*}.$$

We have that $K_{H^{+}}/K$ is an unramified extension at the finite prime divisors of K, $K_H\subseteq K_{H^{+}}$ and

$$Gal(K_{H^{+}}/K)\cong J_K/U_K^{+}{K}^{*}.$$

Let $K/k$ be a finite abelian extension. Then, with the notation given above, if $E=\mathcal{M}K\cap k({\Lambda }_N)$, we have $K_{\mathfrak{gex}}=DK$ for some subfield ${(E_{\mathfrak{ge}}^{\mathcal{H}})}_{\mathfrak{gex}}\subseteq D\subseteq E_{\mathfrak{gex}}$ for some decomposition group $\mathcal{H}$. In most cases, we have $\mathcal{H}=\{Id\}$. In this case, ${(E_{\mathfrak{ge}}^{\mathcal{H}})}_{\mathfrak{gex}}=E_{\mathfrak{gex}}$ and $K_{\mathfrak{gex}}=E_{\mathfrak{gex}}K$.

In this paper, we particularly study the case ${(E_{\mathfrak{ge}}^{\mathcal{H}})}_{\mathfrak{gex}}\ne E_{\mathfrak{gex}}$. A general result is the following.

Proposition 2.

Let $K/k$ be a finite abelian extension and let E be given by (1). Then, if $P_1,\dots ,P_r$ are the finite primes of k ramified in K and

$$e_{P_j}(E_{\mathfrak{ge}}^{\mathcal{H}}|k)=e_{P_j}(E|k)=e_{P_j}(E_{\mathfrak{ge}}|k),$$

for all $1\le j\le r$, it follows that ${(E_{\mathfrak{ge}}^{\mathcal{H}})}_{\mathfrak{gex}}=E_{\mathfrak{gex}}$.

Proof. 

The group of Dirichlet characters associated with $E_{\mathfrak{gex}}$ is

$$Y:=\prod _{P\in R_T^{+}}{X}_P=\prod _{j=1}^r{X}_{P_j},$$

where X is the group of Dirichlet characters associated with E. Each $X_{P_j}$ is cyclic of order $e_{P_j}(E|k)$. The field associated with $X_{P_j}$ is the subfield of $k({\Lambda }_{P_j})$ of degree $e_{P_j}(E|k)$ over k. It follows that

$$[E_{\mathfrak{gex}}:k]=\prod _{j=1}^r{e}_{P_j}(E|k).$$

Let Z be the group of Dirichlet characters associated to ${(E_{\mathfrak{ge}}^{\mathcal{H}})}_{\mathfrak{gex}}$. The only finite primes of k possibly ramified in ${(E_{\mathfrak{ge}}^{\mathcal{H}})}_{\mathfrak{gex}}$ are $P_1,\dots ,P_r$. The group of Dirichlet characters associated to ${(E_{\mathfrak{ge}}^{\mathcal{H}})}_{\mathfrak{gex}}$ is ${\prod }_{j=1}^r{Z}_{P_j}$. By hypothesis, $X_{P_j}=Z_{P_j}$ for all $1\le j\le r$. Therefore ${(E_{\mathfrak{ge}}^{\mathcal{H}})}_{\mathfrak{gex}}=E_{\mathfrak{gex}}$. □

Corollary 2.

We have ${(E_{\mathfrak{ge}}^{\mathcal{H}})}_{\mathfrak{gex}}\ne E_{\mathfrak{gex}}\iff$ there exists $1\le j_0\le r$ such that $e_{P_{j_0}}(E_{\mathfrak{gex}}^{\mathcal{H}}|k)<e_{P_{j_0}}(E|k)=e_{P_{j_0}}(E_{\mathfrak{gex}}|k)=e_{P_{j_0}}(E_{\mathfrak{gex}}|k)$.

Proof. 

It follows from Proposition 2 and from the facts that $E_{\mathfrak{gex}}/E_{\mathfrak{ge}}$ is not ramified at any finite prime and that ${\mathfrak{p}}_{\infty }$ is fully ramified in $E_{\mathfrak{gex}}/E_{\mathfrak{ge}}$. □

In the rest of the section, we will focus on the cases ${(E_{\mathfrak{ge}}^{\mathcal{H}})}_{\mathfrak{gex}}\ne E_{\mathfrak{gex}}$.

Definition 2.

For a finite abelian extension $K/k$, we define $K^{ext}:=E_{\mathfrak{gex}}K$, where E is given by (1).

In this paper, we will prove that it always happens

$$K^{ext}=K_{\mathfrak{gex}}=E_{\mathfrak{gex}}K,$$

where $K/k$ is any finite abelian extension.

Ramírez–Rzedowski–Villa defined $K^{ext}$ for a finite abelian extension $K/k$. In general, we have $K_{\mathfrak{gex}}\subseteq K^{ext}$.

We recall the following theorem from class field theory.

Theorem 3.

Let F be a global function field. Let $N/F$ be a finite abelian extension, and let $B<C_F$ be the subgroup of the idèle class group of F corresponding to N. Then, if ${\mathbb{F}}_q$ is the field of constants of F, ${\mathbb{F}}_{q^{\kappa }}$ is the field of constants of N, where

$$\kappa :=min\{\sigma \in \mathbb{N}\mid \mathit{there}\mathit{exists}\tilde{\overrightarrow{\alpha }}\in B\mathit{such}\mathit{that}\mathrm{deg}\tilde{\overrightarrow{\alpha }}=\sigma \}.$$

(3)

Remark 2.

Hereafter, κ is given by (3).

Remark 3.

Note that if F is any global function field, then the fields of the constants of $F_{\mathfrak{gex}}$ and of $F_{H^{+}}$ are the same. Therefore, in the special case ${(E_{\mathfrak{ge}}^{\mathcal{H}})}_{\mathfrak{gex}}\ne E_{\mathfrak{gex}}$, if we obtain that if the field of constants of $E_{\mathfrak{gex}}K$ is contained in the one of $K_{H^{+}}$ then, because $E_{\mathfrak{gex}}K=E_{\mathfrak{ge}}K$ is an extension of constants of $K_{\mathfrak{ge}}$, say $E_{\mathfrak{gex}}K=K_{\mathfrak{ge}}{\mathbb{F}}_{q^{\mu }}$ and $K_{\mathfrak{ge}}\subseteq K_H\subseteq K_{H^{+}}$ and ${\mathbb{F}}_{q^{\mu }}\subseteq K_{H^{+}}$, it follows that $E_{\mathfrak{gex}}K\subseteq K_{H^{+}}$ and therefore $E_{\mathfrak{gex}}K\subseteq K_{\mathfrak{gex}}$. Since $K_{\mathfrak{gex}}\subseteq E_{\mathfrak{gex}}K$, we obtain $E_{\mathfrak{gex}}K=K_{\mathfrak{gex}}$.

Consider a finite abelian extension $K/k$. The idèle class subgroup B of the idèle class group $C_K$, the idèle class group of K, associated to $K_{H^{+}}$ is

$$\begin{array}{c}B=U_K^{+}{K}^{*}/K^{*}=\left(\prod _{\mathfrak{P}|\infty }ker{\varphi }_{\mathfrak{P}}\times \prod _{\mathfrak{P}\nmid \infty }{U}_{\mathfrak{P}}\right)K^{*}/K^{*},\end{array}$$

where

$$\begin{array}{c}\hfill U_{\mathfrak{P}}:=U_{K_{\mathfrak{P}}},ker{\varphi }_{\mathfrak{P}}:=ker{\varphi }_{K_{\mathfrak{P}}},\mathrm{and}{\varphi }_{\mathfrak{P}}={\varphi }_{\infty }\circ N_{K_{\mathfrak{P}}|k_{\infty }}.\end{array}$$

We denote $N_{\mathfrak{P}}:=N_{K_{\mathfrak{P}}|k_{\infty }}$.

On the one hand, if $\overrightarrow{\alpha }\in U_K^{+}$, then $\overrightarrow{\alpha }={\left({\alpha }_{\mathfrak{P}}\right)}_{\mathfrak{P}}$, ${\alpha }_{\mathfrak{P}}\in U_{\mathfrak{P}}$ for $\mathfrak{P}\nmid \infty$, so that ${\mathrm{deg}}_{\mathfrak{P}}{\alpha }_{\mathfrak{P}}=0$ for $\mathfrak{P}\nmid \infty$. On the other hand, if ${\mathfrak{P}}_1$ and ${\mathfrak{P}}_2$ are two infinite primes, $N_{{\mathfrak{P}}_1}{K}_{{\mathfrak{P}}_1}^{*}=N_{{\mathfrak{P}}_2}{K}_{{\mathfrak{P}}_2}^{*}$. It follows that if we fix an infinite prime $\mathfrak{P}$ of K, then

Lemma 2.

We have

$$\begin{array}{cc}\hfill \kappa :& =min\{\sigma \in \mathbb{N}\mid \mathit{there}\mathit{exists}\overrightarrow{\alpha }\in U_K^{+}\mathit{such}\mathit{that}\mathrm{deg}\overrightarrow{\alpha }=\sigma \}\hfill \\ \hfill & =min\{\sigma \in \mathbb{N}\mid \mathit{there}\mathit{exists}\tilde{\overrightarrow{\alpha }}\in B\mathit{such}\mathit{that}\mathrm{deg}\tilde{\overrightarrow{\alpha }}=\sigma \}\hfill \\ \hfill & =min\{\sigma \in \mathbb{N}\mid \mathit{there}\mathit{exists}x\in K_{\mathfrak{P}}^{*}\mathit{such}\mathit{that}x\in ker{\varphi }_{\mathfrak{P}}\mathit{and}{\mathrm{deg}}_{\mathfrak{P}}x=\sigma \}.\hfill \end{array}$$

5. Cyclic Extensions of Prime Power Degree

In this section, we study the extended genus field $K_{\mathfrak{gex}}$ of a finite cyclic extension $K/k$ of degree $l^n$ with l a prime number and $n\ge 1$. We will assume that the extension $K/k$ is geometric, that is, the field of constants of K is ${\mathbb{F}}_q$. We consider four types of primes l:

(1)

$l=p$, where p is the characteristic of k, the Artin–Schreier–Witt case,

(2)

$l\ne p$ and $l\nmid q-1$,

(3)

$l^n|q-1$, the Kummer case,

(4)

$l^{\rho }|q-1$ with $1\le \rho <n$ and $l^n\nmid q-1$, the “semi-Kummer” case.

5.1. The Artin–Schreier–Witt Case: $l=p$

Since $\mathcal{H}$ is a subgroup of the inertia group of ${\mathfrak{p}}_{\infty }$ in $E/k$ and the order of this last group is a divisor of $q-1$, the order of $\mathcal{H}$ is relatively prime to p. Hence, $\mathcal{H}=\{Id\}$. It follows that ${(E_{\mathfrak{ge}}^{\mathcal{H}})}_{\mathfrak{gex}}=E_{\mathfrak{gex}}$.

5.2. Case $l\ne p$ and $l\nmid q-1$

By the same reason as in Section 5.1, the order of $\mathcal{H}$ is relatively prime to l. Thus $\mathcal{H}=\{Id\}$ and ${(E_{\mathfrak{ge}}^{\mathcal{H}})}_{\mathfrak{gex}}=E_{\mathfrak{gex}}$.

5.3. The Kummer Case: $l^n|q-1$

5.3.1. The Genus Field in the Cyclotomic Case

Let $K=F=k\left(\sqrt[l^n]{D^{*}}\right)$, with $D=P_1^{{\alpha }_1}\dots P_r^{{\alpha }_r}$, be a Kummer cyclic extension of k. Let $X=\langle \chi \rangle$ be the group of Dirichlet characters associated to F. Note that for any $\nu \in \mathbb{N}$ relatively prime to l, the field associated to ${\chi }^{\nu }$ is F since $X=\langle {\chi }^{\nu }\rangle$. The above corresponds to the fact that $F=k\left(\sqrt[l^n]{{(D^{\nu })}^{*}}\right)$.

When $D=P\in R_T^{+}$, we have that the character associated to F is ${\left({\displaystyle \genfrac{}{}{0.5pt}{}{ }{P}}\right)}_{l^n}$, the Legendre symbol that is defined as follows: if P is of degree d, then for any $N\in R_T$ with $P\nmid N$, $NmodP\in {(R_T/\langle P\rangle )}^{*}\cong {\mathbb{F}}_{q^d}^{*}$. Then ${\left({\displaystyle \genfrac{}{}{0.5pt}{}{N}{P}}\right)}_{l^n}$ is defined as the unique element of ${\mathbb{F}}_{q^d}^{*}$ such that $N^{{\displaystyle \frac{q^d-1}{l^n}}}\equiv {\left({\displaystyle \genfrac{}{}{0.5pt}{}{N}{P}}\right)}_{l^n}modP$. We have that ${\left({\displaystyle \genfrac{}{}{0.5pt}{}{ }{P}}\right)}_{l^n}$ is the character associated to $k\left(\sqrt[l^n]{P^{*}}\right)$. Let us denote ${\chi }_P={\left({\displaystyle \genfrac{}{}{0.5pt}{}{ }{P}}\right)}_{l^n}$. Then ${\chi }_P^{\nu }$ is the character associated to $k\left(\sqrt[l^n]{{(P^{\nu })}^{*}}\right)$.

Hence, if ${\chi }_D$ is the character associated to $k\left(\sqrt[l^n]{D^{*}}\right)$, then ${\chi }_D={\prod }_{j=1}^r{\chi }_{P_j}^{{\alpha }_j}$.

Remark 4.

Let l be a prime number different from p, the characteristic of K, such that $l^{\kappa }|q-1$, $\kappa \ge 1$. We have that $-1\in {({\mathbb{F}}_q^{*})}^{l^{\kappa }}$ for all l and all κ except when $l=2$ and $2^{\kappa +1}\nmid q-1$.

Proof. 

If l is odd, ${(-1)}^{l^{\kappa }}=-1$ for all $\kappa \in \mathbb{N}$. Let $l=2$. If $2^{\kappa +1}|q-1$, ${\mathbb{F}}_q$ contains a primitive $2^{\kappa +1}$th root of unity $\xi$. Let $\mu :={\xi }^{2^{\kappa }}$. Then $\mu \ne 1$ and ${\mu }^2=1$. Thus $\mu =-1$.

On the other hand, if $2^{\kappa +1}\nmid q-1$, $\xi \notin {\mathbb{F}}_q^{*}$. Now, if we had $\mu =-1={\rho }^{2^{\kappa }}$ for some $\rho \in {\mathbb{F}}_q^{*}$, then $\rho$ is a primitive $2^{\kappa +1}$th root of unity, contrary to our hypothesis. □

Corollary 3.

For any prime number l such that $l^{\kappa }|q-1$, with $\kappa \in \mathbb{N}$ and $D\in R_T$, we have $k\left(\sqrt[l^{\kappa }]{D^{*}}\right)=k\left(\sqrt[l^{\kappa }]{D}\right)$, except when $\mathrm{deg}D$ is odd, $l=2$, and $2^{\kappa +1}\nmid q-1$. We also have that if $\gamma \in {\mathbb{F}}_q^{*}$ and $\epsilon ={(-1)}^{\mathrm{deg}D}\gamma$, then ${\mathbb{F}}_q\left(\sqrt[l^{\kappa }]{\epsilon }\right)={\mathbb{F}}_q\left(\sqrt[l^{\kappa }]{\gamma }\right)$ with the same exception.

Proof. 

If $\mathrm{deg}D$ is even, ${(-1)}^{\mathrm{deg}D}=1$. If $\mathrm{deg}D$ is odd, ${(-1)}^{\mathrm{deg}D}=-1\in {({\mathbb{F}}_q^{*})}^{l^{\kappa }}$ except when $l=2$ and $2^{\kappa +1}\nmid q-1$. The same argument works for ${\mathbb{F}}_q\left(\sqrt[l^{\kappa }]{\epsilon }\right)$ and ${\mathbb{F}}_q\left(\sqrt[l^{\kappa }]{\gamma }\right)$. □

In general, for a radical extension, we have:

Theorem 4.

Let $F=k\left(\sqrt[s]{\gamma D}\right)$ be a geometric separable extension of k, $\gamma \in {\mathbb{F}}_q^{*}$, and let $D=P_1^{{\alpha }_1}\dots P_r^{{\alpha }_r}\in R_T$. Then

$$\begin{array}{c}{e}_{F/k}(P_j)={\displaystyle \frac{s}{\gcd ({\alpha }_j,s)}},1\le j\le r\mathit{and}\\ e_{\infty }(F|k):=e_{F/k}({\mathfrak{p}}_{\infty })={\displaystyle \frac{s}{\gcd (\mathrm{deg}D,s)}}.\end{array}$$

As a consequence, we obtain the following result for a cyclic cyclotomic Kummer extension $F=k(\sqrt[l^n]{D^{*}})$. Let X be the group of Dirichlet characters associated to F and let $Y={\prod }_{P\in R_T^{+}}{X}_P$ be the group associated to M, the maximal cyclotomic extension of F unramified at the finite primes.

Let $P=P_j$, $X=X_P=\langle {\chi }_P\rangle$, and let $F_P$ be the field associated to $X_P$. Then, $F_P$ is cyclotomic, P is the only ramified prime in $F_P/k$, and P is tamely ramified in $F_P/k$. This implies that $F_P\subseteq k({\Lambda }_P)$ and $Gal(k({\Lambda }_P)/k)\cong C_{q^{d_P}-1}$ with $d_P:=\mathrm{deg}P$. Therefore, $F_P$ is the only field of degree $o({\chi }_P)=:l^{{\beta }_P}$ over k. Since $F_P/k$ is a Kummer extension, it follows that $F_P=k\left(\sqrt[l^{{\beta }_P}]{P^{*}}\right)$.

Theorem 5.

The maximal unramified cyclotomic extension of $F=k\left(\sqrt[l^n]{D^{*}}\right)$ at the finite primes is $M:=k(\sqrt[l^n]{{(P_1^{{\alpha }_1})}^{*}},\dots ,\sqrt[l^n]{{(P_r^{{\alpha }_r})}^{*}})$. In other words,

$$F_{\mathfrak{gex}}=(\sqrt[l^n]{{(P_1^{{\alpha }_1})}^{*}},\dots ,\sqrt[l^n]{{(P_r^{{\alpha }_r})}^{*}}).$$

Proof. 

The field M corresponds to the group of Dirichlet characters $Y={\prod }_{P\in R_T^{+}}{X}_P$ and the field associated to $X_P$ is $F_P=k\left(\sqrt[l^{{\beta }_P}]{P^{*}}\right)$, for each $P\in R_T^{+}$. The result follows. □

Remark 5.

Let $\alpha =l^{a}b$ with $gcd(b,l)=1$ and $a<n$. Then $k\left(\sqrt[l^n]{{(P^{\alpha })}^{*}}\right)=k\left(\sqrt[l^{n-a}]{P^{*}}\right)$ and

$$F_{\mathfrak{gex}}=k(\sqrt[l^{n-a_1}]{P_1^{*}},\dots ,\sqrt[l^{n-a_r}]{P_r^{*}})=F_1\cdots F_r,$$

with $F_j=k\left(\sqrt[l^{n-a_j}]{P_j^{*}}\right)$, $1\le j\le r$.

Another proof of Theorem 5 is obtained using Abhyankar’s Lemma. On the one hand, we have that

$$[M:k]=\prod _{P\in R_T^{+}}|X_P|=\prod _{j=1}^r|X_{P_j}|=\prod _{j=1}^r{e}_{M/k}(P_j)=\prod _{j=1}^r{l}^{n-a_j}.$$

On the other hand, if $F_j=k\left(\sqrt[l^{n-a_j}]{{(P_j)}^{*}}\right)$, from Abyankar’s Lemma, $F{F}_j/F$ is unramified at every finite prime, so $F{F}_1\cdots F_r/F$ is unramified at the finite primes and $F\subseteq F_1\cdots F_r$. Hence $F_1\cdots F_r\subseteq F_{\mathfrak{gex}}$ and $[F_1\cdots F_r:k]=[M:k]$. Therefore, $M=F_1\cdots F_r$.

The following theorem gives the genus field of a cyclic extension.

Theorem 6.

Let $E=k(\sqrt[l^n]{D^{*}})$, with $D=P_1^{{\alpha }_1}\cdots P_r^{{\alpha }_r}$, $1\le {\alpha }_j\le l^n-1$, ${\alpha }_j=b_j{l}^{a_j}$ with $gcd(b_j,l)=1$, $1\le j\le r$, $P_1,\dots ,P_r\in R_T^{+}$ different monic irreducible polynomials with $\mathrm{deg}{P}_j=c_j{l}^{d_j}$, $gcd(c_j,l)=1$, $1\le j\le r$. We order the polynomials $P_1,\dots ,P_r$ so that $0=a_1\le \cdots \le a_r\le n-1$.

Let $E_{\mathfrak{gex}}:=E_1\cdots E_r$ with $E_j=k(\sqrt[l^{n-a_j}]{P_j^{*}})$, $1\le j\le r$. Let

$$\begin{array}{cc}\hfill e_{\infty }(E|k)=l^t\mathit{with}t& =n-min\{n,v_l(\mathrm{deg}D)\},\hfill \\ \hfill & \\ \hfill e_{\infty }(E_{\mathfrak{gex}}|k)=l^m\mathit{with}m& =\underset{1\le j\le r}{max}{v}_l(e_{\infty }(E_j|k))\hfill \\ \hfill & =max\{n-a_j-min\{n-a_j,d_j\}1\le j\le r\}.\hfill \end{array}$$

Let $i_0$, $1\le i_0\le r$, be such that $n-a_{i_0}-min\{n-a_{i_0},d_{i_0}\}=m$ and $n-a_j-d_j<m$ for $j>i_0$. For $m>0$ we have $gcd(\mathrm{deg}{P}_{i_0},l^n)=l^{d_{i_0}}$, and therefore there exist $a,b\in \mathbb{Z}$ such that $a\mathrm{deg}{P}_{i_0}+b{l}^n=l^{d_{i_0}}$. For $j<i_0$, we have $d_{i_0}\le d_j$. Let $z_j:=-a{c}_j{l}^{d_j-d_{i_0}}$. For $j>i_0$, let $y_j\equiv -c_j{c}_{i_0}^{-1}mod{l}^n\in \mathbb{Z}$.

Then

$$E_{\mathfrak{ge}}=F_1\cdots F_r,$$

where $F_j=E_j$ with $1\le j\le r$ if $m=t$, i.e, $E_{\mathfrak{ge}}=E_{\mathfrak{gex}}$, and if $m>t\ge 0$, then

$$\begin{array}{c}{F}_j:=\left\{\begin{array}{cc}k\left(\sqrt[l^{n-a_j}]{P_j{P}_{i_0}^{z_j}}\right)\hfill & \mathit{if}j<i_0,\hfill \\ k\left(\sqrt[l^{d_{i_0}+t}]{P_{i_0}^{*}}\right)\hfill & \mathit{if}j=i_0,\hfill \\ k\left(\sqrt[l^{n-a_j}]{P_j{P}_{i_0}^{y_j{l}^{d_j-d_{i_0}}}}\right)\hfill & \mathit{if}j>i_0\mathit{and}{d}_j\ge d_{i_0},\hfill \\ k\left(\sqrt[l^{n-a_j+d_{i_0}-d_j}]{P_j^{l^{d_{i_0}-d_j}}{P}_{i_0}^{y_j}}\right)\hfill & \mathit{if}j>i_0\mathit{and}{d}_{i_0}>d_j.\hfill \end{array}\right.\end{array}$$

(4)

Remark 6.

When $m=t$, we may also use the description of $K_{\mathfrak{ge}}$ given in the case $m>t$.

5.3.2. The Genus Field in the General Case

The following theorem gives the genus field of a general $l^n$ cyclic extension.

Theorem 7.

Let $K=k(\sqrt[l^n]{\gamma D})\subseteq k{({\Lambda }_D)}_u$, with $\gamma \in {\mathbb{F}}_q^{*}$, $D=P_1^{{\alpha }_1}\dots P_r^{{\alpha }_r}$, $1\le {\alpha }_j\le l^n-1$, ${\alpha }_j=b_j{l}^{a_j}$ with $gcd(b_j,l)=1$, $1\le j\le r$, $P_1,\dots ,P_r\in R_T^{+}$ different polynomials and some $u\in \mathbb{N}$. We order the polynomials $P_1,\dots ,P_r$ so that $0=a_1\le \cdots \le a_r\le n-1$. Let $E=K_u\cap k({\Lambda }_D)$, t as in Theorem 6, and $\alpha =v_l(|\mathcal{H}|)$. Let ${\mathcal{H}}^{\prime }:=\mathcal{H}{\mid }_{E_{\mathfrak{ge}}}$. Then $E_{\mathfrak{ge}}^{{\mathcal{H}}^{\prime }}=F_1\cdots F_{i_0-1}{F}_{i_0+1}\cdots F_r(\sqrt[l^{d_{i_0}+(t-\alpha )}]{P_{i_0}^{*}})$, where $F_j$ are given in (4) for all j. Thus

$$K_{\mathfrak{ge}}=E_{\mathfrak{ge}}^{{\mathcal{H}}^{\prime }}K=\prod _{\stackrel{i=1}{i\ne i_0}}^r{F}_{i}K(\sqrt[l^{d_{i_0}+(t-\alpha )}]{P_{i_0}^{*}}).$$

Further, if $d=\min \{n,v_l(\mathrm{deg}D)\}$, we have

$$|\mathcal{H}|=l^{\alpha }=[{\mathbb{F}}_q(\sqrt[l^n]{{(-1)}^{\mathrm{deg}D}\gamma }):{\mathbb{F}}_q(\sqrt[l^d]{{(-1)}^{\mathrm{deg}D}\gamma })].$$

The general structure of $K_{\mathfrak{gex}}$ when $K/k$ is a finite l–Kummer extension for a prime number l, is given by $K_{\mathfrak{gex}}=DK$ with D a field satisfying ${(E_{\mathfrak{ge}}^{\mathcal{H}})}_{\mathfrak{gex}}\subseteq D\subseteq E_{\mathfrak{gex}}$.

With notations given above, particularly in Theorem 6, we consider first the case $m>t$.

Proposition 3.

If $m>t$ then $i\ge 2$ and there exists $j<i$ such that $m=n-a_j-d_j=n-a_i-d_i$.

Proof. 

First, assume that $i\ge 2$. Suppose that for all $1\le j\le i-1$ we have $n-a_j-d_j<n-a_i-d_i=m$.

We have that for all $j\ne i$, $n-a_i-d_i>n-a_j-min\{n-a_j,d_j\}\ge n-a_j-d_j$. Thus

$$\begin{array}{c}n-a_j-d_j<n-a_i-d_i,\mathrm{so}\mathrm{that}{a}_i+d_i<a_j+d_j\mathrm{for}\mathrm{all}j\ne i.\end{array}$$

(5)

We have

$$\begin{array}{cc}\hfill \mathrm{deg}D& =\sum _{j=1}^r{\alpha }_j\mathrm{deg}{P}_j=\sum _{j=1}^r{b}_j{l}^{a_j}{c}_j{l}^{d_j}\hfill \\ \hfill & =b_i{c}_i{l}^{a_i+d_i}+l^{a_i+d_i+1}(\sum _{\begin{array}{c}j=1\\ j\ne i\end{array}}^r{b}_j{c}_j{l}^{a_j+d_j-a_i-d_i-1}).\hfill \end{array}$$

Hence $v_l(\mathrm{deg}D)=l^{a_i+d_i}$. It follows that

$$\begin{array}{c}n-min\{n,v_l(\mathrm{deg}D)\}\ge n-v_l(\mathrm{deg}D)=n-a_i-d_i=m\mathrm{and}\\ t\ge n-min\{n,v_l(\mathrm{deg}D)\}\ge m\ge t.\end{array}$$

Therefore, $m=t$, contrary to our assumption. Thus, there exists $1\le j\le i-1$ with $n-a_j+d_j=n-a_i-d_i=m$.

The same argument shows that if $i=1$, then $m=t$. □

From Theorems 6 and 7, we have that for all $j\ne i$, we have $e_{P_j}(E_{\mathfrak{ge}}^{\mathcal{H}}|k)=e_{P_j}(K|k)=e_{P_j}(E|k)$. In case there exists $1\le j\le i-1$ such that $n-a_j-d_j=n-a_i-d_i$ we obtain that

$$\begin{array}{c}\begin{array}{cc}\hfill e_{P_i}(E_j|k)& =e_{P_i}(k(\sqrt[l^{n-a_j}]{P_j{P}_i^{-a{c}_j{l}^{d_j-d_i}}})|k)=l^{n-a_j-d_j+d_i}\hfill \\ \hfill & =l^{n-a_i}=e_{P_i}(E|k)=e_{P_i}(K|k).\hfill \end{array}\end{array}$$

Hence

$$\begin{array}{c}\hfill e_{P_i}(E_{\mathfrak{ge}}^{\mathcal{H}}|k)=e_{P_i}(K|k)=e_{P_i}(E|k).\end{array}$$

From the above, the following result is immediate.

Proposition 4.

If there exists $1\le j\le i-1$ such that $n-a_j-d_j=n-a_i-d_i$, in particular when $m>t$, then ${(E_{\mathfrak{ge}}^{\mathcal{H}})}_{\mathfrak{gex}}=E_{\mathfrak{gex}}$.

The first main result on extended genus fields is the following:

Theorem 8.

With the above notations, we have that $K_{\mathfrak{gex}}=E_{\mathfrak{gex}}K$, except possibly in the following case:

(a) 

$K\ne E$,

(b) 

$\mathcal{H}\ne \{Id\}$,

(c) 

$t=m>0$,

(d) 

$m=n-a_i-min\{n-a_i,d_i\}>n-a_j-min\{n-a_j,d_j\}$ for all $j\ne i$.

Proof. 

If $\mathcal{H}=\{Id\}$, the result follows. If $E=K$, K is cyclotomic and therefore $\mathcal{H}=\{Id\}$. If $m>t$, then, from Proposition 3, we have that $e_{P_j}(E_{\mathfrak{ge}}^{\mathcal{H}}|k)=e_{P_j}(E_{\mathfrak{ge}}|k)$ for all $1\le j\le r$ and therefore ${(E_{\mathfrak{ge}}^{\mathcal{H}})}_{\mathfrak{gex}}=E_{\mathfrak{gex}}$. If $n-a_i-min\{n-a_i,d_i\}=n-a_j-min\{n-a_j,d_j\}$ for some $j\ne i$, then $e_{P_j}(E_{\mathfrak{ge}}^{\mathcal{H}}|k)=e_{P_j}(E_{\mathfrak{ge}}|k)$ for all $1\le j\le r$ and therefore ${(E_{\mathfrak{ge}}^{\mathcal{H}})}_{\mathfrak{gex}}=E_{\mathfrak{gex}}$. If $t=0$, we have that ${\mathfrak{p}}_{\infty }$ is unramified in $E/k$ so that $\mathcal{H}=\{Id\}$. □

The above result shows that, outside of the special case given in Theorem 8, we have $K_{\mathfrak{gex}}=K^{ext}$.

5.3.3. The Special Case

We now consider the special case, that is, the exception mentioned in Theorem 8. We will prove that, also in this case, the equality $K_{\mathfrak{gex}}=K^{ext}=E_{\mathfrak{gex}}K$ holds.

Let $K=k\left(\sqrt[l^n]{\gamma D},\right)$ be a geometric separable extension of k, with $\gamma \in {\mathbb{F}}_q^{*}$ and let $D=P_1^{{\alpha }_1}\dots P_r^{{\alpha }_r}\in R_T$, with $P_1,\dots ,P_r\in R_T^{+}$ distinct, $1\le {\alpha }_j\le l^n-1$, $1\le j\le r$. Let ${\alpha }_j=l^{a_j}{b}_j$, $l\nmid b_j$, $\mathrm{deg}{P}_j=c_j{l}^{d_j}$, $l\nmid c_j$. We assume that we have the exception given in Theorem 8. Let $E=k\left(\sqrt[l^n]{D^{*}}\right)$, and $\mathrm{deg}D=l^{\delta }c$ with $l\nmid c$. Then $e_{\infty }(K|k)=l^{n-\delta }=l^t=l^m=l^{n-a_i-d_i}$, so that $\delta =a_i+d_i$.

Since $m=t>0$, we have $m=n-a_i-min\{n-a_i,d_i\}=n-a_i-d_i$ and $n-a_i-d_i>n-a_j-d_j$ for all $j\ne i$. We also have that $\epsilon :={(-1)}^{\mathrm{deg}D}\gamma \notin {({\mathbb{F}}_q^{*})}^l$.

Lemma 3.

We have $E_{\mathfrak{ge}}=E_{\mathfrak{ge}}^{\mathcal{H}}E$. It also holds that $EK/E$ and $EK/K$ are extensions of constants and $EK=E{\mathbb{F}}_q\left(\sqrt[l^n]{\epsilon }\right)=K{\mathbb{F}}_q\left(\sqrt[l^n]{\epsilon }\right)$. That is,

$$EK=E\left(\sqrt[l^n]{\epsilon }\right)=K\left(\sqrt[l^n]{\epsilon }\right).$$

We also have that $E_{\mathfrak{ge}}K/K_{\mathfrak{ge}}$ and $E_{\mathfrak{ge}}K/E_{\mathfrak{ge}}$ are extensions of constants. Furthermore, $E_{\mathfrak{ge}}K=K_{\mathfrak{ge}}{\mathbb{F}}_q\left(\sqrt[l^n]{\epsilon }\right)=K_{\mathfrak{ge}}\left(\sqrt[l^n]{\epsilon }\right)$ and $E_{\mathfrak{ge}}=E_{\mathfrak{ge}}{\mathbb{F}}_q\left(\sqrt[l^n]{\epsilon }\right)=E_{\mathfrak{ge}}\left(\sqrt[l^n]{\epsilon }\right)$.

Proof. 

The extension $E_{\mathfrak{ge}}/E_{\mathfrak{ge}}^{\mathcal{H}}$ is fully ramified at the infinite prime ${\mathfrak{p}}_{\infty }$. Since $E_{\mathfrak{ge}}^{\mathcal{H}}\subseteq E_{\mathfrak{ge}}^{\mathcal{H}}E\subseteq E_{\mathfrak{ge}}$ and since $e_{\infty }(E|k)=e_{\infty }(E_{\mathfrak{ge}}|k)$, it follows that $E_{\mathfrak{ge}}=E_{\mathfrak{ge}}^{\mathcal{H}}E$.

Now, $EK=k\left(\sqrt[n]{\gamma D}\right)k\left(\sqrt[n]{{(1)}^{\mathrm{deg}D}D}\right)=E\left(\sqrt[l^n]{\epsilon }\right)=K\left(\sqrt[l^n]{\epsilon }\right)$.

We also have $E_{\mathfrak{ge}}K=E_{\mathfrak{ge}}^{\mathcal{H}}EK=E_{\mathfrak{ge}}^{\mathcal{H}}KEK=K_{\mathfrak{ge}}K\left(\sqrt[l^n]{\epsilon }\right)=K_{\mathfrak{ge}}\left(\sqrt[l^n]{\epsilon }\right)$. Therefore

$$\begin{array}{c}{E}_{\mathfrak{ge}}K=E_{\mathfrak{ge}}EK=E_{\mathfrak{ge}}\left(\sqrt[l^n]{\epsilon }\right).\end{array}$$

□

Corollary 4.

The field of constants of $E_{\mathfrak{ge}}K$ is ${\mathbb{F}}_q\left(\sqrt[l^n]{\epsilon }\right)$.

Theorem 9.

In the exceptional case given in Theorem 8, we have that $E_{\mathfrak{ge}}=E_{\mathfrak{gex}}$, the field of constants of $K_{\mathfrak{ge}}$ is ${\mathbb{F}}_{q^{{\mathrm{deg}}_K{\mathfrak{p}}_{\infty }}}$, and the field of constants of $E_{\mathfrak{gex}}K$ is ${\mathbb{F}}_q\left(\sqrt[l^n]{\epsilon }\right)$.

Proof. 

Since $m=t$, we have $E_{\mathfrak{ge}}=E_{\mathfrak{gex}}$. □

Later on, we will see that the field of constants of $K_{\mathfrak{ge}}$ is ${\mathbb{F}}_q\left(\sqrt[l^{\delta }]{\gamma }\right)$.

We fix an infinite prime $\mathfrak{P}$ of K, and we denote $K_{\infty }:=K_{\mathfrak{P}}$.

Since $K_{\infty }=k_{\infty }\left(\sqrt[l^n]{\gamma D}\right)$, $\mathrm{deg}D=d=l^{\delta }c$ with $l\nmid c$, we have

$$\begin{array}{cc}\hfill D(T)& =T^d+a_{d-1}{T}^{d-1}+\cdots +a_{1}T+a_0\hfill \\ \hfill & =T^d\left(1+a_{d-1}\left({\displaystyle \frac{1}{T}}\right)+\cdots a_1{\left({\displaystyle \frac{1}{T}}\right)}^{d-1}+a_0{\left({\displaystyle \frac{1}{T}}\right)}^d\right)=T^d{D}_1(1/T).\hfill \end{array}$$

We have that $D_1(1/T)\in U_{\infty }^{(1)}$, and, since l is different from the characteristic, it follows that ${(U_{\infty }^{(1)})}^{l^n}=U_{\infty }^{(1)}$. Therefore

$$K_{\infty }=k_{\infty }\left(\sqrt[l^n]{\gamma D}\right)=k_{\infty }\left(\sqrt[l^n]{\gamma T^d{D}_1(1/T)}\right)=k_{\infty }\left(\sqrt[l^n]{\gamma T^{l^{\delta }c}}\right).$$

Since $gcd(l,c)=1$, there exists $c_1\in \mathbb{Z}$ such that $c{c}_1\equiv -1mod{l}^n$. Thus

$$K_{\infty }=k_{\infty }\left(\sqrt[l^n]{{\gamma }^{c_1}{T}^{l^{\delta c{c}_1}}}\right)=k_{\infty }\left(\sqrt[l^n]{{\gamma }^{c_1}{(1/T)}^{l^{\delta }}}\right)=k_{\infty }\left(\sqrt[l^n]{{\gamma }^{c_1}{\pi }_{\infty }^{l^{\delta }}}\right).$$

Now, $[K_{\infty }:k_{\infty }]=e_{\infty }(K|k)f_{\infty }(K|k)=l^{n-\delta }{\mathrm{deg}}_K\mathfrak{P}$. Set $K_0=k\left(\sqrt[l^{\delta }]{\gamma D}\right)\subseteq K$. We have that $e_{\infty }(K_0|k)={\displaystyle \frac{l^{\delta }}{gcd(\mathrm{deg}D,l^{\delta })}}={\displaystyle \frac{l^{\delta }}{gcd(l^{\delta }c,l^{\delta })}}={\displaystyle \frac{l^{\delta }}{l^{\delta }}}=1$ and $e_{\infty }(K|K_0)=e_{\infty }(K|k)=l^{n-\delta }=[K:K_0]$. Therefore, ${\mathfrak{p}}_{\infty }$ is fully ramified in $K/K_0$.

We have $f_{\infty }(K|k)=f_{\infty }(K_0|k)=f_{\infty }(K_{0,\infty }|k_{\infty })={\mathrm{deg}}_K\mathfrak{P}$ and

$$K_{0,\infty }=k_{\infty }\left(\sqrt[l^{\delta }]{\gamma D}\right)=k_{\infty }\left(\sqrt[l^{\delta }]{\gamma T^d}\right)=k_{\infty }\left(\sqrt[l^{\delta }]{\gamma T^{l^{\delta }c}}\right)=k_{\infty }\left(\sqrt[l^{\delta }]{\gamma }\right).$$

Lemma 4.

The field of constants of $K_{\mathfrak{ge}}$ are ${\mathbb{F}}_q\left(\sqrt[l^{\delta }]{\gamma }\right)$.

Next, we will prove that the field of constants of $E_{\mathfrak{ge}}K=E_{\mathfrak{gex}}K$ and of $K_{H^{+}}$ are the same.

Let

$$\begin{array}{c}{f}_{\infty }(K|k)={\mathrm{deg}}_K\mathfrak{P}=[K_{0,\infty }:k_{\infty }]=[{\mathbb{F}}_q\left(\sqrt[l^{\delta }]{\gamma }\right):{\mathbb{F}}_q]=:l^{\lambda }.\end{array}$$

We also have

$$\begin{array}{c}\hfill EK=E\left(\sqrt[l^n]{\epsilon }\right)=K\left(\sqrt[l^n]{\epsilon }\right).\end{array}$$

Then, $EK/E$ is an extension of constants and, since ${\mathrm{deg}}_E{\mathfrak{p}}_{\infty }=1$, it follows that

$$\begin{array}{c}\hfill f_{\infty }(EK|k)=f_{\infty }(EK|E)=[EK:E]=[{\mathbb{F}}_q\left(\sqrt[l^n]{\epsilon }\right):{\mathbb{F}}_q]=:l^{\nu }.\end{array}$$

Now, $|\mathcal{H}|=f_{\infty }(EK|K)={\displaystyle \frac{f_{\infty }(EK|k)}{f_{\infty }(K|k)}}={\displaystyle \frac{l^{\nu }}{l^{\lambda }}}=l^{\nu -\lambda }=:l^u$. The field of constants of $K_{\mathfrak{ge}}$ are ${\mathbb{F}}_{q^{l^{\lambda }}}$ and the field of constants of $E_{\mathfrak{ge}}K$ are ${\mathbb{F}}_q\left(\sqrt[l^n]{\epsilon }\right)={\mathbb{F}}_{q^{f_{\infty }(EK|k)}}={\mathbb{F}}_{q^{l^{\nu }}}$. We have ${\mathbb{F}}_q\left(\sqrt[l^{\delta }]{\gamma }\right)={\mathbb{F}}_q\left(\sqrt[l^{\delta }]{{\gamma }^{c_1}}\right)={\mathbb{F}}_{q^{l^{\lambda }}}$. Therefore $[{\mathbb{F}}_q\left(\sqrt[l^{\delta }]{{\gamma }^{c_1}}\right):{\mathbb{F}}_q]=l^{\lambda }$. From Theorem 2, we obtain that ${\gamma }^{c_1}\in {({\mathbb{F}}_q^{*})}^{l^{\delta -\lambda }}\backslash {({\mathbb{F}}_q^{*})}^{l^{\delta -\lambda +1}}$.

Let ${\gamma }^{c_1}={\theta }^{l^{\delta -\lambda }}$, with $\theta \in {\mathbb{F}}_q^{*}$ and $\theta \notin {({\mathbb{F}}_q^{*})}^l$. Then

$$K_{\infty }=k_{\infty }\left(\sqrt[l^n]{{\gamma }^{c_1}{\pi }_{\infty }^{l^{\delta }}}\right)=k_{\infty }\left(\sqrt[l^n]{{\theta }^{l^{\delta -\lambda }}{\pi }_{\infty }^{l^{\delta }}}\right)=k_{\infty }\left(\sqrt[l^{n-\delta +\lambda }]{\theta {\pi }_{\infty }^{l^{\lambda }}}\right).$$

The element $\xi :=\sqrt[l^{n-\delta +\lambda }]{\theta {\pi }_{\infty }^{l^{\lambda }}}$ satisfies ${\xi }^{l^{n-\delta +\lambda }}=\theta {\pi }_{\infty }^{l^{\lambda }}$, that is, $\xi$ is a root of $X^{l^{n-\delta +\lambda }}-\theta {\pi }_{\infty }^{l^{\lambda }}\in k_{\infty }\left[X\right]$. Since $[K_{\infty }:k_{\infty }]=l^{n-\delta +\lambda }$, the polynomial

$$X^{l^{n-\delta +\lambda }}-\theta {\pi }_{\infty }^{l^{\lambda }}$$

is irreducible. We also have $K_{0,\infty }=k_{\infty }\left(\sqrt[l^{\delta }]{\gamma }\right)=k_{\infty }{\mathbb{F}}_{q^{l^{\lambda }}}$ and $K/K_0$ is fully ramified at ${\mathfrak{p}}_{\infty }$.

Set $\tilde{\Pi }:=\xi =\sqrt[l^{n-\delta +\lambda }]{\theta {\pi }_{\infty }^{l^{\lambda }}}$. Then ${\tilde{\Pi }}^{l^{n-\delta +\lambda }}=\theta {\pi }_{\infty }^{l^{\lambda }}$ and

$$\begin{array}{cc}\hfill v_{{\mathfrak{P}}_{\infty }}\left({\tilde{\Pi }}^{l^{n-\delta +\lambda }}\right)& =l^{n-\delta +\lambda }{v}_{{\mathfrak{P}}_{\infty }}(\tilde{\Pi })=e_{\infty }(K_{\infty }|k_{\infty })v_{\infty }\left(\theta {\pi }_{\infty }^{l^{\lambda }}\right)\hfill \\ \hfill & =e_{\infty }(K|k)l^{\lambda }=l^{n-\delta }{l}^{\lambda }=l^{n-\delta +\lambda }.\hfill \end{array}$$

Hence, $v_{{\mathfrak{P}}_{\infty }}(\tilde{\Pi })=1$ and $\tilde{\Pi }$ is a prime element of $K_{\infty }$. We also have

$${\mathrm{deg}}_{K_{\infty }}\tilde{\Pi }={\mathrm{deg}}_{K_{\infty }}{\mathfrak{P}}_{\infty }{v}_{{\mathfrak{P}}_{\infty }}(\tilde{\Pi })={\mathrm{deg}}_K{\mathfrak{p}}_{\infty }·1=l^{\lambda }.$$

Now, $\theta \notin {({\mathbb{F}}_q^{*})}^l$. Let ${\zeta }_{l^{n-\delta +\lambda }}$ be a primitive $l^{n-\delta +\lambda }$th root of unity and set $N_{\infty }:=N_{K_{\infty }|k_{\infty }}$. We have

$$\begin{array}{cc}\hfill Irr(\tilde{\Pi },X,k_{\infty })& =X^{l^{n-\delta +\lambda }}-\theta {\pi }_{\infty }^{l^{\lambda }}=\prod _{j=0}^{l^{n-\delta +\lambda }-1}\left(X-{\zeta }_{l^{n-\delta +\lambda }}^j\tilde{\Pi }\right).\hfill \end{array}$$

Thus

$$\begin{array}{cc}\hfill N_{\infty }\tilde{\Pi }& =\prod _{j=0}^{l^{n-\delta +\lambda }-1}\left({\zeta }_{l^{n-\delta +\lambda }}^j\tilde{\Pi }\right)={(-1)}^{l^{n-\delta +\lambda }}\prod _{j=0}^{l^{n-\delta +\lambda }-1}\left(-{\zeta }_{l^{n-\delta +\lambda }}^j\tilde{\Pi }\right)\hfill \end{array}$$

$$\begin{array}{cc}\hfill & ={(-1)}^{l^{n-\delta +\lambda }}\left(-\theta {\pi }_{\infty }^{l^{\lambda }}\right)={(-1)}^{l^{n-\delta +\lambda }+1}\theta {\pi }_{\infty }^{l^{\lambda }}.\hfill \end{array}$$

Now we consider a generic element $Y\in K_{\infty }^{*}$:

$$\begin{array}{c}Y={\tilde{\Pi }}^s\Lambda w,\mathrm{with}s\in \mathbb{Z},\Lambda \in {\mathbb{F}}_{q^{l^{\lambda }}},\mathrm{and}w\in U_{K_{\infty }}^{(1)}.\end{array}$$

Then

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill N_{\infty }{\tilde{\Pi }}^s& ={\left(N_{\infty }\tilde{\Pi }\right)}^s={(-1)}^{(l^{n-\delta +\lambda }+1)s}{\theta }^s{\pi }_{\infty }^{l^{\lambda }s},\hfill \\ \hfill N_{\infty }\Lambda & =N_{K_{0,\infty }|k_{\infty }}\left(N_{K_{\infty }|K_{0,\infty }}\Lambda \right)=N_{K_{0,\infty }|k_{\infty }}({\Lambda }^{l^{l^{n-\delta }}})={\left(N_{K_{0,\infty }|k_{\infty }}\Lambda \right)}^{l^{n-\delta }},\hfill \\ \hfill N_{\infty }w& =v\in U_{\infty }^{(1)}.\hfill \end{array}\end{array}$$

It follows that

$$\begin{array}{cc}\hfill {\varphi }_{{\mathfrak{P}}_{\infty }}(Y)& ={\varphi }_{\infty }(N_{\infty }(Y))={\varphi }_{\infty }\left({(-1)}^{(l^{n-\delta +\lambda }+1)s}{\theta }^s{\pi }_{\infty }^{l^{\lambda }s}{(N_{K_{0,\infty }|k_{\infty }}\Lambda )}^{l^{n-\delta }}v\right)\hfill \\ \hfill & ={(-1)}^{(l^{n-\delta +\lambda }+1)s}{\theta }^s{\left(N_{K_{0,\infty }|k_{\infty }}\Lambda \right)}^{l^{n-\delta }}\hfill \\ \hfill & ={(-\theta )}^s{\left[{(-1)}^{l^{\lambda }}\left(N_{K_{0,\infty }|k_{\infty }}\Lambda \right)\right]}^{l^{n-\delta }}.\hfill \end{array}$$

Therefore, $Y\in ker{\varphi }_{{\mathfrak{P}}_{\infty }}\iff \mathrm{there}\mathrm{exists}\Lambda \in {\mathbb{F}}_{q^{l^{\lambda }}}$ such that

$${(-\theta )}^s{\left[{(-1)}^{l^{\lambda }}\left(N_{K_{0,\infty }|k_{\infty }}\Lambda \right)\right]}^{l^{n-\delta }}=1.$$

Now, $N_{K_{0,\infty }|k_{\infty }}{\mathbb{F}}_{q^{l^{\lambda }}}={\mathbb{F}}_q$, thus $N_{\infty }{\mathbb{F}}_{q^{l^{\lambda }}}^{*}={\left({\mathbb{F}}_q^{*}\right)}^{l^{n-\delta }}$ and $\Lambda \in {\mathbb{F}}_{q^{l^{\lambda }}}$. Hence, $-{\theta }^s\in {({\mathbb{F}}_q^{*})}^{l^{n-\delta }}$.

Next, we break our study in two cases: $n=1$ and $n\ge 2$.

5.3.4. Case $n=1$

This case was considered in Theorem 3.5 [6].

5.3.5. Case $n\ge 2$

We now assume that $n>1$. We always have, since $\theta \notin {({\mathbb{F}}_q^{*})}^l$, that $-\theta \notin {({\mathbb{F}}_q^{*})}^l$ because $n\ge 2$ and therefore $-1\in {({\mathbb{F}}_q^{*})}^l$ (see Remark 4). Hence, $-{\theta }^s\in {({\mathbb{F}}_q^{*})}^{l^{n-\delta }}\iff l^{n-\delta }|s$. That is, $ker{\varphi }_{{\mathfrak{P}}_{\infty }}=\{Y={\tilde{\Pi }}^s\Lambda w\mid l^{n-\delta }|s\}$.

Because $\mathrm{deg}Y=\mathrm{deg}\left({\tilde{\Pi }}^s\Lambda w\right)=\mathrm{deg}\tilde{\Pi }·v_{{\mathfrak{P}}_{\infty }}(Y)=l^{\lambda }·s$, it follows that

$$min\{\kappa \in \mathbb{N}\mid \mathrm{there}\mathrm{exists}\tilde{\overrightarrow{\alpha }}\in B\mathrm{and}\mathrm{deg}\tilde{\overrightarrow{\alpha }}=\kappa \}=l^{n-\delta +\lambda },$$

and that the field of constants of $K_{H^{+}}$ is ${\mathbb{F}}_{q^{l^{n-\delta +\lambda }}}$.

We have that $l^{\lambda }=[{\mathbb{F}}_q\left(\sqrt[l^{\delta }]{\gamma }\right):{\mathbb{F}}_q]$, so that, from Theorem 2, we obtain

$$\begin{array}{c}\gamma \in {({\mathbb{F}}_q^{*})}^{l^{\delta -\lambda }}\backslash {({\mathbb{F}}_q^{*})}^{l^{\delta -\lambda +1}}.\end{array}$$

(6)

On the other hand, the field of constants of $K_{\mathfrak{ge}}E$ is ${\mathbb{F}}_q\left(\sqrt[l^n]{\epsilon }\right)$ and $[{\mathbb{F}}_q\left(\sqrt[l^n]{\epsilon }\right):{\mathbb{F}}_q]=l^{\nu }$. Again, from Theorem 2, we obtain that

$$\epsilon \in {({\mathbb{F}}_q^{*})}^{l^{n-\nu }}\backslash {({\mathbb{F}}_q^{*})}^{l^{n-\nu +1}}.$$

We have $u=\nu -\lambda \ge 1$ so that $\nu \ge \lambda +1$ and $n-\nu \le n-1$. Therefore, ${(-1)}^{\mathrm{deg}D}=\pm 1\in {({\mathbb{F}}_q^{*})}^{l^{n-\nu }}$. Hence, $\gamma ={(-1)}^{\mathrm{deg}D}\epsilon \in {({\mathbb{F}}_q^{*})}^{l^{n-\nu }}$. It follows from (6) that $n-\nu \le \delta -\lambda$ and $n-\delta \le \nu -\lambda$.

Thus, $e_{\infty }(K|k)=l^{n-\delta }|l^{\nu -\lambda }=l^u=|\mathcal{H}|$. Since $|\mathcal{H}||e_{\infty }(K|k)=e_{\infty }(E|k)$, $l^u|l^{n-\delta }$. Therefore, $\nu -\lambda \le n-\delta$, so that $\nu -\lambda =n-\delta$ and $\nu =n-\delta +\lambda$. In particular, $|\mathcal{H}|=l^u=l^{n-\delta }=e_{\infty }(K|k)$.

It follows that the field of constants of $E_{\mathfrak{ge}}K=E_{\mathfrak{gex}}K$ is ${\mathbb{F}}_q\left(\sqrt[l^n]{\epsilon }\right)={\mathbb{F}}_{q^{l^{\nu }}}={\mathbb{F}}_{q^{l^{n-\delta +\lambda }}}$. In short, the field of constants of both $K_{H^{+}}$ and $E_{\mathfrak{ge}}K$, is ${\mathbb{F}}_{q^{l^{n-\delta +\lambda }}}={\mathbb{F}}_{q^{l^{u+\lambda }}}$.

Thus, $E_{\mathfrak{ge}}K\subseteq K_{H^{+}}$ and $E_{\mathfrak{ge}}K\subseteq K_{\mathfrak{gex}}\subseteq E_{\mathfrak{gex}}K=E_{\mathfrak{ge}}K$. Therefore, $E_{\mathfrak{gex}}K\subseteq K_{H^{+}}$. Since $K_{\mathfrak{gex}}\subseteq E_{\mathfrak{gex}}K$, we finally obtain that $K_{\mathfrak{gex}}=E_{\mathfrak{gex}}K$.

Theorem 10.

For Kummer extensions satisfying the special case described in Theorem 8, we have $K_{\mathfrak{gex}}=E_{\mathfrak{gex}}K$.

Corollary 5.

For any cyclic Kummer extension $k\left(\sqrt[l^n]{\gamma D}\right)$ of k, we have $K_{\mathfrak{gex}}=E_{\mathfrak{gex}}K$.

5.4. Semi-Kummer Case: $l^{\rho }|q-1$, $\rho \ge 1$ and $l^n\nmid q-1$

Recall that we only need to consider the case ${(E_{\mathfrak{ge}}^{\mathcal{H}})}_{\mathfrak{gex}}\ne E_{\mathfrak{gex}}$, or, equivalently, there exists a finite prime $P_j$ ramified in $E_{\mathfrak{gex}}/E_{\mathfrak{ge}}^{\mathcal{H}}$. The prime ${\mathfrak{p}}_{\infty }$ is fully ramified in $E_{\mathfrak{gex}}/E_{\mathfrak{ge}}^{\mathcal{H}}$.

We have that $\mathcal{H}\subseteq I_{\infty }(E_{\mathfrak{ge}}/k)$, the inertia group of ${\mathfrak{p}}_{\infty }$ in the extension $E_{\mathfrak{ge}}/k$.

It follows that $Gal(E_{\mathfrak{gex}}/E_{\mathfrak{ge}}^{\mathcal{H}})\subseteq I_{\infty }(E_{\mathfrak{gex}}/k)$, and we have that $I_{\infty }(E_{\mathfrak{gex}}/k)$ is a cyclic group of order a power of l. Set $I:=I_{\infty }(E_{\mathfrak{gex}}/k)$, an l–cyclic group. Furthermore $|I|=e_{\infty }(E_{\mathfrak{gex}}|k)|q-1$.

Using Proposition 1, we obtain:

Proposition 5.

$E_{\mathfrak{gex}}=E_{\mathfrak{ge}}$.

Proof. 

Let $G:=I=I_{\infty }(E_{\mathfrak{gex}}/k)$, $H_1:=I_{P_i}(E_{\mathfrak{gex}}/E_{\mathfrak{ge}}^{\mathcal{H}})$ and $H_2:=Gal(E_{\mathfrak{gex}}/E_{\mathfrak{ge}})$. By hypothesis, we have $H_1\ne \{Id\}$. Set $\Phi :=H_1\cap H_2$ and $F:=E_{\mathfrak{gex}}^{\Phi }$.

Then, $P_i$ is fully ramified in $E_{\mathfrak{gex}}/F$ and $E_{\mathfrak{ge}}\subseteq F$. Therefore $P_i$ is fully ramified and non-ramified in $E_{\mathfrak{gex}}/F$. Hence $F=E_{\mathfrak{gex}}$ and $H_1\cap H_2=\{Id\}$. Finally, it follows that $H_2=\{Id\}$ and that $E_{\mathfrak{gex}}=E_{\mathfrak{ge}}$. □

Lemma 5.

We have that $f_{\infty }(K|k)={\mathrm{deg}}_K{\mathfrak{p}}_{\infty }$.

Proof. 

Since the extension $K/k$ is geometric, and ${\mathrm{deg}}_k{\mathfrak{p}}_{\infty }=1$, we have $f_{\infty }(K|k)=f_{\infty }(K|k){\mathrm{deg}}_k{\mathfrak{p}}_{\infty }={\mathrm{deg}}_K{\mathfrak{p}}_{\infty }$. □

We use the following notation. Let $|\mathcal{H}|:=l^u=f_{\infty }(EK|K)$. Since $\mathcal{H}$ is a quotient of I, it follows that $l^u|l^{\rho }$ because $|I|=e_{\infty }(E_{\mathfrak{gex}}|k)|q^{{\mathrm{deg}}_K{\mathfrak{p}}_{\infty }}-1=q-1$. In particular, $u\le \rho$.

Set $l^{\lambda }:={\mathrm{deg}}_K{\mathfrak{p}}_{\infty }=f_{\infty }(K|k)$. We have

Since e_P(K|k) = e_P(E|k) for all P ∈ $R_T^{+}$ ∪ {∞}, from Abhyankar’s Lemma, we obtain that e_P(EK|E) = 1 for all P ∈ $R_T^{+}$ ∪ {∞}, that is, EK/E is an unramified extension. We will show that it is an extension of constants.

It is easy to see that $[K:k]=[E:k]$. We have, on the one hand

$$\begin{array}{c}[K:E\cap K]={\displaystyle \frac{[K:k]}{[E\cap K:k]}}={\displaystyle \frac{[E:k]}{[E\cap K:k]}}=[E:E\cap K].\end{array}$$

On the other hand

$$\begin{array}{c}\hfill [EK:E]=[K:E\cap K]\mathrm{and}[EK:K]=[E:E\cap K].\end{array}$$

Therefore

$$\begin{array}{c}\hfill [EK:K]=[EK:E]=[E:E\cap K]=[K:E\cap K].\end{array}$$

Because $EK/E$ and $EK/K$ are unramified extensions,

$$e_P(E|E\cap K)=e_P(K|E\cap K)\mathrm{for}\mathrm{all}P\in R_T^{+}\cup \{\infty \}.$$

We have that $E_{\mathfrak{ge}}K/K_{\mathfrak{ge}}=E_{\mathfrak{ge}}^{\mathcal{H}}K$ is an extension of constants of degree $|\mathcal{H}|=l^u=[E_{\mathfrak{ge}}K:K_{\mathfrak{ge}}]$. We also have that the field of constants of $K_{\mathfrak{ge}}$ is ${\mathbb{F}}_{q^{{\mathrm{deg}}_K{\mathfrak{p}}_{\infty }}}$. Hence, the field of constants of $E_{\mathfrak{ge}}K=E_{\mathfrak{gex}}K$ is ${\mathbb{F}}_{q^{\psi }}$, where $\psi ={\mathrm{deg}}_K{\mathfrak{p}}_{\infty }·|\mathcal{H}|=l^{\lambda +u}=f_{\infty }(EK|k)$.

Let us see that the field of constants of $EK$ is also ${\mathbb{F}}_{q^{\psi }}={\mathbb{F}}_{q^{l^{\lambda +u}}}$, the same as that of $E_{\mathfrak{ge}}K$.

Since $K/k$ is tamely ramified, the conductor of constants is the minimum $\eta$ such that $K\subseteq k{({\Lambda }_N)}_{\eta }$. We have that $\eta =td$, where $t=f_{\infty }(K|k)=f_{\infty }(K|J)={\mathrm{deg}}_K{\mathfrak{p}}_{\infty }=l^{\lambda }$, $d=f_{\infty }(EK|K)=f_{\infty }(E_{\mathfrak{ge}}K|K_{\mathfrak{ge}})=|\mathcal{H}|=l^u$, and $J=K\cap {}_{n}k({\Lambda }_N)=K\cap k({\Lambda }_N)=K\cap E$. Therefore

$$\begin{array}{c}\eta =l^{\lambda }·l^u=l^{\lambda +u}=\psi .\end{array}$$

Furthermore, we have

$$\begin{array}{c}\hfill \eta =[K:J]=[K:K\cap E](=[E:K\cap E]=[EK:E]=[EK:K]).\end{array}$$

We have

If $EK\cap k_{\eta }=k_{\sigma }\subseteq k_{\eta }$, then $K\subseteq EK=k_{\sigma }E\subseteq k_{\sigma }k({\Lambda }_N)=k{({\Lambda }_N)}_{\sigma }$. Since $\eta$ is minimum, it follows that $\eta =\sigma$, and $EK={(EK)}_{\eta }=E_{\eta }=K_{\eta }$. Therefore, the field of constants of $EK$ is ${\mathbb{F}}_{q^{\psi }}={\mathbb{F}}_{q^{l^{\lambda +u}}}$.

Proposition 6.

The field of constants of either $E_{\mathfrak{ge}}K$ or $EK$ is ${\mathbb{F}}_{q^{l^{\lambda +u}}}$, where $l^{\lambda }={\mathrm{deg}}_K{\mathfrak{p}}_{\infty }$ and $l^u=|\mathcal{H}|$.

Furthermore, $[E_{\eta }:E]=\eta =\psi =l^{\lambda +u}=[K:J]=[K:E\cap K]=[E:E\cap K]=[EK:K]=[EK:E]$.

We have that $EK/E$ is an extension of constants of degree $\eta$, and the same is true for the extension $EK/K$. The field of constants of $E^{\mathcal{H}}K$ is ${\mathbb{F}}_{q^{l^{\lambda }}}$, that is, the same as for the field $K_{\mathfrak{ge}}=E_{\mathfrak{ge}}^{\mathcal{H}}K$.

Let ${\mathfrak{P}}_{\infty }$ be a prime above ${\mathfrak{p}}_{\infty }$ and denote $K_{\infty }:=K_{{\mathfrak{P}}_{\infty }}$, $k_{\infty }:=k_{{\mathfrak{p}}_{\infty }}$. Let $e_{{\mathfrak{p}}_{\infty }}(K|k)=l^{\tau }|q-1$, that is, $l^{\tau }|l^{\rho }$ and $\tau \le \rho$. We have

$$[K_{\infty }:k_{\infty }]=e_{\infty }(K|k)f_{\infty }(K|k)=l^{\tau }·l^{\lambda }=l^{\tau +\lambda }.$$

Let $k_{\infty }\subseteq F\subseteq K_{\infty }$ be the inertia field of $K_{\infty }/k_{\infty }$, that is, $F:=K_{\infty }^{I_{\infty }(K_{\infty }/k_{\infty })}$ and $[F:k_{\infty }]=l^{\lambda }$. We have that $F/k_{\infty }$ is unramified.

For each local field, there exists a unique unramified extension of each degree. Therefore, $F=k_{\infty }{\mathbb{F}}_{q^{l^{\lambda }}}$, that is, $F/k_{\infty }$ is an “extension of constants” of degree $l^{\lambda }$. More precisely,

$${\mathcal{O}}_{{\mathfrak{P}}_{\infty }}/{\mathfrak{P}}_{\infty }\cong {\mathbb{F}}_{q^{l^{\lambda }}}\mathrm{and}{F}^{*}=\langle {\pi }_F\rangle \times {\mathbb{F}}_{q^{l^{\lambda }}}^{*}\times U_F^{(1)},$$

where ${\pi }_F={\pi }_{\infty }=1/T$ is a prime element of $F^{*}$.

Let $K_{\infty }^{*}=\langle \tilde{\Pi }\rangle \times {\mathbb{F}}_{q^{l^{\lambda }}}^{*}\times U_{K_{\infty }}^{(1)}$ where $\tilde{\Pi }$ is a prime element of $K_{\infty }^{*}$. Now, we have

$$1=v_{K_{\infty }}(\tilde{\Pi })={\displaystyle \frac{e_{\infty }(K|k)}{[K_{\infty }:k_{\infty }]}}{v}_{k_{\infty }}(N_{K_{\infty }/k_{\infty }}(\tilde{\Pi }))={\displaystyle \frac{1}{f_{\infty }(K|k)}}{v}_{k_{\infty }}(N_{K_{\infty }/k_{\infty }}(\tilde{\Pi })).$$

Hence $v_{k_{\infty }}(N_{K_{\infty }/k_{\infty }}(\tilde{\Pi }))=l^{\lambda }$.

We have

It follows that $K_{\infty }/F$ is a Kummer extension, say $K_{\infty }=F(\sqrt[l^{\tau }]{Y})$ for some $Y\in F^{*}=\langle {\pi }_{\infty }\rangle \times {\mathbb{F}}_{q^{l^{\lambda }}}^{*}\times U_F^{(1)}$.

Let $Y={\pi }_{\infty }^s\Lambda w$, with $s\in \mathbb{Z}$, $\Lambda \in {\mathbb{F}}_{q^{l^{\lambda }}}^{*}$, and $w\in U_F^{(1)}$. Since $gcd(l,p)=1$, we have $U_F^{(1)}={(U_F^{(1)})}^{l^{\tau }}$. We write $s=\alpha l^{\tau }+r$, where $0\le r<l^{\tau }$. Then, if $w_0^{l^{\tau }}=w$, we have

$$K_{\infty }=F\left(\sqrt[l^{\tau }]{{\pi }_{\infty }^{\alpha l^{\tau }+r}\Lambda w_0^{l^{\tau }}}\right)=F\left(\sqrt[l^{\tau }]{{\pi }_{\infty }^r\Lambda }\right).$$

Let $r=l^b{r}_0$, with $0\le b<\tau$ and $gcd(l,r_0)=1$. Set $F_1:=F\left(\sqrt[l^b]{{\pi }_{\infty }^{l^b{r}_0}}\Lambda \right)=F\left(\sqrt[l^b]{\Lambda }\right)$. Thus $F_1/F$ is unramified, $F\subseteq F_1\subseteq K_{\infty }$, and $K_{\infty }/F$ is totally ramified. It follows that $F_1=F$ and that $b=0$, that is, $gcd(r,l)=1$.

Therefore, $K_{\infty }=F\left(\sqrt[l^{\tau }]{{\pi }_{\infty }\theta }\right)$ for some $\theta \in {\mathbb{F}}_{q^{l^{\lambda }}}^{*}$. Set $\varphi :=\sqrt[l^{\tau }]{{\pi }_{\infty }\theta }$. Then, ${\varphi }^{l^{\tau }}={\pi }_{\infty }\theta$. Hence

$$l^{\tau }{v}_{K_{\infty }}(\varphi )=v_{K_{\infty }}({\varphi }^{l^{\tau }})=v_{K_{\infty }}({\pi }_{\infty }\theta )=v_{K_{\infty }}({\pi }_{\infty })=e(K_{\infty }|F)v_{\infty }({\pi }_{\infty })=l^{\tau }·1.$$

It follows that $v_{K_{\infty }}(\varphi )=1$. Therefore, we may take $\varphi =\tilde{\Pi }$ as a prime element of F.

Now we consider $E_{\infty }=k_{\infty }\left(\sqrt[l^{\tau }]{{\pi }_{\infty }\mu }\right)$ for some $\mu \in {\mathbb{F}}_q^{*}$. We have

Because $E_{\mathfrak{gex}}\subseteq k({\Lambda }_N)$ is cyclotomic, the field of constants of $E_{H^{+}}$ is also ${\mathbb{F}}_q$.

As before, $\vartheta :=\sqrt[l^{\tau }]{{\pi }_{\infty }\mu }$ is a prime element of $E_{\infty }$. We have

$$\begin{array}{c}{X}^{l^{\tau }}-{\pi }_{\infty }\mu =\prod _{i=0}^{l^{\tau }-1}\left(X-{\zeta }_{l^{\tau }}^i\vartheta \right).\end{array}$$

Hence

$$\begin{array}{c}\hfill \prod _{i=0}^{l^{\tau }-1}\left(-{\zeta }_{l^{\tau }}^i\vartheta \right)={(-1)}^{l^{\tau }}\prod _{i=0}^{l^{\tau }-1}\left({\zeta }_{l^{\tau }}^i\vartheta \right)={(-1)}^{l^{\tau }}{N}_{E_{\infty }/k_{\infty }}\vartheta =-{\pi }_{\infty }\mu .\end{array}$$

Thus

$$\begin{array}{c}\hfill N_{E_{\infty }/k_{\infty }}\vartheta ={(-1)}^{l^{\tau }+1}\mu {\pi }_{\infty }.\end{array}$$

Since the field of constants of $E_{H^{+}}$ is ${\mathbb{F}}_q$, from Theorem 3, there exists an element of degree 1 in $E_{\infty }^{*}$ satisfying

$${\varphi }_{E_{\infty }}(X)={\varphi }_{\infty }(N_{E_{\infty }/k_{\infty }}(X))=1.$$

Set $X={\vartheta }^s\alpha w$ with $s\in \mathbb{Z},\alpha \in {\mathbb{F}}_q^{*},w\in U_{E_{\infty }}^{(1)}$. Since $\mathrm{deg}X=\mathrm{deg}{\pi }_{\infty }{v}_{\infty }(X)=1·s=s$, it follows that $s=1$. Furthermore, $N_{E_{\infty }/k_{\infty }}(w)\in U_{\infty }^{(1)}={\left(U_{\infty }^{(1)}\right)}^{l^{\tau }}$ and $N_{E_{\infty }/k_{\infty }}(\alpha )={\alpha }^{l^{\tau }}$. Therefore

$$1={\varphi }_{E_{\infty }}(X)={\varphi }_{\infty }(N_{E_{\infty }/k_{\infty }}(X))={\varphi }_{\infty }\left(\left({(-1)}^{l^{\tau }+1}\mu {\pi }_{\infty }\right){\alpha }^{l^{\tau }}{u}^{l^{\tau }}\right)={(-1)}^{l^{\tau }+1}\mu {\alpha }^{l^{\tau }},$$

where $u\in U_{\infty }^{(1)}$. Therefore, $-\mu ={(-\alpha )}^{-l^{\tau }}\in {\left({\mathbb{F}}_q^{*}\right)}^{l^{\tau }}$ and, since $\tau <n$, it follows that $-1\in {\left({\mathbb{F}}_q^{*}\right)}^{l^{\tau }}$. Thus $\mu \in {\left({\mathbb{F}}_q^{*}\right)}^{l^{\tau }}$ and $E_{\infty }=k_{\infty }\left(\sqrt[l^{\tau }]{{\pi }_{\infty }\mu }\right)=k_{\infty }\left(\sqrt[l^{\tau }]{{\pi }_{\infty }}\right)$.

We obtain

$$\begin{array}{c}{E}_{\infty }{K}_{\infty }=K_{\infty }{k}_{\infty }\left(\sqrt[l^{\tau }]{{\pi }_{\infty }}\right)=F\left(\sqrt[l^{\tau }]{{\pi }_{\infty }\theta },\sqrt[l^{\tau }]{{\pi }_{\infty }}\right)=F\left(\sqrt[l^{\tau }]{{\pi }_{\infty }\theta },\sqrt[l^{\tau }]{\theta }\right)=K_{\infty }\left(\sqrt[l^{\tau }]{\theta }\right).\end{array}$$

The field of constants of $EK$ is ${\mathbb{F}}_{q^{l^{\lambda +u}}}$, therefore, $[{\mathbb{F}}_{q^{l^{\lambda }}}\left(\sqrt[l^{\tau }]{\theta }\right):{\mathbb{F}}_{q^{l^{\lambda }}}]=l^u$. From Theorem 2 we have that $\theta \in {\left({\mathbb{F}}_{q^{l^{\lambda }}}^{*}\right)}^{l^{\tau -u}}\backslash {\left({\mathbb{F}}_{q^{l^{\lambda }}}^{*}\right)}^{l^{\tau -u+1}}$.

In short, $K_{\infty }=F\left(\sqrt[l^{\tau }]{{\pi }_{\infty }\theta }\right)$ with $\theta \in {\left({\mathbb{F}}_{q^{l^{\lambda }}}^{*}\right)}^{l^{\tau -u}}\backslash {\left({\mathbb{F}}_{q^{l^{\lambda }}}^{*}\right)}^{l^{\tau -u+1}}$, $F=k_{\infty }{\mathbb{F}}_{q^{l^{\lambda }}}$, and $\tilde{\Pi }=\sqrt[l^{\tau }]{\theta {\pi }_{\infty }}$.

The irreducible polynomial of $\tilde{\Pi }$ over F is $X^{l^{\tau }}-\theta {\pi }_{\infty }\in F\left[X\right]$. Then $X^{l^{\tau }}-\theta {\pi }_{\infty }={\prod }_{j=0}^{l^{\tau }-1}(X-{\zeta }_{l^{\tau }}^j\tilde{\Pi })$ and

$$\begin{array}{cc}\hfill N_{K_{\infty }/F}\tilde{\Pi }& =\prod _{j=0}^{l^{\tau }-1}{\zeta }_{l^{\tau }}^j\tilde{\Pi }={(-1)}^{l^{\tau }}\prod _{j=0}^{l^{\tau }-1}(-{\zeta }_{l^{\tau }}^j\tilde{\Pi })={(-1)}^{l^{\tau }}(-\theta {\pi }_{{\pi }_{\infty }})={(-1)}^{l^{\tau }+1}\theta {\pi }_{\infty },\hfill \\ \hfill N_{K_{\infty }/k_{\infty }}\tilde{\Pi }& =N_{F/k_{\infty }}(N_{K_{\infty }/F}\tilde{\Pi })=N_{F/k_{\infty }}({(-1)}^{l^{\tau }+1}\theta {\pi }_{\infty })\hfill \\ \hfill & ={(-1)}^{(l^{\tau }+1)l^{\lambda }}(N_{F/k_{\infty }}\theta ){\pi }_{\infty }^{l^{\lambda }}.\hfill \end{array}$$

Now, $N_{F/k_{\infty }}{\mathbb{F}}_{q^{l^{\lambda }}}^{*}={\mathbb{F}}_q^{*}$. Therefore, $N_{F/k_{\infty }}\theta \in {\left({\mathbb{F}}_q^{*}\right)}^{l^{\tau -u}}\backslash {\left({\mathbb{F}}_q^{*}\right)}^{l^{\tau -u+1}}$.

Let us see the norm of an arbitrary element X of $K_{\infty }^{*}$. Let $X={\tilde{\Pi }}^s\Lambda \omega$ with $s\in \mathbb{Z}$, $\Lambda \in {\mathbb{F}}_{q^{l^{\lambda }}}^{*}$, and $\omega \in U_{K_{\infty }}^{(1)}$. Then

$$\begin{array}{c}\begin{array}{cc}\hfill N_{K_{\infty }/k_{\infty }}\omega & ={\omega }_0\in U_{\infty }^{(1)},\hfill \\ \hfill N_{K_{\infty }/k_{\infty }}{\tilde{\Pi }}^s& ={(-1)}^{(l^{\tau }+1)l^{\lambda }s}{\xi }^s{\pi }_{\infty }^{l^{\lambda }s}\mathrm{with}\xi =N_{F/k_{\infty }}\theta \in {\left({\mathbb{F}}_q^{*}\right)}^{l^{\tau -u}}\backslash {\left({\mathbb{F}}_q^{*}\right)}^{l^{\tau -u+1}},\hfill \\ \hfill N_{K_{\infty }/k_{\infty }}\Lambda & =N_{F/k_{\infty }}(N_{K_{\infty }/F}\Lambda )=N_{F/k_{\infty }}{\Lambda }^{l^{\tau }}={(N_{F/k_{\infty }}\Lambda )}^{l^{\tau }},\hfill \\ \hfill & \mathrm{and}{N}_{K_{\infty }/k_{\infty }}{\mathbb{F}}_{q^{l^{\lambda }}}^{*}={({\mathbb{F}}_q^{*})}^{l^{\tau }}.\hfill \end{array}\end{array}$$

Therefore

$$\begin{array}{c}\hfill N_{K_{\infty }/k_{\infty }}X={(-1)}^{(l^{\tau }+1)l^{\lambda }s}{\xi }^s{\pi }_{\infty }^{l^{\lambda }s}{\left(N_{F/k_{\infty }}\Lambda \right)}^{l^{\tau }}{\omega }_0,\end{array}$$

and

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill {\varphi }_{K_{\infty }}(X)& ={\varphi }_{\infty }(N_{K_{\infty }/k_{\infty }}(X))={(-1)}^{(l^{\tau }+1)l^{\lambda }s}{\xi }^s{\left(N_{F/k_{\infty }}\Lambda \right)}^{l^{\tau }}\hfill \\ \hfill & ={\left({(-1)}^{l^{\lambda }}\xi \right)}^s{\left[{(-1)}^{l^{\lambda }s}(N_{F/k_{\infty }}\Lambda )\right]}^{l^{\tau }}.\hfill \end{array}\end{array}$$

Now

$$\begin{array}{c}\hfill X\in ker{\varphi }_{K_{\infty }}\iff {\left({(-1)}^{l^{\lambda }}\xi \right)}^s{\left[{(-1)}^{l^{\lambda }s}(N_{F/k_{\infty }}\Lambda )\right]}^{l^{\tau }}=1.\end{array}$$

In other words,

$$X\in ker{\varphi }_{K_{\infty }}\iff \mathrm{there}\mathrm{exists}\Lambda \in {\mathbb{F}}_{q^{l^{\lambda }}}^{*}\mathrm{such}\mathrm{that}{\left(N_{F/k_{\infty }}\Lambda \right)}^{l^{\tau }}={\left(\pm {\xi }^{-1}\right)}^s.$$

Since $l^{\rho }|q-1$ with $\rho \ge 1$ and $l^n\nmid q-1$, we have $n\ge 2$. Thus $-1\in {({\mathbb{F}}_q^{*})}^{l^{\tau -u+1}}$. Therefore, ${\xi }^s\in {({\mathbb{F}}_q^{*})}^{l^{\tau }}$. Since $\xi \in {\left({\mathbb{F}}_q^{*}\right)}^{l^{\tau -u}}\backslash {\left({\mathbb{F}}_q^{*}\right)}^{l^{\tau -u+1}}$, it follows that $l^u|s$ and that $l^u$ is the minimum positive integer with this property. For such X, we have

$${\mathrm{deg}}_{K_{\infty }}X={\mathrm{deg}}_{K_{\infty }}{\mathfrak{P}}_{\infty }{v}_{K_{\infty }}(X)=l^{\lambda }·l^u=l^{\lambda +u}.$$

Therefore, the field of constants of $K_{H^{+}}$ is ${\mathbb{F}}_{l^{\lambda +u}}$, the same as for $E_{\mathfrak{gex}}K$.

We have obtained our first main result:

Theorem 11.

Let $K/k$ be a geometric cyclic extension of degree $l^n$ with l a prime number and $n\ge 1$. Then, if E is given by(1), we have

$$\begin{array}{c}{K}_{\mathfrak{gex}}=E_{\mathfrak{gex}}K.\end{array}$$

6. General Finite Abelian Extensions

The following theorem is the main key to obtain the extended genus field of a finite abelian extension.

Lemma 6.

Let $E_1$ and $E_2$ be two finite cyclotomic extensions of k, and let $E=E_1{E}_2$. Then $E_{\mathfrak{gex}}={(E_1)}_{\mathfrak{gex}}{(E_2)}_{\mathfrak{gex}}$.

As a consequence, we obtain our final main result.

Theorem 12.

Let $K/k$ be any geometric finite abelian extension. Then, if E is given in (1) , we have that

$$K_{\mathfrak{gex}}=E_{\mathfrak{gex}}K.$$

Proof. 

Let $K=K_1\cdots K_s$, where each $K_j/k$ is a cyclic extension of prime power degree. Let $E=E_1\cdots E_s$ with each $E_j$ given in (1). Then, from Lemma 6, we obtain

$$E_{\mathfrak{gex}}={(E_1)}_{\mathfrak{gex}}\cdots {(E_s)}_{\mathfrak{gex}}.$$

Therefore, from Theorem 11, it follows that

$$K_{\mathfrak{gex}}\subseteq E_{\mathfrak{gex}}K={(E_1)}_{\mathfrak{gex}}{K}_1\cdots {(E_s)}_{\mathfrak{gex}}{K}_s={(K_1)}_{\mathfrak{gex}}\cdots {(K_s)}_{\mathfrak{gex}}\subseteq K_{\mathfrak{gex}}.$$

Hence $K_{\mathfrak{gex}}=E_{\mathfrak{gex}}K$. □

We obtain some consequences from Theorem 12. We consider $K/k$ a geometric abelian finite extension. We have $K_{\mathfrak{gex}}=E_{\mathfrak{gex}}K$ and $K_{\mathfrak{ge}}=E_{\mathfrak{ge}}^{\mathcal{H}}K$. Let $K\subseteq {}_{n}k{({\Lambda }_N)}_m$, $\mathcal{M}=L_n{k}_m$, and $E=\mathcal{M}K\cap k({\Lambda }_N)$. For any finite abelian extension $L/J$ and any prime $\mathfrak{p}$ of J, we denote by $e_{\mathfrak{p}}^{*}(L|J)$ the tame ramification index of the prime $\mathfrak{p}$ in the extension $L/J$, namely, if $e_{\mathfrak{p}}(L|J)=p^{\beta }\alpha$ with $gcd(\alpha ,p)=1$, then $e_{\mathfrak{p}}^{*}(L|J)=\alpha$. The set of tame ramification indexes is multiplicative.

Lemma 7.

We have $e_{\infty }(E|k)=e_{\infty }^{*}(K|k)$.

Proof. 

First, note that if $k\subseteq J\subseteq \mathcal{M}$, then $e_{\infty }^{*}(\mathcal{M}|J)=e_{\infty }^{*}(J|k)=1$. Now we consider

Since $e_{\infty }^{*}(K\mathcal{M}|K)|e_{\infty }^{*}(\mathcal{M}|K\cap \mathcal{M})$, it follows that $e_{\infty }^{*}(K\mathcal{M}|K)=e_{\infty }^{*}(\mathcal{M}|K\cap \mathcal{M})=1$. Therefore, $e_{\infty }^{*}(K|K\cap \mathcal{M})=e_{\infty }^{*}(K\mathcal{M}|\mathcal{M})$.

We have

It follows that

$$\begin{array}{cc}\hfill e_{\infty }^{*}(K\mathcal{M}|k)& =e_{\infty }^{*}(K\mathcal{M}|\mathcal{M})=e_{\infty }^{*}(K|K\cap \mathcal{M})=e_{\infty }^{*}(E|k)=e_{\infty }(E|k).\hfill \end{array}$$

Now

$$\begin{array}{cc}\hfill e_{\infty }^{*}(K|K\cap \mathcal{M})& =e_{\infty }^{*}(K|K\cap \mathcal{M})e_{\infty }^{*}(K\cap \mathcal{M}|k)=e_{\infty }^{*}(K|k)=e_{\infty }(E|k).\hfill \end{array}$$

□

Theorem 13.

We have $[K_{\mathfrak{gex}}:K_{\mathfrak{ge}}]=[E_{\mathfrak{gex}}:E_{\mathfrak{ge}}^{\mathcal{H}}]=[E_{\mathfrak{gex}}:E_{\mathfrak{ge}}]·|\mathcal{H}|$. In particular $[K_{\mathfrak{gex}}:K_{\mathfrak{ge}}]|q-1$. It also holds

$$f_{\infty }(K_{\mathfrak{gex}}:K_{\mathfrak{ge}})=|\mathcal{H}|,e_{\infty }(K_{\mathfrak{gex}}:K_{\mathfrak{ge}})=[E_{\mathfrak{gex}}:E_{\mathfrak{ge}}].$$

Furthermore, the field of constants of $K_{\mathfrak{ge}}$ is ${\mathbb{F}}_{q^{{\mathrm{deg}}_K{\mathfrak{p}}_{\infty }}}$, ${\mathrm{deg}}_K{\mathfrak{p}}_{\infty }=f_{\infty }(K|k)$, and the field of constants of both $K_{\mathfrak{gex}}$ and $K_{H^{+}}$ is ${\mathbb{F}}_{q^{|\mathcal{H}|{\mathrm{deg}}_K{\mathfrak{p}}_{\infty }}}$, and we have $|\mathcal{H}|{\mathrm{deg}}_K{\mathfrak{p}}_{\infty }=f_{\infty }(EK|k)$.

Proof. 

We have

It follows that $[K_{\mathfrak{gex}}:K_{\mathfrak{ge}}]=[E_{\mathfrak{gex}}:E_{\mathfrak{ge}}^{\mathcal{H}}]|q-1$.

Now, since $E_{\mathfrak{ge}}K/{(E_{\mathfrak{ge}}K)}^{\mathcal{H}}=K_{\mathfrak{ge}}$ is an extension of constants of degree, in fact, $|\mathcal{H}|=f_{\infty }(E_{\mathfrak{ge}}K|K_{\mathfrak{ge}})$, we will see that the extension $K_{\mathfrak{gex}}/E_{\mathfrak{ge}}K$ is totally ramified.

We have that $e_{\infty }(E_{\mathfrak{ge}}K|k)=e_{\infty }(K|k)$. Hence $e_{\infty }^{*}(E_{\mathfrak{ge}}K|E_{\mathfrak{ge}})=1$. Similarly, we obtain $e_{\infty }^{*}(E_{\mathfrak{gex}}K|E_{\mathfrak{gex}})=1$.

Therefore, $e_{\infty }(E_{\mathfrak{gex}}K|E_{\mathfrak{ge}}K)=e_{\infty }(E_{\mathfrak{gex}}|E_{\mathfrak{ge}})=[E_{\mathfrak{gex}}:E_{\mathfrak{ge}}]$ and $K_{\mathfrak{gex}}/E_{\mathfrak{ge}}K$ is totally ramified.

Since ${\mathrm{deg}}_k{\mathfrak{p}}_{\infty }=1$ and $K/k$ is geometric, we obtain that $f_{\infty }(K|k)={\mathrm{deg}}_K{\mathfrak{p}}_{\infty }$. We know that the field of constants of $K_{\mathfrak{ge}}$ is ${\mathbb{F}}_{q^{{\mathrm{deg}}_K{\mathfrak{p}}_{\infty }}}$.

Finally, $E_{\mathfrak{ge}}K/E_{\mathfrak{ge}}^{\mathcal{H}}K$ is an extension of constants of degree $|\mathcal{H}|=f_{\infty }(EK|K)$. Hence, the field of constants of both $K_{\mathfrak{gex}}$ and $K_{H^{+}}$ is ${\mathbb{F}}_{q^{{\mathrm{deg}}_K{\mathfrak{p}}_{\infty }·|\mathcal{H}|}}={\mathbb{F}}_{q^{f_{\infty }(EK|k)}}$. □

7. Discussion

In this paper, it is shown that the notion of the extended genus field of an abelian finite extension of the field of rational functions given by B. Anglès and J.F. Jaulent by means of class field theory is the same as the one given by E. Ramírez-Ramírez, M. Rzedowski-Calderón, and G. Villa-Salvador using Dirichlet characters.