Rationality of Three-Dimensional Quotients by Monomial Actions

TL;DR

This paper proves that for certain finite 2-groups acting monomially on a three-variable rational function field over a field with specific properties, the fixed field remains rational, extending understanding of field invariants.

Contribution

It establishes the rationality of fixed fields under monomial actions of finite 2-groups on three-variable rational function fields, under particular field conditions.

Findings

Fixed field $K(x,y,z)^G$ is rational over $K$ for the specified group actions.

The result applies to fields with characteristic not 2 and containing square roots of all elements.

Provides applications demonstrating the theorem's utility.

Abstract

Let $G$ be a finite 2-group and $K$ be a field satisfying that (i) $\fn{char}K\ne 2$, and (ii) $\sqrt{a}\in K$ for any $a\in K$. If $G$ acts on the rational function field $K(x,y,z)$ by monomial $K$-automorphisms, then the fixed field $K(x,y,z)^G$ is rational (= purely transcendental) over $K$. Applications of this theorem will be given.

Equations

$$\sigma (x_j)=a_j(\sigma )\prod _{i=1}^n{x}_i^{m_{ij}}$$

$$\left(\begin{array}{c}\sigma (x_1)\\ \sigma (x_2)\\ ⋮\\ \sigma (x_n)\end{array}\right)=A(\sigma )\cdot \left(\begin{array}{c}{x}_1\\ x_2\\ ⋮\\ x_n\end{array}\right)+B(\sigma )$$

$$u=\frac{x-\frac{a}{x}}{xy-\frac{ab}{xy}},v=\frac{y-\frac{b}{y}}{xy-\frac{ab}{xy}}.$$

$$x+(a/x)=(-b{u}^2+a{v}^2+1)/v,$$

$$y+(b/y)=(b{u}^2-a{v}^2+1)/u,$$

$$xy+(ab/(xy))=(-b{u}^2-a{v}^2+1)/(uv),$$

$$\frac{x-\frac{a}{x}}{\frac{bx}{y}-\frac{ay}{x}}=\frac{u}{b{u}^2-a{v}^2}.$$

$$u=\frac{x-\frac{a}{x}}{xy-\frac{ab}{xy}},v=\frac{y-\frac{b}{y}}{xy-\frac{ab}{xy}}.$$

$$\sigma (\alpha )=-\alpha ,\sigma (x)=x,\sigma (y)=b(x^2-c)/y$$

$$\Phi :⟨x_1,\dots ,x_n⟩$$

$$M$$

$$\sigma$$

$$\tau$$

$$\sigma :(x_1,x_2,x_3)\mapsto (a{x}_1,b/x_3,c/x_2)$$

$$(x_1,x_2,x_3)\mapsto (a{x}_1,b/x_2,c/x_3)$$

$$(x_1,x_2,x_3)\mapsto (a/x_1,b{x}_2,c{x}_3)$$

$$(x_1,x_2,x_3)\mapsto (a/x_1,b{x}_3,c{x}_2)$$

$$\sigma :(x_1,x_2,x_3)\mapsto (a_1/x_1,a_2/x_2,a_3/x_3)$$

$$\sigma :(x_1,x_2,x_3)\mapsto \left((\sqrt{-1}{)}^i{x}_1,x_3,a/x_2\right)$$

$$\sigma :(x_1,x_2,x_3)\mapsto (a/x_1,x_3,b/x_2)$$

$$u=\frac{x_2-\frac{b}{x_2}}{x_2{x}_3-\frac{b^2}{x_2{x}_3}},v=\frac{x_3-\frac{b}{x_3}}{x_2{x}_3-\frac{b^2}{x_2{x}_3}}.$$

$$\sigma:u\mapsto\frac{x_{3}-\frac{b}{x_{3}}}{\frac{bx_{3}}{x_{2}}-\frac{bx_{2}}{x_{3}}},~{}v\mapsto\frac{-\big{(}x_{2}-\frac{b}{x_{2}}\big{)}}{\frac{bx_{3}}{x_{2}}-\frac{bx_{2}}{x_{3}}}.$$

$$\frac{-\big{(}x_{2}-\frac{b}{x_{2}}\big{)}}{\frac{bx_{3}}{x_{2}}-\frac{bx_{2}}{x_{3}}}=\frac{u}{bu^{2}-bv^{2}}$$

$$\sigma :(w,x_1,v)\mapsto (-1/w,a/x_1,1/(cv))$$

$$y_{1}=x_{1},~{}~{}~{}Y_{2}=\big{(}\sqrt{-1}-w\big{)}\big{/}\big{(}\sqrt{-1}+w\big{)},~{}~{}~{}y_{2}=[x_{1}-(a/x_{1})]Y_{2}.$$

$$y_3=v(x_1-(a/x_1)+\sqrt{-1}{Y}_2)/(x_1-(a/x_1)+Y_2).$$

$$\sigma :(x_1,x_2,x_3)\mapsto (a{x}_1,x_3,b{x}_1/x_2).$$

$$\begin{array}[]{ll}X_{1}=(y_{1}+y_{3})/2,&X_{2}=[y_{2}-(y_{1}y_{3}/y_{2})]/2,\\ X_{3}=(y_{1}-y_{3})/(2\sqrt{-1}),&X_{4}=[y_{2}+(y_{1}y_{3}/y_{2})]/(2\sqrt{-1}).\end{array}$$

$$\sigma :(Y_1,Y_2,Y_3)\mapsto (-Y_2/Y_3^2,-Y_1/Y_3^2,-1/Y_3).$$

$$\sigma :(x_1,x_2,x_3)\mapsto (c/x_1,x_3,b{x}_1/x_2)$$

$$u=\frac{y_1-\frac{A}{y_1}}{y_1{y}_2-\frac{AB}{y_1{y}_2}},v=\frac{y_2-\frac{B}{y_2}}{y_1{y}_2-\frac{AB}{y_1{y}_2}}.$$

$$\sigma$$

Full text

Rationality of Three-Dimensional Quotients by Monomial Actions

Ming-chang Kang*(1)* and Yuri  G. Prokhorov*(2)*

Abstract. Let $G$ be a finite 2-group and $K$ be a field satisfying that (i) $\mathop{{\rm char}\vphantom{\dim}}K\neq 2$, and (ii) $\sqrt{a}\in K$ for any $a\in K$. If $G$ acts on the rational function field $K(x,y,z)$ by monomial $K$-automorphisms, then the fixed field $K(x,y,z)^{G}$ is rational (= purely transcendental) over $K$. Applications of this theorem will be given.

*(1)*Department of Mathematics and Taida Institute of Mathematical Sciences, National Taiwan University, Taipei, Taiwan; E-mail: kang@math.ntu.edu.tw

*(2)*Department of Algebra, Faculty of Mathematics and Mechanics, Moscow State University, Moscow, Russia; E-mail: prokhorov@gmail.com

††

The first-named author was partially supported by National Center for Theoretic Sciences (Taipei office). Parts of the work in this paper were done while he visited Department of Mathematics, Yunnan Normal University, Kunming.

The second-named author was partially supported by the Russian Foundation for Basic Research (grants No  06-01-72017-MNTIa, 08-01-00395-a) and Leading Scientific Schools (grants No  NSh-1983.2008.1, NSh-1987.2008.1).

Mathematics Subject Classification (2000). Primary 12F12, 13A50, 14E08.

Keywords. Rationality problem, Noether’s problem, multiplicative group actions.

§1. Introduction

Let $K$ be any field and $K(x_{1},\ldots,x_{n})$ be the rational function field of $n$ variables over $K$. A $K$-automorphism $\sigma$ of $K(x_{1},\ldots,x_{n})$ is said to be a monomial automorphism if

[TABLE]

where $(m_{ij})_{1\leq i,j\leq n}\in GL_{n}(\mathbbm{Z})$ and $a_{j}(\sigma)\in K\backslash\{0\}$. If $a_{j}(\sigma)=1$ for all $1\leq j\leq n$, then $\sigma$ is called a purely monomial automorphism. A group action on $K(x_{1},\ldots,x_{n})$ by monomial $K$-automorphisms is also called a multiplicative group action. Multiplicative actions are crucial in solving rationality problems for linear group actions, e.g. Noether problem (see, for examples, [Sw, CHKP]).

However, the rationality problem of a multiplicative action on $K(x_{1},\ldots,x_{n})$ is rather intricate. We recall some previously known results first.

Theorem 1.1** (Hajja [Ha1, Ha2]).**

Let $K$ be any field and $K(x,y)$ be the rational function field of two variables over $K$. If $G$ is a finite group acting on the rational function field $K(x,y)$ by monomial $K$-automorphisms, then the fixed field $K(x,y)^{G}$ is rational $(=$ purely transcendental$)$ over $K$.

Theorem 1.2** (Hajja, Kang, Hoshi and Rikuna [HK1, HK2, HR]).**

Let $K$ be any field and $K(x,y,z)$ be the rational function field of three variables over $K$. If $G$ is a finite group acting on $K(x,y,z)$ by purely monomial $K$-automorphisms, then the fixed field $K(x,y,z)^{G}$ is rational over $K$.

In the above Theorem 1.2, it is impossible to replace the assumption “purely monomial $K$-automorphisms” by the weaker assumption “monomial $K$-automorphisms” as the fixed field $K(x,y,z)^{\langle\sigma\rangle}$ is not rational over $K$ where the monomial $K$-automorphism $\sigma$ is defined by $\sigma:x\mapsto y\mapsto z\mapsto-1/(xyz)$ (see the last paragraph of [Ha1]). However, if we assume that $K$ is an algebraically closed field, then we can get an affirmative result for monomial group actions. In fact, the main result of this article is the following theorem.

Theorem 1.3**.**

Let $G$ be a finite $2$-group and $K$ be a field satisfying that

1. (i)

$\mathop{{\rm char}\vphantom{\dim}}K\neq 2$, and 2. (ii)

$\sqrt{a}\in K$* for any $a\in K$.*

If $G$ acts on the rational function field $K(x,y,z)$ by monomial $K$-automorphisms, then the fixed field $K(x,y,z)^{G}$ is rational over $K$.

In the above theorem, it is assumed that $\mathop{{\rm char}\vphantom{\dim}}K\neq 2$. We may as well show that $K(x,y,z)^{G}$ is rational over $K$ when $\mathop{{\rm char}\vphantom{\dim}}K=2$ and $K$ satisfies the assumption that $\sqrt{a}\in K$ for any $a\in K$, by using the techniques in [HK1, HK2] and the method developed in Section 3 of this paper. Because we intend to highlight the main techniques in the proof of Theorem 1.3 and because we try to shorten the length of this article, we omit the proof of the case when $\mathop{{\rm char}\vphantom{\dim}}K=2$.

We will give another remark on Theorem 1.3. It is impossible to generalize Theorem 1.3 to the case of rational function fields of four variables. In fact, if $G\simeq C_{2}^{3}$, there is a monomial action of $G$ on $\mathbbm{C}(x_{1},x_{2},x_{3},x_{4})$ so that the fixed field $\mathbbm{C}(x_{1},x_{2},x_{3},x_{4})^{G}$ is not retract rational over $\mathbbm{C}$; in particular, it is not rational over $\mathbbm{C}$ [CHKK, Example 5.11].

An application of Theorem 1.3 is the following theorem.

Theorem 1.4**.**

Let $G$ be a finite $2$-group and $K$ be a field satisfying that

1. (i)

$\mathop{{\rm char}\vphantom{\dim}}K\neq 2$, and 2. (ii)

$\sqrt{a}\in K$* for any $a\in K$.*

Suppose that $\rho:G\to GL(V)$ is a linear representation of $G$ over $K$ such that either $\dim_{K}V\leq 5$ or $\rho$ is the direct sum of three $2$-dimensional irreducible representations. Then both the fixed field $K(V)^{G}$ and the quotient $P(V)/G$ are rational over $K$.

We remark that, if we assume that the field $K$ is algebraically closed, the conclusion of the above Theorem 1.3 is still valid for any finite group $G$, i.e. no assumption about $G$ being a $2$-groups is necessary. We refer the reader to [Pr, Section 5] for some key ideas of the proof. We choose to publish parts of the general result because of two reasons. First of all, the proof of the general case is rather long and complicated; we had better publish this result in two separate papers. Second, the result for 2-groups will be used in a forthcoming paper on Noether’s problem and Bogomolov multipliers (i.e. the unramified Brauer groups) for groups of order 64 [CHKK].

We will point out that Theorem 1.4 provides a quick proof for the special case when $K$ is algebraically closed of the following theorem.

Theorem 1.5** (Chu, Hu, Kang and Prokhorov [CHKP]).**

Let $G$ be a group of order $32$ with exponent $e$, and $K$ be a field satisfying

(i)* $\mathop{{\rm char}\vphantom{\dim}}K=2$, or*

(ii)* $\mathop{{\rm char}\vphantom{\dim}}K\neq 2$ and $K$ contains a primitive $e$-th root of unity,*

Then $K(x_{g}:g\in G)^{G}$ is rational over $K$.

Note that Theorem 1.5 was proved using the classification of groups of order $32$, i.e. the structures and representations of these groups provided by the data base of GAP [CHKP]. If we choose to avoid using the classification of groups of order $32$, we still have a proof, thanks to Theorem 1.4, but we need the stronger assumptions that $\mathop{{\rm char}\vphantom{\dim}}K\neq 2$ and $\sqrt{a}\in K$ for any $a\in K$ (e.g. $K$ is algebraically closed).

We will organize this paper as follows. In Section 2 we recall several preliminary results which will be used subsequently. Section 3 contains the complete proof of Theorem 1.3; the proof uses only elementary methods and will be more accessible to most readers. Another proof of Theorem 1.3 for the case when the ground field $K$ is algebraically closed and $\mathop{{\rm char}\vphantom{\dim}}K=0$ will be given in Section 4. This new proof is shorter than that in Section 3, more conceptual, and no classification of finite subgroups of $GL_{3}(\mathbbm{Z})$ [Ta] is required; but the price is that some machinery in algebraic geometry is used. The proof of Theorem 1.4 will be given in Section 5. As an application of it, we will give a quick proof of Theorem 1.5 provided that $K$ is algebraically closed.

Standing Notations. Throughout this paper, $K$ is a field, $K(x_{1},\ldots,x_{n})$ or $K(x,y,z)$ is the rational function field over $K$. We will denote by $\zeta_{n}$ a primitive $n$-th root of unity. Whenever we write $\zeta_{n}\in K$, it is understood that either $\mathop{{\rm char}\vphantom{\dim}}K=0$ or $\mathop{{\rm char}\vphantom{\dim}}K>0$ with $\mathop{{\rm char}\vphantom{\dim}}K\nmid n$.

If $G$ is a finite group, the exponent of $G$ is $\mathop{{\rm lcm}\vphantom{\dim}}\{\mathop{{\rm ord}\vphantom{\dim}}(g):g\in G\}$ where $\mathop{{\rm ord}\vphantom{\dim}}(g)$ is the order of an element $g\in G$. The cyclic group of order $n$ and the dihedral group of order $2n$ will be denoted by $C_{n}$ and $D_{n}$ respectively.

If $G$ is a finite group acting on $K(x_{1},\ldots,x_{n})$ by $K$-automorphism, the actions of $G$ are called monomial actions if for any $\sigma\in G$, any $1\leq j\leq n$, $\sigma\cdot x_{j}=a_{j}(\sigma)\cdot\prod_{1\leq i\leq n}x_{i}^{m_{ij}}$ where $m_{ij}\in\mathbbm{Z}$ and $a_{j}(\sigma)\in K\backslash\{0\}$. The actions are called purely monomial actions if they are monomial actions satisfying $a_{j}(\sigma)=1$ for any $\sigma\in G$, any $1\leq j\leq n$. In the rational function field $K(x_{1},\ldots,x_{n})$, an element of the form $x_{1}^{\lambda_{1}}x_{2}^{\lambda_{2}}\cdots x_{n}^{\lambda_{n}}$ where each $\lambda_{i}\in\mathbbm{Z}$ will be called a power product in $x_{1},\ldots,x_{n}$; we refrain from calling it a monomial to avoid possible confusion with the monomial action and with $a\cdot x_{1}^{\lambda_{1}}\cdots x_{n}^{\lambda_{n}}$ where $a\in K$.

§2. Preliminaries

In this section we recall several results which will be used in the proof of Theorem 1.3.

Theorem 2.1** ([HK3, Theorem 1]).**

Let $G$ be a finite group acting on $L(x_{1},\ldots,x_{n})$, the rational function field of $n$ variables over a field $L$. Suppose that

(i)* for any $\sigma\in G$, $\sigma(L)\subset L$,*

(ii)* the restriction of the action of $G$ to $L$ is faithful, and*

(iii)* for any $\sigma\in G$,*

[TABLE]

where $A(\sigma)\in GL_{n}(L)$ and $B(\sigma)$ is an $n\times 1$ matrix over $L$.

Then there exist elements $z_{1},\ldots,z_{n}\in L(x_{1},\ldots,x_{n})$ which are algebraically independent over $L$ and $L(x_{1},\ldots,x_{n})=L(z_{1},\ldots,z_{n})$ so that $\sigma(z_{i})=z_{i}$ for any $\sigma\in G$, any $1\leq i\leq n$.

Theorem 2.2** ([AHK, Theorem 3.1]).**

Let $L$ be any field, $L(x)$ the rational function field of one variable over $L$, and $G$ a finite group acting on $L(x)$. Suppose that, for any $\sigma\in G$, $\sigma(L)\subset L$ and $\sigma(x)=a_{\sigma}\cdot x+b_{\sigma}$ where $a_{\sigma},b_{\sigma}\in L$ and $a_{\sigma}\neq 0$. Then $L(x)^{G}=L^{G}(f)$ for some polynomial $f\in L[x]$. In fact, if $m=\min\{\deg g(x):g(x)\in L[x]^{G}\setminus L^{G}\}$, any polynomial $f\in L[x]^{G}$ with $\deg f=m$ satisfies the property $L(x)^{G}=L^{G}(f)$.

Theorem 2.3** ([CHK, Theorem 2.3]).**

Let $K$ be any field, $a,b\in K\backslash\{0\}$ and $\sigma:K(x,y)\to K(x,y)$ be a $K$-automorphism of the rational function field $K(x,y)$ defined by $\sigma(x)=a/x$, $\sigma(y)=b/y$. Then $K(x,y)^{\langle\sigma\rangle}=K(u,v)$ where

[TABLE]

Moreover,

[TABLE]

and

[TABLE]

Proof..

The proof of Formula (2.1) was given in [CHK, p.156]. ∎

Theorem 2.4** ([Ka1, Theorem 2.4]).**

Let $K$ be any field. Define a $K$-automorphism $\sigma$ on the rational function field $K(x,y)$ by $\sigma(x)=a/x$, $\sigma(y)=\{b_{1}[x+(a/x)]+b_{2}\}/y$ where $a,b_{1},b_{2}\in K$ with $ab_{1}\neq 0$. Then $K(x,y)^{\langle\sigma\rangle}=K(u,v)$ where

[TABLE]

with $b=b_{1}[x+(a/x)]+b_{2}$.

Theorem 2.5** ([HKO, Theorem 6.7; HK2, Theorem 2.4]).**

Let $F$ be a field with $\mathop{{\rm char}\vphantom{\dim}}F\neq 2$, $E=F(\alpha)$ be a field extension of $F$ defined by $\alpha^{2}=a\in F\backslash\{0\}$. Let $E(x,y)$ be the rational function field of two variables over $E$ and $\sigma$ be an $F$-automorphism on $E(x,y)$ defined by

[TABLE]

where $b,c\in F$ with $b\neq 0$. Then $E(x,y)^{\langle\sigma\rangle}$ is rational over $F$ if and only if the Hilbert symbol $(a,b)_{2}$ is trivial in the field $F(\sqrt{ac})$.

Theorem 2.6** ([Ka4, Theorem 1.4]).**

Let $K$ be a field and $G$ be a finite group. Assume that

(i)* $G$ contains an abelian normal subgroup $H$ so that $G/H$ is cyclic of order $n$;*

(ii)* $\mathbbm{Z}[\zeta_{n}]$ is a unique factorization domain; and*

(iii)* $\zeta_{e}\in K$ where $e$ is the exponent of $G$.*

If $G\to GL(V)$ is any finite-dimensional linear representation of $G$ over $K$, then $K(V)^{G}$ is rational over $K$.

Definition 2.7**.**

Let $G$ be a finite group acting on $K(x_{1},\ldots,x_{n})$ by monomial $K$-automorphisms. Let $z_{1},z_{2},\ldots,z_{n}\in K(x_{1},\ldots,x_{n})$ satisfy

(i)* $\mathop{{\rm trdeg}\vphantom{\dim}}_{K}K(z_{1},\ldots,z_{n})=n$,*

(ii)* for any $\sigma\in G$, any $1\leq j\leq n$, $\sigma\cdot z_{j}=c_{j}(\sigma)\cdot\prod_{1\leq i\leq n}z_{i}^{m_{ij}}$ where $m_{ij}\in\mathbbm{Z}$ and $c_{j}(\sigma)\in K\backslash\{0\}$.*

We will define a group homomorphism $\rho_{\mathop{\underline{z}}}:G\to GL_{n}(\mathbbm{Z})$ by $\rho_{\mathop{\underline{z}}}(\sigma)=(m_{ij})_{1\leq i,j\leq n}$ if $\sigma\cdot z_{j}$ is given by (ii). Note that $\mathop{\underline{z}}$ denotes the ordered transcendental basis $(z_{1},\ldots,z_{n})$;* if $\mathop{\underline{z}}$ is understood from the context, we will simply write $\rho$ for $\rho_{\mathop{\underline{z}}}$.*

Lemma 2.8**.**

Let $K$ be any field and $G$ be a finite group acting on $K(x_{1},\ldots,x_{n})$ by monomial $K$-automorphisms. Then there is a normal subgroup $H$ of $G$ so that

(i)* $K(x_{1},\ldots,x_{n})^{H}=K(z_{1},\ldots,z_{n})$ for some power products $z_{1},\ldots,z_{n}$ in $x_{1},\ldots,x_{n}$,*

(ii)* $G/H$ acts on $K(z_{1},\ldots,z_{n})$ by monomial $K$-automorphisms, and*

(iii)* $\rho_{\mathop{\underline{z}}}:G/H\to GL_{n}(\mathbbm{Z})$ is injective.*

Proof..

Induction on the order of $G$.

Without loss of generality, we may assume that $G$ acts faithfully on $K(x_{1},\ldots,x_{n})$. Define $H_{0}=\mathop{{\rm Ker}\vphantom{\dim}}\{\rho_{\mathop{\underline{x}}}:G\to GL_{n}(\mathbbm{Z})\}$. For any $\tau\in H_{0}$, $\tau(x_{i})=a_{i}(\tau)\cdot x_{i}$ for some $a_{i}(\tau)\in K\backslash\{0\}$. In particular, $H_{0}$ is an abelian subgroup of $G$. Choose $\tau_{1},\ldots,\tau_{m}\in H_{0}$ so that $H_{0}$ is generated by $\tau_{1},\ldots,\tau_{m}$.

Define $\langle x_{1},\ldots,x_{n}\rangle$ to be the multiplicative subgroup of $K(x_{1},\ldots,x_{n})\backslash\{0\}$ generated by $x_{1},\ldots,x_{n}$, i.e. $\langle x_{1},\ldots,x_{n}\rangle=\{M=x_{1}^{\lambda_{1}}x_{2}^{\lambda_{2}}\cdots x_{n}^{\lambda_{n}}:\lambda_{i}\in\mathbbm{Z}$ for $1\leq i\leq n\}$. Define a group homomorphism

[TABLE]

where $K^{\times}=K\backslash\{0\}$ and $M=x_{1}^{\lambda_{1}}\cdots x_{n}^{\lambda_{n}}$ with $\lambda_{i}\in\mathbbm{Z}$. Note that the image of $\Phi$ is a finite group. It is not difficult to find that $\mathop{{\rm Ker}\vphantom{\dim}}(\Phi)=\langle M_{1},\ldots,M_{n}\rangle$ where $M_{1},\ldots,M_{n}$ are $n$ power products in $x_{1},\ldots,x_{n}$. Moreover, $K(x_{1},\ldots,x_{n})^{H_{0}}=K(M_{1},\ldots,M_{n})$.

We will show that $G$ (equivalently, $G/H_{0}$) acts on $K(M_{1},\ldots,M_{n})$ by monomial $K$-automorphisms. For any $\sigma\in G$, any $1\leq i\leq n$, $\sigma(M_{i})=c_{i}(\sigma)\cdot N_{i}$ where $N_{i}=x_{1}^{v_{1}}\cdots x_{n}^{v_{n}}$ with $v_{i}\in\mathbbm{Z}$ and $c_{i}(\sigma)\in K\backslash\{0\}$. To show that each $N_{i}$ is a power product in $M_{1},M_{2},\ldots,M_{n}$, it suffices to show that $N_{i}\in K(x_{1},\ldots,x_{n})^{H_{0}}=K(M_{1},\ldots,M_{n})$. From the identity $\sigma(M_{i})=c_{i}(\sigma)\cdot N_{i}$, we find that, for any $\tau\in H_{0}$, $c_{i}(\sigma)\cdot\tau(N_{i})=\tau\sigma(M_{i})=\sigma\cdot(\sigma^{-1}\tau\sigma)(M_{i})=\sigma(M_{i})$ because $\sigma^{-1}\tau\sigma\in H_{0}$. Thus $\tau(N_{i})=N_{i}$ for any $\tau\in H_{0}$.

Now consider $\rho_{\mathop{\underline{M}}}:G/H_{0}\to GL_{n}(\mathbbm{Z})$. By induction hypothesis, we can find $z_{1}$, $\ldots$, $z_{n}\in K(M_{1},\ldots,M_{n})$ so that (i) $z_{1},\ldots,z_{n}$ are power products in $M_{1},\ldots,M_{n}$, (ii) $K(z_{1},\ldots,z_{n})=K(M_{1},\ldots,M_{n})^{H/H_{0}}$ for some normal subgroup $H/H_{0}$ in $G/H_{0}$ (where $H\supset H_{0}$), (iii) $G/H$ acts monomially on $K(z_{1},\ldots,z_{n})$, (iv) $\rho_{\mathop{\underline{z}}}:G/H\to GL_{n}(\mathbbm{Z})$ is injective. ∎

§3. Proof of Theorem 1.3

We will prove Theorem 1.3 in this section. Because of Lemma 2.8, it suffices to consider the situation when $\rho:G\to GL_{3}(\mathbbm{Z})$ is injective. This condition will remain in force throughout this section.

Unless otherwise specified, we will assume, in this section, that $G$ is a finite 2-group acting on $K(x_{1},x_{2},x_{3})$ by monomial $K$-automorphisms, and $K$ is a field satisfying the conditions (i) $\mathop{{\rm char}\vphantom{\dim}}K\neq 2$, and (ii) $\sqrt{a}\in K$ for any $a\in K$. For example, in the following Theorem 3.4 it is understood that $K$ satisfies the above two conditions although there is no mention about the assumptions on $K$ in the statement of Theorem 3.4. On the other hand, in Theorem 3.3, $K$ satisfies the weaker conditions that $\mathop{{\rm char}\vphantom{\dim}}K\neq 2$ and $\sqrt{-1}\in K$, which will be stated explicitly.

Since $\rho:G\to GL_{3}(\mathbbm{Z})$ is injective, $G$ is isomorphic to a finite subgroup of $GL_{3}(\mathbbm{Z})$ as an abstract group. It is known that, up to conjugation in $GL_{3}(\mathbbm{Z})$, there are precisely 73 non-isomorphic finite subgroups in $GL_{3}(\mathbbm{Z})$ [Ta, HK2, p.807]. We will denote by $W_{i}(j)$ the group $W_{i}$ which appears on page $j$ of Tahara’s paper [Ta]. The 2-groups in $GL_{3}(\mathbbm{Z})$ are the following 36 groups $G$,

1. (I)

$G\simeq C_{2}:W_{i}(173)$ where $1\leq i\leq 5$; 2. (II)

$G\simeq C_{4}:W_{i}(174)$ where $1\leq i\leq 4$; and

$G\simeq C_{2}\times C_{2}:W_{i}(174)$ where $5\leq i\leq 15$; 3. (III)

$G\simeq C_{4}\times C_{2}:W_{i}(187)$ where $1\leq i\leq 2$;

$G\simeq C_{2}\times C_{2}\times C_{2}:W_{i}(187)$ where $3\leq i\leq 6$; and

$G\simeq D_{4}:W_{i}(187)$ where $7\leq i\leq 14$; 4. (IV)

$G\simeq D_{4}\times C_{2}:W_{1}(194)$ and $W_{2}(195)$.

Convention. Suppose that $\sigma,\tau\in G$. We will adopt the convention

[TABLE]

to indicate the fact that $\sigma$ and $\tau$ are $K$-automorphisms on $K(x_{1},x_{2},x_{3})$ defined by $\sigma(x_{1})=\epsilon_{1}x_{1}$, $\sigma(x_{2})=1/x_{2}$, $\sigma(x_{3})=1/x_{3}$, $\tau(x_{1})=1/x_{1}$, $\tau(x_{2})=\epsilon_{2}/x_{2}$, $\tau(x_{3})=\epsilon_{3}/x_{3}$.

Theorem 3.1**.**

Let $G=W_{i}(173)$ for $1\leq i\leq 4$ and $K$ be a field with $\mathop{{\rm char}\vphantom{\dim}}K\neq 2$. Then $K(x_{1},x_{2},x_{3})^{G}$ is rational over $K$.

Proof..

First we will “normalize” the coefficients of $\sigma(x_{i})$ where $G=\langle\sigma\rangle$.

When $G=W_{4}(173)$, the action of $G=\langle\sigma\rangle$ is given by

[TABLE]

for some $a,b,c\in K\backslash\{0\}$. Since $G\simeq C_{2}$, it follows that $a=\pm 1$. Define $y_{2}=x_{2}$, $y_{3}=b/x_{3}$. Then $K(x_{2},x_{3})=K(y_{2},y_{3})$ and $\sigma:(y_{2},y_{3})\mapsto(y_{3},by_{2}/c)$. Since $G\simeq C_{2}$, we find that $b/c=1$. Apply Theorem 2.1. We find that $K(x_{1},x_{2},x_{3})^{G}=K(y_{2},y_{3},x_{1})^{G}=K(y_{2},y_{3})^{G}(u)$ for some $u$ with $\sigma(u)=u$. Note that $K(y_{2},y_{3})^{G}=K(y_{2}+y_{3},(y_{2}-y_{3})^{2})$ is rational over $K$.

When $G=W_{1}(173)$, the action of $G$ is given by

[TABLE]

for some $a,b,c\in K\backslash\{0\}$. Apply Theorem 2.1 and reduce the question to $K(x_{2},x_{3})^{G}$. Apply Theorem 1.1. Done.

When $G=W_{2}(173)$ and $W_{3}(173)$, the actions of $G$ are given by

[TABLE]

and

[TABLE]

respectively where $a,b,c\in K\backslash\{0\}$.

Apply Theorem 2.1 to both cases. The questions are reduced to $K(x_{1})^{G}$. Note that $K(x_{1})^{G}$ is rational by Lüroth’s Theorem.

∎

Theorem 3.2**.**

Let $K$ be a field with $\mathop{{\rm char}\vphantom{\dim}}K\neq 2$ and $G=W_{5}(173)$, i.e. the action of $G=\langle\sigma\rangle$ is given by

[TABLE]

for some $a_{1},a_{2},a_{3}\in K\backslash\{0\}$. Then $K(x_{1},x_{2},x_{3})^{G}$ is rational over $K$ if and only if $[K(\sqrt{a_{1}},\sqrt{a_{2}},\sqrt{a_{3}}):K]\leq 4$.

Proof..

If $[K(\sqrt{a_{1}},\sqrt{a_{2}},\sqrt{a_{3}}):K]=8$, it is Saltman who shows that $K(x_{1},x_{2},x_{3})^{G}$ is not retract rational over $K$ [Sa] (note that $K$ is an infinite field in this situation). In particular, $K(x_{1},x_{2},x_{3})^{G}$ is not rational over $K$. See [Ka2] for a generalization of Saltman’s Theorem.

If $[K(\sqrt{a_{1}},\sqrt{a_{2}},\sqrt{a_{3}}):K]\leq 4$, we may assume that $a_{3}\in a_{1}^{\epsilon_{1}}a_{2}^{\epsilon_{2}}K^{2}$ for some $\epsilon_{1},\epsilon_{2}\in\{0,1\}$. First consider the case $a_{3}=a_{1}a_{2}b^{2}$ for some $b\in K\backslash\{0\}$. Define $y=x_{3}/(bx_{1}x_{2})$. Then $K(x_{1},x_{2},x_{3})=K(x_{1},x_{2},y)$ and $\sigma(y)=1/y$. Define $z=(1-y)/(1+y)$. Then $\sigma(z)=-z$. It follows that $K(x_{1},x_{2},x_{3})^{G}=K(x_{1},x_{2},z)^{G}=K(x_{1},x_{2})^{G}(u)$ for some $u$ with $\sigma(u)=u$ by Theorem 2.1. Note that $K(x_{1},x_{2})^{G}$ is rational by Theorem 1.1. The other cases $a_{3}\in a_{1}K^{2}$, $a_{3}\in a_{2}K^{2}$ and $a_{3}\in K^{2}$ can be proved similarly. ∎

Theorem 3.3**.**

Let $G=W_{i}(174)$ for $1\leq i\leq 4$ and $K$ satisfy the conditions that $\mathop{{\rm char}\vphantom{\dim}}K\neq 2$ and $\sqrt{-1}\in K$. Then $K(x_{1},x_{2},x_{3})^{G}$ is rational over $K$.

Proof..

Case 1. $G=W_{1}(174)$.

The action of $G=\langle\sigma\rangle$ is given by

[TABLE]

for some integer $i$ and some $a\in K\backslash\{0\}$. Apply Theorem 2.1. We find that $K(x_{1}$, $x_{2},x_{3})^{G}=K(x_{2},x_{3})^{G}(u)$ for some $u$ with $\sigma(u)=u$. By Theorem 1.1, $K(x_{2},x_{3})^{G}$ is rational over $K$.

Case 2. $G=W_{2}(174)$.

The action of $G=\langle\sigma\rangle$ is given by

[TABLE]

for some $a,b\in K\backslash\{0\}$. Define $u$ and $v$ by

[TABLE]

By Theorem 2.3, we get $K(x_{1},x_{2},x_{3})^{\langle\sigma^{2}\rangle}=K(x_{1},u,v)$. Note that

[TABLE]

Note that

[TABLE]

by Theorem 2.3 again.

Define $w=u/v$. Then $K(x_{1},u,v)=K(w,x_{1},v)$ and

[TABLE]

where $c=b[w-(1/w)]$.

Define

[TABLE]

Then $K(w,x_{1},v)=K(y_{1},y_{2},v)$ and $\sigma(y_{1})=a/y_{1}$, $\sigma(y_{2})=y_{2}$, $\sigma(v)=1/(cv)$.

It is not difficult to verify that c=b[w-(1/w)]=2\sqrt{-1}b\big{(}x_{1}-(a/x_{1})+\sqrt{-1}Y_{2}\big{)}\big{(}x_{1}-(a/x_{1})-\sqrt{-1}Y_{2}\big{)}\big{/}[\big{(}x_{1}-(a/x_{1})+Y_{2}\big{)}\big{(}x_{1}-(a/x_{1})-Y_{2}\big{)}].

Define

[TABLE]

It follows that $\sigma(y_{3})=d/y_{3}$ where $d=1/(2\sqrt{-1}b)$. Now we find that $K(w,x_{1},v)^{\langle\sigma\rangle}$ $=K(y_{1},y_{2},y_{3})^{\langle\sigma\rangle}=K(y_{2})(y_{1},y_{3})^{\langle\sigma\rangle}$ is rational over $K(y_{2})$ by Theorem 2.3.

Case 3. $G=W_{3}(174)$.

The action of $G=\langle\sigma\rangle$ is given by

[TABLE]

for some $a,b\in K\backslash\{0\}$. Define $y_{1}=x_{2}$, $y_{2}=x_{3}$, $y_{3}=bx_{1}/x_{2}$. Then $K(x_{1},x_{2},x_{3})=K(y_{1},y_{2},y_{3})$ and $\sigma:(y_{1},y_{2},y_{3})\mapsto(y_{2},y_{3},\epsilon y_{1}y_{3}/y_{2})$. Since $\mathop{{\rm ord}\vphantom{\dim}}(\sigma)=4$, it follows that $\epsilon^{2}=1$, i.e. $\epsilon=\pm 1$.

Case 3.1. $\epsilon=1$.

The action $\sigma:(y_{1},y_{2},y_{3})\mapsto(y_{2},y_{3},y_{1}y_{3}/y_{2})$ is a purely monomial automorphism. Thus $K(y_{1},y_{2},y_{3})^{G}$ is rational over $K$ by Theorem 1.2.

Case 3.2. $\epsilon=-1$.

We find that $\sigma:(y_{1},y_{2},y_{3})\mapsto(y_{2},y_{3},-y_{1}y_{3}/y_{2})$.

Define

[TABLE]

Then $K(y_{1},y_{2},y_{3})=K(X_{1},X_{2},X_{3},X_{4})$ with the relation $X_{1}^{2}+X_{2}^{2}+X_{3}^{2}+X_{4}^{2}=0$. Note that $\sigma:X_{1}\leftrightarrow X_{2}$, $X_{3}\mapsto X_{4}\mapsto-X_{3}$. Thus $\sigma^{2}:X_{1}\mapsto X_{1}$, $X_{2}\mapsto X_{2}$, $X_{3}\mapsto-X_{3}$, $X_{4}\mapsto-X_{4}$.

Define $Y_{3}=X_{3}/X_{4}$, $Y_{1}=X_{1}Y_{3}$, $Y_{2}=X_{2}Y_{3}$, $Y_{4}=X_{3}^{2}$. Then $K(X_{1},X_{2},X_{3},X_{4})^{\langle\sigma^{2}\rangle}=K(Y_{1},Y_{2},Y_{3},Y_{4})$ with the relation $Y_{1}^{2}+Y_{2}^{2}+Y_{4}(1+Y_{3}^{2})=0$. Hence $Y_{4}\in K(Y_{1},Y_{2},Y_{3})$. It follows that $K(y_{1},y_{2},y_{3})^{\langle\sigma\rangle}=K(Y_{1},Y_{2},Y_{3})^{\langle\sigma\rangle}$. Note that

[TABLE]

Define $z_{1}=Y_{1}/Y_{3}$, $z_{2}=Y_{2}/Y_{3}$. Then $K(Y_{1},Y_{2},Y_{3})=K(z_{1},z_{2},Y_{3})$ and $\sigma:(z_{1},z_{2}$, $Y_{3})\mapsto(z_{2},z_{1},-1/Y_{3})$.

Apply Theorem 2.1. We find that $K(Y_{1},Y_{2},Y_{3})^{\langle\sigma\rangle}=K(z_{1},z_{2},Y_{3})^{\langle\sigma\rangle}=K(Y_{3})^{\langle\sigma\rangle}(u_{1},u_{2})$ for some $u_{1}$, $u_{2}$ with $\sigma(u_{1})=u_{1}$, $\sigma(u_{2})=u_{2}$.

Case 4. $G=W_{4}(174)$.

The action of $G=\langle\sigma\rangle$ is given by

[TABLE]

for some $b,c\in K\backslash\{0\}$.

Define $y_{1}=x_{2}$, $y_{2}=x_{3}$, $y_{3}=bx_{1}/x_{2}$. Then $K(x_{1},x_{2},x_{3})=K(y_{1},y_{2},y_{3})$ and $\sigma:(y_{1},y_{2},y_{3})\mapsto(y_{2},y_{3},a/(y_{1}y_{2}y_{3}))$ for some $a\in K\backslash\{0\}$.

Note that $\sigma^{2}:(y_{1},y_{2},y_{3})\mapsto(y_{3},a/(y_{1}y_{2}y_{3}),y_{1})$.

Define $t=y_{1}y_{3}$. Then $K(y_{1},y_{2},y_{3})=K(t,y_{1},y_{2})$ and $\sigma^{2}:(t,y_{1},y_{2})\mapsto(t,t/y_{1},a/(ty_{2}))$. Define $A=t$, $B=a/t$, $u$ and $v$ as follows

[TABLE]

By Theorem 2.3, we find that $K(t,y_{1},y_{2})^{\langle\sigma^{2}\rangle}=K(t,u,v)$. Note that

[TABLE]

Note that

[TABLE]

by Theorem 2.3.

Define $w=u/v$. Then $K(t,u,v)=K(t,w,v)$ and $\sigma:(t,w,v)\mapsto(a/t,-1/w,C/v)$ where $C=w/(Bw^{2}-A)$.

Define $Z_{1}=({\sqrt{-1}-w})/({\sqrt{-1}+w})$, $z_{2}=w/t$. Then $\sigma(Z_{1})=-Z_{1}$, $\sigma(z_{2})=-1/(az_{2})$.

Define $z_{1}=Z_{1}[z_{2}+1/(az_{2})]$. Then $\sigma:(z_{1},z_{2},v)\mapsto(z_{1},-1/(az_{2}),C/v)$ where $C=w/(Bw^{2}-A)=1/[(aw/t)-(t/w)]=1/[az_{2}-(1/z_{2})]$.

Define $z_{3}=v[az_{2}-(1/z_{2})]$. Then $K(t,u,v)=K(t,w,v)=K(z_{1},z_{2},z_{3})$ and $\sigma:(z_{1},z_{2},z_{3})\mapsto(z_{1},-1/(az_{2}),D/z_{3})$ where $D=az_{2}-(1/z_{2})=a[z_{2}+\sigma(z_{2})]$. By Theorem 2.4 (instead of Theorem 2.3 !), $K(z_{1},z_{2},z_{3})^{\langle\sigma\rangle}$ is rational over $K(z_{1})$.

∎

Theorem 3.4**.**

Let $G=W_{i}(174)$ for $5\leq i\leq 15$. Then $K(x_{1},x_{2},x_{3})^{G}$ is rational over $K$.

Proof..

Case 1. $G=W_{5}(174)$.

The action of $G=\langle\sigma,\tau\rangle$ is given by

[TABLE]

for some $a_{1},a_{2},a_{3},b_{2},b_{3}\in K\backslash\{0\}$ and $\epsilon=\pm 1$.

Since $\sqrt{a_{1}},\sqrt{a_{2}},\sqrt{a_{3}}\in K$, we may define $y_{i}=x_{i}/\sqrt{a_{i}}$ for $1\leq i\leq 3$. It follows that $\sigma:(y_{1},y_{2},y_{3})\mapsto(\epsilon y_{1},c_{2}/y_{2},c_{3}/y_{3})$ and $\tau:(y_{1},y_{2},y_{3})\mapsto(1/y_{1},1/y_{2},1/y_{3})$ for some $c_{2},c_{3}\in K\backslash\{0\}$. Since $\sigma\tau=\tau\sigma$, it follows that $c_{2}^{2}=c_{3}^{2}=1$.

If $c_{2}c_{3}=1$, we define $y_{4}=y_{2}/y_{3}$. It follows that $\sigma(y_{4})=\tau(y_{4})=1/y_{4}$. Define $y_{5}=(1-y_{4})/(1+y_{4})$. Then $\sigma(y_{5})=\tau(y_{5})=-y_{5}$. Thus $K(y_{1},y_{2},y_{3})^{G}=K(y_{1},y_{2},y_{5})^{G}=K(y_{1},y_{2})^{G}(y_{0})$ for some $y_{0}$ with $\sigma(y_{0})=\tau(y_{0})=y_{0}$ by Theorem 2.1. Now $K(y_{1},y_{2})^{G}$ is rational by Theorem 1.1.

If $c_{2}c_{3}=-1$, we may assume that $c_{2}=-1$ and $c_{3}=1$. In this situation, define $y_{4}=y_{3}$. The arguments in the preceding paragraph remain valid. Done.

It is can be shown that $K(x_{1},x_{2},x_{3})^{G}$ is rational when $G=W_{5}(174)$ and $K$ is any field with $\mathop{{\rm char}\vphantom{\dim}}K\neq 2$, i.e. the assumption that $\sqrt{a}\in K$ for any $a\in K$ can be waived. But we omit the proof here.

Case 2. $G=W_{6}(174)$.

The action of $G=\langle\sigma,\tau\rangle$ is given by

[TABLE]

for some $a_{1},a_{2},a_{3},a_{4}\in K\backslash\{0\}$ and $\epsilon_{1},\epsilon_{3}\in\{1,-1\}$.

Define $y_{i}=x_{i}/\sqrt{a_{i}}$ for $1\leq i\leq 3$. Using the relation $\sigma\tau=\tau\sigma$, we get

[TABLE]

where $\epsilon_{1},\epsilon_{2},\epsilon_{3}\in\{1,-1\}$.

If at least one of $\epsilon_{1},\epsilon_{2},\epsilon_{3}$ is 1, say $\epsilon_{3}=1$, define $y_{4}=(1-y_{3})/(1+y_{3})$. Then we get $\sigma(y_{3})=-y_{3}$ and $\sigma(y_{3})=y_{3}$. Then we may apply Theorem 2.1 and get $K(y_{1},y_{2},y_{3})^{G}=K(y_{1},y_{2},y_{4})^{G}=K(y_{1},y_{2})^{G}(y_{0})$ with $\sigma(y_{0})=\tau(y_{0})=y_{0}$.

It remains to consider the situation $\epsilon_{1}=\epsilon_{2}=\epsilon_{3}=-1$.

Define $z_{1}$, $u$ and $v$ by

[TABLE]

Then $K(y_{1},y_{2},y_{3})^{\langle\sigma\rangle}=K(z_{1},u,v)$ by Theorem 2.3. Define $w=u/v$. Use the same technique as in the proof of Case 2 of Theorem 3.3. We find that

[TABLE]

where $A=[y_{2}-(1/y_{2})]^{2}=[(u^{2}+w^{2}-u^{2}w^{2})^{2}/(u^{2}w^{2})]-4$, $B=-w^{2}/(w^{2}-1)$. Note that $\tau(A)=A$.

Define $z_{2}=u(w-1)/w$. Then $K(z_{1},u,v)=K(z_{1},z_{2},w)$ and $\tau(z_{2})=-1/z_{2}$. It is not difficult to verify that

[TABLE]

Define $t=(\sqrt{-1}-z_{2})/(\sqrt{-1}+z_{2})$. Then $K(z_{1},u,w)=K(t,w,z_{1})$ and

[TABLE]

where $A=2\sqrt{-1}[w^{2}(1+t^{2})-4wt+2(1+t^{2})]/(1-t^{2})$.

Define $x=tw$, $y=z_{1}(1-t)/e$ where $e\in K$ satisfies $e^{2}=2\sqrt{-1}$. We will use Theorem 2.5 to show that $K(t,x,y)^{\langle\tau\rangle}$ is rational over $K(t^{2})$; in particular, $K(t,x,y)^{\langle\tau\rangle}$ is rational over $K$.

Note that $\tau(t)=-t$, $\tau(x)=x$, $\tau(y)=B/y$ where

[TABLE]

In applying Theorem 2.5, define $a=t^{2}$, $F=K(t^{2})$, $b=(1+t^{2})/t^{2}=(1+a)/a$ and $c=-(2a+2a^{3})/(1+a)^{2}$. It is known that $K(t,x,y)^{\langle\tau\rangle}=F(t)(x,y)^{\langle\tau\rangle}$ is rational over $F$ if and only if the Hilbert symbol $(a,b)_{2}$ is trivial in the field $F(\sqrt{ac})$.

We claim that $F(\sqrt{ac})$ is a rational function field of one variable over $K$. Note that $F(\sqrt{ac})=F(\sqrt{1+a^{2}})$ because $\sqrt{-2}\in K\subset F$. Write $p=\sqrt{1+a^{2}}$, $s=p-a$. Then $F(\sqrt{1+a^{2}})=K(a,p)$. From the relation $(p-a)(p+a)=1$, we find that $p+a=1/(p-a)=1/s\in K(s)$. Thus $a=(1-s^{2})/(2s)\in K(s)$. Hence $F(\sqrt{1+a^{2}})=K(a,p)=K(s)$.

Now we find that $(a,b)_{2}=(a,(1+a)/a)_{2}=(a,a(1+a))_{2}=(a,-(1+a))_{2}=(a,1+a)_{2}$ because $-1\in K^{2}$. To show that $(a,1+a)_{2}=0$ is equivalent to find a solution in $F(\sqrt{ac})=K(s)$ of the equation

[TABLE]

Substitute $a=(1-s^{2})/(2s)$ into the above equation. We get

[TABLE]

Note that $X=\sqrt{-1}$, $Y=1$ is a solution we need. Hence the result.

It can be shown that the field $F(t)(x,y)^{\langle\tau\rangle}$ is the function field of certain conic bundle of $\mathbbm{P}^{1}$ defined over $F$. Thus we may as well use Iskovskikh’s theory of conic bundles [Is] to show that $F(t)(x,y)^{\langle\tau\rangle}$ is rational over $F$. See also [Ka3].

Case 3. $G=W_{7}(174)$.

The action of $G={\langle\sigma,\tau\rangle}$ is given by

[TABLE]

where $a_{2},a_{3},b\in K\backslash\{0\}$ and $\epsilon,\epsilon_{1},\epsilon_{2}\in\{1,-1\}$.

Apply Theorem 2.1. The question is reduced to the rationality of $K(x_{2},x_{3})^{G}$. Done.

Case 4. $G=W_{8}(174)$.

The action of $G={\langle\sigma,\tau\rangle}$ is given by

[TABLE]

where $\epsilon_{1},a_{2},a_{3},b_{1},b_{2},b_{3}\in K\backslash\{0\}$.

Define $y_{1}=x_{1}/\sqrt{b_{1}}$, $y_{2}=x_{2}/\sqrt{a_{2}}$, $y_{3}=x_{3}/\sqrt{a_{3}}$. We find that

[TABLE]

where $\epsilon_{1},\epsilon_{2},\epsilon_{3}\in\{1,-1\}$.

The situation is the same as in Case 1. Hence we omit the proof.

Case 5. $G=W_{9}(174)$.

The action of $G={\langle\sigma,\tau\rangle}$ is given by

[TABLE]

where $a_{i},b_{j}\in K\backslash\{0\}$.

Apply Theorem 2.1. The question is reduced to the rationality of $K(x_{2},x_{3})^{G}$.

Case 6. $G=W_{10}(174)$.

The action of $G={\langle\sigma,\tau\rangle}$ is given by

[TABLE]

where $a_{i},b_{j}\in K\backslash\{0\}$.

Define $y=x_{1}/\sqrt{a_{1}}$. Then $\sigma(y)=1/y$ and $\tau(y)=\pm y$.

If $\tau(y)=-y$, we will apply Theorem 2.1 and reduce the question to the rationality of $K(y)^{G}$.

If $\tau(y)=y$, define $z=(1-y)/(1+y)$. Then $\sigma(z)=-z$ and $\tau(z)=z$. Apply Theorem 2.1 and reduced the question to the rationality of $K(x_{2},x_{3})^{G}$.

Case 7. $G=W_{11}(174)$.

After the “normalization” of the coefficients, the action of $G={\langle\sigma,\tau\rangle}$ is given by

[TABLE]

where $b\in K\backslash\{0\}$ and $\epsilon=\pm 1$.

Define $y_{i}=x_{i}/\sqrt{b}$ for $i=2,3$. Then $\sigma:y_{2}\leftrightarrow y_{3}$ and $\tau:y_{2}\mapsto 1/y_{2}$, $y_{3}\mapsto 1/y_{3}$.

If $\tau(x_{1})=-x_{1}$, we may apply Theorem 2.1 and consider the rationality of $K(x_{1})^{G}$.

If $\tau(x_{1})=x_{1}$, define $y_{1}=(1-x_{1})/(1+x_{1})$. We get $\sigma(y_{1})=-y_{1}$ and $\tau(y_{1})=y_{1}$. Apply Theorem 2.1 and consider the rationality of $K(y_{2},y_{3})^{G}$. Done.

Case 8. $G=W_{12}(174)$.

The action of $G={\langle\sigma,\tau\rangle}$ is given by

[TABLE]

where $b_{1},b_{2},b_{3}\in K\backslash\{0\}$.

Define $y_{1}=x_{1}$, $y_{2}=x_{2}/\sqrt{b_{2}}$, $y_{3}=x_{3}/\sqrt{b_{2}}$. Use the fact that $\sigma\tau=\tau\sigma$. We find that

[TABLE]

where $\epsilon=\pm 1$.

If $\epsilon=1$, apply Theorem 1.2. Thus $K(y_{1},y_{2},y_{3})^{G}$ is rational.

It remains to consider the case that $\epsilon=-1$.

Define $t=y_{2}/y_{3}$. Then $K(y_{1},y_{2},y_{3})=K(t,y_{1},y_{2})$ and

[TABLE]

Define

[TABLE]

By Theorem 2.3 we find that $K(t,y_{1},y_{2})^{\langle\tau\rangle}=K(t,u,w)$ and

[TABLE]

where $A=w^{2}/(1+w^{2})$ (the computation is similar to Case 4).

Define $s=[(1-t)/(1+t)][u-(A/u)]$. Then $K(t,u,w)=K(s,u,w)$ and $\sigma(s)=s$. Thus we may reduce the question to the rationality of $K(w)(u)^{\langle\sigma\rangle}$, which is easy by Lüroth’s Theorem. Hence the result.

Case 9. $G=W_{13}(174)$.

The action of $G={\langle\sigma,\tau\rangle}$ is given by

[TABLE]

where $a_{1},a_{2},a_{3}\in K\backslash\{0\}$.

Define $x_{0}=x_{2}/x_{3}$. Then $K(x_{1},x_{2},x_{3})=K(x_{0},x_{1},x_{2})$ and

[TABLE]

Apply Theorem 2.2. It suffices to show that $K(x_{0},x_{1})^{G}$ is rational. But this follows from Theorem 1.1.

Case 10. $G=W_{14}(174)$.

The action of $G={\langle\sigma,\tau\rangle}$ is given by

[TABLE]

where $a_{1},a_{2},a_{3}\in K\backslash\{0\}$.

Since $\tau^{2}=1$, we find that $a_{3}=a_{1}a_{2}$. From $\sigma\tau(x_{2})=\tau\sigma(x_{2})$, we find that $a_{2}=a_{3}$. Thus $a_{1}=1$. In summary, we have that

[TABLE]

where $a\in K\backslash\{0\}$.

Define $y_{i}=x_{i}/\sqrt{a}$ for $i=2,3$. Then $\sigma:y_{2}\leftrightarrow y_{3}$, $\tau:y_{2}\mapsto x_{1}/y_{3}$, $y_{3}\mapsto 1/(x_{1}y_{2})$. Thus $G$ acts on $K(x_{1},y_{2},y_{3})$ by purely monomial $K$-automorphisms. Hence we may apply Theorem 1.2.

Case 11. $G=W_{15}(174)$.

The action of $G={\langle\sigma,\tau\rangle}$ is given by

[TABLE]

Define $x_{0}=x_{2}/x_{3}$. The proof is almost the same as that of Case 9. $G=W_{13}(174)$.

∎

Theorem 3.5**.**

Let $G=W_{i}(187)$ for $1\leq i\leq 2$. Then $K(x_{1},x_{2},x_{3})^{G}$ is rational over $K$.

Proof..

Case 1. $G=W_{1}(187)$.

The action of $G={\langle\sigma,\tau\rangle}$ is given by

[TABLE]

where $b,c_{1},c_{2},c_{3}\in K\backslash\{0\}$ and $i\in\{0,1,2,3\}$.

Since $\sigma\tau=\tau\sigma$, we find that $i=0$ or 2, $c_{2}=c_{3}$, $(b/c_{2})^{2}=1$. In summary, we have

[TABLE]

where $a_{1},a_{2}\in K\backslash\{0\}$ and $\epsilon_{1},\epsilon_{2}\in\{1,-1\}$.

Define $y_{1}=x_{1}/\sqrt{a_{1}}$, $y_{2}=x_{2}/\sqrt{a_{2}}$, $y_{3}=x_{3}/\sqrt{a_{2}}$. We get that

[TABLE]

where $\epsilon_{1},\epsilon_{2}\in\{1,-1\}$.

If $\epsilon_{1}=1$, define $z_{1}=(1-y_{1})/(1+y_{1})$. Then $\sigma(z_{1})=z_{1}$, $\tau(z_{1})=-z_{1}$. Apply Theorem 2.1 and reduce the question to $K(y_{2},y_{3})^{G}$.

If $\epsilon_{1}=-1$ and $\epsilon_{2}=1$, define $z_{2}=(1-y_{2})/(1+y_{2})$, $z_{3}=(1-y_{3})/(1+y_{3})$. Then $\sigma(z_{2})=z_{3}$, $\sigma(z_{3})=-z_{2}$, $\tau(z_{2})=-z_{2}$, $\tau(z_{3})=-z_{3}$. Apply Theorem 2.1 and reduce the question to $K(y_{1})^{G}$.

It remains to consider the situation when $\epsilon_{1}=\epsilon_{2}=-1$.

Note that $\sigma^{2}(y_{1})=y_{1}$, $\sigma^{2}(y_{2})=-1/y_{2}$, $\sigma^{2}(y_{3})=-1/y_{3}$. Define

[TABLE]

By Theorem 2.3 we find that $K(y_{1},y_{2},y_{3})^{\langle\sigma^{2}\rangle}=K(y_{1},u,v)=K(y_{1},w,v)$ and

[TABLE]

where $A=w/(w^{2}-1)$.

Define $z_{1}=y_{1}+(1/y_{1})$, $z_{2}=v[y_{1}-(1/y_{1})]$. Then $K(y_{1},w,v)^{\langle\tau\rangle}=K(z_{1},z_{2},w)$ and

[TABLE]

where $B=(z_{1}^{2}-4)/[(1/w)-w]$.

Define $t=z_{1}[w+(1/w)]$, $z_{3}=z_{2}[(1/w)-w]/(z_{1}-2)$. Then $K(z_{1},z_{2},w)=K(t,w,z_{3})$ and

[TABLE]

Thus $K(t,w,z_{3})^{\langle\sigma\rangle}$ is rational over $K(t)$ by Theorem 2.4 (instead of Theorem 2.3).

Case 2. $G=W_{2}(187)$.

The action of $G={\langle\sigma,\tau\rangle}$ is given by

[TABLE]

where $a_{i},b_{j}\in K\backslash\{0\}$.

Define $y_{1}=x_{2}$, $y_{2}=x_{3}$, $y_{3}=a_{2}x_{1}/x_{2}$. Using the relation $\sigma^{4}=1$, $\sigma\tau=\tau\sigma$, we get

[TABLE]

where $\epsilon\in\{1,-1\}$ and $a\in K\backslash\{0\}$.

Define $z_{i}=y_{i}/\sqrt{a}$ for $1\leq i\leq 3$. Then we have

[TABLE]

where $\epsilon\in\{1,-1\}$.

If $\epsilon=1$, the action of $G$ is a purely monomial action. Apply Theorem 1.2.

It remains to consider the situation when $\epsilon=-1$.

Define $t=z_{1}z_{3}$. Then $K(z_{1},z_{2},z_{3})=K(t,z_{1},z_{2})$ and

[TABLE]

Note that $\sigma^{2}(t)=t$, $\sigma^{2}(z_{1})=t/z_{1}$, $\sigma^{2}(z_{2})=-t/z_{2}$. Define

[TABLE]

By Theorem 2.3 we find that $K(t,z_{1},z_{2})^{\langle\sigma^{2}\rangle}=K(t,u,v)=K(t,w,v)$ and

[TABLE]

where $A=-w/[t(w^{2}+1)]$.

Define $u_{1}=t+(1/t)$, $u_{2}=w[t-(1/t)]$, $u_{3}=v(1+t)$. Then $K(t,w,v)^{\langle\tau\rangle}=K(u_{1},u_{2}$, $u_{3})$ and

[TABLE]

where $B=A(1-t^{2})=w(t^{2}-1)/[t(w^{2}+1)]=u_{2}(u_{1}^{2}-4)/[u_{2}^{2}+(u_{1}^{2}-4)]$.

Define $v_{1}=u_{1}$, $v_{2}=u_{2}/(u_{1}-2)$, $v_{3}=u_{3}$. Then

[TABLE]

where $B=(v_{1}+2)v_{2}/[v_{2}^{2}+((v_{1}+2)/(v_{1}-2))]$.

Define $v_{4}=(\sqrt{-1}-v_{2})/(\sqrt{-1}+v_{2})$. Then $\sigma(v_{4})=-v_{4}$. It is not difficult to verify that

[TABLE]

Define $s=v_{1}^{2}$, $\alpha=v_{1}$, $x=v_{1}v_{4}$, $y=e\cdot v_{3}(v_{4}^{2}+v_{1}v_{4}+1)/[(v_{1}-2)(v_{4}-1)]$ where $e\in K$ satisfying $e^{2}=4\sqrt{-1}$. Note that $K(u_{1},u_{2},u_{3})=K(\alpha,x,y)$ and we find that

[TABLE]

where $f(x)=\frac{1}{s}\{[x+(s/2)]^{2}-[(s^{2}-4s)/4]\}$.

We will apply Theorem 2.5 (note that Formula (3.2) looks very similar to Formula (3.1) in the proof of Case 2 in Theorem 3.4). In the notation of Theorem 2.5, $F=K(s)$ and $E=F(\alpha)$. It remains to show that the Hilbert symbol $(s,1/s)_{2}$ is trivial in $K(s,\sqrt{s-4})$. Since $-1\in K^{2}$, we find that $(s,1/s)_{2}=(s,s)_{2}=(s,-s)_{2}=0$. Hence $K(\alpha,x,y)^{\langle\sigma\rangle}=K(s)(\alpha,x,y)^{\langle\sigma\rangle}$ is rational over $K(s)$.

∎

Theorem 3.6**.**

Let $G=W_{i}(187)$ for $3\leq i\leq 6$. Then $K(x_{1},x_{2},x_{3})^{G}$ is rational over $K$.

Proof..

Case 1. $G=W_{3}(187)$.

The action of $G=\langle\sigma,\tau,\lambda\rangle$ is given by

[TABLE]

where $\epsilon_{1},\epsilon_{2},\epsilon_{3}\in\{1,-1\}$, $a_{1},a_{2},a_{3}\in K\backslash\{0\}$ (for the actions of $\sigma$ and $\tau$, compare with the actions of $\sigma$ and $\tau$ for $G=W_{6}(174)$ in the proof of Theorem 3.4).

Since $\sigma\lambda=\lambda\sigma$ and $\tau\lambda=\lambda\tau$, we find that $a_{1}^{2}=a_{2}^{2}=a_{3}^{2}=1$.

We will consider only the case $\epsilon_{1}=-1$, because the case $\epsilon_{1}=1$ is easier and can be proved similarly.

Define $y_{1}=x_{1}$ and

[TABLE]

for $i=2,3$. Thus we get

[TABLE]

where $\epsilon_{2},\epsilon_{3},a_{1},a_{2},a_{3}\in\{1,-1\}$.

Define $z_{1}=y_{1}/y_{2}$, $z_{2}=y_{2}/y_{3}$, $z_{3}=y_{3}^{2}$. Then $K(y_{1},y_{2},y_{3})^{\langle\sigma\rangle}=K(z_{1},z_{2},z_{3})$. It is not difficult to see that anyone of $\tau(z_{i})$, $\lambda(z_{j})$ where $1\leq i,j\leq 3$ is of the form

[TABLE]

where $\epsilon\in\{1,-1\}$ and $b_{1},b_{2},b_{3}\in\mathbbm{Z}$. Hence $\langle\tau,\lambda\rangle$ acts on $K(z_{1},z_{2},z_{3})$ by monomial $K$-automorphisms. Since $\langle\tau,\lambda\rangle\simeq C_{2}\times C_{2}$, we find that $K(z_{1},z_{2},z_{3})^{\langle\tau,\lambda\rangle}$ is rational over $K$ by Theorem 3.4. Done.

Case 2. $G=W_{4}(187)$.

The action of $G=\langle\sigma,\tau,\lambda\rangle$ is given by

[TABLE]

where $a_{i},b_{j},c_{k}\in K\backslash\{0\}$.

Define $y_{i}=x_{i}/\sqrt{c_{i}}$ for $1\leq i\leq 3$. It is easy to see that

[TABLE]

where $\epsilon_{1},\epsilon_{2},\epsilon_{3},e_{1},e_{2},e_{3}\in\{1,-1\}$ and $e_{2}=e_{3}$.

Define $z_{i}=(1-y_{i})/(1+y_{i})$ for $1\leq i\leq 3$. Consider $K(z_{1},z_{2},z_{3})^{\langle\lambda\rangle}$ and then proceed as in the above Case 1. The details are left to the reader.

Case 3. $G=W_{5}(187)$.

The action of $G=\langle\sigma,\tau,\lambda\rangle$ is given by

[TABLE]

where $\epsilon\in\{1,-1\}$, $a_{1},a_{2},a_{3}\in K\backslash\{0\}$ (for the actions of $\sigma$ and $\tau$, compare with those for $G=W_{12}(174)$ in the proof of Theorem 3.4).

Since $G\simeq C_{2}\times C_{2}\times C_{2}$, it is easy to see that $a_{2}=a_{3}$ and $a_{1},a_{2}\in\{1,-1\}$.

Define $y_{1}=x_{1}$, $y_{2}=x_{2}$, $y_{3}=x_{2}/x_{3}$. We find that

[TABLE]

where $\epsilon,\epsilon_{1},\epsilon_{2}\in\{1,-1\}$.

Define

[TABLE]

By Theorem 2.3, we get $K(y_{1},y_{2},y_{3})^{\langle\tau\rangle}=K(u,v,y_{3})$ and

[TABLE]

where $A=w^{2}/(w^{2}-\epsilon)$.

Define $z_{1}=w$, $z_{2}=(1-y_{3})/(1+y_{3})$, $z_{3}=2uy_{3}/(1+y_{3})$. Then $K(u,w,y_{3})=K(z_{1},z_{2},z_{3})$ and

[TABLE]

where $B=A(1-z_{2}^{2})=z_{1}^{2}(1-z_{2}^{2})/(z_{1}^{2}-\epsilon)$.

Case 3.1. $\epsilon_{2}=-1$ and $\epsilon\epsilon_{1}\epsilon_{2}=-1$.

Define $u_{1}=z_{1}z_{2}$, $u_{2}=z_{2}^{2}$, $u_{3}=z_{2}z_{3}$. Then $K(z_{1},z_{2},z_{3})^{\langle\lambda\rangle}=K(u_{1},u_{2},u_{3})$ and

[TABLE]

where $C=u_{1}^{2}u_{2}(1-u_{2})/(\epsilon-u_{1}^{2})$.

If $\epsilon=1$, then $K(u_{1},u_{2},u_{3})^{\langle\sigma\rangle}=K(u_{1},u_{2},u_{3}+(C/u_{3}))$.

If $\epsilon=-1$, define $u_{4}=u_{3}(1-\sqrt{-1}u_{1})/u_{1}$. Then we get that

[TABLE]

Thus $\sigma$ acts on $K(u_{2})(u_{1},u_{4})$ as a monomial automorphism over $K(u_{2})$. Hence $K(u_{1},u_{2},u_{4})^{\langle\sigma\rangle}=K(u_{2})(u_{1},u_{4})^{\langle\sigma\rangle}$ is rational over $K(u_{2})$.

Case 3.2. $\epsilon_{2}=-1$ and $\epsilon\epsilon_{1}\epsilon_{2}=1$.

Define $u_{1}=z_{1}$, $u_{2}=z_{2}^{2}$, $u_{3}=z_{2}z_{3}$. The details are left to the reader.

Case 3.3. $\epsilon_{2}=1$.

This case is easier and the proof is omitted.

Case 4. $G=W_{6}(187)$.

The action of $G=\langle\sigma,\tau,\lambda\rangle$ is given by

[TABLE]

where $a_{1},a_{2},a_{3}\in K\backslash\{0\}$ (for the action of $\sigma$ and $\tau$, see $G=W_{14}(174)$ in the proof of Theorem 3.4).

Since $G\simeq C_{2}\times C_{2}\times C_{2}$, it follows that $a_{1}^{2}=1$, $a_{2}=a_{3}$, and $a_{2}^{2}=a_{1}$.

Define $y_{1}=x_{1}x_{2}/x_{3}$, $y_{2}=x_{2}$, $y_{3}=x_{3}$. Then we have

[TABLE]

where $\epsilon\in\{1,-1\}$ and $a^{2}=\epsilon$.

Define

[TABLE]

By Theorem 2.3, we find that $K(y_{1},y_{2},y_{3})^{\langle\tau\rangle}=K(y_{1},u,v)$ and

[TABLE]

Thus $\langle\sigma,\lambda\rangle$ acts on $K(y_{1},u,v)$ by monomial $K$-automorphisms. By Theorem 3.4, $K(y_{1},u,v)^{\langle\sigma,\lambda\rangle}$ is rational over $K$.

∎

Theorem 3.7**.**

Let $G=W_{i}(187)$ for $7\leq i\leq 14$. Then $K(x_{1},x_{2},x_{3})^{G}$ is rational over $K$.

Proof..

Case 1. $G=W_{7}(187)$.

The action of $G=\langle\sigma,\tau\rangle$ is given by

[TABLE]

where $b,b_{1},b_{2},b_{3}\in K\backslash\{0\}$ and $j\in\{0,1,2,3\}$ (for the action of $\sigma$, see Case 1. $G=W_{1}(187)$ in the proof of Theorem 3.5).

Define $y_{1}=x_{1}/\sqrt{b_{1}}$, $y_{2}=x_{2}/\sqrt{b}$, $y_{3}=b_{2}x_{3}/\sqrt{b}$. Then we have

[TABLE]

for some $a\in K\backslash\{0\}$ and $j\in\{0,1,2,3\}$.

Since $\tau\sigma\tau^{-1}=\sigma^{-1}$, we find that $a^{2}=1$.

For $2\leq i\leq 3$, define $z_{i}=y_{i}/\sqrt{-1}$ if $a=-1$, and define $z_{i}=y_{i}$ if $a=1$. Thus we get $K(x_{1},x_{2},x_{3})=K(y_{1},z_{2},z_{3})$ and

[TABLE]

where $\epsilon\in\{1,-1\}$ and $j\in\{0,1,2,3\}$.

Case 1.1. $\epsilon=1$.

Define $u_{i}=(1-z_{i})/(1+z_{i})$ for $2\leq i\leq 3$. We get

[TABLE]

If $j\in\{1,3\}$, apply Theorem 2.1 and reduce the question to $K(y_{1})^{G}$.

If $j\in\{0,2\}$, define $w_{2}=u_{2}/u_{3}$, $w_{3}=u_{2}^{2}$. Then $K(y_{1},u_{2},u_{3})^{\langle\sigma^{2}\rangle}=K(y_{1},w_{2},w_{3})$. Note that $\langle\sigma,\tau\rangle$ acts on $K(y_{1},w_{2},w_{3})$ by monomial $K$-automorphisms and $\langle\sigma,\tau\rangle\simeq C_{2}\times C_{2}$ as a group of automorphisms on $K(y_{1},w_{2},w_{3})$. Thus we may apply Theorem 3.4.

Case 1.2. $\epsilon=-1$ and $j=0$.

Define $z_{1}=(1-y_{1})/(1+y_{1})$. Then $\sigma(z_{1})=z_{1}$, $\tau(z_{1})=-z_{1}$. Apply Theorem 2.1 and it suffices to consider the rationality of $K(z_{2},z_{3})^{G}$.

Case 1.3. $\epsilon=-1$ and $j=2$.

Define

[TABLE]

By Theorem 2.3, $K(y_{1},z_{2},z_{3})^{\langle\sigma^{2}\rangle}=K(y_{1},u,v)$ and

[TABLE]

Define $w_{1}=u+v$, $w_{2}=u-v$. Then we find that

[TABLE]

We get a monomial group action. Thus $K(y_{1},w_{2},w_{3})^{\langle\sigma,\tau\rangle}$ is rational by Theorem 3.4.

Case 1.4. $\epsilon=-1$ and $j\in\{1,3\}$. Write $\eta=(\sqrt{-1})^{j}$.

Define $u_{2}=z_{2}/\eta$, $u_{3}=-z_{3}/\eta$. We get

[TABLE]

Define $v_{2}=(1-u_{2})/(1+u_{2})$, $v_{3}=(1-u_{3})/(1+u_{3})$. We find that $K(y_{1},z_{2},z_{3})=K(y_{1},v_{2},v_{3})$ and

[TABLE]

Define $w_{1}=y_{1}v_{2}$, $w_{2}=v_{2}/v_{3}$, $w_{3}=v_{2}^{2}$. Then $K(y_{1},v_{2},v_{3})^{\langle\sigma^{2}\rangle}=K(w_{1},w_{2},w_{3})$ and $\langle\sigma,\tau\rangle$ acts on $K(w_{1},w_{2},w_{3})$ by monomial $K$-automorphisms because

[TABLE]

Thus we may apply Theorem 3.4.

Case 2. $G=W_{8}(187)$.

The action of $G=\langle\sigma,\tau\rangle$ is given by

[TABLE]

where $a_{i},b_{j}\in K\backslash\{0\}$.

Apply Theorem 2.1 and reduce the question to $K(x_{2},x_{3})^{G}$.

Case 3. $G=W_{9}(187)$.

The action of $G=\langle\sigma,\tau\rangle$ is given by

[TABLE]

where $b_{2},b_{3},c_{1},c_{2},c_{3}\in K\backslash\{0\}$.

Define $y_{1}=x_{1}$, $y_{2}=x_{2}$, $y_{3}=b_{2}/x_{3}$. Use the fact that $G\simeq D_{4}$. We find that

[TABLE]

where $a\in K\backslash\{0\}$ and $\epsilon_{1},\epsilon_{2}\in\{1,-1\}$.

Define $z_{i}=y_{i}/\sqrt{a}$ for $2\leq i\leq 3$. Then $K(x_{1},x_{2},x_{3})=K(y_{1},z_{2},z_{3})$ and

[TABLE]

where $\epsilon_{1},\epsilon_{2}\in\{1,-1\}$.

Define

[TABLE]

By Theorem 2.3, $K(y_{1},z_{2},z_{3})^{\langle\sigma^{2}\rangle}=K(y_{1},u,v)$ and

[TABLE]

Define $w_{2}=u+v$, $w_{3}=u-v$. Then

[TABLE]

We may apply Theorem 3.4 to assert that $K(y_{1},w_{2},w_{3})^{\langle\sigma,\tau\rangle}$ is rational over $K$.

Case 4. $G=W_{10}(187)$.

The action of $G=\langle\sigma,\tau\rangle$ is given by

[TABLE]

where $a_{2},a_{3},b_{1},b_{2},b_{3}\in K\backslash\{0\}$.

Define $y_{1}=x_{1}$, $y_{2}=\sqrt{a_{3}}x_{2}/\sqrt{a_{2}}$, $y_{3}=(\sqrt{a_{2}}\cdot\sqrt{a_{3}})/x_{3}$. Use the relations that $\tau^{2}=1$ and $\tau\sigma\tau^{-1}=\sigma^{-1}$. We get

[TABLE]

where $\epsilon_{1},\epsilon_{2}\in\{1,-1\}$.

Define

[TABLE]

By Theorem 2.3 we find that $K(y_{1},y_{2},y_{3})^{\langle\sigma^{2}\rangle}=K(y_{1},u,v)$ and

[TABLE]

Define $z_{2}=u+v$, $z_{3}=u-v$. Then $\langle\sigma,\tau\rangle$ acts on $K(y_{1},u,v)=K(y_{1},z_{2},z_{3})$ by monomial $K$-automorphisms because

[TABLE]

Thus we may apply Theorem 3.4.

Case 5. $G=W_{11}(187)$.

The action of $G=\langle\sigma,\tau\rangle$ is given by

[TABLE]

where $a_{1},a_{3},b_{2},b_{3}\in K\backslash\{0\}$.

Use the relations that $\tau^{4}=\tau^{2}=1$ and $\tau\sigma\tau^{-1}=\sigma^{-1}$. We find that

[TABLE]

where $\epsilon\in\{1,-1\}$, $a_{3},b\in K\backslash\{0\}$.

Define $y_{1}=x_{1}$, $y_{2}=x_{2}/\sqrt{b}$, $y_{3}=\sqrt{b}/x_{3}$ and use that relation $\tau\sigma\tau^{-1}(y_{3})=\sigma^{-1}(y_{3})$. We find that

[TABLE]

where $\epsilon\in\{1,-1\}$ and $a^{2}=\epsilon$.

Note that $\sigma^{2}:(y_{1},y_{2},y_{3})\mapsto(y_{1},y_{1}/(ay_{2}),1/(ay_{1}y_{3}))$.

Define

[TABLE]

By Theorem 2.3, $K(y_{1},y_{2},y_{3})^{\langle\sigma^{2}\rangle}=K(y_{1},u,v)$ (note that $a^{2}=\epsilon$) and

[TABLE]

Case 5.1. $\epsilon=-1$.

Define $z_{2}=u+y_{1}v$, $z_{3}=u-y_{1}v$. Then $K(y_{1},u,v)=K(y_{1},z_{2},z_{3})$ and

[TABLE]

Apply Theorem 3.4.

Case 5.2. $\epsilon=1$.

Define $z_{2}$ and $z_{3}$ as in Case 1. The details are omitted.

Case 6. $G=W_{12}(187)$.

The action of $G=\langle\sigma,\tau\rangle$ is given by

[TABLE]

where $a_{1},a_{2},a_{3}\in K\backslash\{0\}$ and $\epsilon\in\{1,-1\}$.

Since $\tau^{4}=1$ and $\tau\sigma\tau^{-1}=\sigma^{-1}$, we find that $a_{1}=\epsilon$, $a_{2}^{2}=1$.

Define $y_{1}=x_{1}$, $y_{2}=x_{2}/\sqrt{a_{3}}$, $y_{3}=x_{3}/\sqrt{a_{3}}$. We get that

[TABLE]

where $\epsilon_{1},\epsilon_{2}\in\{1,-1\}$.

Define $z_{1}=y_{1}$, $z_{2}=y_{2}$, $z_{3}=1/y_{3}$. Then we have

[TABLE]

Define

[TABLE]

Then proceed by the same way as in the previous Case 5. $G=W_{11}(187)$. Note that $\sigma:(u,v)\mapsto(-\epsilon_{1}\epsilon_{2}z_{1}^{2}v/(u^{2}-\epsilon_{1}z_{1}^{2}v^{2})$, $\epsilon_{2}u/(u^{2}-\epsilon_{1}z_{1}^{2}v^{2}))$. Thus, if $\epsilon_{1}=-1$, we should take the change of variables $u\pm\sqrt{-1}z_{1}v$, instead of the “usual” change of variables $u\pm z_{1}v$. The details are left to the reader.

Case 7. $G=W_{13}(187)$.

The action of $G=\langle\sigma,\tau\rangle$ is given by

[TABLE]

where $a_{3},b_{1},b_{2},b_{3}\in K\backslash\{0\}$.

Use the relations $\tau^{2}=1$ and $\tau\sigma\tau^{-1}=\sigma^{-1}$. We find that $b_{1}=1$ and $b_{2}=b_{3}\in\{1$, $-1\}$. Define $y_{1}=x_{1}$, $y_{2}=\sqrt{a_{3}}x_{2}$, $y_{3}=x_{3}/\sqrt{a_{3}}$. Then we get

[TABLE]

where $\epsilon\in\{1,-1\}$.

If $\epsilon=1$, apply Theorem 1.2.

If $\epsilon=-1$, note that $\sigma^{2}:(y_{1},y_{2},y_{3})\mapsto(y_{1},y_{1}/y_{2},y_{1}/y_{3})$. Define

[TABLE]

The proof is almost the same as the previous Case 5. $G=W_{11}(187)$ by taking the change of variables $z_{2}=u+v$, $z_{3}=u-v$ because $\sigma:(u,v)\mapsto(-v/(u^{2}-v^{2}),u/(u^{2}-v^{2}))$ and $\tau:(u,v)\mapsto(-y_{1}v,-y_{1}u)$.

Case 8. $G=W_{14}(187)$.

The action of $G=\langle\sigma,\tau\rangle$ is given by

[TABLE]

where $a_{3},b\in K\backslash\{0\}$, $\epsilon\in\{1,-1\}$.

Define $y_{1}=x_{1}$, $y_{2}=\sqrt{a_{3}}x_{2}$, $y_{3}=x_{3}/\sqrt{a_{3}}$. Use the relation $\tau\sigma\tau^{-1}=\sigma^{-1}$. We get

[TABLE]

where $a^{2}=\epsilon\in\{1,-1\}$.

Define $u$ and $v$ by Formula (3.3) in the previous Case 7. $G=W_{13}(187)$. Note that $\sigma:(y_{1},u,v)\mapsto(1/y_{1},-v/(u^{2}-v^{2}),u/(u^{2}-v^{2}))$, $\tau:(y_{1},u,v)\mapsto(\epsilon y_{1},av,\epsilon au)$.

Discuss the situations $\epsilon=1$ and $\epsilon=-1$ separately. The proof is almost the same as in the Case 7. $G=W_{13}(187)$.

∎

Theorem 3.8**.**

Let $G=W_{1}(194),W_{2}(195)$. Then $K(x_{1},x_{2},x_{3})^{G}$ is rational over $K$.

Proof..

Case 1. $G=W_{1}(194)$.

The action of $G=\langle\sigma,\tau,\lambda\rangle$ is given by

[TABLE]

where $j\in\{0,1,2,3\}$, $\epsilon\in\{1,-1\}$, $a_{1},a_{2},a_{3}\in K\backslash\{0\}$ (for the actions of $\sigma$ and $\tau$, see Case 1. $G=W_{7}(187)$ in the proof of Theorem 3.7).

Use the relations $\sigma\lambda=\lambda\sigma$ and $\tau\lambda=\lambda\tau$. We find that $j=0$ or 2, $a_{2}=a_{3}\in\{1,-1\}$. We write these actions as follows

[TABLE]

where $\epsilon_{1},\epsilon_{2},\epsilon_{3}\in\{1,-1\}$ and $a\in K\backslash\{0\}$.

Define

[TABLE]

By Theorem 2.3 we find that $K(x_{1},x_{2},x_{3})^{\langle\sigma^{2}\rangle}=K(x_{1},u,v)$ and

[TABLE]

Define $y_{2}=u+v$, $y_{3}=u-v$. Then we get

[TABLE]

Thus we may apply Theorem 3.6 and conclude that $K(x_{1},y_{2},y_{3})^{\langle\sigma,\tau,\lambda\rangle}$ is rational over $K$.

Case 2. $G=W_{2}(195)$.

The action of $G=\langle\sigma,\tau,\lambda\rangle$ is given by

[TABLE]

where $a,b,c_{1},c_{2},c_{3}\in K\backslash\{0\}$, $\epsilon\in\{1,-1\}$ (see Case 5. $G=W_{11}(187)$ in the proof of Theorem 3.7).

Define $y_{1}=x_{1}$, $y_{2}=x_{2}/\sqrt{b}$, $y_{3}=\sqrt{b}/x_{3}$ and use the relations $\tau\sigma\tau^{-1}=\sigma^{-1}$, $\sigma\lambda=\lambda\sigma$, $\tau\lambda=\lambda\tau$. We find that

[TABLE]

where $\epsilon,\epsilon_{1},\epsilon_{2}\in\{1,-1\}$ and $a^{2}=\epsilon_{1}$.

We will “copy” the proof of Case 5. $G=W_{11}(187)$ in Theorem 3.7 without changing anything. Note that $\lambda:(u,v)\mapsto(\epsilon\epsilon_{2}au/y_{1},\epsilon\epsilon_{2}avy_{1})$.

Following the proof of $G=W_{11}(187)$, we define $z_{2}=u+y_{1}v$, $z_{3}=u-y_{1}v$. Since $\lambda(y_{1})=\epsilon_{1}/y_{1}$, we find that $\lambda:(z_{2},z_{3})\mapsto(\epsilon\epsilon_{2}az_{2}/y_{1},\epsilon\epsilon_{2}az_{3}/y_{1})$ if $\epsilon_{1}=1$; while $\lambda:(z_{2},z_{3})\mapsto(\epsilon\epsilon_{2}az_{3}/y_{1},\epsilon\epsilon_{2}az_{2}/y_{1})$ if $\epsilon_{1}=-1$.

Thus we may discuss various cases when $\epsilon,\epsilon_{1}=\pm 1$ and prove that $K(y_{1},z_{2},z_{3})^{\langle\sigma,\tau,\lambda\rangle}$ is rational over $K$ by Theorem 3.6.

∎

Proof of Theorem 1.3..

By all the theorems proved in this section, i.e. Theorems 3.1 – 3.8, the proof of Theorem 1.3 is completed. ∎

§4. Another proof of Theorem 1.3

In this section we will give an alternative, short proof of Theorem 1.3 in the case where the ground field $K$ is algebraically closed and has characteristic [math]. The proof is geometric and does not require computations. Moreover it does not use the classification of finite subgroups in $GL_{3}(\mathbbm{Z})$ [Ta].

Throughout this section we assume that $G$ is a finite $2$-group and $K$ is an algebraically closed field of characteristic [math].

The following Lemma 4.1, Lemma 4.3, Corollary 4.2 and Corollary 4.4 are meant to replace a classification of finite subgroups of $GL_{2}(\mathbbm{Z})$ and $GL_{3}(\mathbbm{Z})$. (In fact, finite subgroups of $GL_{2}(\mathbbm{Z})$ are listed in [Ha1, Ha2] and those of $GL_{3}(\mathbbm{Z})$ are listed in [Ta].)

Lemma 4.1 is an easy exercise in the representation theory of finite groups.

Lemma 4.1**.**

Let $G$ be a finite $2$-group. Then any representation $G\to GL_{3}(\mathbbm{Q})$ is reducible.

Corollary 4.2**.**

Let $G$ be a finite $2$-group. Then any integral representation $G\to GL_{3}(\mathbbm{Z})$ has a subrepresentation of $\mathbbm{Z}$-rank $1$.

Now let $G$ be a finite $2$-group acting monomially on $K(x,y,z)$. Let $\rho:G\to GL_{3}(\mathbbm{Z})$ be the integral representation introduced in Definition 2.7. By Corollary 4.2 there is a $1$-dimensional subrepresentation $\rho_{1}$. Denote the quotient representation by $\rho_{2}:G\to GL_{2}(\mathbbm{Z})$. In a suitable basis in $\mathbbm{Z}^{3}$ any element of $\rho(G)$ can be written as

[TABLE]

and then

[TABLE]

We will denote by $M$ the lattice $\mathbbm{Z}^{3}$. Thus the group $G$ acts on $M$. By Lemma 2.8 we may assume that the representation $\rho$ is faithful on $M$.

Lemma 4.3**.**

Any $2$-subgroup in $GL_{2}(\mathbbm{Z})$ is conjugated to a subgroup of the group generated by two matrices $A$ and $B$ $($and isomorphic to $D_{4})$

[TABLE]

Outline of the proof.

For any matrix $C$ in the group we have $\mathop{{\rm tr}\vphantom{\dim}}C=0$ or $\pm 1$ and $\det C=\pm 1$. Since $C$ is an element of order $2^{k}$, its minimum polynomial divides $t^{2^{k}}-1$. This immediately implies $\mathop{{\rm tr}\vphantom{\dim}}C=0$ and $C^{4}=E$. Hence $C$ is conjugate to either $A$ or $B$, or $\pm E$. On the other hand, the order of the group is at most $8$ by the famous Minkowski theorem (see, for example, [Ta, page 169]). ∎

Let $N\subset G$ be the maximal subgroup that acts trivially on $L$. It is easy to see that so is the restriction $\rho_{2}|_{N}$.

Corollary 4.4**.**

In the above notation, there are only the following possibilities:

[TABLE]

By Lemma 4.3, for any $g\in G$, the action of $g$ on $K(x,y,z)$ can be written as follows:

[TABLE]

where $a_{i}\in\{1,-1\}$, $\alpha_{i}\in K^{\times}$, and $\Psi_{i}\in K(x)^{\times}$ are some monomials, i.e. elements of the form $cx^{n}$ where $c\in K^{\times}$ and $n$ is an integer.

**Remark **.

Define $L:=K(x)$. We may regard $K(x,y,z)$ as the function field of the projective surface $\mathbbm{P}^{1}_{L}\times\mathbbm{P}^{1}_{L}$ over $L$ and the induced action of $G$ on $\mathbbm{P}^{1}_{L}\times\mathbbm{P}^{1}_{L}$ is semi-linear. Here $y$ and $z$ are non-homogeneous coordinates on the first and second factors respectively.

We use some facts about del Pezzo surfaces. Good references are [Ma1, Ma2, MT]. Recall that over an algebraically closed field there are two types of del Pezzo surfaces of degree $8$: the Hirzebruch surface $\mathbbm{F}_{1}$ and $\mathbbm{P}^{1}\times\mathbbm{P}^{1}$. A form $S$ of $\mathbbm{F}_{1}$ over any field $L$ is not minimal, that is, it contains a unique $(-1)$-curve. Contracting it we get a Brauer-Severi variety $S^{\prime}$ having an $L$-point. Hence, $S^{\prime}\simeq\mathbbm{P}^{2}_{L}$. A form of $\mathbbm{P}^{1}\times\mathbbm{P}^{1}$ over a field $L$ is called a minimal del Pezzo surface of degree $8$. Taking into account the remark after Corollary 4.4, one find that Theorem 1.3 is a consequence of Proposition 4.5 below.

Proposition 4.5**.**

Let $L$ be a field of characteristic [math] and $G$ be a finite $2$-group. Let $S$ a minimal del Pezzo surface of degree $8$ defined over $L$. Suppose $G$ acts on both $L$ and $S$ which is regarded as an $L^{G}$-scheme, and let $N$ be the kernel of the homomorphism $G\to\operatorname{Aut}L$. Assume that $N$ is given as in (4.1). If $L^{G}$ is a $C_{1}$-field, then $S/G$ is rational over $L^{G}$.

For the definition $C_{i}$-fields, see [Gr, page 3; DKT, page 1256]. An algebraic function field of one variable over an algebraically closed field is a $C_{1}$-field by Tsen’s Theorem [Gr, page 22; DKT, Theorem 5.8, page 1259].

Proof..

Our proof is by induction on the order of $G$.

If $G=\{1\}$, then $S$ has an $L$-point (because $L$ is a $C_{1}$- field) and so $S$ is a rational quadric in $\mathbbm{P}^{3}$.

From now on we assume that $n:=|G|>1$ and the assertion holds for all groups $G_{1}$ with $|G_{1}|<n$.

Let $\bar{L}$ be the algebraic closure. Set $\bar{S}:=S\otimes\bar{L}$. The group $N$ acts linearly on $S$ over $L$ and on $\bar{S}$ over $\bar{L}$. In particular, $N$ acts on $\mathop{{\rm Pic}\vphantom{\dim}}(\bar{S})\simeq\mathbbm{Z}\oplus\mathbbm{Z}$.

First we assume that $|N|>2$. By Formula (4.1) there is an element $g\in N$ of order $4$ and $\tau:=g^{2}$ is contained in $Z(G)\cap N$, where $Z(G)$ is the center of $G$. Indeed, in this case, $\tau$ is the only element of order $2$ in $Z(N)\supset Z(G)\cap N$. Denote $N_{0}:=\{1,\tau\}$ and $F:=S/N_{0}$. By our assumptions, $F$ is defined over $L$ and $G/N_{0}$ acts naturally on $F$ by semi-linear automorphisms. Thus $S/G\simeq F/(G/N_{0})$. In a suitable (non-homogeneous) coordinate system in $\bar{F}\simeq\mathbbm{P}^{1}_{\bar{L}}\times\mathbbm{P}^{1}_{\bar{L}}$ over $\bar{L}$ there are only the following possibilities for the action:

1. (1)

$\tau:x\longmapsto y,\quad y\longmapsto x$; 2. (2)

$\tau:x\longmapsto x,\quad y\longmapsto-y$; 3. (3)

$\tau:x\longmapsto-x,\quad y\longmapsto-y$.

The element $\tau=g^{2}$ acts trivially on $\mathop{{\rm Pic}\vphantom{\dim}}(\bar{S})\simeq\mathbbm{Z}\oplus\mathbbm{Z}$. Thus the case (1) is impossible. In the case (2) we have $\bar{F}\simeq\mathbbm{P}^{1}_{\bar{L}}\times\mathbbm{P}^{1}_{\bar{L}}$, so $F$ a minimal del Pezzo surface of degree $8$. We reduce the problem to an action of a smaller group $G_{1}=G/N$ on $F$.

Consider the case (3). Let $\pi:S\to F$ be the quotient morphism. Over the field $\bar{L}$, the group $N$ acts freely in codimension one on $S$ and has exactly 4 fixed points. These points give us 4 singular points on $F$ of type $A_{1}$ and $F$ has no other singularities. In particular, this implies that the divisor $K_{F}$ is Cartier. Since the map $\pi:S\to F$ is étale in codimension one, $K_{S}=\pi^{*}K_{F}$. Hence $-K_{F}$ is ample and $K_{F}^{2}=K_{S}^{2}/2=4$. This means that $F$ is a del Pezzo surface of degree 4 having 4 singular points of type $A_{1}$. According to the classification [CT], there are exactly 4 lines $l_{1},\dots,l_{4}$ on $F$. Let $\mu:\tilde{F}\to F$ be the minimal resolution and let $\tilde{l}_{i}$ be the proper transform of $l_{i}$. Then $\tilde{F}$ is a weak del Pezzo surface in the sense that its anti-canonical divisor $-K_{\tilde{F}}$ is nef and big (such surfaces are also called generalized del Pezzo surfaces [CT]). Moreover, $K_{\tilde{F}}^{2}=K_{F}^{2}=4$. Again by [CT, Prop. 6.1] $\tilde{l}_{1},\dots,\tilde{l}_{4}$ are disjointed $(-1)$-curves (and $\tilde{F}$ contains no other $(-1)$-curves). Consider their contraction $\varphi:\tilde{F}\to F^{\prime}$. Here $K_{F^{\prime}}^{2}=K_{\tilde{F}}^{2}+4=8$. Further, $-K_{F^{\prime}}=\varphi_{*}(-K_{\tilde{F}})$ is nef and big, i.e., $F^{\prime}$ is a weak del Pezzo surface of degree 8. Moreover, every $(-2)$-curve $C$ on $\tilde{F}$ meets two $\varphi$-exceptional curves $\tilde{l}_{i}$. Hence the surface $F^{\prime}$ contains no curves with negative self-intersection numbers. Therefore, $F^{\prime}$ is a minimal del Pezzo surface of degree 8. Finally, both the birational maps $\mu:\tilde{F}\rightarrow F^{\prime}$ and $\varphi:\tilde{F}\rightarrow F$ are canonically defined, so they are $G/N$-equivariant. Again we reduce the problem to an action of $G_{1}=G/N$ on $F^{\prime}$ with $|G_{1}|<|G|$.

Now assume that $|N|=2$, so $N=\{1,\tau\}$. As above, $S/G\simeq F/(G/N_{0})$ and in a suitable coordinate system in $\bar{F}\simeq\mathbbm{P}^{1}_{\bar{L}}\times\mathbbm{P}^{1}_{\bar{L}}$ over $\bar{L}$ there are only the possibilities (1), (2) and (3) listed above. In cases (2) and (3) we may argue as above. It remains to consider the case (1). Then the quotient $\bar{F}=\bar{S}/N$ is isomorphic to $\mathbbm{P}^{2}_{\bar{L}}$. Thus $F$ is a del Pezzo surface of degree $9$ (a Brauer-Severi variety). In this situation, $G_{1}=G/N$ acts on $L$ effectively. Then by Lemma 4.6 below the quotient $S/G\simeq\mathbbm{P}^{2}_{L}/(G/N)$ is $L$-rational.

Finally, if $N=\{1\}$, i.e., the action of $G$ on $L$ is effective, we may also apply Lemma 4.6 to finish the proof. ∎

Lemma 4.6**.**

Let $\Bbbk$ be a field of characteristic [math] and let $S$ be a del Pezzo surface of degree $d\geq 5$ over $\Bbbk$. Suppose a finite group $G$ acts effectively on both $\Bbbk$ and $S$ which is regarded as a $\Bbbk^{G}$-scheme. If $\Bbbk^{G}$ is a $C_{1}$-field, then $S/G$ is rational over $\Bbbk^{G}$.

Proof..

We claim that the natural $G$-equivariant morphism $S\rightarrow S/G\times_{\Bbbk^{G}}\Bbbk$ is an isomorphism. Since it is a local property, it suffices to show that, if $G$ acts on an affine domain $R$ over a field $\Bbbk$ so that $\Bbbk$ is left invariant by the action of $G$ and $G$ acts faithfully both on $\Bbbk$ and $R$, then $R^{G}\otimes_{\Bbbk^{G}}\Bbbk\simeq R$. This follows from the Galois descent lemma [KMRT, Lemma 18.1, page 279]. Note that the Galois descent lemma is essentially another form of Theorem 2.1.

Now consider the following diagram

[TABLE]

Since $S/G\times_{\Bbbk^{G}}\Bbbk\simeq S$, it follows that $S/G$ is a minimal del Pezzo surface of degree $d\geq 5$. By [Ma2, Chapter 4, §7] it is sufficient to show that $S/G$ has a $\Bbbk^{G}$-point. This holds for an arbitrary $\Bbbk$ if $d=5$ [SD] and for $C_{1}$-fields if $d\geq 6$ [Ma1]. Thus $S/G$ is $\Bbbk^{G}$-rational. ∎

§5. Applications

Proof of Theorem 1.4.

By 2.2, we may assume that $V$ contains no $1$-dimensional subrepresentation. By [Se, pages 53 and 65], every irreducible representation of $G$ is an induced representation from some 1-dimensional representation of $H$ for some subgroup $H$ of $G$; thus it is a monomial representation. It follows that $V$ is isomorphic to (i) an irreducible 4-dimensional representation, (ii) the direct sum of two irreducible 2-dimensional representations, or (iii) the direct sum of three 2-dimensional irreducible representations.

Case 1. $V$ is an irreducible 4-dimensional representation.

Since $V$ is induced from a 1-dimensional representation, for any $\sigma\in G$, $\sigma(e_{i})=a_{i}(\sigma)\cdot e_{j}$ for some basis $e_{1}$, $e_{2}$, $e_{3}$, $e_{4}$ of $V$ where $a_{i}(\sigma)\in K\backslash\{0\}$. If $x_{1}$, $x_{2}$, $x_{3}$, $x_{4}$ denotes the dual basis of $e_{1}$, $e_{2}$, $e_{3}$, $e_{4}$, then $K(V)=K(x_{1},x_{2},x_{3},x_{4})$ and $\sigma\cdot x_{i}=b_{i}(\sigma)\cdot x_{j}$ for any $\sigma\in G$ where $b_{i}(\sigma)\in K\backslash\{0\}$.

Define $y_{i}=x_{i}/x_{4}$ for $1\leq i\leq 3$. Then $\sigma(x_{4})=t_{\sigma}\cdot x_{4}$ where $t_{\sigma}\in K(y_{1},y_{2},y_{3})$ for any $\sigma\in G$. Apply Theorem 2.2. We find that $K(x_{1},x_{2},x_{3},x_{4})^{G}=K(y_{1},y_{2},y_{3})^{G}(f)$ for some $f$ satisfying $\sigma(f)=f$ for any $\sigma\in G$. Note that $G$ acts on $K(y_{1},y_{2},y_{3})$ by monomial $K$-automorphisms. Apply Theorem 1.3. We find that $K(y_{1},y_{2},y_{3})^{G}$ is rational over $K$. Thus the quotient $P(V)/G$ is also rational.

Case 2. $V$ is a direct sum of irreducible 2-dimensional representations and certain 1-dimensional representations.

Apply Theorem 2.2 and Theorem 1.1. The details are omitted.

Case 3. $V\simeq V_{1}\oplus V_{2}\oplus V_{3}$ where each $V_{i}$ is the representation space of an irreducible $2$-dimensional representation of $G$. It follows that $K(V)=K(x_{1},x_{2},y_{1},y_{2},z_{1},z_{2})$ and $G$ acts on elements of the set $\{x_{1},x_{2}\}$ (resp. $\{y_{1},y_{2}\}$, $\{z_{1},z_{2}\}$) by permutations up to non-zero elements in $K$. Define $x=x_{1}/x_{2},y=y_{1}/y_{2},z=z_{1}/z_{2}$. Apply Theorem 2.2 repeatedly. We reduce the question to the rationality of $K(x,y,z)^{G}$. Apply Theorem 1.3.

∎

For the proof of Theorem 1.5, we need two standard results in group theory, whose proof will be omitted.

Lemma 5.1**.**

Let $p$ be a prime number, $G$ be a finite $p$-group. Denote by $Z(G)$ the center of $G$. Let $H$ be a normal subgroup of $G$. If $H\supsetneq\{1\}$, then $H\cap Z(G)\supsetneq\{1\}$.

Lemma 5.2**.**

Let $G$ be a finite $2$-group of exponent $e$, and $K$ be a field containing a primitive $e$-th root of unity. For each non-negative integer $i$, let $s_{i}$ be the number of non-equivalent $2^{i}$-dimensional irreducible representations of $G$ over $K$. Then $\sum_{i}s_{i}4^{i}=|G|$.

Proof of Theorem 1.5 when $K$ is algebraically closed..

Let $G$ be a group of order 32. If $G$ is abelian, then $K(G)$ is rational by Fischer’s Theorem [Sw, Theorem 6.1]. Thus we will assume that $G$ is non-abelian from now on. Let $Z(G)$ be the center of $G$, and $[G,G]$ be the commutator subgroup of $G$.

If $\mathop{{\rm char}\vphantom{\dim}}K=2$, then $K(G)$ is rational by Kuniyoshi’s Theorem [Ku]. Thus we will assume that $\mathop{{\rm char}\vphantom{\dim}}K\neq 2$ and $K$ is algebraically closed from now on.

Case 1. $|Z(G)|\geq 8$.

Since $|G/Z(G)|\leq 4$, $G/Z(G)$ is isomorphic to $C_{2}\times C_{2}$. Thus there is an abelian subgroup $H$ so that $Z(G)\subset H\subset G$ and $[G:H]=2$. Apply Theorem 2.6. We find that $K(G)$ is rational.

Case 2. $Z(G)\simeq C_{2}$ or $C_{4}$.

Let $\tau\in Z(G)$ be the unique element of order 2 in $Z(G)$.

We claim that there is an irreducible representation $\rho:G\to GL(V)$ over $K$, which is faithful.

Assume the validity of this claim. Note that $\dim_{K}V\leq 4$ by Lemma 5.2. Hence $K(V)^{G}$ is rational over $K$ by Theorem 1.4. Since $K(G)$ is rational over $K(V)^{G}$ by Theorem 2.1, we find that $K(G)$ is rational.

Now we will prove this claim. Suppose not, i.e. every irreducible representation $\rho:G\to GL(W)$ is not faithful. Then $\mathop{{\rm Ker}\vphantom{\dim}}(\rho)\cap Z(G)\supsetneq\{1\}$ by Lemma 5.1. Thus $\tau\in\mathop{{\rm Ker}\vphantom{\dim}}(\rho)\cap Z(G)$. We conclude that $\rho(\tau)$ is trivial for all irreducible representations of $G$. But the regular representation of $G$ is faithful and is the direct sum of these irreducible representations of $G$. This leads to a contradiction.

Case 3. $Z(G)\simeq C_{2}\times C_{2}$.

We will show that $G$ has a faithful 4-dimensional representation, which is either irreducible or a direct sum of two 2-dimensional irreducible representations. In any case, the rationality of $K(G)$ can be proved as in Case 2.

Step 1. If $G$ has a faithful 4-dimensional representation, we are done. From now on, we will assume that, if $G$ has a 4-dimensional irreducible representation, then it is not faithful.

We will prove that $G$ has no 4-dimensional irreducible representation at all.

Suppose that $\rho:G\to GL_{4}(K)$ is irreducible. Then $\mathop{{\rm Ker}\vphantom{\dim}}(\rho)\supsetneq\{1\}$. Thus $\rho$ gives rise to a faithful irreducible 4-dimensional representation $G/\mathop{{\rm Ker}\vphantom{\dim}}(\rho)\to GL_{4}(K)$. Note that $|G/\mathop{{\rm Ker}\vphantom{\dim}}(\rho)|\leq 16$ which is impossible by Lemma 5.2.

Step 2. We will show that $G$ has at least four non-equivalent 2-dimensional irreducible representation.

Let $G$ has $p$ 1-dimensional irreducible representations and has $s$ 2-dimensional irreducible representations. Then $p+4s=|G|=32$ and $p=|G/[G,G]|$ which is a divisor of 16. The only solutions of $(p,s)$ are $(p,s)=(4,7)$, $(8,6)$, $(16,4)$. Thus $s\geq 4$.

Step 3. From Step 2, let $\rho_{i}:G\to GL_{2}(K)$ be all the 2-dimensional irreducible representations where $1\leq i\leq s$ with $s\geq 4$.

We claim that, there is some $\rho_{i}$ so that $\mathop{{\rm Ker}\vphantom{\dim}}(\rho_{i})\not\supset Z(G)$.

Otherwise, assume that $\mathop{{\rm Ker}\vphantom{\dim}}(\rho_{i})\supset Z(G)$ for all $1\leq i\leq s$. Note that the kernel of any 1-dimensional representation contains $[G,G]$. Hence $[G,G]\cap(\bigcap_{1\leq i\leq s}\mathop{{\rm Ker}\vphantom{\dim}}(\rho_{i}))\supset[G,G]\cap Z(G)\supsetneq\{1\}$ by Lemma 5.1. It follows that there is some element $g\in G$, $g\neq 1$, but $g$ belongs to the kernel of any irreducible representation of $G$. This is impossible by the same arguments as in the last paragraph of Case 2.

Step 4. By Step 3, we may assume that $\mathop{{\rm Ker}\vphantom{\dim}}(\rho_{1})\not\supset Z(G)$. Write $\mathop{{\rm Ker}\vphantom{\dim}}(\rho_{1})\cap Z(G)=\langle\lambda\rangle$ where $\lambda$ is an element of order two in $Z(G)\simeq C_{2}\times C_{2}$.

Define $H=\bigcap_{1\leq i\leq s}\mathop{{\rm Ker}\vphantom{\dim}}(\rho_{i})$. We will prove that $H=\{1\}$.

Suppose that $H\supsetneq\{1\}$. But Lemma 5.1, $H\cap Z(G)\supsetneq\{1\}$. But $H\cap Z(G)\subset\mathop{{\rm Ker}\vphantom{\dim}}(\rho_{1})\cap Z(G)=\langle\lambda\rangle$. Hence $H\cap Z(G)=\langle\lambda\rangle$. Thus the group $G/\langle\lambda\rangle$ has at least four 2-dimensional irreducible representations. If the number of 1-dimensional irreducible representations of $G/\langle\lambda\rangle$ is $q$, then $q+4\cdot 4\leq|G/\langle\lambda\rangle|$ by Lemma 5.2. Since $|G/\langle\lambda\rangle|=16$, we get a contradiction.

Step 5. Since $H=\{1\}$, we find that $\lambda\notin H$. We may assume that $\lambda\notin\mathop{{\rm Ker}\vphantom{\dim}}(\rho_{2})$.

Define $H_{0}=\mathop{{\rm Ker}\vphantom{\dim}}(\rho_{1})\cap\mathop{{\rm Ker}\vphantom{\dim}}(\rho_{2})$. If $H_{0}\supsetneq\{1\}$, then $H_{0}\cap Z(G)\supsetneq\{1\}$ by Lemma 5.1. Hence $H_{0}\cap Z(G)\subset\mathop{{\rm Ker}\vphantom{\dim}}(\rho_{1})\cap Z(G)=\langle\lambda\rangle$. Thus $H_{0}\cap Z(G)=\langle\lambda\rangle$, which is impossible because $\lambda\notin\mathop{{\rm Ker}\vphantom{\dim}}(\rho_{2})$.

Step 6. Since $\{1\}=H_{0}=\mathop{{\rm Ker}\vphantom{\dim}}(\rho_{1})\cap\mathop{{\rm Ker}\vphantom{\dim}}(\rho_{2})$, the direct sum of $\rho_{1}$ and $\rho_{2}$ is a faithful representation of $G$. Done.

∎

**Remark **.

As we note in Section 1, the above proof is valid only under the assumption that $\sqrt{a}\in K$ for any $a\in K$ instead of the weaker assumption that $\zeta_{e}\in K$ (which is given in Theorem 1.5). Consider the following proposition : Let G be a non-abelian group of order $32$ and let $e$ be the exponent of $G$. If $K$ is a field containing $\zeta_{e}$ and $G\rightarrow GL(V)$ is an irreducible representation over $K$ where $dimV=4$, then $K(V)^{G}$ is rational over $K$. Note that the above proposition is valid if we assume the classification of groups of order $32$, because there are exactly seven such groups (they are groups $(42;49),(43;50),(44;43),(45;44),(46;6),(47;7),(48;8)$ in the notation of [CHKP, page 3025]) and all the $4$-dimensional irreducible representations can be described explicitly. On the other hand, at present we don’t have a proof of the above proposition if we try to avoid using the classification of groups of order $32$. With this proposition in hand, the above proof of Theorem 1.5 may be adapted to the situation assuming only that $\zeta_{e}\in K$.