A Note on Extended Genus Fields of Kummer Extensions of Global Rational Function Fields

Abstract

We consider the generalization of the extended genus field of a prime degree cyclic Kummer extension of a global rational function field obtained by R. Clement in 1992 to general Kummer extensions. We observe that the same approach of Clement works in general.

1. Introduction

In the case of number fields, both the genus and the extended genus fields of a number field K are canonically defined as the maximal extension of K of the form $KF$ contained in the Hilbert class field and in the extended Hilbert class field, respectively, where F is an abelian extension of $\mathbb{Q}$. The Hilbert class field (resp. extended Hilbert class field) of K is defined as the maximal unramified (resp. maximal unramified at the finite primes) abelian extension of K.

In the case of global function fields, the situation is quite different. There are several possible reasonable definitions of the Hilbert class field of a function field K. The main reason is that, in the number field case, the Hilbert class field is a finite extension of K. In the case of a function field K, the extensions of constants of K are unramified so that the maximal unramified abelian extension of K is of an infinite degree.

The first one to consider genus fields, or more precisely extended genus fields, of function fields was R. Clement in [1]. She considered a cyclic Kummer extension K of $k:={\mathbb{F}}_q\left(T\right)$, a global rational function field of prime degree l, where $l|q-1$. She used the ideas H. Hasse developed in [2], where he considered a quadratic extension K of the field of rational numbers $\mathbb{Q}$. He presented the extended Hilbert class field of K and provided the extended genus field of $K/\mathbb{Q}$, using class field theory.

For a quadratic number field K of $\mathbb{Q}$, the extended Hilbert class field $K_{H^{+}}$ of K is the abelian extension such that the fully decomposed primes of K in $K_{H^{+}}$ are precisely the non-zero principal prime ideals of K generated by a totally positive element. In the case of a quadratic field K of $\mathbb{Q}$, an element is totally positive if and only if its norm in $\mathbb{Q}$ is the square of a real number. In other words, the norm group in the idèle group $J_K$ of K is $K^{*}(\Delta \times {\prod }_p{U}_p)$, where p runs through the prime numbers, $U_p$ denotes the group of units of ${\mathbb{Q}}_p$ and $\Delta =\left\{({\alpha }_1,\dots ,{\alpha }_t)\in {\prod }_{i=1}^t{K}_{{\mathfrak{p}}_i}\mid {\prod }_{i=1}^t{\mathrm{N}}_{K_{{\mathfrak{p}}_i}/\mathbb{R}}\left({\alpha }_i\right)\in {{\mathbb{R}}^{*}}^2\right\}$ where ${\left\{{\mathfrak{p}}_i\right\}}_{i=1}^t$ is the set of infinite primes in K ($t=1$ or 2 and $K_{{\mathfrak{p}}_i}\in \{\mathbb{R},\mathbb{C}\}$).

For a cyclic Kummer extension $K/k$ of prime degree l, where $k={\mathbb{F}}_q\left(T\right)$ and $l|q-1$, the analogue is to consider as the norm group of $K_{H^{+}}$ the subgroup $K^{*}(\Delta \times {\prod }_{\mathfrak{p}\nmid \infty }{U}_{\mathfrak{p}})$ of the idèle group $J_K$ of K, where $\Delta =\{({\alpha }_1,\dots ,{\alpha }_t)\in {\prod }_{i=1}^t{K}_{{\mathfrak{p}}_i}\mid {\prod }_{i=1}^t{\mathrm{N}}_{K_{{\mathfrak{p}}_i}/k_{\infty }}\left({\alpha }_i\right)\in k_{\infty }^{*l}\}$ and $U_{\mathfrak{p}}$ denotes the group of units of $K_{\mathfrak{p}}$. This is the approach taken by Clement.

In [3], E. O. Curiel-Anaya, M. R. Maldonado-Ramírez and the first author use this approach to extend the result to general cyclic Kummer extensions.

In this note, we will see that this approach still works for general Kummer extensions K of k and obtain a version of the extended genus field for this kind of extension. More precisely, let $K/k$ be given by

$$K=k\left(\sqrt[n]{{\gamma }_1{D}_1,\dots ,{\sqrt[n]{\gamma }}_s{D}_s}\right).$$

Let $P_1,\dots ,P_r$ be the primes of k that ramify in K, with ramification indices $e_1,\dots ,e_r$. Then, the main result of this paper, Theorem 1, states that the extended genus field of $K/k$ is given by

$$\Gamma ={\mathbb{F}}_{q^n}\left(\sqrt[e_1]{P_1},\dots ,\sqrt[e_r]{P_r}\right).$$

In the last section, we compare three definitions of extended genus fields for these extensions and present some examples.

2. General Kummer Extensions of $\mathit{k}$

Let $k:={\mathbb{F}}_q\left(T\right)$ and let K be a Kummer extension of k of exponent $n|q-1$. Then,

$$\mathrm{Gal}(K/k)\cong C_{n_1}\times \cdots \times C_{n_s},$$

where, in general, $C_m$ denotes the cyclic group of order m and $n_s|\cdots |n_2|n_1=n$.

We have

$$\begin{array}{c}K=k\left(\sqrt[n_1]{{\gamma }_1{D}_1},\dots ,\sqrt[n_s]{{\gamma }_s{D}_s}\right)=K_1\cdots K_s,\end{array}$$

(1)

where $K_i=k\left(\sqrt[n_i]{{\gamma }_i{D}_i}\right)$, the constant ${\gamma }_i$ is different from zero, that is, ${\gamma }_i\in {\mathbb{F}}_q^{*}$, $D_i\in R_T:={\mathbb{F}}_q\left[T\right]$ is a monic polynomial and $\mathrm{Gal}(K_i/k)\cong C_{n_i}$, $1\le i\le s$.

Let $P_1,\dots ,P_r\in R_T^{+}:=\{P\in R_T\mid P\mathrm{is}\mathrm{a}\mathrm{monic}\mathrm{irreducible}\mathrm{polynomial}\}$ be the primes of k ramified in K with ramification indices $e_1,\dots ,e_r$, respectively. Let

$$\Delta :=\{({\alpha }_1,\dots ,{\alpha }_t)\in \prod _{i=1}^t{K}_{{\mathfrak{p}}_i}\mid \prod _{i=1}^t{\mathrm{N}}_{K_{{\mathfrak{p}}_i}/k_{\infty }}\left({\alpha }_i\right)\in k_{\infty }^{*n}\},$$

where ${\left\{{\mathfrak{p}}_i\right\}}_{i=1}^t$ are the infinite primes in K, that is, the primes above ${\mathcal{P}}_{\infty }$ ($=\infty$), and the infinite prime of k and $k_{\infty }:=k_{{\mathcal{P}}_{\infty }}$.

Let ${\mathcal{O}}_K:=\{\alpha \in K\mid v_{\mathfrak{p}}\left(\alpha \right)\ge 0\mathrm{for}\mathrm{all}\mathfrak{p}\nmid \infty \}$. Then, ${\mathcal{O}}_K$ is a Dedekind domain, so that the prime ideals of ${\mathcal{O}}_K$ correspond canonically to the finite primes of K. We denote the idèle group of K by $J_K$.

Definition 1.

The extended Hilbert class field $K_{H^{+}}$ of K is the finite abelian extension of K such that the finite primes of K fully decomposed in $K_{H^{+}}$ are precisely the principal non-zero ideals generated by an element whose norm from K to k is an n-th power in $k_{\infty }$.

The extended genus field Γ of K with respect to k is the maximal abelian extension of k contained in $K_{H^{+}}$.

Proposition 1.

We have that $[J_K:K^{*}\left(\Delta \times {\prod }_{\mathfrak{p}\nmid \infty }{U}_{\mathfrak{p}}\right)]$ is finite.

Proof. 

The same proof in [1] (Proposition 1.1) works in this case. We present the proof for the sake of completeness. Let $A:=\Delta \times {\prod }_{\mathfrak{p}\nmid \infty }{U}_{\mathfrak{p}}$ and $B:={\prod }_{\mathfrak{p}|\infty }{K}_{\mathfrak{p}}^{*}\times {\prod }_{\mathfrak{p}\nmid \infty }{U}_{\mathfrak{p}}$. We have

$$\begin{array}{c}{K}^{*}\left(\Delta \times \prod _{\mathfrak{p}\nmid \infty }{U}_{\mathfrak{p}}\right)=K^{*}A\subseteq K^{*}B=K^{*}\left(\prod _{\mathfrak{p}|\infty }{K}_{\mathfrak{p}}^{*}\times \prod _{\mathfrak{p}\nmid \infty }{U}_{\mathfrak{p}}\right),\\ J_K/K^{*}B=J_K/K^{*}\left(\prod _{\mathfrak{p}|\infty }{K}_{\mathfrak{p}}^{*}\times \prod _{\mathfrak{p}\nmid \infty }{U}_{\mathfrak{p}}\right)\underset{\psi }{\overset{\cong }{\to }}Cl\left({\mathcal{O}}_K\right)\end{array}$$

where the isomorphism $\psi$ is the one induced by the natural map $\theta :J_K⟶D_K$ given by $\theta \left(\overrightarrow{\alpha }\right)={\prod }_{\mathfrak{p}}{\mathfrak{p}}^{{\alpha }_{\mathfrak{p}}}$, $D_K$ denotes the divisor group of K, and the ideal class group $Cl\left({\mathcal{O}}_K\right)$ is isomorphic to $D_K/D_{\infty }{P}_K$, where $D_{\infty }$ is the free group generated by the infinite primes and $P_K$ is the group of principal divisors. We have that $Cl\left({\mathcal{O}}_K\right)$ is finite.

Now, $A\subseteq B$. Therefore,

$$\begin{array}{c}\begin{array}{cc}\hfill [K^{*}B:K^{*}A]& ={\displaystyle \frac{[B:A]}{[K^{*}\cap B:K^{*}\cap A]}}={\displaystyle \frac{[{\prod }_{\mathfrak{p}|\infty }{K}_{\mathfrak{p}}^{*}\times {\prod }_{\mathfrak{p}\nmid \infty }{U}_{\mathfrak{p}}:\Delta \times {\prod }_{\mathfrak{p}\nmid \infty }{U}_{\mathfrak{p}}]}{[K^{*}\cap B:K^{*}\cap A]}}\hfill \\ \hfill & ={\displaystyle \frac{[{\prod }_{\mathfrak{p}|\infty }{K}_{\mathfrak{p}}^{*}:\Delta ]}{[U_K:U_K^{(+)}]}},\hfill \end{array}\end{array}$$

where

$$\begin{array}{c}\hfill U_K:=K^{*}\cap B=\{x\in K^{*}\mid v_{\mathfrak{p}}\left(x\right)=0\mathrm{for}\mathrm{all}\mathfrak{p}\nmid \infty \}={\mathcal{O}}_K^{*}(\mathrm{units}\mathrm{of}{\mathcal{O}}_K)\end{array}$$

and

$$\begin{array}{c}\hfill U_K^{(+)}:=\{x\in U_K\mid {\mathrm{N}}_{K/k}\left(x\right)=\prod _{\mathfrak{p}|\infty }{\mathrm{N}}_{K_{\mathfrak{p}}/k_{\infty }}\left(x\right)\in k_{\infty }^{*n}\}.\end{array}$$

Now, $n|q-1$ so that $\gcd (n,p)=1$, where p is the characteristic of k. Therefore, ${\left(U_{\mathfrak{p}}^{\left(1\right)}\right)}^n=U_{\mathfrak{p}}^{\left(1\right)}$. Since $k_{\infty }^{*}=\langle {\pi }_{\infty }\rangle \times {\mathbb{F}}_q^{*}\times U_{\infty }^{\left(1\right)}$, where ${\pi }_{\infty }=1/T$ is a uniformizing parameter, we have $k_{\infty }^{*n}=\langle {\pi }_{\infty }^n\rangle \times {{\mathbb{F}}_q^{*}}^n\times {\left(U_{\infty }^{\left(1\right)}\right)}^n=\langle {\pi }_{\infty }^n\rangle \times {{\mathbb{F}}_q^{*}}^n\times U_{\infty }^{\left(1\right)}$. Thus, $\left|{\displaystyle \frac{k_{\infty }^{*}}{k_{\infty }^{*n}}}\right|=n^2$. Consider

$$\begin{array}{c}\phi :\prod _{\mathfrak{p}|\infty }{K}_{\mathfrak{p}}^{*}⟶{\displaystyle \frac{k_{\infty }^{*}}{k_{\infty }^{*n}}},\mathrm{given}\mathrm{by}\phi \left({\left({\alpha }_{\mathfrak{p}}\right)}_{\mathfrak{p}|\infty }\right)=\left(\prod _{\mathfrak{p}|\infty }{\mathrm{N}}_{K_{\mathfrak{p}}/k_{\infty }}\left({\alpha }_{\mathfrak{p}}\right)\mathrm{mod}{k}_{\infty }^{*n}\right).\end{array}$$

Then $\ker \phi =\Delta$ and $\phi$ induces an injection

$$\begin{array}{c}\hfill \tilde{\phi }:{\displaystyle \frac{\left({\prod }_{\mathfrak{p}|\infty }{K}_{\mathfrak{p}}^{*}\right)}{\Delta }}\hookrightarrow {\displaystyle \frac{k_{\infty }^{*}}{k_{\infty }^{*n}}}.\end{array}$$

Hence, $[B:A]|n^2$ and finally, $[J_K:K^{*}A]<\infty$. □

We have that $K^{*}\left(\Delta \times {\prod }_{\mathfrak{p}\nmid \infty }{U}_{\mathfrak{p}}\right)$ is an open subgroup of finite index of $J_K$. Let L be the class field corresponding to $K^{*}\left(\Delta \times {\prod }_{\mathfrak{p}\nmid \infty }{U}_{\mathfrak{p}}\right)$. Then, ${\mathrm{N}}_{L/K}{J}_L=K^{*}\left(\Delta \times {\prod }_{\mathfrak{p}\nmid \infty }{U}_{\mathfrak{p}}\right)$ and $\mathrm{Gal}(L/K)\cong {\displaystyle \frac{J_K}{K^{*}\left(\Delta \times {\prod }_{\mathfrak{p}\nmid \infty }{U}_{\mathfrak{p}}\right)}}$.

Proposition 2.

Let $\mathfrak{q}$ be a finite prime of K, that is, $\mathfrak{q}$ is a non-zero prime ideal of ${\mathcal{O}}_K$. Then, $\mathfrak{q}$ decomposes fully in L if and only if $\mathfrak{q}$ is a principal ideal with $\mathfrak{q}=\langle \delta \rangle$ and ${\mathrm{N}}_{K/k}\left(\delta \right)\in k_{\infty }^{*n}$.

Proof. 

From class field theory, we have that $\mathfrak{q}$ decomposes fully in $L/K$ if and only if $\theta {\lceil K_{\mathfrak{q}}^{*}\rceil }_{\mathfrak{q}}\subseteq K^{*}\left(\Delta \times {\prod }_{\mathfrak{p}\nmid \infty }{U}_{\mathfrak{p}}\right)$, where $\theta {\lceil x\rceil }_{\mathfrak{q}}:=(\dots ,1,x,1,\dots )$ for $x\in K_{\mathfrak{q}}^{*}$ (see [4] (Corolario 17.6.196)) if and only if for all $x\in K_{\mathfrak{q}}^{*}$ there exist ${\beta }_x\in K^{*}$ and ${\overrightarrow{\alpha }}_x\in \Delta \times {\prod }_{\mathfrak{p}\nmid \infty }{U}_{\mathfrak{p}}$ with $\theta {\lceil x\rceil }_{\mathfrak{q}}={\beta }_x{\overrightarrow{\alpha }}_x$. Thus, ${\overrightarrow{\alpha }}_x=(\dots ,{\beta }_x^{-1},\dots ,{\beta }_x^{-1},{\beta }_x^{-1}x,{\beta }_x^{-1},\dots )$. This is equivalent to $v_{\mathfrak{p}}\left({\beta }_x\right)=v_{\mathfrak{p}}\left(x\right)=0$ for all $\mathfrak{p}\ne \mathfrak{q}$ and $\mathfrak{p}\nmid \infty$ and ${\mathrm{N}}_{K/k}\left({\beta }_x^{-1}\right)={\prod }_{\mathfrak{p}|\infty }{\mathrm{N}}_{K_{\mathfrak{p}}/k_{\infty }}\left({\beta }_x^{-1}\right)\in k_{\infty }^{*n}$.

Assume $\mathfrak{q}$ decomposes fully in L. Let $x\in K_{\mathfrak{q}}^{*}$. The principal ideal in ${\mathcal{O}}_K$ generated by ${\beta }_x$ satisfies $\langle {\beta }_x\rangle ={\mathfrak{q}}^{n_x}={\beta }_x{\mathcal{O}}_K$, where $n_x=v_q\left(x\right)\in \mathbb{Z}$. In particular, for $x\in K_{\mathfrak{q}}^{*}$ with $v_{\mathfrak{q}}\left(x\right)=1$, we have $\langle {\beta }_x\rangle =\mathfrak{q}$, ${\beta }_x\in K^{*}$. Hence, $\mathfrak{q}$ is a principal ideal, $\mathfrak{q}=\langle {\beta }_x\rangle$ and ${\mathrm{N}}_{K/k}\left({\beta }_x\right)\in k_{\infty }^{*n}$.

Conversely, if $\mathfrak{q}=\langle \delta \rangle$ satisfies ${\mathrm{N}}_{K/k}\left(\delta \right)\in k_{\infty }^{*n}$, then, for $x\in K_{\mathfrak{q}}^{*}$, if $v_{\mathfrak{q}}\left(x\right)=m=v_{\mathfrak{q}}\left({\delta }^m\right)$, we obtain that $x{\delta }^{-m}\in U_{\mathfrak{q}}$ and that $\theta {\lceil x\rceil }_{\mathfrak{q}}=(\dots ,1,x,1,\dots )={\beta }_x{\overrightarrow{\alpha }}_x$ with ${\beta }_x={\delta }^m$ and ${\overrightarrow{\alpha }}_x=(\dots ,{\delta }^{-m},{\delta }^{-m}x,{\delta }^{-m},\dots )$. Therefore, $v_{\mathfrak{p}}\left({\beta }_x\right)=v_{\mathfrak{p}}\left({\delta }^m\right)=0$ for all $\mathfrak{p}\ne \mathfrak{q}$ and $\mathfrak{p}\nmid \infty$, and ${\mathrm{N}}_{K/k}\left({\beta }_x^{-m}\right)={\left({\mathrm{N}}_{K/k}{\delta }^{-1}\right)}^m\in k_{\infty }^{*n}$. Hence, $\mathfrak{q}$ decomposes fully in $L/K$. □

Corollary 1.

The extended class field $K_{H^{+}}$ of K is the class field corresponding to $K^{*}\left(\Delta \times {\prod }_{\mathfrak{q}\nmid \infty }{U}_{\mathfrak{q}}\right)$ and a finite prime $\mathfrak{q}$ of K decomposes fully in $K_{H^{+}}$ if and only if $\mathfrak{q}$ is a principal ideal and the norm from K to k of a generator belongs to $k_{\infty }^{*n}$.

Proposition 3.

The field of constants of $K_{H^{+}}$ is ${\mathbb{F}}_{q^n}$.

Proof. 

We have (see [4] (Teorema 17.6.192)) that the field of constants of $K_{H^{+}}$ is ${\mathbb{F}}_{q^d}$ where $d=\min \{\mathrm{deg}\overrightarrow{\alpha }\mid \overrightarrow{\alpha }\in \Delta \times {\prod }_{\mathfrak{p}\nmid \infty }{U}_{\mathfrak{p}}\mathrm{and}\mathrm{deg}\overrightarrow{\alpha }>0\}$. We will see that $d=n$.

On the one hand, since ${\prod }_{\mathfrak{p}|\infty }{\mathrm{N}}_{K_{\mathfrak{p}}/k_{\infty }}{K}_{\mathfrak{p}}^{*}=k_{\infty }^{*n}$, there exists $\overrightarrow{\alpha }\in \Delta \times {\prod }_{\mathfrak{p}\nmid \infty }{U}_{\mathfrak{p}}$ with $v_{\infty }\left({\prod }_{\mathfrak{p}|\infty }{\mathrm{N}}_{K_{\mathfrak{p}}/k_{\infty }}\left({\alpha }_{\mathfrak{p}}\right)\right)=v_{\infty }\left({\pi }_{\infty }^n\right)=n$. On the other hand, let $\overrightarrow{\alpha }\in \Delta \times {\prod }_{\mathfrak{p}\nmid \infty }{U}_{\mathfrak{p}}$. Then, $v_{\mathfrak{p}}\left({\alpha }_{\mathfrak{p}}\right)=0$ for all $\mathfrak{p}\nmid \infty$ and $\xi ={\prod }_{\mathfrak{p}|\infty }{\mathrm{N}}_{K_{\mathfrak{p}}/k_{\infty }}\left({\alpha }_{\mathfrak{p}}\right)\in k_{\infty }^{*n}$. Therefore (see [4] (Proposición 17.3.38)),

$$\begin{array}{cc}\hfill \mathrm{deg}\overrightarrow{\alpha }& =\sum _{\mathfrak{p}}\mathrm{deg}\mathfrak{p}·v_{\mathfrak{p}}\left({\alpha }_{\mathfrak{p}}\right)=\sum _{\mathfrak{p}|\infty }\mathrm{deg}\mathfrak{p}·v_{\mathfrak{p}}\left({\alpha }_{\mathfrak{p}}\right)=\sum _{\mathfrak{p}|\infty }{f}_{\mathfrak{p}}{v}_{\mathfrak{p}}\left({\alpha }_{\mathfrak{p}}\right)\hfill \\ \hfill & =\sum _{\mathfrak{p}|\infty }{v}_{\infty }\left({\mathrm{N}}_{K_{\mathfrak{p}}/k_{\infty }}\left({\alpha }_{\mathfrak{p}}\right)\right)=v_{\infty }\left(\prod _{\mathfrak{p}|\infty }{\mathrm{N}}_{K_{\mathfrak{p}}/k_{\infty }}\left({\alpha }_{\mathfrak{p}}\right)\right)=v_{\infty }\left(\xi \right)=na\hfill \end{array}$$

for some $a\in \mathbb{Z}$. Hence, $d=n$. □

Proposition 4.

Every abelian extension $L/k$ such that $K\subseteq L$, $L/K$ is unramified at the finite primes and such that $\mathrm{Gal}(L/k)$ is of exponent a divisor of n is contained in $K_{H^{+}}$.

Proof. 

Since $K_{H^{+}}$ is the class field of $\Delta \times {\prod }_{\mathfrak{p}\nmid \infty }{U}_{\mathfrak{p}}$ (Corollary 1), we have that $K_{H^{+}}/K$ is unramified at the finite primes.

Let $L/k$ be an abelian extension with $K\subseteq L$, such that $L/K$ is unramified at the finite primes and such that $\mathrm{Gal}(L/K)$ is an exponent and a divisor of n.

Let ${\mathrm{N}}_{L/K}{J}_L=K^{*}\mathcal{H}\subseteq J_K$ for some $\mathcal{H}$. If $\mathfrak{q}$ is a finite prime, then $\mathfrak{q}$ is unramified in $L/K$ so that $U_{\mathfrak{q}}\subseteq \mathcal{H}$ (see [4] (Corolario 17.6.196)). If $\overrightarrow{\alpha }\in {\prod }_{\mathfrak{p}\nmid \infty }{U}_{\mathfrak{p}}$, and ${(\_,E/F)}_{\mathfrak{p}}$ denotes the Artin reciprocity map, we have (see [4] (Teorema 17.6.149))

$$(\overrightarrow{\alpha },L/K)=\prod _{\mathfrak{p}|\infty }(1,L_{\mathfrak{P}}/K_{\mathfrak{p}})\times \prod _{\mathfrak{p}\nmid \infty }({\alpha }_{\mathfrak{p}},L_{\mathfrak{P}}/K_{\mathfrak{p}})=1,$$

where $\mathfrak{P}$ is a prime in L above $\mathfrak{p}$. It follows that $\overrightarrow{\alpha }\in \mathcal{H}$ and ${\prod }_{\mathfrak{p}\nmid \infty }{U}_{\mathfrak{p}}\subseteq \mathcal{H}$.

If $\overrightarrow{\alpha }\in \Delta$, we have the commutative diagram

where $\iota$ is the natural embedding. We have that ${\mathrm{N}}_{K/k}(\Delta )=\{(\delta ,1,\dots ,1,\dots )\mid \delta \in k_{\infty }^{*n}\}$. Since $\mathrm{Gal}(L/k)$ is of exponent n, we have that ${\mathrm{N}}_{K/k}(\Delta )\subseteq \ker (\_,L/k)$ so that $\Delta \subseteq \ker (\_,L/K)=K^{*}\mathcal{H}$. Hence, $\Delta \times {\prod }_{\mathfrak{p}\nmid \infty }{U}_{\mathfrak{p}}\subseteq K^{*}\mathcal{H}$ and $L\subseteq K_{H^{+}}$. □

Proposition 5.

The extension $K_{H^{+}}/k$ is a Galois extension.

Proof. 

It is the same as the one in [1]. It follows from the fact that $\sigma \left(\Delta \times {\prod }_{\mathfrak{p}\nmid \infty }{U}_{\mathfrak{p}}\right)=\Delta \times {\prod }_{\mathfrak{p}\nmid \infty }{U}_{\mathfrak{p}}$ for each k-embedding $\sigma$ of $K_{H^{+}}$ in an algebraic closure $\overline{k}$ of k. □

The following result is the analogue of Lemma 1.5 and Proposition 1.6 in [1]. The same proof works in our more general setting.

Proposition 6.

Let $I_K$ be the group of fractional ideals of ${\mathcal{O}}_K$ and $P_K$ the subgroup of principal ideals. Let $J_K^{(+)}=\{\overrightarrow{\alpha }\in J_K\mid {\left({\alpha }_{\mathfrak{p}}\right)}_{\mathfrak{p}|\infty }\in \Delta \}$, $K^{(+)}=K^{*}\cap J_K^{(+)}$ and $P_K^{(+)}=\{\langle \beta \rangle \in P_K\mid \beta \in K^{(+)}\}$. Then,

$$\begin{array}{c}{\displaystyle \frac{J_K}{K^{*}\left(\Delta \times {\prod }_{\mathfrak{p}\nmid \infty }{U}_{\mathfrak{p}}\right)}}\cong {\displaystyle \frac{J_K^{(+)}}{K^{(+)}\left(\Delta \times {\prod }_{\mathfrak{p}\nmid \infty }{U}_{\mathfrak{p}}\right)}}\cong {\displaystyle \frac{I_K}{P_K^{(+)}}}=:Cl{\left({\mathcal{O}}_K\right)}^{(+)}.\end{array}$$

Next, we recall a result from class field theory.

Proposition 7.

Let $E/F$ be a finite extension of global fields. Let N be an open subgroup of finite index of $J_E$ such that $E^{*}\subseteq N$. Let $E_N$ be the class field of N, that is, ${\mathrm{N}}_{E_N/E}\left(J_{E_N}\right)=N$. Let $F_0$ be the maximal abelian extension of F contained in $E_N$. Then, the norm group of $F_0$ in $J_F$ is $F^{*}{\mathrm{N}}_{E/F}\left(N\right)$.

Proof. 

The norm ${\mathrm{N}}_{E/F}:J_E⟶J_F$ is an open map so that ${\mathrm{N}}_{E/F}\left(N\right)$ is an open subgroup of $J_F$. Since N is of finite index in $J_E$, we have that $F^{*}{\mathrm{N}}_{E/F}\left(N\right)$ is of finite index in $J_F$.

To show that $F_0$ is the class field of $F^{*}{\mathrm{N}}_{E/F}\left(N\right)$, it suffices to show that the group ${\mathrm{N}}_{E/F}\left(N\right)$ is contained in any open subgroup of finite index in $J_F$ such that its class field $F^{\prime }$ is contained in $E_N$.

Let $N^{\prime }$ be an open subgroup of finite index in $J_E$ such that its class field is $E{F}^{\prime }\subseteq E_N$. Since $N\subseteq N^{\prime }$, it follows that ${\mathrm{N}}_{E/F}\left(N\right)\subseteq {\mathrm{N}}_{E/F}\left(N^{\prime }\right)$. We have the commutative diagram (see [4] (Teorema 17.6.137))

Therefore, ${\mathrm{N}}_{E/F}\left(N\right)$ is contained in the subgroup of $J_F$ corresponding to $F^{\prime }$. □

As a direct consequence of Proposition 7, we have

Corollary 2.

The extended genus field Γ of $K/k$ is the class field of $k^{*}{\mathrm{N}}_{K/k}\left(\Delta \times {\prod }_{\mathfrak{p}\nmid \infty }{U}_{\mathfrak{p}}\right)$ and

$$\begin{array}{c}[\Gamma :k]=\left|\mathrm{Gal}(\Gamma /k)\right|=\left[J_k:k^{*}{\mathrm{N}}_{K/k}\left(\Delta \times \prod _{\mathfrak{p}\nmid \infty }{U}_{\mathfrak{p}}\right)\right].\end{array}$$

To obtain $\Gamma$, we have to compute ${\mathrm{N}}_{K/k}\left(\Delta \times {\prod }_{\mathfrak{p}\nmid \infty }{U}_{\mathfrak{p}}\right)$.

First, we prove two lemmas.

Lemma 1.

We have $J_k=k^{*}\left(k_{\infty }^{*}\times {\prod }_{P\in R_T^{+}}{U}_P\right)$.

Proof. 

Let $\overrightarrow{\beta }={\left({\beta }_{\infty },{\beta }_P\right)}_{P\in R_T^{+}}\in J_k$. Let $Q_1,\dots ,Q_t\in R_T^{+}$ be the finite primes such that $c_i=v_{Q_i}\left({\beta }_{Q_i}\right)\ne 0$. We have that $v_P\left({\beta }_P\right)=0$ for all $P\notin \{Q_1,\dots ,Q_t\}$. Let $f\in k^{*}$ be defined by $f={\prod }_{i=1}^t{Q}_i^{c_i}$ ($f=1$ if $t=0$). Then, $f^{-1}\overrightarrow{\beta }\in k_{\infty }^{*}\times {\prod }_{P\in R_T^{+}}{U}_P$. The result is obtained.

Another proof is as follows: the class field F of $k^{*}\left(k_{\infty }^{*}\times {\prod }_{P\in R_T^{+}}{U}_P\right)$ is unramified over k and ${\mathcal{P}}_{\infty }$ decomposes fully. Hence, F is an extension of constants and the infinite prime (of degree one) decomposes fully in F. Therefore, $F=k$. □

Lemma 2.

Let $\mathfrak{p}$ be a finite prime in K. Let P be the irreducible polynomial in $R_T$ corresponding to the prime in k below $\mathfrak{p}$.

(1)

If $\mathfrak{p}$ is unramified, then ${\mathrm{N}}_{K_{\mathfrak{p}}/k_P}\left(U_{\mathfrak{p}}\right)=U_P$.

(2)

If $\mathfrak{p}$ is ramified with ramification index $e_P$, then ${\mathrm{N}}_{K_{\mathfrak{p}}/k_P}\left(U_{\mathfrak{p}}\right)={\mathbb{F}}_P^{*e_P}\times U_P^{\left(1\right)}$, where ${\mathbb{F}}_P$ is the field of constants of $k_P$.

Proof. 

For any extension $E/F$ of local fields, we have $e_{E/F}=[U_F:{\mathrm{N}}_{E/F}\left(U_E\right)]$ (see [4] (Corolario 17.3.39)). If $E/F$ is tamely ramified, ${\mathrm{N}}_{E/F}\left(U_E^{\left(1\right)}\right)=U_F^{\left(1\right)}$.

(1)

If $\mathfrak{p}$ is unramified, then $e=1=[U_P:{\mathrm{N}}_{K_{\mathfrak{p}}/k_P}\left(U_{\mathfrak{p}}\right)]$. Hence, ${\mathrm{N}}_{K_{\mathfrak{p}}/k_P}\left(U_{\mathfrak{p}}\right)=U_P$.

(2)

Since $K_{\mathfrak{p}}/k_P$ is tamely ramified, we have ${\mathrm{N}}_{K_{\mathfrak{p}}/k_P}\left(U_{\mathfrak{p}}^{\left(1\right)}\right)=U_P^{\left(1\right)}$. Therefore,

$${\mathrm{N}}_{K_{\mathfrak{p}}/k_P}\left(U_{\mathfrak{p}}\right)={\mathrm{N}}_{K_{\mathfrak{p}}/k_P}\left({\mathbb{F}}_q^{*}\times U_{\mathfrak{p}}^{\left(1\right)}\right)={\mathrm{N}}_{K_{\mathfrak{p}}/k_P}\left({\mathbb{F}}_q^{*}\right)\times {\mathrm{N}}_{K_{\mathfrak{p}}/k_P}\left(U_{\mathfrak{p}}^{\left(1\right)}\right)={{\mathbb{F}}_q^{*}}^e\times U_P^{\left(1\right)}.$$

□

Proposition 8.

We have

$$[\Gamma :k]=\left[J_k:k^{*}{\mathrm{N}}_{K/k}\left(\Delta \times \prod _{\mathfrak{p}\nmid \infty }{U}_{\mathfrak{p}}\right)\right]=n\prod _{i=1}^r{e}_i$$

where $e_i$ is the ramification index of $P_i$ in $K/k$, $1\le i\le r$.

Proof. 

From Lemmas 1 and 2, we obtain

$$\begin{array}{cc}\hfill & \left[J_k:k^{*}{\mathrm{N}}_{K/k}\left(\Delta \times \prod _{\mathfrak{p}\nmid \infty }{U}_{\mathfrak{p}}\right)\right]\hfill \\ \hfill & =\left[k^{*}\left(\Delta \times \prod _{\mathfrak{p}\nmid \infty }{U}_{\mathfrak{p}}\right):k^{*}\left(k_{\infty }^{*n}\times \prod _{i=1}^r\left({\mathbb{F}}_{P_i}^{*e_i}\times U_{P_i}^{\left(1\right)}\right)\times \prod _{\begin{array}{c}P\in R_T^{+}\\ P\notin \{P_1,\dots ,P_r\}\end{array}}{U}_P\right)\right].\hfill \end{array}$$

Set $C:=\Delta \times {\prod }_{\mathfrak{p}\nmid \infty }{U}_{\mathfrak{p}}$ and $D:=k_{\infty }^{*n}\times {\prod }_{i=1}^r\left({\mathbb{F}}_{P_i}^{*e_i}\times U_{P_i}^{\left(1\right)}\right)\times {\prod }_{\begin{array}{c}P\in R_T^{+}\\ P\notin \{P_1,\dots ,P_r\}\end{array}}{U}_P$.

Then,

$$\begin{array}{c}[k^{*}C:k^{*}D]={\displaystyle \frac{[C:D]}{[k^{*}\cap C:k^{*}\cap D]}}\end{array}$$

and

$$\begin{array}{c}\hfill {\displaystyle \frac{C}{D}}\cong {\displaystyle \frac{k_{\infty }^{*}}{k_{\infty }^{*n}}}\times \prod _{i=1}^r{\displaystyle \frac{{\mathbb{F}}_{P_i}^{*}\times U_{P_i}^{\left(1\right)}}{{\mathbb{F}}_{P_i}^{*e_i}\times U_{P_i}^{\left(1\right)}}}\times \prod _{\begin{array}{c}P\in R_T^{+}\\ P\notin \{P_1,\dots ,P_r\}\end{array}}{\displaystyle \frac{U_P}{U_P}}\cong C_n^2\times \prod _{i=1}^r{C}_{e_i}.\end{array}$$

Consider now $\overrightarrow{\alpha }\in k^{*}\cap C$. Then ${\alpha }_P\in U_P$ for all $P\in R_T^{+}$. Thus $v_P\left({\alpha }_P\right)=0$ for all $P\in R_T^{+}$. It follows that $v_{\infty }\left({\alpha }_{\infty }\right)=0$ and that $\overrightarrow{\alpha }=x\in {\mathbb{F}}_q^{*}$. Thus, $k^{*}\cap C={\mathbb{F}}_q^{*}$. Similarly $k^{*}\cap D={{\mathbb{F}}_q^{*}}^n$. Therefore $[k^{*}\cap C:k^{*}\cap D]=n$. It follows that $[k^{*}C:k^{*}D]={\displaystyle \frac{n^2\times {\prod }_{i=1}^r{e}_i}{n}}=n·{\prod }_{i=1}^r{e}_i$. □

We have the main result of this note.

Theorem 1.

Let K be given by (1). Then, the extended genus field of $K/k$ is

$$\Gamma ={\mathbb{F}}_{q^n}\left(T,\sqrt[e_1]{P_1},\sqrt[e_2]{P_2},\dots ,\sqrt[e_r]{P_r}\right).$$

Proof. 

Let $M={\mathbb{F}}_{q^n}\left(T,\sqrt[e_1]{P_1},\sqrt[e_2]{P_2},\dots ,\sqrt[e_r]{P_r}\right)$. Then, $[M:k]=n·{\prod }_{i=1}^r{e}_i$. From Proposition 8, we obtain that $[\Gamma :k]=[M:k]$, and from Proposition 4, we obtain that $M\subseteq K_{H^{+}}$. Since M is abelian and $\Gamma$ is the maximal abelian extension of k contained in $K_{H^{+}}$, it follows that $M\subseteq \Gamma$. Hence, $\Gamma =M$. □

Remark 1.

We observe that the extended genus field Γ of a Kummer extension $K/k$ is also a Kummer extension and furthermore that the extended genus field of $\Gamma /k$ is Γ itself.

3. Comparison Among Various Extended Genus Fields

Let K be given by (1). We compare three different definitions of extended genus fields.

First, we denote by $K_{\mathfrak{gex},\mathrm{clement}}$ the extended genus field obtained in this work, that is, $K_{\mathfrak{gex},\mathrm{clement}}=\Gamma ={\mathbb{F}}_{q^n}\left(T,\sqrt[e_1]{P_1},\sqrt[e_2]{P_2},\dots ,\sqrt[e_r]{P_r}\right)$.

In [5], the extended genus field of K was defined as $K_{\mathfrak{gex},\mathrm{rarzvi}}=E_{\mathfrak{gex},\mathrm{rarzvi}}K$, where we have $E=k\left(\sqrt[n_1]{{(-1)}^{\mathrm{deg}{D}_1}{D}_1},\sqrt[n_2]{{(-1)}^{\mathrm{deg}{D}_2}{D}_2},\dots ,\sqrt[n_1]{{(-1)}^{\mathrm{deg}{D}_s}{D}_s}\right)$ and then

$$\begin{array}{c}{E}_{\mathfrak{gex},\mathrm{rarzvi}}=k\left(\sqrt[e_1]{{(-1)}^{\mathrm{deg}{P}_1}{P}_1},\sqrt[e_2]{{(-1)}^{\mathrm{deg}{P}_2}{P}_2},\dots ,\sqrt[e_r]{{(-1)}^{\mathrm{deg}{P}_r}{P}_r}\right).\end{array}$$

Therefore

$$\begin{array}{c}\hfill K_{\mathfrak{gex},\mathrm{rarzvi}}={\mathbb{F}}_q\left(T,\sqrt[e_1]{{ϵ}_1},\sqrt[e_2]{{ϵ}_2},\dots ,\sqrt[e_r]{{ϵ}_r}\right)\left(\sqrt[e_1]{P_1},\sqrt[e_2]{P_2},\dots ,\sqrt[e_r]{P_r}\right),\end{array}$$

where ${ϵ}_i={(-1)}^{\mathrm{deg}{P}_i}{\gamma }_i$, $1\le i\le r$. Since $e_i|n$, we have ${\mathbb{F}}_q\left(\sqrt[e_1]{{ϵ}_1},\sqrt[e_2]{{ϵ}_2},\dots ,\sqrt[e_r]{{ϵ}_r}\right)\subseteq {\mathbb{F}}_{q^n}$. Therefore, $K_{\mathfrak{gex},\mathrm{rarzvi}}\subseteq K_{\mathfrak{gex},\mathrm{clement}}$.

Next, using the definition of extended genus field given by B. Anglès and J.-F. Jaulent in [6], it was obtained in Theorem 5.1 from [7] that $K_{\mathfrak{gex},\mathrm{angjau}}=FK$ for some subfield F of $E_{\mathfrak{gex},\mathrm{rarzvi}}$. Therefore, $K_{\mathfrak{gex},\mathrm{angjau}}\subseteq K_{\mathfrak{gex},\mathrm{rarzvi}}$.

Finally, we have

$$K_{\mathfrak{gex},\mathrm{angjau}}\subseteq K_{\mathfrak{gex},\mathrm{rarzvi}}\subseteq K_{\mathfrak{gex},\mathrm{clement}}.$$

We also remark that while $K_{\mathfrak{gex},\mathrm{clement}}$ is defined just for Kummer extensions $K/k$, $K_{\mathfrak{gex},\mathrm{rarzvi}}$ and $K_{\mathfrak{gex},\mathrm{angjau}}$ are defined for any separable finite extension $K/k$.

Example 1.

(1) 

Let $K:=k\left(\sqrt[l]{\gamma P}\right)$ where l is a prime number such that $l|q-1$, $P\in R_T^{+}$ with $\mathrm{deg}P<l$ and $\gamma \in {\mathbb{F}}_q^{*}\setminus {\left({\mathbb{F}}_q^{*}\right)}^l$. Then, $K_{\mathfrak{gex},\mathrm{angjau}}=K_{\mathfrak{gex},\mathrm{rarzvi}}$ (see [7] ([Example 5.2)).

(2) 

Let l be a prime number such that $l^2|q-1$ and let $K=k\left(\sqrt[l^2]{\gamma D}\right)$ where $\gamma \in {\mathbb{F}}_q^{*}$, $D=P_1^{{\alpha }_1}\cdots P_r^{{\alpha }_r}\in R_T$. Then, $K_{\mathfrak{gex},\mathrm{angjau}}=K_{\mathfrak{gex},\mathrm{rarzvi}}$. (see [7] (Example 5.3)).

(3) 

Now, let l be a prime number such that $l|q-1$ and let $K:=k\left(\sqrt[l]{{(-1)}^{\mathrm{deg}P}P}\right)$ where $P\in R_T^{+}$ and $\mathrm{deg}P$ denotes the degree of P. Then, $K_{\mathfrak{gex},\mathrm{rarzvi}}=K⫋K_l=K_{\mathfrak{gex},\mathrm{clement}}$, where $K_l=K{\mathbb{F}}_{q^l}$ denotes the extension of constants of degree l of K.

Remark 2.

It is very likely that, when $K/k$ is abelian, we have $K_{\mathfrak{gex},\mathrm{angjau}}=K_{\mathfrak{gex},\mathrm{rarzvi}}$. In fact, in all known examples, as well as for elementary abelian extensions [8] (Theorem 4.2), this is the case.

4. Discussion

In this paper, it is shown that the approach of R. Clement to the extended genus field of a Kummer cyclic extension of a global rational function field can be extended to a general Kummer extension of a global rational function field. It remains the study of the case of abelian extensions that are not necessarily Kummer. Furthermore, the case where the ground field is not k has not been studied.