Infinite families of Artin–Schreier function fields with any prescribed class group rank

Abstract

We study the Galois module structure of the class groups of the Artin–Schreier extensions K over k of extension degree p, where $k:={\mathbb F}_q(T)$ is the rational function field and p is a prime number. The structure of the p-part $Cl_K(p)$ of the ideal class group of K as a finite G-module is determined by the invariant ${\lambda }_n$, where $G:=\operatorname {\mathrm {Gal}}(K/k)=\langle {\sigma } \rangle $ is the Galois group of K over k, and ${\lambda }_n = \dim _{{\mathbb F}_p}(Cl_K(p)^{({\sigma }-1)^{n-1}}/Cl_K(p)^{({\sigma }-1)^{n}})$. We find infinite families of the Artin–Schreier extensions over k whose ideal class groups have guaranteed prescribed ${\lambda }_n$-rank for $1 \leq n \leq 3$. We find an algorithm for computing ${\lambda }_3$-rank of $Cl_K(p)$. Using this algorithm, for a given integer $t \ge 2$, we get infinite families of the Artin–Schreier extensions over k whose ${\lambda }_1$-rank is t, ${\lambda }_2$-rank is $t-1$, and ${\lambda }_3$-rank is $t-2$. In particular, in the case where $p=2$, for a given positive integer $t \ge 2$, we obtain an infinite family of the Artin–Schreier quadratic extensions over k whose $2$-class group rank (resp. $2^2$-class group rank and $2^3$-class group rank) is exactly t (resp. $t-1$ and $t-2$). Furthermore, we also obtain a similar result on the $2^n$-ranks of the divisor class groups of the Artin–Schreier quadratic extensions over k.

1 Introduction

There have been active studies on the structure of the class groups of number fields and function fields; for instance, we refer to [1–5, 6, 8, 10, 11, 13–16, 19–25]. For studying the structure of class groups, the following methods have been used: genus theory [1, 3, 6], Rédei matrix [2, 15, 23], and Conner and Hurrelbrink’s exact hexagon [5, 13].

The Galois module structure of the class groups of cyclic extensions over the rational function field $k:={\mathbb F}_q(T)$ has been studied in [2, 8, 14, 19], where ${\mathbb F}_q$ is a finite field of order q. We need to introduce the following definitions for description of the previous developments. Let K be a cyclic extension over k of extension degree prime p. We denote the ideal class group of K by $Cl_K$ and that of divisor class group by $J_K$ . Let $G := \mathrm{Gal}(K/k)$ be the Galois group of K over k. Then $Cl_K$ and $J_K$ are finite G-modules. Let ${\sigma }$ be a generator of G and ${\mathbb Z}_p$ the ring of p-adic integer. The structures of $Cl_K(p)$ and $J_K(p)$ as finite modules over the discrete valuation ring ${\mathbb Z}_p[{\sigma }]/(1 + {\sigma } + \cdots + {\sigma }^{p-1}) \simeq {\mathbb Z}_p[\zeta _p]$ are determined by the following ranks:

$$ \begin{align*} {\lambda}_n &:= \dim_{{\mathbb F}_p}(Cl_K(p)^{({\sigma}-1)^{n-1}}/Cl_K(p)^{({\sigma}-1)^n}) \quad \mbox{and}\\ {\mu}_n &:= \dim_{{\mathbb F}_p}(J_K(p)^{({\sigma}-1)^{n-1}}/J_K(p)^{({\sigma}-1)^n}), \end{align*} $$

where $Cl_K(p)$ (resp. $J_K(p)$ ) is the p-Sylow subgroup of $Cl_K$ (resp. $J_K$ ) and $\zeta _p$ is a primitive pth root of unity.

We point out that in particular, when $p = 2$ , the rank ${\lambda }_n$ of $Cl_K$ is exactly equal to the $2^n$ -rank of $Cl_K$ and the rank ${\mu }_n$ of $J_K$ gives the $2^n$ -rank of $J_K$ , where the $2^n$ -rank of $Cl_K$ is defined as $\dim _{{\mathbb F}_2}(Cl_K^{2^{n-1}} /Cl_K^{2^n})$ and similarly for $J_K$ . This is because ${\sigma }$ acts $-1$ on $Cl_K$ , which implies that the rank ${\lambda }_n$ of the finite module $Cl_K$ over ${\mathbb Z}[\zeta _2] = {\mathbb Z}$ is exactly the $2^n$ -rank of $Cl_K$ , and similarly it also holds for $J_K$ .

There are exactly two kinds of cyclic extensions of prime extension degree over the rational function field k: Kummer extension and Artin–Schreier extension. For Kummer extensions L over k, Anglés and Jaulent [1] (resp. Wittmann [19]) studied the ${\lambda }_1$ -rank (resp. ${\lambda }_2$ -rank) of the ideal class groups of L and the authors of this paper [22] studied the ${\lambda }_3$ -rank of the ideal class groups of L. Furthermore, for Artin–Schreier extensions over k, there also have been some studies on the computation of ${\lambda }_1$ and ${\lambda }_2$ for their ideal class groups [2, 8]. However, there has been no result yet on finding infinite families of Artin–Schreier extensions over k whose ideal class groups have guaranteed prescribed ${\lambda }_n$ -rank of the ideal class group of Artin–Schreier extension for $1 \leq n \leq 3$ . This is one of the motivations of our paper.

In this paper, we study the Galois module structure of the class groups of the Artin–Schreier extensions K over k of extension degree p, where $k:={\mathbb F}_q(T)$ is the rational function field of characteristic p and p is a prime number. The structure of the p-part $Cl_K(p)$ of the ideal class group of K as a finite G-module is determined by the invariant ${\lambda }_n$ , where $G:=\operatorname {\mathrm {Gal}}(K/k)=\langle {\sigma } \rangle $ . In detail, first of all, for a given positive integer t, we obtain infinite families of K over k whose ${\lambda }_1$ -rank of $Cl_K$ is t and ${\lambda }_n$ -rank of $Cl_K$ is zero for $n \ge 2$ , depending on the ramification behavior of the infinite place $\infty $ of k (Theorems 3.2–3.4). We then find infinite families of the Artin–Schreier extensions over k whose ideal class groups have guaranteed prescribed ${\lambda }_n$ -rank for n up to 3. We find an algorithm for computing ${\lambda }_3$ -rank of $Cl_K(p)$ . Using this algorithm, for a given integer $t \ge 2$ , we get infinite families of the Artin–Schreier extensions over k whose ${\lambda }_1$ -rank is t, ${\lambda }_2$ -rank is $t-1$ , and ${\lambda }_3$ -rank is $t-2$ (Theorem 5.1). In particular, in the case where $p=2$ , for a given positive integer $t \ge 2$ , we obtain an infinite family of the Artin–Schreier quadratic extensions over k which have 2-class group rank exactly t, $2^2$ -class group rank $t-1$ , and $2^3$ -class group rank $t-2$ (Corollary 5.3). Furthermore, we also obtain a similar result on the $2^n$ -ranks of the divisor class groups of the Artin–Schreier quadratic extensions over k for n up to 3 (Corollary 5.4). Finally, in Tables 1 and 2, we give some implementation results for explicit infinite families using Theorems 3.2–3.4 and 5.1. These implementation results are done by MAGMA.

We remark that as a main tool for computation of ${\lambda }_3$ , we use an analogue of Rédei matrix. We emphasize that there is no number field analogue for the Artin–Schreier extensions over k, while there is a number field analogue for Kummer extensions over k.

2 Preliminaries

Let q be a power of a prime number p, and let $k := {\mathbb F}_q(T)$ be the rational function field. The prime divisor of k corresponding to $(1/T)$ is called the infinite place and denoted by $\infty $ . Let $K/k$ be a cyclic extension of degree p. Then $K/k$ is an Artin–Schreier extension: that is, $K = k({\alpha })$ , where ${\alpha }^p-{\alpha } = D$ , $D \in k$ , and that D cannot be written as $x^p-x$ for any $x \in k$ . Conversely, for any $D \in k$ and D cannot be written as $x^p-x$ for any $x \in k$ , $k({\alpha })/k$ is a cyclic extension of degree p, where ${\alpha }^p-{\alpha }=D$ .

For $D, D' \in k$ , let $K_1:=k({\alpha })$ and $K_2:=k({\beta })$ be two Artin–Schreier extensions over k with ${\alpha }^p - {\alpha } = D$ and ${\beta }^p-{\beta } = D'$ , respectively. Two Artin–Schreier extensions $K_1$ and $K_2$ are equal if and only if they satisfy the following relations [8, p. 256]:

$$ \begin{align*} &{\alpha} \quad \rightarrow \quad x{\alpha} + B_0 = {\beta},\\[2pt] &D \quad \rightarrow \quad xD+(B_0^p-B_0) = D',\\[2pt] &x \in {\mathbb F}_p^\times, ~B_0 \in k. \end{align*} $$

Thus, D can be normalized to satisfy the following conditions:

(2.1) $$ \begin{align} &D = \sum_{i=1}^{m} \frac{Q_i}{P_i^{r_i}} + f(T),\\ &(P_i, Q_i) = 1, ~p \nmid r_i \mbox{ for } 1 \leq i \leq t,\nonumber \\ &p \nmid \deg f(T) \; \mbox{ if } \deg {f(T)} \geq 1, \mbox{ and} \nonumber\\ & {f(T)} =0 \; \mbox{ if } f(T) \in {\mathbb F}_q \mbox{ with } \operatorname{\mathrm{Tr}}_{{\mathbb F}_q/{\mathbb F}_p} (f) = 0, \nonumber \end{align} $$

where $P_i$ is a monic irreducible polynomial in ${\mathbb F}_q[T]$ , $Q_i$ , $f(T) \in {\mathbb F}_q[T]$ , and $\deg {Q_i} < \deg {P_i^{r_i}}$ for $1 \leq i \leq t$ ; the last condition follows from noting that if $f(T)=c$ in ${\mathbb F}_q^\times $ with $\operatorname {\mathrm {Tr}}_{{\mathbb F}_q/{\mathbb F}_p}(c) = 0$ , then there exists $b \in {\mathbb F}_q^\times $ such that $b^p-b = c$ .

Throughout this paper, let be the Artin–Schreier extension over k of extension degree p, where $x^p-x = D_m$ has no root in k, is a root of $x^p-x=D_m$ , and the normalized $D_m$ satisfies (2.1). We note that all the finite places of k which are totally ramified in K are $P_1, \dots , P_t$ . In the following lemma, we state the ramification behavior of the infinite place $\infty $ of k in K.

Lemma 2.1 [8, p. 256]

Let be the Artin–Schreier extension over k of extension degree p, where and $D_m$ is defined in (2.1). Then we have the followings.

1. (i) The infinite place $\infty $ of k is totally ramified in K if and only if $\deg {f(T)} \geq 1$ .

2. (ii) The infinite place $\infty $ of k is inert in K if and only if $f(T) = c \in {\mathbb F}_q^\times $ , where $x^p-x-c$ is irreducible over ${\mathbb F}_q$ .

3. (iii) The infinite place $\infty $ of k splits completely in K if and only if $f(T) = 0$ .

For descriptions of ${\lambda }_1$ and ${\lambda }_2$ , we use the notion of the Hasse symbol which is first introduced in [7].

Definition 2.1 [8, p. 257]

Let be the Artin–Schreier extension over k of extension degree p, where for some $D_m \in k$ . Let P be a finite place of k which is unramified in K, and let $\left (\frac {K/k}{P}\right )$ be the Artin symbol of P. Then , where $\left \{\frac {D_m}{P}\right \}$ is defined as follows:

$$ \begin{align*}\left \{\frac{D_m}{P}\right \} = \operatorname{\mathrm{Tr}}_{(\mathcal{O}_K/P)/{\mathbb F}_p}(D_m \mbox{ mod } P);\end{align*} $$

$\operatorname {\mathrm {Tr}}_{(\mathcal {O}_K/P)/{\mathbb F}_p}$ denotes the trace function from $\mathcal {O}_K/P$ to ${\mathbb F}_p$ and $\mathcal {O}_K$ is the integral closure of K. We call $\left \{\frac {\cdot }{\cdot }\right \}$ the Hasse symbol.

Lemma 2.2 [8]

Let be the Artin–Schreier extension over k of extension degree p, where , which is defined in (2.1). Then we have the followings.

1. (i) ${\lambda }_1 = \left \{ \begin {array}{ll} m & \mbox {if}\ \deg f(T) \ge 1\ \mbox { or}\\&\ \ \ \ f(T) = c \in {\mathbb F}_q^\times, \mbox { where}\ x^p-x=c \in {\mathbb F}_q^\times \ \mbox {is irreducible over}\ {\mathbb F}_q,\\[3pt] m-1 & \mbox {if}\ f(T) = 0. \\ \end {array} \right.$

2. (ii) We have ${\lambda }_2 = {\lambda }_1 - \operatorname {\mathrm {rank}}(R)$ , where $R = [r_{ij}]$ is a matrix over ${\mathbb F}_p$ defined by

   $$ \begin{align*}r_{ij} = \left\{ \begin{array}{ll} \left\{\frac{Q_j/P_j^{r_j}}{P_i} \right\}, & \mbox{for } 1 \leq i \neq j \leq m,\\[4pt] -\left(\sum_{j=1, i \neq j}^{m} r_{ij} + \left\{\frac{f}{P_i}\right\} \right), & \mbox{for } 1 \leq i=j \leq m. \end{array} \right.\end{align*} $$

3. We call such matrix R as the Rédei matrix.

We recall that the Hilbert class field $H_K$ of K is the maximal unramified abelian extension of K where the infinite places of k split completely in K. The genus field $\mathcal {G}_K$ of K is the maximal subextension $K \subseteq \mathcal {G}_K \subseteq H_K$ which is abelian over k. In Lemma 2.3, we state a description of the genus field of the Artin–Schreier extension.

Lemma 2.3 [8, Theorem 4.1]

Let be the Artin–Schreier extension over k of extension degree p, where $D_m$ is defined in (2.1) and is a root of $x^p-x=D_m$ . Let ${\alpha }_i\ ($ resp. ${\beta })$ be a root of $x^p-x = Q_i/P_i^{r_i}$ for $1 \leq i \leq m\ ($ resp. $x^p-x = f(T))$ in $\overline {k}$ . Then the genus field $\mathcal {G}_{K}$ of K is $\mathcal {G}_{K} = k({\alpha }_1, \dots , {\alpha }_m, {\beta }).$

We now introduce explicit criteria for determining whether a place of k is totally ramified or not in the Artin–Schreier extension K.

Lemma 2.4 [18, Proposition 3.7.8]

Let $K=k(y)$ be the Artin–Schreier extension over k of extension degree p, where $y^p-y=u$ for some $u \in k$ . For a place P of k, we define the integer $m_P$ by

$$ \begin{align*}m_P := \left\{ \begin{array}{ll} m, & \mbox{if there is an element}\ z \in k\ \mbox{satisfying} \\ & \upsilon_P(u-(z^p-z)) = -m < 0\ \mbox{and}\ m \not \equiv 0\quad \pmod p, \\ -1, & \mbox{if}\ \upsilon_P(u-(z^p-z)) \geq 0\ \mbox{for some}\ z \in k. \end{array} \right.\end{align*} $$

Then we have the followings.

1. (i) P is totally ramified in $K/k$ if and only if $m_P> 0$ .

2. (ii) P is unramified in $K/k$ if and only if $m_P = -1$ .

Lemma 2.5 [17, Proposition 14.1]

Let K be a function field over the rational function field $k={\mathbb F}_q(T)$ , and let $\infty $ be the infinite place of k. Denote the ideal class group (resp. the divisor class group) of K by $Cl_K$ (resp. $J_K$ ) and S be a set of places of K lying over $\infty $ . Then

$$ \begin{align*}0 \rightarrow \mathcal{D}^0_K(S)/\mathcal{P}_K(S) \rightarrow J_K \rightarrow Cl_K \rightarrow {\mathbb Z}/d{\mathbb Z} \rightarrow 0\end{align*} $$

is an exact sequence, where $\mathcal {D}_K^{0}(S)$ is the divisor group with support only in S whose degree is zero, $\mathcal {P}_K(S)$ is a principal divisor with support only in S, and d is the greatest common divisor of the elements in $\{\deg P : P \in S \}$ .

Using Lemma 2.5, we can easily obtain the following corollary, which gives relation between the ideal class group of K and the divisor class group of K, where K is the Artin–Schreier function field over k.

Lemma 2.6 Let K be the Artin–Schreier extension over k with extension degree p, and let all the notations be the same as in Lemma 2.5. Then we have the following.

1. (i) If $\infty $ is totally ramified in K, then $\mathcal {D}_K^0(S)$ is trivial and $d = 1$ ; thus,

   $$ \begin{align*}0 \rightarrow J_K \rightarrow Cl_K \rightarrow 0\end{align*} $$
   is exact.

2. (ii) If $\infty $ is inert in K, then $\mathcal {D}_K^0(S)$ is trivial and $d = p$ ; therefore,

   $$ \begin{align*}0 \rightarrow J_K \rightarrow Cl_K \rightarrow {\mathbb Z}/p{\mathbb Z} \rightarrow 0\end{align*} $$
   is an exact sequence.

3. (iii) If $\infty $ splits completely in K, then $d = 1$ ; thus,

   $$ \begin{align*}0 \rightarrow \mathcal{D}_K^{0}(S)/\mathcal{P}_K(S) \rightarrow J_K \rightarrow Cl_K \rightarrow 0\end{align*} $$
   is exact.

3 Infinite families of Artin–Schreier function fields with any prescribed class group ${\lambda }$ -rank

In this section, for any positive integer t, we find infinite families of Artin–Schreier function fields K over k whose ${\lambda }$ -rank of the ideal class group $Cl_K$ of K is t and ${\lambda }_n$ -rank of $Cl_K$ is zero for $n \ge 2$ , depending on the ramification behavior of the infinite place $\infty $ of k. Theorem 3.2 deals with the case where the infinite place $\infty $ of k is totally ramified in K and Theorem 3.3 (resp. Theorem 3.4) treats the case where the infinite place $\infty $ of k splits completely (resp. $\infty $ is inert) in K.

We first give the following lemma, which shows the property of the trace over finite fields. This lemma plays a key role in the proofs of Theorems 3.2–3.4.

Lemma 3.1 Let h be a monic irreducible polynomial in ${\mathbb F}_q[T]$ and $\mathfrak {h} := q^{\deg h}$ . Let g be a nonzero element in ${\mathbb F}_q[T]$ , and let $\tilde {g} \in {\mathbb F}_{\mathfrak {h}}$ be $\phi \circ \pi (g)$ , where

$$ \begin{align*}g \in {\mathbb F}_q[T] \xrightarrow{\hspace{0.2cm}\pi\hspace{0.2cm}} \pi(g) \in {\mathbb F}_q[T]/{\langle} h {\rangle} \xrightarrow{\hspace{0.2cm}\phi\hspace{0.2cm}} {\mathbb F}_{\mathfrak{h}}.\end{align*} $$

Then we have $\operatorname {\mathrm {Tr}}_{{\mathbb F}_{\mathfrak {h}}/{\mathbb F}_q} \widetilde {g} = 0$ if and only if the following holds:

1. (i) If $\deg g =0$ , then $q \mid \deg h$ .

2. (ii) If $\deg g \geq 1$ , then $g \equiv b(T)^q-b(T)\ \pmod {h}$ for some $b(T) \in {\mathbb F}_q[T]$ .

Proof We note that ${\mathbb F}_{\mathfrak {h}} \simeq {\mathbb F}_q[T]/{\langle } h {\rangle }$ since h is an irreducible polynomial over ${\mathbb F}_q$ .

First, assume that $\deg g = 0$ : that is, g is an element of ${\mathbb F}_q^\times $ , and so $g = \tilde {g}$ . Then we have the following:

$$ \begin{align*}\operatorname{\mathrm{Tr}}_{{\mathbb F}_{\mathfrak{h}}/{\mathbb F}_q} \tilde{g} = 0 \text{ if and only if } q \mid \deg h;\end{align*} $$

this is because $\operatorname {\mathrm {Tr}}_{{\mathbb F}_{\mathfrak {h}}/{\mathbb F}_q} \tilde {g} = \tilde {g}\cdot \deg h$ in ${\mathbb F}_q$ .

Now, we consider the case where $\deg g \geq 1$ . Assume that $g \equiv b(T)^q-b(T)\ \pmod h$ . Then we have

$$ \begin{align*}\tilde{g} = \phi \circ \pi (g) = \phi ((b(T))^q-(b(T))) = \phi(b(T))^q - \phi (b(T)) = \tilde{b}^q - \tilde{b},\end{align*} $$

where $\tilde {b} := \phi (b(T)) \in {\mathbb F}_{\mathfrak {h}}$ . Therefore, the result follows immediately by [12, Theorem 2.25]. Conversely, now assume that $\operatorname {\mathrm {Tr}}_{{\mathbb F}_{\mathfrak {h}/{\mathbb F}_q}}(\tilde {g}) = 0$ : that is, there exists some $\tilde {b}\in {\mathbb F}_{\mathfrak {h}}$ such that $\tilde {g} = \tilde {b}^q - \tilde {b}$ . Let $b(T) := \phi ^{-1}(\tilde {b})$ ; there exists such $b(T) \in {\mathbb F}_q[T]$ since $\phi $ is isomorphism. Thus, we get

$$ \begin{align*}g= \pi^{-1} \circ \phi^{-1}(\tilde{g}) = \pi^{-1} \circ \phi^{-1}(\tilde{b}^q - \tilde{b}) = \pi^{-1}((b(T))^q-(b(T)));\end{align*} $$

this implies that $g \equiv b(T)^q - b(T)\ \pmod h$ .

Theorem 3.2 Let t be a positive integer. Let

be the Artin–Schreier extension over the rational function field $k = {\mathbb F}_q(T)$ of extension degree p, where

satisfies (2.1). Assume that the infinite place $\infty $ of k is totally ramified in K; equivalently, $\deg f(T) \ge 1$ with $p \nmid \deg f(T)$ . We further assume that the followings hold:

1. (i) $p \nmid \deg P_i$ for any i with $1 \leq i \leq t$ .

2. (ii) $f(T) \equiv \mathfrak {c}_i\ \pmod {P_i}$ , where $\mathfrak {c}_i \in {\mathbb F}_q^\times $ such that $\operatorname {\mathrm {Tr}}_{{\mathbb F}_q/{\mathbb F}_p}(\mathfrak {c}_i) \ne 0$ for any i with $1 \leq i \leq t$ .

3. (iii) $Q_j \equiv {P_j}^{r_j}(b_i(T)^q-b_i(T))\ \pmod {P_i}$ for any i with $1 \leq i \ne j \leq t$ , where $b_i(T)$ is a polynomial in ${\mathbb F}_q[T]$ .

Then the ${\lambda }_1$ -rank of the ideal class group $Cl_K$ of K and ${\mu }_1$ -rank of the divisor class group $J_K$ of K are t. Moreover, for $n \ge 2$ , the ${\lambda }_n$ -rank of $Cl_K$ and the ${\mu }_n$ -rank of $J_K$ are zero.

In particular, for the case when $p=2$ , the $2$ -class groups $Cl_K(2)$ and $J_K(2)$ are elementary abelian $2$ -groups: that is, isomorphic to $({\mathbb Z}/2{\mathbb Z})^{t}$ .

Proof We note that by Lemma 2.6, the ideal class group of K and the divisor class group of K are isomorphic; thus, ${\lambda }_n = {\mu }_n$ for $n \ge 1$ . Since ${\lambda }_n$ is a decreasing sequence as n grows ( ${\lambda }_{n-1}$ and ${\lambda }_n$ may have the same value), it suffices to show the following:

(3.1) $$ \begin{align} {\lambda}_1 = t \quad \mbox{and} \quad {\lambda}_2 = 0. \end{align} $$

By Lemma 2.2, we can easily get ${\lambda }_1 = t$ . Thus, we will show that the rank of R is t, where R is the Rédei matrix over ${\mathbb F}_p$ which is defined in Lemma 2.2.

Let $f(T)$ be a polynomial in ${\mathbb F}_q[T]$ which satisfies condition (ii). For convenience, let ${\delta }_i:=\deg P_i$ for $1 \leq i \leq t$ . Then we have the following:

$$ \begin{align*}\operatorname{\mathrm{Tr}}_{{\mathbb F}_{q^{{\delta}_i}}/{\mathbb F}_q}(f\quad \pmod {P_i}) = \operatorname{\mathrm{Tr}}_{{\mathbb F}_{q^{{\delta}_i}}/{\mathbb F}_q} \mathfrak{c}_i = {\delta}_i \mathfrak{c}_i;\end{align*} $$

the last equality follows from the fact that $\mathfrak {c}_i \in {\mathbb F}_q^\times $ . Thus, by the definition of the Hasse symbol, we obtain

(3.2) $$ \begin{align} \left\{ \frac{f(T)}{P_i} \right\} = \operatorname{\mathrm{Tr}}_{{\mathbb F}_q/{\mathbb F}_p} (\operatorname{\mathrm{Tr}}_{{\mathbb F}_{q^{{\delta}_i}}/{\mathbb F}_q} (f\quad \pmod {P_i})) = \operatorname{\mathrm{Tr}}_{{\mathbb F}_q/{\mathbb F}_p} ({\delta}_i\mathfrak{c}_i) = {\delta}_i\operatorname{\mathrm{Tr}}_{{\mathbb F}_q/{\mathbb F}_p} \mathfrak{c}_i \ne 0; \end{align} $$

for the last equality, we use conditions (i) and (ii).

Now, let $Q_j$ ( $1 \leq j \leq t$ ) be a polynomial in ${\mathbb F}_q[T]$ which satisfies condition (iii). Then, for $1 \leq i \ne j \leq t$ , we have

$$ \begin{align*}Q_j{\overline{P_j}}^{r_j} \equiv b_i(T)^q - b_i(T)\quad \pmod {P_i},\end{align*} $$

where $P_j\overline {P_j} \equiv 1\ \pmod {P_i}$ . We note that $\overline {P_j}$ always exist since $P_i$ and $P_j$ are relative prime in ${\mathbb F}_q[T]$ . Then, by Lemma 3.1, we obtain $\operatorname {\mathrm {Tr}}_{{\mathbb F}_{{\delta }_i}/{\mathbb F}_q} (Q_j{\overline {P_j}}^{r_j}\ \pmod {P_i}) = 0$ , where ${\delta }_i:=\deg P_i$ . Thus, we obtain

(3.3) $$ \begin{align} \left\{ \frac{Q_j/{P_j}^{r_j}}{P_i} \right\} = \operatorname{\mathrm{Tr}}_{{\mathbb F}_q/{\mathbb F}_p}(\operatorname{\mathrm{Tr}}_{{\mathbb F}_{q^{{\delta}_i}}/{\mathbb F}_q} (Q_j\overline{P_j}^{r_j}\quad \pmod {P_i}) = \operatorname{\mathrm{Tr}}_{{\mathbb F}_q/{\mathbb F}_p} 0 = 0. \end{align} $$

Therefore, we get a $t \times t$ Rédei matrix $R = [r_{ij}]$ over ${\mathbb F}_p$ as follows:

(3.4) $$ \begin{align} {\small{R = \left[ \begin{array}{cccc} r_{11} & 0 & \cdots & 0\\ 0 & r_{22} & \cdots & 0\\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0& \cdots & r_{tt} \\ \end{array} \right]}}, \end{align} $$

where $r_{ii} = \left \{ \frac {f(T)}{P_i} \right \} \ne 0$ in ${\mathbb F}_p$ for every $1 \leq i \leq t$ . We can easily check that the rank of R is t; therefore, we get ${\lambda }_2 = {\lambda }_1 - \operatorname {\mathrm {rank}}(R) = 0$ .

For the case where $p=2$ , the $2^n$ -rank of $Cl_K$ and that of $J_K$ are exactly $\lambda _n$ and ${\mu }_n$ , respectively; therefore, $Cl_K(2) \simeq J_K(2) \simeq ({\mathbb Z}/2{\mathbb Z})^t$ .

Theorem 3.3 Let t be a positive integer. Let

be the Artin–Schreier extension over the rational function field $k = {\mathbb F}_q(T)$ of extension degree p, where

satisfies (2.1). Assume that the infinite place $\infty $ splits completely in K; equivalently, $f(T)=0$ . We further assume that the followings hold:

1. (i) $p \nmid \deg P_i$ for any i with $1 \leq i \leq t+1$ .

2. (ii) $Q_t \equiv \mathfrak {c}_iP_t^{r_t}\ \pmod {P_i}$ , where $\mathfrak {c}_i \in {\mathbb F}_q^\times $ such that $\operatorname {\mathrm {Tr}}_{{\mathbb F}_q/{\mathbb F}_p}(\mathfrak {c}_i) \ne 0$ for any i with ${1 \leq i \leq t}$ .

3. (iii) $Q_j \equiv P_j^{r_j}(b_i(T)^q-b_i(T))\ \pmod {P_i}$ for any $1 \leq i \leq t+1$ , $1 \leq j \leq t$ , $i \ne j$ , where $b_i(T) \in {\mathbb F}_q[T]$ .

Then the ${\lambda }_1$ -rank of the ideal class group $Cl_K$ of K is t. Moreover, for $n \ge 2$ , the ${\lambda }_n$ -rank of $Cl_K$ is zero.

In particular, for the case when $p=2$ , the $2$ -class group $Cl_K(2)$ is an elementary abelian $2$ -group: that is, isomorphic to $({\mathbb Z}/2{\mathbb Z})^{t}$ .

Proof As in Theorem 3.2, we will show (3.1). The fact that ${\lambda }_1 = t$ comes immediately from Lemma 2.2. Thus, it is sufficient to show that $\lambda _2=0$ : that is, $\operatorname {\mathrm {rank}}(R) = {\lambda }_1 = t$ , where R is the Rédei matrix of K defined in Lemma 2.2.

Let $D_i:=\frac {Q_i}{P_i^{r_i}}$ for $1 \leq i \leq t+1$ . Using the same reasoning as in Theorem 3.2, we get $\left \{D_t/P_i \right \} \ne 0$ for every $1 \leq i \leq t$ ; we note that we use conditions (i) and (ii). Thus, the $i(t+1)$ th entry of R is nonzero for $1 \leq i \leq t$ . By condition (iii), we obtain $\left \{ D_j/ P_i \right \} =0$ from Lemma 3.1; this implies that the $ij$ th entries of R are all zero for $1 \leq i \leq t+1$ and $1 \leq j \leq t$ with $i \ne j$ .

Therefore, we obtain a $(t+1) \times (t+1)$ matrix $R = [r_{ij}]$ over ${\mathbb F}_p$ as follows:

$$ \begin{align*}{\small{R = \left[ \begin{array}{ccccl} -r_{1,t+1} & 0 & \cdots & 0 & r_{1,t+1}\\ 0 & -r_{2,t+1} & \cdots & 0 & r_{2,t+1}\\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0& \cdots & -r_{t,t+1} & r_{t,t+1}\\ 0 & 0 & \cdots & 0 & 0 \\ \end{array} \right]}},\end{align*} $$

where $r_{i,t+1} \ne 0$ in ${\mathbb F}_p$ for every $1 \leq i \leq t$ . Thus, the result follows immediately.

For the case where $p=2$ , since ${\lambda }_n$ gives the full $2^n$ -rank of $Cl_K$ , we obtain that $Cl_K(2) \simeq ({\mathbb Z}/2{\mathbb Z})^t$ .

Theorem 3.4 Let t be a positive integer. Let

be the Artin–Schreier extension over the rational function field $k = {\mathbb F}_q(T)$ of extension degree p, where

satisfies (2.1). Assume that $\infty $ is inert in K; equivalently, $f(T) = c \in {\mathbb F}_q^\times $ , where $x^p-x-c$ is irreducible over ${\mathbb F}_q$ . We further assume that the followings hold: for some $\mathfrak {c} \in {\mathbb F}_q$ ,

1. (i) $p \nmid \deg {P_i}$ for every $1 \leq i \leq t$ .

2. (ii) $Q_j \equiv P_j^{r_j}(b_i(T)^q-b_i(T))$ for any i with $1 \leq i \ne j \leq t$ , where $b_i(T) \in {\mathbb F}_q[T]$ .

Then the ${\lambda }_1$ -rank of the ideal class group $Cl_K$ of K is t. Moreover, for $n \ge 2$ , the ${\lambda }_n$ -rank of $Cl_K$ is zero.

In particular, for the case when $p=2$ , then $Cl_K(2)$ is isomorphic to $({\mathbb Z}/2{\mathbb Z})^t$ and $J_K(2)$ is isomorphic to $({\mathbb Z}/2{\mathbb Z})^{t-1}$ .

Proof We can simply get ${\lambda }_1 = t$ by Lemma 2.2; we now show that ${\lambda }_2 = 0$ , which implies that the rank of the Rédei matrix R is t. As usual, set $D_i:=\frac {Q_i}{P_i^{r_i}}$ . Using Lemma 3.1, we obtain $\left \{ D_j/ P_i \right \} = 0$ for every $1 \leq i \ne j \leq t$ . Now, we compute $\left \{ c/P_i\right \}$ for $1 \leq i \leq t$ , where $c \in {\mathbb F}_q^\times $ . Let ${\delta }_{i}$ be the degree of $P_i$ . By the definition of Hasse norm, we have

(3.5) $$ \begin{align} \left\{ \frac{c}{P_i} \right\} &=& \operatorname{\mathrm{Tr}}_{{\mathbb F}_q/{\mathbb F}_p} \operatorname{\mathrm{Tr}}_{{{\mathbb F}_q^{{\delta}_i}}/{\mathbb F}_q}(c\quad \pmod {P_i}) = \operatorname{\mathrm{Tr}}_{{\mathbb F}_{q}/{\mathbb F}_p} ({\delta}_i c) = {\delta}_i\operatorname{\mathrm{Tr}}_{{\mathbb F}_q/{\mathbb F}_p} (c). \end{align} $$

We note that $\operatorname {\mathrm {Tr}}_{{\mathbb F}_q/{\mathbb F}_p}(c) \ne 0$ since $x^p-x-c$ is irreducible over ${\mathbb F}_q$ . Therefore, (3.5) is nonzero; we use condition (i). Using the definition of the Rédei matrix R in Lemma 2.2, we get a $t \times t$ matrix $R = [r_{ij}]$ over ${\mathbb F}_p$ which is given in (3.4). Hence, the desired result follows.

For the case where $p=2$ , the 2-class group of $Cl_K$ is isomorphic to $({\mathbb Z}/2{\mathbb Z})^t$ by the fact that ${\lambda }_n$ gives the full $2^n$ -rank of $Cl_K$ . By Lemma 2.6, the remaining result follows.

4 Computing the ${\lambda }_3$ -rank of class groups of Artin–Schreier function fields

In this section, Algorithm 1 presents an explicit method for computing the ${\lambda }_3$ -rank of the ideal class groups of Artin–Schreier extensions K over k. In Theorem 4.3, we provide a proof for Algorithm 1. In particular, we obtain an explicit method for determining the exact $2^3$ -rank of the ideal class groups of Artin–Schreier quadratic extensions over k (Corollary 4.4).

The following lemma plays a crucial role for the proof of Theorem 4.3.

Lemma 4.1 Let be the Artin–Schreier extension over k of extension degree p, where $D_m(T) = \sum _{i=1}^{m} \frac {Q_i}{P_i^{r_i}} + f(T)$ is defined as (2.1) and is a root of $x^p-x = D_m$ . For $1 \leq i \leq m$ , let ${\alpha }_i$ be a root of $x^p-x = D_i := Q_i/P_i^{r_i}$ and let ${\gamma }_i$ be a root of the following equation in $\overline {k}$ :

$$ \begin{align*} \mathbf{X}^p-\mathbf{X} = \mathcal{D}_i := \frac{{{\alpha}_i}^2 P_i^{r_i}}{Q_i}. \end{align*} $$

Then $k({\alpha }_i,{\gamma }_i)/k({\alpha }_i)$ is unramified, where all the infinite places of $k({\alpha }_i)$ split completely in $k({\alpha }_i, {\gamma }_i)$ .

Proof We first show that $k({\alpha }_i, {\gamma }_i)/k({\alpha }_i)$ is an unramified extension. Let ${\mathfrak p}_i \in k({\alpha }_i)$ be a place which lies above a finite place P of k. We note that it suffices to show the following by Lemma 2.4:

(4.1) $$ \begin{align} v_{{\mathfrak p}_i}(\mathcal{D}_i) = 2v_{{\mathfrak p}_i}({\alpha}_i) + v_{{\mathfrak p}_i}(P_i^{r_i})-v_{{\mathfrak p}_i}(Q_i) \geq 0. \end{align} $$

We consider the following three possible cases: $P = P_i$ for $1 \leq i \leq m$ , P divides $Q_i \in {\mathbb F}_q[T]$ , and $(P, P_i)=(P,Q_i)=1$ . Using a valuation property, we can easily show the following, where n is a positive integer.

(4.2) $$ \begin{align} &\mbox{If}\ v_{{\mathfrak p}_i}({\alpha}_i^n-{\alpha}_i) < 0,\ \mbox{then}\ v_{{\mathfrak p}_i}({\alpha}_i^n-{\alpha}_i) = nv_{{\mathfrak p}_i}({\alpha}_i) < 0. \end{align} $$

(4.3) $$ \begin{align} &\mbox{If}\ v_{{\mathfrak p}_i}({\alpha}_i^n-{\alpha}_i)> 0,\ \mbox{then}\ v_{{\mathfrak p}_i}({\alpha}_i) \geq 0.\qquad\qquad\qquad\ \ \end{align} $$

(4.4) $$ \begin{align} &\!\!\!\mbox{If}\ v_{{\mathfrak p}_i}({\alpha}_i^n-{\alpha}_i) = 0,\ \mbox{then}\ v_{{\mathfrak p}_i}({\alpha}_i^n-{\alpha}_i) = v_{{\mathfrak p}_i}({\alpha}_i) = 0. \end{align} $$

We denote the ramification index of $\mathfrak {p}_i$ over P in $k({\alpha }_i)/k$ by $e(\mathfrak {p}_i|P)$ and the residue class field degree of $\mathfrak {p}_i$ over P by $f(\mathfrak {p}_i|P)$ .

(i) Suppose that $P = P_i$ . Then we have $e({{\mathfrak p}_i}|P) = e({{\mathfrak p}_i}|P_i) = p$ since $P_i$ is the only totally ramified finite place for $k({\alpha }_i)/k$ . Therefore, we have $v_{{\mathfrak p}_i}({\alpha }_i^p-{\alpha }_i) = v_{{\mathfrak p}_i}(Q_i/P_i^{r_i}) = -pr_i < 0$ ; this implies that $v_{{\mathfrak p}_i}({\alpha }_i) = -r_i$ by (4.2). Therefore, (4.1) holds true.

(ii) Suppose that P divides $Q_i$ in ${\mathbb F}_q[T]$ . Under the given assumption, we have $e({{\mathfrak p}_i}|P) = 1$ ; this is because $(P,P_i) = 1$ as $(P_i, Q_i)=1$ and $P_i$ is the only totally ramified finite place for $k({\alpha }_i)/k$ . Consequently, we have

$$ \begin{align*}v_{{{\mathfrak p}_i}}({\alpha}_i^p-{\alpha}_i) = v_{{{\mathfrak p}_i}}(Q_i/P_i^{r_i}) = v_{P}(Q_i/P_i^{r_i}) = v_P(Q_i)> 0;\end{align*} $$

thus, $v_{{{\mathfrak p}_i}}({\alpha }_i) \geq 0$ by (4.3). Assuming that $v_{{{\mathfrak p}_i}}({\alpha }_i)=0$ , we obtain

(4.5) $$ \begin{align} v_{P}(N_{k({\alpha}_i)/k} ({\alpha}_i)) = f({{\mathfrak p}_i}|P)v_{{{\mathfrak p}_i}}({\alpha}_i) = 0. \end{align} $$

However, since $v_{{{\mathfrak p}_i}}(N_{k({\alpha }_i)/k} ({\alpha }_i)) = v_{{\mathfrak p}_i}(Q_i/P_i^{r_i})>0$ (4.5) cannot happen. Therefore, we have $v_{{\mathfrak p}_i}(\mathcal {D}_i) = 2v_P(Q_i) - v_P(Q_i)> 0$ and (4.1) follows; we use the fact that $v_{{\mathfrak p}_i}({\alpha }_i) = v_P(Q_i)> 0$ . As a result, ${{\mathfrak p}_i}$ is unramified in $k({\alpha }_i,{\gamma }_i)$ .

(iii) Suppose that $(P,P_i) = (P,Q_i)= 1$ . In this case, we get $v_{{\mathfrak p}_i}({\alpha }_i) = 0$ by (4.4) since $v_{{\mathfrak p}_i}({\alpha }_i^p-{\alpha }_i)=0$ . Therefore, (4.1) follows immediately.

Now, it remains to show that all the infinite places of $k({\alpha }_i)$ split completely in $k({\alpha }_i,{\gamma }_i)$ . Let ${\mathfrak p}_{\infty }$ (resp. ${\mathfrak P}_{\infty }$ ) be a place of $k({\alpha }_i)$ (resp. $k({\alpha }_i, {\gamma }_i)$ ) lying above the infinite place ${\infty }$ of k (resp. ${\mathfrak p}_{\infty }$ ). We first note that $v_{{\mathfrak p}_{\infty }}({\alpha }_i^p-{\alpha }_i) = v_{{\mathfrak p}_{\infty }}(Q_i/P_i^{r_i})> 0$ ; thus, $v_{{\mathfrak p}_{\infty }}({\alpha }_i)\geq 0$ by (4.3). By a similar computation method as in (4.5), we obtain $v_{{\mathfrak p}_{\infty }}({\alpha }_i)> 0$ , and therefore $v_{{\mathfrak p}_{\infty }}({\alpha }_i) = v_{{\mathfrak p}_{\infty }}({\alpha }_i^p-{\alpha }_i) = \deg {P_i^{r_i}} - \deg {Q_i}$ . Hence, we get

$$ \begin{align*}v_{{\mathfrak p}_{\infty}}(\mathcal{D}_i) = 2v_{{\mathfrak p}_{\infty}}({\alpha}_i) + v_{{\mathfrak p}_{\infty}}(P_i^{r_i})-v_{{\mathfrak p}_{\infty}}(Q_i) = 2(\deg{P_i^{r_i}} - \deg {Q_i}) - \deg{P_i^{r_i}} + \deg {Q_i}> 0;\end{align*} $$

from this fact and by Lemma 2.4, we can conclude that ${\mathfrak p}_{\infty }$ is unramified in $k({\alpha }_i,{\gamma }_i)/k({\alpha }_i)$ .

Now, it is enough to show that $f({\mathfrak P}_{\infty }|{\mathfrak p}_{\infty })$ is 1. For the proof, we assume that $f({\mathfrak P}_{\infty }|{\mathfrak p}_{\infty }) = p$ . We first note that

(4.6) $$ \begin{align} N_{k({\alpha}_i,{\gamma}_i)/k({\alpha}_i)}({\gamma}_i) = {\gamma}_i^p-{\gamma}_i = {\alpha}_i^2P_i^{r_i}/Q_i. \end{align} $$

On the other hand, we have

(4.7) $$ \begin{align} v_{{\mathfrak p}_{\infty}}(N_{k({\alpha}_i,{\gamma}_i)/k({\alpha}_i)} ({\gamma}_i)) = f({\mathfrak P}_{\infty}|{\mathfrak p}_{\infty})v_{{\mathfrak P}_{\infty}}({\gamma}_i) = pv_{{\mathfrak P}_{\infty}}({\gamma}_i). \end{align} $$

Also, we can obtain

(4.8) $$ \begin{align} pv_{{\mathfrak P}_{\infty}}({\gamma}_i) = v_{{\mathfrak p}_{\infty}}({\gamma}_i^p-{\gamma}_i) = v_{{\mathfrak P}_{\infty}}({\gamma}_i^p-{\gamma}_i), \end{align} $$

by combining (4.6) with (4.7). Furthermore, since $v_{{\mathfrak p}_{\infty }}({\gamma }_i^p-{\gamma }_i) = pv_{{\mathfrak P}_{\infty }}({\gamma }_i)> 0$ , we have

(4.9) $$ \begin{align} pv_{{\mathfrak P}_{\infty}}({\gamma}_i) = \min\{pv_{{\mathfrak P}_{\infty}}({\gamma}_i) ,v_{{\mathfrak P}_{\infty}}({\gamma}_i)\} = v_{{\mathfrak P}_{\infty}}({\gamma}_i), \end{align} $$

which is a contradiction. Therefore, the infinite place of $k({\alpha }_i)$ splits completely in $k({\alpha }_i,{\gamma }_i)$ .

Lemma 4.2 Let K be the Artin–Schreier extension over k of extension degree p. Let $H_{K}$ be the Hilbert class field of K, and let $\mathcal {G}_{K}$ be the genus field of ${K}$ . Let $\mathcal {H}$ be a fixed field of a subgroup of $\mathrm{Gal}(H_K/\mathcal {G}_K)$ which is isomorphic to $Cl_K^{({\sigma }-1)^2}$ . Then $Cl_{K}(p)^{({\sigma }-1)}/Cl_{K}(p)^{({\sigma }-1)^{2}}$ is isomorphic to $\mathrm{Gal}(\mathcal {H}/\mathcal {G}_{K})$ ; thus, we can define the following composite map:

(4.10) $$ \begin{align} \Psi: Cl_{K}(p)^G \cap Cl_{K}(p)^{({\sigma}-1)} \rightarrow Cl_{K}(p)^{({\sigma}-1)}/Cl_{K}(p)^{({\sigma}-1)^{2}} \xrightarrow{\simeq} \mathrm{Gal}(\mathcal{H}/\mathcal{G}_{K}), \end{align} $$

where the first map is induced by the inclusion map.

Then ${\lambda }_3$ is equal to ${\lambda }_2-\operatorname {\mathrm {rank}}(\mathcal {R})$ , where $\mathcal {R}$ is a matrix representing $\Psi $ over ${\mathbb F}_p$ and ${\lambda }_2$ is obtained by Lemma 2.2.

Proof We note that $\mathrm{Gal}(H_K/K) \simeq Cl_K$ and $\mathrm{Gal}(\mathcal {G}_K/K) \simeq Cl_K(p)/Cl_K(p)^{({\sigma }-1)} \simeq Cl_K/Cl_K^{({\sigma }-1)}$ [19, pp. 328–329]; therefore, $\mathrm{Gal}(H_K/\mathcal {G}_K) \simeq Cl_K^{({\sigma }-1)}$ . By the Galois correspondence, we have isomorphisms $\mathrm{Gal}(\mathcal {H}/\mathcal {G}_K) \simeq Cl_K^{({\sigma }-1)}/Cl_K^{({\sigma }-1)^2}$ and $Cl_K^{({\sigma }-1)}/Cl_K^{({\sigma }-1)^2} \simeq Cl_K(p)^{({\sigma }-1)}/Cl_K(p)^{({\sigma }-1)^2}$ ; thus, we have the isomorphism $Cl_K(p)^{({\sigma }-1)}/Cl_K(p)^{({\sigma }-1)^2} \xrightarrow {\simeq } \operatorname {\mathrm {Gal}}(\mathcal {H}/\mathcal {G}_K)$ .

Let $\Psi $ be the map defined as in (4.10). Then we have

$$ \begin{align*} |\operatorname{\mathrm{Ker}} (\Psi)| = |Cl_K(p)^G \cap Cl_K(p)^{({\sigma}-1)^2}|. \end{align*} $$

We claim that for any positive integer n,

(4.11) $$ \begin{align} |Cl_K(p)^G \cap Cl_K(p)^{({\sigma}-1)^{n-1}}| = |Cl_K(p)^{({\sigma}-1)^{n-1}}/Cl_K(p)^{({\sigma}-1)^{n}}|. \end{align} $$

We consider a short exact sequence

$$ \begin{align*}0 \rightarrow Cl_K(p)^G \cap Cl_K(p)^{({\sigma}-1)^{n-1}} \xrightarrow{\imath} Cl_K(p)^{({\sigma}-1)^{n-1}} \xrightarrow{{\sigma}-1} Cl_K(p)^{({\sigma}-1)^{n}} \rightarrow 0,\end{align*} $$

where $\imath $ denotes an inclusion map. Then $Cl_K(p)^{({\sigma }-1)^n}$ is isomorphic to

$$ \begin{align*}Cl_K(p)^{({\sigma}-1)^{n-1}}/\mbox{Im}(\imath) = Cl_K(p)^{({\sigma}-1)^{n-1}} / Cl_K(p)^G \cap Cl_K(p)^{({\sigma}-1)^{n-1}}.\end{align*} $$

Therefore, we have the following:

$$ \begin{align*} |Cl_K(p)^{({\sigma}-1)^n}| = \frac{|Cl_K(p)^{({\sigma}-1)^{n-1}}|}{|Cl_K(p)^G \cap Cl_K(p)^{({\sigma}-1)^{n-1}}|}. \end{align*} $$

We can rewrite this as

$$ \begin{align*} |Cl_K(p)^G \cap Cl_K(p)^{({\sigma}-1)^{n-1}}| = \frac{|Cl_K(p)^{({\sigma}-1)^{n-1}}|}{|Cl_K(p)^{({\sigma}-1)^{n}}|} = |Cl_K(p)^{({\sigma}-1)^{n-1}}/Cl_K(p)^{({\sigma}-1)^{n}}|; \end{align*} $$

hence, (4.11) follows.

Therefore, we compute as follows:

$$ \begin{align*} {\lambda}_3 &= \mbox{dim}_{{\mathbb F}_p}(Cl_K(p)^{({\sigma}-1)^{2}}/Cl_K(p)^{({\sigma}-1)^{3}}) = \dim_{{\mathbb F}_p}(Cl_K(p)^{G}/Cl_K(p)^{({\sigma}-1)^{2}})\\ &= \dim_{{\mathbb F}_p}(\operatorname{\mathrm{Ker}}(\Psi)) = \dim_{{\mathbb F}_p}(Cl_K(p)^G \cap Cl_K(p)^{{({\sigma}-1)}}) - \dim_{{\mathbb F}_p}(\mbox{Im} (\Psi))\\ &= \dim_{{\mathbb F}_p}(Cl_K(p)^{({\sigma}-1)}/Cl_K(p)^{({\sigma}-1)^{2}}) - \dim_{{\mathbb F}_p}(\mbox{Im} (\Psi)) = {\lambda}_2 - \dim_{{\mathbb F}_p}(\mbox{Im} (\Psi))\\ &= {\lambda}_2 - \mbox{rank}(\mathcal{R}), \end{align*} $$

where $\mathcal {R}$ is a matrix representing $\Psi $ over ${\mathbb F}_p$ and ${\lambda }_2$ is obtained by Lemma 2.2. We note that the second equality and the fifth one hold by (4.11) with $n =3$ and 2, respectively.

Theorem 4.3 Let K be the Artin–Schreier extension over the rational function field k of extension degree p. Then the ${\lambda }_3$ -rank of the ideal class group of K can be computed by Algorithm 1.

Proof By Lemma 4.2, we have ${\lambda }_3 = {\lambda }_2 - \operatorname {\mathrm {rank}}(\mathcal {R})$ , where $\mathcal {R}$ is a matrix representing $\Psi $ which is defined as in (4.10). Therefore, it is sufficient to compute the matrix $\mathcal {R}$ in an explicit way for computation of ${\lambda }_3$ . We describe how to compute the matrix $\mathcal {R}$ as follows.

Let $\mathcal {I}:=\{1 \leq i \leq m \mid \mbox {the}\ i\mbox {th row vector of}\ R\ \mbox {is zero}\} = \{s_1, \ldots , s_{{\lambda }_2}\}$ , where $s_i < s_j$ for $1 \leq i < j \leq {\lambda }_2$ . For simplicity, we set $\mathcal {P}_i:=P_{s_i}$ and $\mathcal {F}_i = Q_{s_i}/P_{s_i}^{r_{s_i}}$ for $1 \leq i \leq {\lambda }_2$ . Let $\mathcal {D}_i := {\mathfrak {a}_i}^2/\mathcal {F}_i$ , and let ${\gamma }_i$ be a root of $\mathbf {X}^p - \mathbf {X} = \mathcal {D}_i$ in $\overline {k}$ , where $\overline {k}$ is the algebraic closure of k and $\mathfrak {a}_i$ is the root of $x^p-x = \mathcal {F}_i$ in $\overline {k}$ .

Let $L:= k({\alpha }_1, \ldots , {\alpha }_m)$ be a subfield of the genus field $\mathcal {G}_{K}$ defined as the following, where $\mathcal {G}_{K}$ is given in Lemma 2.3.

We now show that $\mathcal {G}_{K}({\gamma }_i)$ is a subfield of $H_{K}$ for $1 \leq i \leq {\lambda }_2$ . We point out that $\mathcal {G}_{K}({\gamma }_i)/\mathcal {G}_{K}$ is an abelian extension by the fact that it is the Artin–Schreier function field. It suffices to show that $\mathcal {G}_{K}({\gamma }_i)/\mathcal {G}_{K}$ is an unramified extension and all the infinite places of $\mathcal {G}_{K}$ split completely in $\mathcal {G}_{K}({\gamma }_i)$ . By Lemma 4.1, $k({\alpha }_i,{\gamma }_i)/k({\alpha }_i)$ is an unramified extension and all the infinite places of $k({\alpha }_i)$ split completely in $k({\alpha }_i,{\gamma }_i)$ . Thus, $L({\gamma }_i)/L$ is an unramified extension; hence, $\mathcal {G}_{K}({\gamma }_i)/\mathcal {G}_{K}$ is an unramified extension.

Now, we show that all the infinite places of $\mathcal {G}_{K}$ split completely in $\mathcal {G}_{K}({\gamma }_i)$ . Every infinite place of $k({\alpha }_i)$ splits completely in $k({\alpha }_i,{\gamma }_i)$ as shown above and all the infinite places of L split completely in $L({\gamma }_i)$ . Also, all the infinite places split completely in $L/k({\alpha }_i)$ by Lemma 2.1. Consequently, all the infinite places of L split completely in the compositum $L({\gamma }_i)$ of L and $k({\alpha }_i,{\gamma }_i)$ .

Let $\mathcal {P}_\infty $ be a place of L which lies above the infinite place $\infty $ of k and $\mathcal {P}'$ a place of $\mathcal {G}_{K}$ which lies above $\mathcal {P}_\infty $ . We consider the following two possible cases: $\mathcal {P}_\infty $ splits completely in $\mathcal {G}_{K}$ or $\mathcal {P}_\infty $ is totally ramified or inert in $\mathcal {G}_{K}$ . We note that the result follows immediately in the former case; thus, it is sufficient to consider the latter case where there is exactly one place lying above $\mathcal {P}_\infty $ in $\mathcal {G}_{K}$ , the number of places in $\mathcal {G}_{K}({\gamma }_i)$ which lie above $\mathcal {P}'$ is exactly p; this is because the infinite places split completely in $L({\gamma }_i)/L$ . Therefore, $\mathcal {P}'$ splits completely in $\mathcal {G}_{K}({\gamma }_i)$ , and the result holds.

We have $\mathcal {H} = \mathcal {G}_{K}({\gamma }_1, \ldots , {\gamma }_{{\lambda }_2})$ since $\mathcal {G}_{K}({\gamma }_i) \subseteq H_{K}$ and $[\mathcal {H}:\mathcal {G}_{K}] = p^{{\lambda }_2}$ . We get

$$ \begin{align*}\left( \frac{\mathcal{H}/\mathcal{G}_{K}}{{\mathfrak p}_i} \right)({\gamma}_j) = {\gamma}_j + \left\{ \frac{\mathcal{D}_j}{\mathcal{P}_i} \right\},\end{align*} $$

where ${\mathfrak p}_i$ is a place of $\mathcal {G}_{K}$ lying above $\mathcal {P}_i$ for $1 \leq i \leq {\lambda }_2$ by the action of the Artin map in the Artin–Schreier function field. Therefore, we determine $\mathcal {R}=[\mathfrak {r}_{ij}]= \left \{\frac {\mathcal {D}_j}{\mathcal {P}_i}\right \}$ .

This process is implemented in Algorithm 1. Steps (1) and (2) of Algorithm 1 give the process of computing ${\lambda }_1$ , ${\lambda }_2$ , and the Rédei matrix R. Step (3) explains the case where ${\lambda }_2=0$ and then the algorithm stops. If $0<{\lambda }_2 <{\lambda }_1$ , then we go to Step (4.1), and if ${\lambda }_2 = {\lambda }_1$ , then we proceed with Step (4.2). Steps (5.1) and (5.2) explain the process of finding $\mathcal {D}_i$ for $1 \leq i \leq {\lambda }_2$ . In Step (6), we determine a matrix $\mathcal {R}$ over ${\mathbb F}_p$ , and finally we obtain ${\lambda }_3 = {\lambda }_2 - \mbox {rank}(\mathcal {R})$ in Step (7).

Corollary 4.4 Let K be the Artin–Schreier quadratic extension over k, and let the ${\lambda }_3$ -rank of $Cl_K$ be computed by Algorithm 1. Then the $2^3$ -rank of $Cl_K$ is exactly ${\lambda }_3$ : that is, $Cl_K(2)$ has a subgroup isomorphic to $({\mathbb Z}/2^3{\mathbb Z})^{{\lambda }_3}$ .

Proof This follows immediately from the fact that ${\lambda }_n$ is exactly equal to the full $2^n$ -rank of $Cl_K$ and Theorem 4.3.

Remark 4.5 For readers, focusing on the case: $p=2$ , we first briefly explain the analogy between Rédei symbols (the 4-rank of the class groups) and the 8-rank of the class groups in the quadratic field case (for more details, see [9]). Then we describe the analogy between Artin–Schreier quadratic extensions over k and quadratic extensions over ${\mathbb Q}$ for computation of ${\lambda }_3$ .

Let F be a quadratic extension over ${\mathbb Q}$ , and let $Cl_F$ be the ideal class group of F. Let $r_4$ (resp. $r_8$ ) be the $2^2$ -rank (resp. $2^3$ -rank) of $Cl_F$ . Let H be the Hilbert class field of F, and let $H_n$ be the unramified abelian subextension of H such that $\operatorname {\mathrm {Gal}}(H_n/F) \simeq Cl_F/Cl_F^n$ for $n=2,4$ .

Basically, a strategy for computing the $2^2$ -rank (resp. $2^3$ -rank) is explicitly finding a subextension $H_2$ (resp. $H_4$ ) of the Hilbert class field of F whose Galois group is isomorphic to $\operatorname {\mathrm {Gal}}(Cl_F/Cl_F^2)$ (resp. $\operatorname {\mathrm {Gal}}(Cl_F^2/Cl_F^4)$ ).

Define two maps as follows:

$$ \begin{align*} &R_4 : {\mathbb F}_2^t \rightarrow Cl_F[2] \xrightarrow{\varphi} Cl_F/Cl_F^2 \xrightarrow{\simeq} \operatorname{\mathrm{Gal}}(H_2/F) \rightarrow \operatorname{\mathrm{Gal}}(H_2/{\mathbb Q}) = \prod_{i=1}^t \operatorname{\mathrm{Gal}}({\mathbb Q}(\sqrt{d_i})/{\mathbb Q}),\\ &R_8 : \operatorname{\mathrm{Ker}} R_4 \rightarrow Cl_F[2] \cap Cl_F^2 \xrightarrow{\psi} Cl_F^2/Cl_F^4 \xrightarrow{\simeq} \operatorname{\mathrm{Gal}}(H_4/H_2) = \prod_{i=1}^{r_4}\operatorname{\mathrm{Gal}}(H_2(\sqrt{{\alpha}_i})/H_2) \rightarrow {\mathbb F}_2^{r_4}, \end{align*} $$

where t is the number of finite primes of ${\mathbb Q}$ which are ramified in F, $Cl_F[2]$ is the 2-torsion part of $Cl_F$ , and the maps $\varphi $ and $\psi $ are induced by the inclusion maps. For computation of $r_4$ and $r_8$ , we find appropriate $d_i$ ( $1 \leq i \leq t$ ) and ${\alpha }_i$ ( $1\leq i \leq r_4)$ . Then we have

$$ \begin{align*}r_4 = t-\dim_{{\mathbb F}_2}R_4 \quad \text{and} \quad r_8=r_4-\dim_{{\mathbb F}_2}R_8.\end{align*} $$

To show the analogy between Artin–Schreier quadratic extensions over k and quadratic extensions over ${\mathbb Q}$ for computation of ${\lambda }_3$ ( $2^3$ -rank), let K be the Artin–Schreier quadratic extension over k. Then the map $R_8$ corresponds to the map $\Psi $ defined in (4.10):

$$ \begin{align*} \Psi: Cl_{K}(2)^G \cap Cl_{K}^{2} \rightarrow Cl_{K}^{2}/Cl_{K}^{4} \xrightarrow{\simeq} \mathrm{Gal}(\mathcal{H}/\mathcal{G}_{K}). \end{align*} $$

Then we have ${\lambda }_3 = {\lambda }_2 - \operatorname {\mathrm {rank}}\mathcal {R}$ , where $\mathcal {R}$ is a matrix over ${\mathbb F}_2$ representing the map $\Psi $ . We recall that ${\lambda }_3$ is the $2^3$ -rank of $Cl_K$ .

5 An infinite family of Artin–Schreier function fields with higher ${\lambda }_n$ -rank

In this section, we find an infinite family of Artin–Schreier function fields which have prescribed ${\lambda }_n$ -rank of the ideal class group for $1 \leq n \leq 3$ . In Theorem 5.1, for any positive integer $t \ge 2$ , we obtain an infinite family of Artin–Schreier extensions over k whose ${\lambda }_1$ -rank is t, ${\lambda }_2$ -rank is $t-1$ , and ${\lambda }_3$ -rank is $t-2$ . Then Corollary 5.3 shows the case where $p=2$ , for a given positive integer $t \ge 2$ , we obtain an infinite family of the Artin–Schreier quadratic extensions over k whose $2$ -class group rank (resp. $2^2$ -class group rank and $2^3$ -class group rank) is exactly t (resp. $t-1$ and $t-2$ ). Furthermore, we also obtain a similar result on the $2^n$ -ranks of the divisor class groups of the Artin–Schreier quadratic extensions over k in Corollary 5.4.

Throughout this section, we define $D_m$ as follows.

Notation 1 Let $D_m := \sum _{i=1}^m D_i +f(T)$ be defined in (2.1) with $D_i = Q_i/P_i^{r_i}$ , where $m, P_i, Q_i$ , and $f(T)$ satisfy one of the followings:

1. (i) $m = \left \{ \begin {array}{ll} t, & \text {if}\ \deg f(T) \geq 1\\[3pt] & \text { or}\ f(T) = c \in {\mathbb F}_q^\times \ \text {such that}\ x^p-x=c\ \text {is irreducible over}\ {\mathbb F}_q,\\[5pt] t+1, & \text {if}\ f(T) = 0. \end {array} \right.$

2. (ii) $Q_j \equiv P_j^{r_j}(b_i(T)^q-b_i(T))\ \pmod {P_i}$ for any $1 \leq i \ne j \leq m$ except $(i,j) = (1,2)$ , where $b_i(T) \in {\mathbb F}_q[T]$ .

3. (iii) If $\deg f(T) \geq 1$ , then $f(T) \equiv P_j^{r_j}(b_i(T)^q-b_i(T))\ \pmod {P_i}$ , where $b_i(T) \in {\mathbb F}_q[T]$ for any $1 \leq i \leq m$ .

4. (iv) If $f(T) \in {\mathbb F}_q^\times $ , then $q \mid \deg P_i$ for any i with $1 \leq i \leq m$ .

5. (v) $Q_j^{-1} \equiv P_j^{r_j}(b_i(T)^q-b_i(T))\ \pmod {P_i}$ , where $b_i(T) \in {\mathbb F}_q[T]$ and $Q_j^{-1}$ denotes the inverse of $Q_j$ modulo $P_i$ for any $1 \leq i \ne j \leq m$ except $(i,j) \ne (1,2)$ .

Theorem 5.1 For a given positive integer $t \ge 2$ , there is an infinite family of Artin–Schreier extensions over k whose ${\lambda }_1$ -rank is t, ${\lambda }_2$ -rank is $t-1$ , and ${\lambda }_3$ -rank is $t-2$ .

Let be the Artin–Schreier function field over k of extension degree p, where $D_m$ is defined in Notation 1 and is a root of $x^p-x = D_m$ . Then the ideal class group $Cl_K$ of K has ${\lambda }_1 = t$ , ${\lambda }_2 = t-1$ , and ${\lambda }_3 = t-2$ .

Remark 5.2 Let ${\mathbb F}_q$ be a finite field of order q, t be a given integer, and $f(T) \in {\mathbb F}_q$ . By condition (i), $m = t+1$ . By condition (ii), we can choose monic irreducible polynomials $P_i \in {\mathbb F}_q[T]$ whose degrees are divisible by p. We note that conditions (iii) and (iv) can be interpreted as

(5.1) $$ \begin{align} \left\{ \frac{D_j}{P_i} \right\} = \left\{ \frac{Q_j^{-1}}{P_i} \right\}=0; \end{align} $$

by the surjectivity of the trace map, there always exist $D_j$ and $Q_j^{-1}$ which satisfy (5.1). Since our choice of $P_i$ ’s are infinite, we have an infinite family of Artin–Schreier extensions which satisfy the conditions in Theorem 5.1.

Proof of Theorem 5.1

Recall that ${\lambda }_2 = {\lambda }_1 - \operatorname {\mathrm {rank}}(R)$ and ${\lambda }_3 = {\lambda }_2 - \operatorname {\mathrm {rank}}(\mathcal {R})$ , where R (resp. $\mathcal {R}$ ) is a matrix over ${\mathbb F}_p$ defined in Lemma 2.2 (resp. Algorithm 1). We need to show that

(5.2) $$ \begin{align} {\lambda}_1 = t, \qquad {\lambda}_2 = t-1, \qquad {\lambda}_3 = t-2; \end{align} $$

this is equivalent to $\operatorname {\mathrm {rank}}(R)= \operatorname {\mathrm {rank}}(\mathcal {R}) = 1$ .

We divide into the following three cases: $\deg f(T) \geq 1$ , $\deg f(T) = 0$ , and $f(T) = c$ , where $x^p-x-c$ is irreducible over ${\mathbb F}_q$ .

Case I. $\deg f(T) \geq 1$ : that is, the infinite place of k is totally ramified in K.

Since $\deg f(T) \geq 1$ , we have $m=t$ by condition (i); this implies that ${\lambda }_1 = m = t$ by Lemma 2.2. For computing ${\lambda }_2$ , we compute every entry of the Rédei matrix R: that is, the Hasse norm $\{D_j/P_i\}$ and $\{f(T)/P_i\}$ for $1 \leq i \ne j \leq m$ . Using Lemma 3.1 and condition (ii), we can easily obtain that $\left \{ \frac {D_2}{P_1} \right \} \ne 0$ and $\left \{ \frac {D_j}{P_i} \right \} = 0$ for any $1 \leq i \ne j\leq m$ except $(i, j) \ne (1,2)$ . Furthermore, we get $\left \{ \frac {f}{P_i} \right \} = 0$ for any $1 \leq i \leq m$ by condition (iii). Therefore, the Rédei matrix R can be written as ${\small {R = \left [ \begin {array}{cccc} p-1 & 1 & \cdots & 0\\ 0 & 0 & \cdots & 0\\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0& \cdots & 0 \\ \end {array} \right ]}};$ thus, ${\lambda }_2 = {\lambda }_1 - \operatorname {\mathrm {rank}}(R) = t-1$ . Lastly, we compute ${\lambda }_3$ of K using Algorithm 1 and Theorem 4.3. Using the definition of a matrix $\mathcal {R}$ which is given in Algorithm 1, it suffices to compute $\left \{\frac {1/Q_j}{P_i} \right \}$ for $1 \leq i \ne j \leq m$ . By the same reasoning as in the computation of R, we get ${\lambda }_3 = {\lambda }_2 - \operatorname {\mathrm {rank}}(\mathcal {R}) = t-2$ . Therefore, (5.2) follows.

Case II. $\deg f(T) = 0$ : that is, the infinite place of k splits completely in K, which is a real extension.

We can easily obtain ${\lambda }_1=t$ by using Lemma 2.2 and the condition $m=t+1$ . For computing ${\lambda }_2$ , we compute every entry of the Rédei matrix R: that is, the value of Hasse norm $\{D_j/P_i\}$ for $1 \leq i \ne j \leq m$ . By the definition of Hasse norm which is defined in Definition 2.1, we get $\{D_2/P_1\} \ne 0$ and $\{D_j/P_i\} = 0$ , where $1 \leq i \ne j \leq m$ except $(i,j) = (1,2)$ . As in Case 1, the rank of Rédei matrix is one: that is, ${\lambda }_2 = {\lambda }_1 - \operatorname {\mathrm {rank}}(R) = t-1$ . Lastly, we compute ${\lambda }_3$ of K; by the same computation method as in Case I, we have ${\lambda }_3 = {\lambda }_2 - \operatorname {\mathrm {rank}}(\mathcal {R}) = t-2$ . Therefore, (5.2) follows.

Case III. $f(T) = c \in {\mathbb F}_q^\times $ , where $x^p-x-c$ is irreducible over ${\mathbb F}_q$ : that is, the infinite place of k is inert in K.

Under this assumption, K is an imaginary extension; so, $m = t$ . We claim that (5.2) holds for this case. We can simply get ${\lambda }_1 = t$ by Lemma 2.2 and we also obtain $\left \{ D_j/ P_i \right \} = 0$ for every $1 \leq i \ne j \leq t = m$ except $(i,j) = (1,2)$ by using the same reasoning as in Case I. Now, we compute the value of $\left \{ c/P_i\right \}$ for $1 \leq i \leq t=m$ , where $c \in {\mathbb F}_q^\times $ . We have

$$ \begin{align*} \left\{ \frac{c}{P_i} \right\} = \operatorname{\mathrm{Tr}}_{{\mathbb F}_{q}/{\mathbb F}_p} (\operatorname{\mathrm{Tr}}_{{\mathbb F}_{\mathfrak{d}}/{\mathbb F}_q} c) = \operatorname{\mathrm{Tr}}_{{\mathbb F}_q/{\mathbb F}_p} (c \deg {P_i}) = \deg {P_i} (\operatorname{\mathrm{Tr}}_{{\mathbb F}_q/{\mathbb F}_p} c); \end{align*} $$

the second equation holds since c is a nonzero element of ${\mathbb F}_q$ and the last equation holds by the property of a trace map over a finite field. We get $\deg {P_i}(\operatorname {\mathrm {Tr}}_{{\mathbb F}_q/{\mathbb F}_p} c) = 0$ in ${\mathbb F}_p$ by Lemma 3.1 by the assumption that $q \mid \deg {P_i}$ for every $1 \leq i \leq m$ ; therefore, (3.5) is zero in ${\mathbb F}_p$ . Hence, ${\lambda }_2 = t-1$ . By the same reasoning as in Case I, ${\lambda }_3 = t-2$ and we have (5.2).

Corollary 5.3 Let be the Artin–Schreier quadratic function field over k of extension degree $2$ , where $D_m$ is defined in Notation $1$ and is a root of $x^2-x = D_m$ .

For any positive integer $t \ge 2$ , there is an infinite family of Artin–Schreier quadratic extensions over k whose $2$ -class group rank is exactly t, $2^2$ -class group rank is $t-1$ , and $2^3$ -class group rank is $t-2$ .

In particular, $Cl_K(2)$ contains a subgroup isomorphic to $({\mathbb Z}/2^n{\mathbb Z})^{t-n+1}$ for $1 \leq n \leq 3$ .

Proof We note that ${\lambda }_n$ is exactly equal to the full $2^n$ -rank ( $1 \leq n \leq 3$ ) of the ideal class group $Cl_K$ of K; therefore, the result follows immediately from Theorem 5.1.

Corollary 5.4 For a given positive integer t, let be the Artin–Schreier quadratic function field over k, where $D_m = \sum _{i=1}^m Q_i/P_i^{r_i} + f(T)$ such that $P_i, Q_i, f(T)$ , and m satisfy the conditions (i)–(v) in Notation 1. Let $J_K$ be the divisor class group of K. Then we have the following infinite family of Artin–Schreier quadratic extensions.

1. (i) For $t \ge 2$ , if $\deg f(T) \ge 1$ (equivalently, $\infty $ is totally ramified in K), then the $2^n$ -class group rank of $J_K$ is exactly equal to $t+1-n$ for $1 \leq n \leq 3$ .

2. (ii) For $t \ge 2$ , if $f(T) =0$ (equivalently, $\infty $ splits completely in K), then the $2^n$ -class group rank of $J_K$ is exactly either $t+1-n$ or $t+2-n$ for $1 \leq n \leq 3$ .

3. (iii) For $t \ge 3$ , if $f(T) \in {\mathbb F}_q^\times $ (equivalently, $\infty $ is inert in K), then the $2^n$ -class group rank of $J_K$ is exactly either $t+1-n$ or $t-n$ for $1 \leq n \leq 3$ .

Proof Since $D_m$ satisfies the conditions (i)–(v) in Notation 1, the ideal class group $Cl_K$ of K has ${\lambda }_1$ -rank t, ${\lambda }_2$ -rank $t-1$ , and ${\lambda }_3$ -rank $t-2$ .

We first assume that $\deg f(T) \geq 1$ : that is, the infinite place $\infty $ of k is totally ramified in K. Then the ideal class group $Cl_K$ of K is isomorphic to the divisor class group $J_K$ of K by Lemma 2.6. Thus, by Lemma 5.3, the $2^n$ -rank of the divisor class group $J_K$ of K is $t+1-n$ for n up to 3; thus, (i) follows.

Next, suppose that $f(T) = 0$ . This is the case where the infinite place $\infty $ of k splits completely in K. Then, by Lemma 2.6, we note that $J_K/R$ is isomorphic to $Cl_K$ , where R denotes the group $\mathcal {D}_K^0(S)/\mathcal {P}_K(S)$ . By the fact the group R is a cyclic group, the $2^n$ -rank of the divisor class group $J_K$ is either $t+1-n$ or $t+2-n$ for n up to 3.

Finally, we assume that $f(T) \in {\mathbb F}_q^\times $ : the case where $\infty $ is inert in K. Then, by the exact sequence given in Lemma 2.6(ii), we get $|Cl_K| = 2|J_K|$ . Since $Cl_K(2)$ contains a subgroup isomorphic to $({\mathbb Z}/2^n{\mathbb Z})^{t-n+1}$ for $1 \le n \le 3$ , $J_K(2)$ contains a subgroup isomorphic to $({\mathbb Z}/2^n{\mathbb Z})^{t-n+1}$ or $({\mathbb Z}/2^n{\mathbb Z})^{t-n}$ for $1 \leq n \le 3$ ; therefore, (iii) holds.

Remark 5.5 We briefly mention that the ${\lambda }_2$ -rank is connected to the embedding problem. For instance, in the quadratic number field $F = {\mathbb Q}(\sqrt {d})$ , the solvability of the conics $X^2 = aY^2 + \frac {d}{a}Z^2$ yields unramified cyclic quartic extensions of F. The solvability of this conic is related to the $\lambda _2$ -rank of $Cl_F$ , which is computed by the Rédei matrix in terms of Legendre symbols. Then the embedding problem for F is not solvable. On the other hand, in our context, the embedding problem for Artin–Schreier extensions K over k is solvable and every finite place of k is wildly ramified in K.

Table 1: Infinite families of Artin–Schreier extensions $K=k({\alpha }_D)$ over k whose ${\lambda }_1$ -rank of the ideal class groups is t and ${\lambda }_n$ -rank is zero for $n\ge 2$ , where ${\alpha }_D^p-{\alpha }_D = D$ .

Table 2: Infinite families of Artin–Schreier extensions $K=k({\alpha }_D)$ over k whose ${\lambda }_1$ -rank of the ideal class groups is t, ${\lambda }_2$ -rank is $t-1$ , and ${\lambda }_3$ -rank is $t-2$ , where ${\alpha }_D^p-{\alpha }_D = D$ .

6 Implementation results

In this section, as implementation results, we explicitly present concrete infinite families of Artin–Schreier extensions over k whose ideal class groups have guaranteed prescribed ${\lambda }_n$ -rank of the ideal class group for $1 \leq n \leq 3$ . In Table 1, for a given positive integer t, we obtain explicit families of Artin–Schreier extensions K over k whose ${\lambda }_1$ -rank of the ideal class group $Cl_K$ is t and ${\lambda }_n$ -rank is zero for $n \ge 2$ , depending on the ramification behavior of the infinite place $\infty $ of k (Theorems 3.2–3.4). Furthermore, in Table 2, for a given integer $t \ge 2$ , we get explicit families of Artin–Schreier extensions over k whose ${\lambda }_1$ -rank of the ideal class groups is t, ${\lambda }_2$ -rank is $t-1$ , and ${\lambda }_3$ -rank is $t-2$ (Theorem 5.1). In the tables, we denote ${\mathbb Z}/m{\mathbb Z}$ by ${\mathbb Z}_m$ for a positive integer m.