Let $k=\mathbb{F}_q(T)$ be the rational function field with constant field $\mathbb{F}_q$, the finite field with $q$ elements, where $q$ is a power of an odd prime number. The set $R=\mathbb{F}_q[T]$ of all the polynomials of $T$ over $\mathbb{F}_q$ is called the integral domain of $k$. A finite extension of $k$ is called an algebraic function field. Let $l\leq 19$ be a prime number such that $l|(q-1)$. The function fields $k(\sqrt[l]{(D(T)})$ (where $D(T)$ are not the $l$-th power of any polynomial) are $l$-th cyclic function fields. Artin studied the case $l=2$ systematically in [1]. By the discussing of ambiguous ideal classes Zhang (see [2]) explicitly expressed the 2-rank of the class group of $k(\sqrt{(D(T)})$ and gave a necessary and sufficient condition for the class number to be odd. Zhang's result was used by Ma and Feng (see [3]) to study the ideal class groups of imaginary quadratic function fields. They obtained a condition for the ideal class groups having exponent $\le 2$.

Here we study the general $l$-th function fields $K=k(\sqrt[l]{D})$, where $2 < l\leq 19$. Denote the Galois group of $K/k$ as $\text{Gal}(K/k)=\langle \sigma \rangle=\{1, \sigma, \sigma^2, \cdots, \sigma^{l-1}\}$. The integral closure of $R$ in $K$ (denoted as ${\cal O}_K$) is called the integral domain of $K$. The invertible elements (units) of ${\cal O}_K$ constitute a group $U_K$, the unit group of $K$ or ${\cal O}_K$.

$K$ is called a real $l$-th function field if $D$ is monic and the degree of the polynomial $D$ is a multiple of $l$, otherwise, we call $K$ a imaginary function field. In the real case, $U_K={\mathbb{F}}_q^\times \times V_K$, where $V_K$ is a free abelian group with rank $l-1$. A set of generators of it is called a basic system of units. Let $U_K=\mathbb{F}_q^\times \times V_K, $ where $V_K$ is a free abelian group with rank $l-1$. If there is an element $\varepsilon$ of $U_K$ satisfying

$ V_K=\langle\varepsilon, \varepsilon^\sigma, \varepsilon^{\sigma^2}, \cdots, \varepsilon^{\sigma^{l-2}}\rangle, $

then $\varepsilon$ is called a Minkowski unit of $K$. It is known (see [4]) that has Minkowski unit if $l$ is a prime number $\le 19$. Here and afterward, we assume that $l$ is a prime number $\le 19$.

Similar to number fields, the set $\mathcal{I}(K)$ of fractional ideals of $K$ is a group with respect to the multiplication of ideals. All the principal fractional ideals constitute a subgroup $\mathcal{P}(K)$ of $\mathcal{I}(K)$, the so called principal ideal subgroup. The quotient group $H(\cal{O}_K)=\mathcal{P}(K)/\mathcal{I}(K)$ is called the ideal class group of $K$. An ideal class containing ideal $\mathfrak{a}$ is denoted as $[\mathfrak{a}]$. It is a classical result that the ideal class group of $K$ is a finite abelian group. From the study of the properties and the constructions of the ambiguous ideal classes we'll prove the following theorems.

Theorem 1.1  Let $K=k(\sqrt[l]{D})$ with $D=a P_1(T)^{\alpha_1}P_2(T)^{\alpha_2}\cdots P_s(T)^{\alpha_s}$, where $a\in \mathbb{F}_q^{\times}$, $P_1(T), P_2(T), \cdots, P_s(T)$ are irreducible polynomials in $\mathbb{F}_q[T]$ and $1\leq \alpha_1, \alpha_2, \cdots, \alpha_s < l$. Then we have

$ \mbox{Rank}_l H(\mathcal{O}_K)\geq\left\{\begin{array}{l} s-2, \ \mbox{if}\ K\ \mbox{is real and }\ \text{N}\varepsilon=1; \\ s-1, \ \mbox{otherwise}. \\ \end{array} \right. $

Theorem 1.2  Suppose that $K=k(\sqrt[l]{D})$ is a real $l$-th cyclic function field and $\text{N}\varepsilon=1$. If $D=X^l-gY^l$ (where $X, Y\in R$, $g$ is a generator of $\mathbb{F}_q^\times$), then the ideal class group of $K$ has an ambiguous class not containing any ambiguous ideal.

Definition 2.1  An ideal $\mathfrak{a}$ of $K$ is called an ambiguous ideal of $K$ if $\mathfrak{a}^\sigma=\mathfrak{a}$. An ideal class $[\mathfrak{a}]$ of $K$ is called ambiguous if $[\mathfrak{a}]^\sigma=[\mathfrak{a}]$.

An ambiguous class is of order $1$ or $l$: if an ambiguous ideal class $[\mathfrak{a}]\neq [1]$, then

$ \begin{array}{rcl} [\mathfrak{a}]^l&=&[\mathfrak{a}]\cdot[\mathfrak{a}]^\sigma\cdots[\mathfrak{a}]^\sigma\\ &=&[\mathfrak{a}]\cdot[\mathfrak{a}]^{\sigma}\cdots[\mathfrak{a}]^{\sigma^2}\\ &=&\cdots\cdots\\ &=&[\mathfrak{a}]\cdot[\mathfrak{a}]^\sigma\cdots[\mathfrak{a}]^{\sigma^{l-1}}\\ &=&[\text{N}\mathfrak{a}]\\ &=&[1].\end{array} $

Assume that $D=P_1^{\alpha _1}P_2^{\alpha _2}\cdots P_s^{\alpha _s}$, then the principal ideal $(P_i)$ of $k$ factors as $(P_i)=\mathfrak{P}_{i}^{l}\ (i=1, 2, \cdots, s)$ in $K$. It is obvious that the ideals

$ \prod\limits_{i=1}\limits^{s}\mathfrak{P}_i^{s_i} \ (s_i\in\{0, 1, \cdots, l-1\})\quad \mbox{and}\quad (1), \quad (\sqrt[l]{D_i}) \ (i\in\{1, \cdots, l-1\}) $

are all ambiguous ideals of $K$, here $D_i$ is the $l$-free part of $D^i$. In addition, if $K$ is a real $l$-th cyclic function field and its Minkowski unit $\varepsilon$ satisfies $\text{N}_{K/k}\varepsilon=1$, from Hilbert theorem 90, there is an element $\gamma \in \mathcal{O}_K$ such that $\varepsilon=\gamma/\gamma^\sigma$. Under this condition,

$ (\gamma^i)\ \mbox{and} \ (\gamma^i\sqrt[l]{D_j})\ (i, j\in\{0, 1, \cdots, l-1\}) $

are also ambiguous ideals of $K$.

In fact, we have listed all the ambiguous ideals of $K$:

Lemma 2.2  If an ambiguous ideal $\mathcal{A}$ of $K$ does not have rational factors, it must be of the form

$ \mathcal{A}=\prod\limits_{i=1}\limits^{s}\mathfrak{P}_i^{s_i}, \quad \mbox{where}\ \mathfrak{P}_i|P_i, \ s_i\in\{0, 1, \cdots, l-1\}. $

Proof  Let $\mathcal{A}$ be an unprincipal ambiguous ideal without rational factors. It factors as

$ \mathcal{A}=\prod\limits_{i\in I}\mathfrak{P}_i^{a_i}\prod\limits_{j\in J}\mathfrak{Q}_j^{b_j}, $

where $\mathfrak{P}_i$'s are ramified prime ideals and $\mathfrak{Q}_j$'s are splitting ones. From the definition of 'ambiguous', we know that

$ \mathcal{A}={{\mathcal{A}}^{\sigma }}=\prod\limits_{i\in I}{\mathfrak{P}_{i}^{{{a}_{i}}}}\prod\limits_{j\in J}{\bar{\mathfrak{Q}}_{j}^{{{b}_{j}}}}, $

where $\bar{\mathfrak{Q}}={{\mathfrak{Q}}^{\sigma }}$. The uniqueness of factorization leads to $b_j=0, \forall j\in J$.

Lemma 2.3  A principal ambiguous ideal without rational factors must be of the following forms:

(1) $(\sqrt[l]{D_i}) \ (i\in\{0, 1, \cdots, l-1\})$;

(2) if $K$ is a real $l$-th cyclic function field and its Minkowfski unit $\varepsilon$ satisfies $\text{N}_{K/k}\varepsilon=1, $ then $\varepsilon$ can be written as $\varepsilon=\gamma/\gamma^\sigma$. Whence $(\gamma^i)\ \mbox{and} \ (\gamma^i\sqrt[l]{D_j})\ (i\in\{0, 1, \cdots, l-1\}, j\in\{1, \cdots, l-1\})$ are principal ambiguous ideals of $K$.

Proof  If $(z)$ is a principal ambiguous ideal without rational factors, where $z\in \mathcal{O}_K$, then $(z)=(z)^\sigma$. This means that $z/z^\sigma=u\in U_K, $ and $\text{N} u=1$.

Case 1  Suppose that $K$ is imaginary. Then $u\in \mathbb{F}_q^\times=\langle g\rangle$ and there exists a positive integer $m$ such that $u=g^{\frac {q-1}{l} m} \ (m \in \{1, \cdots, l-1\})$. Without lost of generality, we can choose that $\sqrt[l]{D}^\sigma =g^{\frac {1-q}{l}}\sqrt[l]{D}$. So we have

$ \sqrt[l]{D_m}^{\sigma} =g^{\frac {1-q}{l}m}\sqrt[l]{D_m}=u^{-1}\sqrt[l]{D_m}. $

It follows that

$ z/\sqrt[l]{D_m} =z^\sigma u/\sqrt[l]{D_m}=\left(z/\sqrt[l]{D_m}\right)^\sigma, $

and so $z/\sqrt[l]{D_m}\in k$. Because $(z)$ does not have rational factors, we have $(z)=(\sqrt[l]{D_m})$.

Case 2  Assume that $K$ is a real function field and $\text{N}(\varepsilon)=g^m\neq 1$ (where $\varepsilon$ is the Minkowski unit of $K$). Suppose, without lost of generality, let $m\in \{1, 2, \cdots, l-1\}$. Then

$ u=c\varepsilon_0^{a_0}\varepsilon_1^{a_1}\cdots\varepsilon_{l-2}^{a_{l-2}} , \ \mbox{where} \ c\in \mathbb{F}_q^\times;\quad \varepsilon_i=\varepsilon^{\sigma^i}, a_i\in \mathbb{Z} \ (i=0, 1, \cdots, l-2). $

Take norms of both side, we get $1=\text{N} u=c^lg^{(a_0+a_1+\cdots+a_{l-2})m}.$ Hence $l|(a_0+a_1+\cdots+a_{l-2})$.

Set

$\begin{array}{lcllcl} b_{l-2} = m(a_0+a_1+\cdots+a_{l-2})/l, \\ \beta_0 = b_{l-2}, \;\;\;\;\;\;\;\; \;\;\;\;\;\;\;\;b_0 = a_0-\beta_0; \\ \beta_1 = b_{l-2}-b_0, \;\;\;\;\;\;\;\;\;b_1 = a_1-\beta_1; \\ \cdots \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\cdots \\ \beta_{l-3} = b_{l-2}-b_{l-4}, \;\;\;\; b_{l-3} = a_{l-3}-\beta_{l-3}; \\ \beta_{l-2} = b_{l-2}-b_{l-3}. \end{array} $

Then we have

$ u=c\varepsilon_0^{b_0}\varepsilon_1^{b_1}\cdots\varepsilon_{l-2}^{b_l-2} \varepsilon_0^{\beta_0}\varepsilon_1^{\beta_1}\cdots\varepsilon_{l-2}^{\beta_l-2}. $

Let

$ \eta_1=\varepsilon_0^{b_0}\varepsilon_1^{b_1}\cdots\varepsilon_{l-2}^{b_{l-2}}, \quad\quad \eta_2=\varepsilon_0^{\beta_0}\varepsilon_1^{\beta_1}\cdots\varepsilon_{l-2}^{\beta_{l-2}}. $

We know from $\text{N}\varepsilon=g^m$ that $\varepsilon_{l-1}=g^m\varepsilon_0^{-1}\varepsilon_1^{-1}\cdots\varepsilon_{l-2}^{-1}$. It implies that

$ \eta_1^\sigma=\varepsilon_1^{b_0}\varepsilon_2^{b_1}\cdots\varepsilon_{l-1}^{b_{l-2}}\\ =\varepsilon_1^{b_0}\varepsilon_2^{b_1}\cdots\varepsilon_{l-2}^{b_{l-3}}\cdot g^{mb_{l-2}}\varepsilon_0^{-b_{l-2}}\varepsilon_1^{-b_{l-2}}\cdots\varepsilon_{l-2}^{-b_{l-2}}\\ =g^{mb_{l-2}}\varepsilon_0^{-b_{l-2}}\varepsilon_1^{b_0-b_{l-2}}\cdots\varepsilon_{l-2}^{b_{l-3}-b_{l-2}}\\ =g^{mb_{l-2}}\varepsilon_0^{-\beta_0}\varepsilon_1^{-\beta_1}\cdots\varepsilon_{l-2}^{-\beta_{l-2}}\\ =g^{mb_{l-2}}\eta_2^{-1}. $

That is

$ \begin{eqnarray*}&& z/z^\sigma=c\eta_1\eta_2=cg^{mb_{l-2}}\eta_1\eta_1^{-\sigma}, \\ && z/\eta_1=cg^{mb_{l-2}}(z^\sigma/\eta_1^\sigma)=c'(z/\eta_1)^\sigma, \quad \mbox{ where } \mbox { } c'\in \mathbb{F}_q^\times.\end{eqnarray*} $

Similar to Case 1, there is a $s\in \{0, 1, \cdots, l-1\}$ such that

$ (z)=(z/\eta _1)= \left(\sqrt[l]{D}^s\right)\ . $

Case 3   $K$ is a real function field and $\text{N}(\varepsilon)=1$. From Hilbert theorem 90, we know that there exists a $\gamma\in \mathcal{O}_K$ such that $\varepsilon=\gamma/\gamma^\sigma$. Denote $\gamma^{\sigma^i}=\gamma_i$, we have

$ \varepsilon_i=\varepsilon^{\sigma_i}= \gamma^{\sigma^i}/\gamma^{\sigma^{i+1}}=\gamma_i/\gamma_{i+1}. $

Hence

$ z/z^\sigma=u=c\varepsilon_0^{a_0}\varepsilon_1^{a_1}\cdots \varepsilon_{l-2}^{a_{l-2}}, \quad \mbox{where} \; c^l=1. $

Set $\tilde{\gamma}=\gamma_0^{a_0}\gamma_1^{a_1}\cdots\gamma_{l-2}^{a_{l-2}}.$ It follows that

$ z/\tilde{\gamma}=c(z/\tilde{\gamma})^\sigma, $

and so

$ (z)=(\gamma^s)\quad \mbox{or} \quad \left(\gamma^s\sqrt[l]{D}^t\right)\; \mbox{where } s, t\in \{1, 2, \cdots, l-1\}. $

Proof of Theorem 1.1  From the above lemmas the number $A$ of ideal classes containing ambiguous ideals of $K=(\sqrt[l]{aP_1(T)^{\alpha_1}P_2(T)^{\alpha_2}\cdots P_s(T)^{\alpha_s}})$ satisfies

$ A=\left\{ \begin{array}{ll} l^{s-2}&\mbox{ if }\ K \ \mbox{ is real and } \ \text{N}\varepsilon=1; \\ l^{s-1}&\mbox{ otherwise }. \\ \end{array}\right. $

This means

$ \mbox{Rank}_l H(\mathcal{O}_K)\geq\left\{\begin{array}{l} s-2, \ \mbox{ if }\ K\ \mbox{ is real and }\ \text{N}\varepsilon=1; \\ s-1, \ \mbox{otherwise}. \end{array} \right. $

Let's count all the ambiguous ideal classes of $K$.

Lemma 2.4  Denote the ideal class group of $K$ as $H(\mathcal{O}_K)$. Let $H(\mathcal{O}_K)^G$ express its subgroup consists of all the ambiguous ideal classes. Then

$ \left|H(\mathcal{O}_K)^G\right|=\frac{l^{s+\delta-1}} {(\mathbb{F}_q^\times:\text{N}_{K/k}K^\times \bigcap\mathbb{F}_q^\times)}. $

where $\delta=\left\{\begin{array}{c} 0, \ \mbox{ if }\ K \ \mbox{ is real ;}\\ 1, \ \mbox{ if }\ K \ \mbox{ is imaginary}. \end{array}\right. $

Proof  For simplicity, we denote the ideal group, the principal ideal group and the ideal class group of field $L$ as $I_L, \ P_L, \ {\rm and} \ C_L\ $ respectively. With our field $K$ we have exact sequence (see [5])

$ 0\longrightarrow P_K\longrightarrow I_K\longrightarrow C_K\longrightarrow 0. $

Because $H^1(\text{G}, I_K)=\oplus_\wp H^1(\text{G}_\wp, \mathbb{Z})$ and $ H^1(\text{G}_\wp, \mathbb{Z})=1$, we have $H^1(\text{N}, I_K)=1.$ That means that the exact sequence

$ 0\longrightarrow P_K^G\longrightarrow I_K^G\longrightarrow C_K^G \longrightarrow H^1(\text{G}, P_K) \longrightarrow 0 $

holds. It is the same to say that we have the following short exact sequence

$ 0\longrightarrow I_K^G/P_K^G \longrightarrow C_K^G \longrightarrow H^1(\text{G}, P_K) \longrightarrow 0. $

It follows that

$ \left|C_K^G\right|=(I_K^G:P_K^G)\cdot \# H^1(\text{G}, P_K). $

But we have $I_K^G\supset P_K^G\supset P_K$, so

$ (I_K^G:P_K^G)=(I_K^G:P_k)/(P_K^G:P_k)=(I_K^G:I_k)/(P_K^G:P_k) =e_0(K)/(P_K^G:P_k), $

where $e_0(K)=l^s$ is the product of the ramified indices of all the ramified prime ideals of $K$. It is seen from Hilbert theorem 90 that $H^1(\text{G}, K^\times)=0$. From the short exact sequence

$ 0\longrightarrow U_K\longrightarrow K^\times\longrightarrow P_K\longrightarrow 0 $

we obtain the long exact sequence

$ 0\longrightarrow U_k\longrightarrow k^\times\longrightarrow P_K^G \longrightarrow H^1(G, U_K)\longrightarrow 0. $

Moreover, we conclude that

$ 0\longrightarrow P_k\longrightarrow P_K^G\longrightarrow H^1(G, U_K)\longrightarrow 0 $

are exact. It means that

$ (P_K^G:P_k)=\# H^1(G, U_K)=\# H^0(G, U_K)/Q(G, U_K). $

where the Herband quotient

$ Q(G, U_K)=\frac{1}{l}\prod\limits_{v\in S_\infty}e_v f_v =\frac{1}{l}e_\infty f_\infty=\left\{\begin{array}{c} \frac{1}{l}, \ \mbox{ if }\ K\ \mbox{ is real };\\ 1, \ \mbox{ if }\ K\ \mbox{ is imaginary }. \end{array} \right. $

Thus we have

$ (P_K^G:P_k)=\frac{l\cdot\#H^0(G, U_K)}{e_\infty f_\infty} =\left\{\begin{array}{c} l\cdot\#H^0(G, U_K), \ \mbox{ if }\ K\ \mbox{ is real };\\ \#H^0(G, U_K), \ \mbox{ if }\ K\ \mbox{ is imaginary }. \end{array} \right. $

On the other hand, we have exact hexagon

$ \begin{array}{ccccc} &H^0(U_K)&\longrightarrow&H^0(K^\times)& \\ \nearrow&&&&\searrow\\ H^1(P_K) &&&&H^0(P_K) \\ \nwarrow&&&&\swarrow\\ &H^1(K^\times)&\longleftarrow&H^1(U_K)& \end{array} $

and get exact sequence

$ 0=H^1(K^\times)\longrightarrow H^1(P_K)\longrightarrow H^0(U_K) \longrightarrow H^0(K^\times). $

So

$ \# H^1(P_K)=\# \ker (U_k/\text{N}_{K/k}U_K\longrightarrow k^\times/\text{N}_{K/k}) =(\text{N}_{K/k}(K^\times)\cap U_k:\text{N}_{K/k}(U_K)). $

But $U_k \supset(\text{N}_{K/k}(K^\times)\bigcap U_k)\supset \text{N}_{K/k}(U_K)$, hence

$ \# H^1(G, P_K)=\# (U_k:\text{N}_{K/k}(U_K))/(U_k:\text{N}_{K/k}(K^\times)\bigcap U_k). $

It implies that

$ \left|H(\mathcal{O}_K)^G\right|=(I_K^G:P_K^G)\cdot\# H^1(G, P_K) =\frac{e_0e_\infty f_\infty}{l\cdot (\mathbb{F}_q^\times:\text{N}_{K/k}K^\times \bigcap\mathbb{F}_q^\times)}. $

Proof of Theorem 1.2  If $K$ is real, then

$ \left|H(\mathcal{O}_K)^G\right| =\frac{l^{s-1}}{(\mathbb{F}_q^\times:\text{N}_{K/k}K^\times \bigcap\mathbb{F}_q^\times)}. $

If $D=X^l-gY^l$, then

$ \eta =\frac {X}{Y}-\frac {\sqrt D}{Y}\in K^* \ {\rm and}\ N\eta=g. $

So $N_{K/k}K^*\bigcap \mathbb{F}_q^\times=\mathbb{F}_q^\times\ $ and $\ (H(O_K)^G)=l^{s-1}.$

On the other hand, if $N\varepsilon =1$, then there are $l^{s-2}$ ambiguous ideals. Thus we can see, in this case, there is an ambiguous ideal class not containing any ambiguous ideal.