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Let $G$ be a finite group, $k$ be a field and $G\to GL(V_{\rm reg})$ be the regular representation of $G$ over $k$. Then $G$ acts naturally on the rational function field $k(V_{\rm reg})$ by $k$-automorphisms. Define $k(G)$ to be the fixed field $k(V_{\rm reg})^G$. Noether's problem asks whether $k(G)$ is rational (resp. stably rational) over $k$. When $k=\bQ$ and $G$ contains a normal subgroup $N$ with $G/H\simeq C_8$ (the cyclic group of order $8$), Jack Sonn proves that $\bQ(G)$ is not stably rational over $\bQ$, which is a non-abelian extension of a theorem of Endo-Miyata, Voskresenskii, Lenstra and Saltman for the abelian Noether's problem $\bQ(C_8)$. Using the method of cohomological invariants, we are able to generalize Sonn's theorem as follows. Theorem. Let $G$ be a finite group and $N$ $\lhd$ $G$ such that $G/N\simeq C_{2^n}$ with $n\geq 3$. If $k$ is a field satisfying that ${\rm char}\,k=0$ and $k(\zeta_{2^n})/k$ is not a cyclic extension where $\zeta_{2^n}$ is a primitive $2^n$-th root of unity, then $k(G)$ is not stably rational (resp. not retract rational) over $k$. \end{abstract}, Comment: Theorem 1.4 and its proof in Section 4 are new. Some minor corrections are added

Let $G$ be a subgroup of $S_{n}$, the symmetric group of degree $n$. For any field $k$, $G$ acts naturally on the rational function field $k(x_{1},\cdots,x_{n})$ via $k$-automorphisms defined by $\sigma\cdot x_{i}:=x_{\sigma\cdot i}$ for any $\sigma\in G$ and $1\leq i\leq n$. In this article, we will show that if $G$ is a solvable transitive subgroup of $S_{14}$ and $\text{char}(k)=7$, then the fixed subfield $k(x_{1},\cdots,x_{14})^{G}$ is rational (i.e., purely transcendental) over $k$. In proving the above theorem, we rely on the Kuniyoshi-Gasch\"utz Theorem or some ideas in its proof.

Let $k$ be a field with char $k\neq 2$ and $k$ be not algebraically closed. Let $a\in k\setminus k^2$ and $L=k(\sqrt{a})(x,y)$ be a field extension of $k$ where $x,y$ are algebraically independent over $k$. Assume that $\sigma$ is a $k$-automorphism on $L$ defined by \[ \sigma: \sqrt{a}\mapsto -\sqrt{a},\ x\mapsto \frac{b}{x},\ y\mapsto \frac{c(x+\frac{b}{x})+d}{y} \] where $b,c,d \in k$, $b\neq 0$ and at least one of $c,d$ is non-zero. Let $L^{\langle\sigma\rangle}=\{u\in L:\sigma(u)=u\}$ be the fixed subfield of $L$. We show that $L^{\langle\sigma\rangle}$ is isomorphic to the function field of a certain surface in $P^4_k$ which is given as the intersection of two quadrics. We give criteria for the $k$-rationality of $L^{\langle\sigma\rangle}$ by using the Hilbert symbol. As an appendix of the paper, we also give an alternative geometric proof of a part of the result which is provided to the authors by J.-L. Colliot-Th\'el\`ene., Comment: To appear in Manuscripta Math. The main theorems (old Theorem 1.7 and Theorem 1.8) incorporated into (new) Theorem 1.8. Section 3 and Section 4 interchanged

Let $k$ be a field and $G$ be a finite group acting on the rational function field $k(x_g : g\in G)$ by $k$-automorphisms defined as $h(x_g)=x_{hg}$ for any $g,h\in G$. We denote the fixed field $k(x_g : g\in G)^G$ by $k(G)$. Noether's problem asks whether $k(G)$ is rational (= purely transcendental) over $k$. It is well-known that if $C(G)$ is stably rational over $C$, then all the unramified cohomology groups $H_[nr}^i(C(G),Q/Z)=0$ for $i \ge 2$. Hoshi, Kang and Kunyavskii [HKK] showed that, for a $p$-group of order $p^5$ ($p$: an odd prime number), $H_[nr}^2(C(G),Q/Z)\neq 0$ if and only if $G$ belongs to the isoclinism family $\Phi_{10}$. When $p$ is an odd prime number, Peyre [Pe3] and Hoshi, Kang and Yamasaki [HKY1] exhibit some $p$-groups $G$ which are of the form of a central extension of certain elementary abelian $p$-group by another one with $H_[nr}^2(C(G),Q/Z)=0$ and $H_[nr}^3(C(G),Q/Z)\neq 0$. However, it is difficult to tell whether $H_[nr}^3(C(G),Q/Z)$ is non-trivial if $G$ is an arbitrary finite group. In this paper, we are able to determine $H_[nr}^3(C(G),Q/Z)$ where $G$ is any group of order $p^5$ with $p=3, 5, 7$. Theorem 1. Let $G$ be a group of order $3^5$. Then $H_[nr}^3(C(G),Q/Z)\neq 0$ if and only if $G$ belongs to $\Phi_7$. Theorem 2. If $G$ is a group of order $3^5$, then the fixed field $C(G)$ is rational if and only if $G$ does not belong to $\Phi_{7}$ and $\Phi_{10}$. Theorem 3. Let $G$ be a group of order $5^5$ or $7^5$. Then $H_[nr}^3(C(G),Q/Z)\neq 0$ if and only if $G$ belongs to $\Phi_6$, $\Phi_7$ or $\Phi_{10}$. Theorem 4. If $G$ is the alternating group $A_n$, the Mathieu group $M_{11}$, $M_{12}$, the Janko group $J_1$ or the group $PSL_2(F_q)$, $SL_2(F_q)$, $PGL_2(F_q)$ (where $q$ is a prime power), then $H_[nr}^d(C(G),Q/Z)=0$ for any $d\ge 2$. Besides the degree three unramified cohomology groups, we compute also the stable cohomology groups., Comment: 63 pages. Some typos are modified. To appear in J. Algebra

Let $k$ be a field, $G$ be a finite group, $k(x(g):g\in G)$ be the rational function field with the variables $x(g)$ where $g\in G$. The group $G$ acts on $k(x(g):g\in G)$ by $k$-automorphisms where $h\cdot x(g)=x(hg)$ for all $h,g\in G$. Let $k(G)$ be the fixed field defined by $k(G):=k(x(g):g\in G)^G=\{f\in k(x(g):g\in G): h\cdot f=f$ for all $h\in G\}$. Noether's problem asks whether the fixed field $k(G)$ is rational (= purely transcendental) over $k$. Let $m$ and $n$ be positive integers and assume that there is an integer $t$ such that $t\in (\bm{Z}/m\bm{Z})^\times$ is of order $n$. Define a group $G_{m,n}:=\langle\sigma,\tau:\sigma^m=\tau^n=1,\tau^{-1}\sigma\tau=\sigma^t\rangle$ $\simeq C_m \rtimes C_n$. We will find a sufficient condition to guarantee that $k(G)$ is rational over $k$. As a result, it is shown that, for any positive integer $n$, the set $S:=\{p: p$ is a prime number such that $\bm{C}(G_{p,n})$ is rational over $\bm{C} \}$ is of positive Dirichlet density; in particular, $S$ is an infinite set.

The finite subgroups of $GL_4(\bm{Z})$ are classified up to conjugation in \cite{BBNWZ}; in particular, there exist $710$ non-conjugate finite groups in $GL_4(\bm{Z})$. Each finite group $G$ of $GL_4(\bm{Z})$ acts naturally on $\bm{Z}^{\oplus 4}$; thus we get a faithful $G$-lattice $M$ with ${\rm rank}_\bm{Z} M=4$. In this way, there are exactly $710$ such lattices. Given a $G$-lattice $M$ with ${\rm rank}_\bm{Z} M=4$, the group $G$ acts on the rational function field $\bm{C}(M):=\bm{C}(x_1,x_2,x_3,x_4)$ by multiplicative actions, i.e. purely monomial automorphisms over $\bm{C}$. We are concerned with the rationality problem of the fixed field $\bm{C}(M)^G$. A tool of our investigation is the unramified Brauer group of the field $\bm{C}(M)^G$ over $\bm{C}$. A formula of the unramified Brauer group ${\rm Br}_u(\bm{C}(M)^G)$ for the multiplicative invariant field was found by Saltman in 1990. However, to calculate ${\rm Br}_u(\bm{C}(M)^G)$ for a specific multiplicatively invariant field requires additional efforts, even when the lattice $M$ is of rank equal to $4$. Theorem 1. Among the $710$ finite groups $G$, let $M$ be the associated faithful $G$-lattice with ${\rm rank}_\bm{Z} M=4$, there exist precisely $5$ lattices $M$ with ${\rm Br}_u(\bm{C}(M)^G)\neq 0$. In these situations, $B_0(G)=0$ and thus ${\rm Br}_u(\bm{C}(M)^G)\subset H^2(G,M)$. The {\rm GAP IDs} of the five groups $G$ are {\rm (4,12,4,12), (4,32,1,2), (4,32,3,2), (4,33,3,1), (4,33,6,1)} in {\rm \cite{BBNWZ}} and in {\rm \cite{GAP}}. Theorem 2. There exist $6079$ finite subgroups $G$ in $GL_5(\bm{Z})$. Let $M$ be the lattice with rank $5$ associated to each group $G$. Among these lattices precisely $46$ of them satisfy the condition ${\rm Br}_u(\bm{C}(M)^G)\neq 0$. A similar result for lattices of rank $6$ is found also., Comment: A new section (Section 3) is added. Several footnotes are added also. The webe page of Yamasaki at Kyoto University is available so that the reader may use the algorithm of this paper

Let $p$ be a prime number and $\zeta_p$ be a primitive $p$-th root of unity in $\bm{C}$. Let $k$ be a field and $k(x_0,\ldots,x_{p-1})$ be the rational function field of $p$ variables over $k$. Suppose that $G=\langle\sigma\rangle \simeq C_p$ acts on $k(x_0,\ldots,x_{p-1})$ by $k$-automorphisms defined as $\sigma:x_0\mapsto x_1\mapsto\cdots\mapsto x_{p-1}\mapsto x_0$. Denote by $P$ the set of all prime numbers and define $P_0=\{p\in P:\bm{Q}(\zeta_p)$ is of class number one$\}$. Theorem. If $k$ is an algebraic number field and $p\in P\backslash (P_0\cup P_k)$, then $k(x_0,\ldots,x_{p-1})^G$ is not stably rational over $k$ where $P_k=\{p\in P: p$ is ramified in $k\}$., Comment: Theorem 1.3 and Lemma 2.7 are new

Let $R$ be a commutative ring, $\pi$ be a finite group, $R\pi$ be the group ring of $\pi$ over $R$. Theorem 1. If $R$ is a commutative artinian ring and $\pi$ is a finite group. Then the Cartan map $c:K_0(R\pi)\to G_0(R\pi)$ is injective. Theorem 2. Suppose that $R$ is a Dedekind domain with $\fn{char}R=p>0$ and $\pi$ is a $p$-group. Then every finitely generated projective $R\pi$-module is isomorphic to $F \oplus \c{A}$ where $F$ is a free module and $\c{A}$ is a projective ideal of $R\pi$. Moreover, $R$ is a principal ideal domain if and only if every finitely generated projective $R\pi$-module is isomorphic to a free module. Theorem 3. Let $R$ be a commutative noetherian ring with total quotient ring $K$, $A$ be an $R$-algebra which is a finitely generated $R$-projective module. Suppose that $I$ is an ideal of $R$ such that $R/I$ is artinian. Let $\{\c{M}_1,\ldots,\c{M}_n\}$ be the set of all maximal ideals of $R$ containing $I$. Assume that the Cartan map $c_i: K_0(A/\c{M}_iA)\to G_0(A/\c{M}_iA)$ is injective for all $1\le i\le n$. If $P$ and $Q$ are finitely generated $A$-projective modules with $KP\simeq KQ$, then $P/IP\simeq Q/IQ$., Comment: Theorem 4.3 and THeorem 4.4 are new

Let $p$ be an odd prime number. Peyre shows that there is a group $G$ of order $p^{12}$ such that $H_{nr}^3(\bm{C}(G), \bm{Q}/\bm{Z})$ is non-trivial. Using Peyre's method, we are able to prove that the same conclusion is true for some groups of order $p^9$.

Let D_n be the dihedral group of order 2n where n \ge 2, 1 \to R \to F \to D_n \to 1 be a free presentation of D_n. R^{ab}:=R/[R,R] becomes a \bm{Z}[D_n]-lattice. We will study the module structure and the rationality problem of R^{ab}., Comment: The title is changed slightly. The name of one of the authors, Akinari Hoshi, is corrected. Several places are improved at the suggestion of a referee. To appear in Journal of Algebra. arXiv admin note: text overlap with arXiv:1312.6738

Theorem. Let $\pi$ be a finite group of order $n$, $R$ be a Dedekind domain satisfying that (i) $\fn{char}R=0$, (ii) every prime divisor of $n$ is not invertible in $R$, and (iii) $p$ is unramified in $R$ for any prime divisor $p$ of $n$. Then all the flabby (resp.\ coflabby) $R\pi$-lattices are invertible if and only if all the Sylow subgroups of $\pi$ are cyclic. The above theorem was proved by Endo and Miyata when $R=\bm{Z}$ \cite[Theorem 1.5]{EM}. As applications of this theorem, we give a short proof and a partial generalization of a result of Torrecillas and Weigel \cite[Theorem A]{TW}, which was proved using cohomological Mackey functors.

Let $K/k$ be a finite Galois extension and $\pi = \fn{Gal}(K/k)$. An algebraic torus $T$ defined over $k$ is called a $\pi$-torus if $T\times_{\fn{Spec}(k)} \fn{Spec}(K)\simeq \bm{G}_{m,K}^n$ for some integer $n$. The set of all algebraic $\pi$-tori defined over $k$ under the stably isomorphism form a semigroup, denoted by $T(\pi)$. We will give a complete proof of the following theorem due to Endo and Miyata \cite{EM5}. Theorem. Let $\pi$ be a finite group. Then $T(\pi)\simeq C(\Omega_{\bm{Z}\pi})$ where $\Omega_{\bm{Z}\pi}$ is a maximal $\bm{Z}$-order in $\bm{Q}\pi$ containing $\bm{Z}\pi$ and $C(\Omega_{\bm{Z}\pi})$ is the locally free class group of $\Omega_{\bm{Z}\pi}$, provided that $\pi$ is isomorphic to the following four types of groups : $C_n$ ($n$ is any positive integer), $D_m$ ($m$ is any odd integer $\ge 3$), $C_{q^f}\times D_m$ ($m$ is any odd integer $\ge 3$, $q$ is an odd prime number not dividing $m$, $f\ge 1$, and $(\bm{Z}/q^f\bm{Z})^{\times}=\langle \bar{p}\rangle$ for any prime divisor $p$ of $m$), $Q_{4m}$ ($m$ is any odd integer $\ge 3$, $p\equiv 3 \pmod{4}$ for any prime divisor $p$ of $m$)., Comment: To appear in Asian J. Math. ; the title is changed

Let $k$ be any field, $G$ be a finite group. Let $G$ act on the rational function field $k(x_g:g\in G)$ by $k$-automorphisms defined by $h\cdot x_g=x_{hg}$ for any $g,h\in G$. Denote by $k(G)=k(x_g:g\in G)^G$ the fixed field. Noether's problem asks, under what situations, the fixed field $k(G)$ will be rational (= purely transcendental) over $k$. According to the data base of GAP there are $10$ isoclinism families for groups of order $243$. It is known that there are precisely $3$ groups $G$ of order $243$ (they consist of the isoclinism family $\Phi_{10}$) such that the unramified Brauer group of $\bm{C}(G)$ over $\bm{C}$ is non-trivial. Thus $\bm{C}(G)$ is not rational over $\bm{C}$. We will prove that, if $\zeta_9 \in k$, then $k(G)$ is rational over $k$ for groups of order $243$ other than these $3$ groups, except possibly for groups belonging to the isoclinism family $\Phi_7$., Comment: arXiv admin note: text overlap with arXiv:1201.5555 by other authors

Let $p$ be an odd prime number, $D_p$ be the dihedral group of order $2p$, $h_p$ and $h^+_p$ be the class numbers of $\bm{Q}(\zeta_p)$ and $\bm{Q}(\zeta_p+ \zeta_p^{-1})$ respectively. Theorem. $h_p^+=1$ if and only if, for any field $k$ admitting a $D_p$-extension, all the algebraic $D_p$-tori over $k$ are stably rational. A similar result for $h_p=1$ and $C_p$-tori is valid also., Comment: A 2-page appendix is added

Let $G$ be a subgroup of $S_6$, the symmetric group of degree 6. For any field $k$, $G$ acts naturally on the rational function field $k(x_1,...,x_6)$ via $k$-automorphisms defined by $\sigma\cdot x_i=x_{\sigma(i)}$ for any $\sigma\in G$, any $1\le i\le 6$. Theorem. The fixed field $k(x_1,...,x_6)^G$ is rational (=purely transcendental) over $k$, except possibly when $G$ is isomorphic to $PSL_2(\bm{F}_5)$, $PGL_2(\bm{F}_5)$ or $A_6$. When $G$ is isomorphic to $PSL_2(\bm{F}_5)$ or $PGL_2(\bm{F}_5)$, then $\bm{C}(x_1,...,x_6)^G$ is $\bm{C}$-rational and $k(x_1,...,x_6)^G$ is stably $k$-rational for any field $k$. The invariant theory of wreath products will be investigated also.

Let $G$ be a subgroup of $S_n$, the symmetric group of degree $n$. For any field $k$, $G$ acts naturally on the rational function field $k(x_1,x_2,\ldots,x_n)$ via $k$-automorphisms defined by $\sigma\cdot x_i=x_{\sigma(i)}$ for any $\sigma\in G$, any $1\le i\le n$. Theorem. If $n\le 5$, then the fixed field $k(x_1,\ldots,x_n)^G$ is purely transcendental over $k$. We will show that $\bm{C}(x_1,\ldots,x_7)^G$ is also purely transcendental over $\bm{C}$ if $G$ is any transitive subgroups of $S_7$ other than $A_7$; a similar result is valid for solvable transitive subgroups of $S_{11}$.

The Bogomolov multiplier of a finite group $G$ is defined as the subgroup of the Schur multiplier consisting of the cohomology classes vanishing after restriction to all abelian subgroups of $G$. This invariant of $G$ plays an important role in birational geometry of quotient spaces $V/G$. We show that in many cases the vanishing of the Bogomolov multiplier is guaranteed by the rigidity of $G$ in the sense that it has no outer class-preserving automorphisms., Comment: 9 pages

Let $K$ be any field and $G$ be a finite group. Let $G$ act on the rational function field $K(x_g: \ g \in G)$ by $K$-automorphisms defined by $g \cdot x_h=x_{gh}$ for any $g, \ h \in G$. Denote by $K(G)$ the fixed field $K(x_g: \ g \in G)^G$. Noether's problem asks whether $K(G)$ is rational (=purely transcendental) over $K$. We will give a brief survey of Noether's problem for abelian groups and dihedral groups, and will show that $\Bbb Q(D_n)$ is rational over $\Bbb Q$ for $n \le 10$.

Let $G$ be a finite group. The Bogomolov multiplier $B_0(G)$ is constructed as an obstruction to the rationality of $\bm{C}(V)^G$ where $G\to GL(V)$ is a faithful representation over $\bm{C}$. We prove that, for any finite groups $G_1$ and $G_2$, $B_0(G_1\times G_2)\xrightarrow{\sim} B_0(G_1)\times B_0(G_2)$ under the restriction map. If $G=N\rtimes G_0$ with $\gcd\{|N|,|G_0|\}=1$, then $B_0(G)\xrightarrow{\sim} B_0(N)^{G_0}\times B_0(G_0)$ under the restriction map. For any integer $n$, we show that there are non-direct-product $p$-groups $G_1$ and $G_2$ such that $B_0(G_1)$ and $B_0(G_2)$ contain subgroups isomorphic to $(\bm{Z}/p \bm{Z})^n$ and $\bm{Z}/p^n \bm{Z}$ respectively. On the other hand, if $k$ is an infinite field and $G=N\rtimes G_0$ where $N$ is an abelian normal subgroup of exponent $e$ satisfying that $\zeta_e\in k$, we will prove that, if $k(G_0)$ is retract $k$-rational, then $k(G)$ is also retract $k$-rational provided that certain "local" conditions are satisfied; this result generalizes two previous results of Saltman and Jambor \cite{Ja}.

Let $k$ be any field, $G$ be a finite group acting on the rational function field $k(x_g:g\in G)$ by $h\cdot x_g=x_{hg}$ for any $h,g\in G$. Define $k(G)=k(x_g:g\in G)^G$. Noether's problem asks whether $k(G)$ is rational (= purely transcendental) over $k$. A weaker notion, retract rationality introduced by Saltman, is also very useful for the study of Noether's problem. We prove that, if $G$ is a Frobenius group with abelian Frobenius kernel, then $k(G)$ is retract $k$-rational for any field $k$ satisfying some mild conditions. As an application, we show that, for any algebraic number field $k$, for any Frobenius group $G$ with Frobenius complement isomorphic to $SL_2(\bm{F}_5)$, there is a Galois extension field $K$ over $k$ whose Galois group is isomorphic to $G$, i.e. the inverse Galois problem is valid for the pair $(G,k)$. The same result is true for any non-solvable Frobenius group if $k(\zeta_8)$ is a cyclic extension of $k$.

Let $k$ be any field, $G$ be a finite group acing on the rational function field $k(x_g:g\in G)$ by $h\cdot x_g=x_{hg}$ for any $h,g\in G$. Define $k(G)=k(x_g:g\in G)^G$. Noether's problem asks whether $k(G)$ is rational (= purely transcendental) over $k$. It is known that, if $\bm{C}(G)$ is rational over $\bm{C}$, then $B_0(G)=0$ where $B_0(G)$ is the unramified Brauer group of $\bm{C}(G)$ over $\bm{C}$. Bogomolov showed that, if $G$ is a $p$-group of order $p^5$, then $B_0(G)=0$. This result was disproved by Moravec for $p=3,5,7$ by computer calculations. We will prove the following theorem. Theorem. Let $p$ be any odd prime number, $G$ be a group of order $p^5$. Then $B_0(G)\ne 0$ if and only if $G$ belongs to the isoclinism family $\Phi_{10}$ in R. James's classification of groups of order $p^5$., Comment: arXiv admin note: substantial text overlap with arXiv:1109.2966

Let $k$ be any field, $G$ be a finite group acting on the rational function field $k(x_g : g\in G)$ by $h\cdot x_g=x_{hg}$ for any $h,g\in G$. Define $k(G)=k(x_g : g\in G)^G$. Noether's problem asks whether $k(G)$ is rational (= purely transcendental) over $k$. It is known that, if $\bC(G)$ is rational over $\bC$, then $B_0(G)=0$ where $B_0(G)$ is the unramified Brauer group of $\bC(G)$ over $\bC$. Bogomolov showed that, if $G$ is a $p$-group of order $p^5$, then $B_0(G)=0$. This result was disproved by Moravec for $p=3,5,7$ by computer computing. We will give a theoretic proof of the following theorem (i.e. by the traditional bare-hand proof without using computers). Theorem. Let $p$ be any odd prime number. Then there is a group $G$ of order $p^5$ satisfying $B_0(G)\neq 0$ and $G/[G,G] \simeq C_p \times C_p$. In particular, $\bC(G)$ is not rational over $\bC$.

Let $G$ be a finite group and $k$ be a field. Let $G$ act on the rational function field $k(x_g:g\in G)$ by $k$-automorphisms defined by $g\cdot x_h=x_{gh}$ for any $g,h\in G$. Noether's problem asks whether the fixed field $k(G)=k(x_g:g\in G)^G$ is rational (i.e. purely transcendental) over $k$. Theorem 1. If $G$ is a group of order $2^n$ ($n\ge 4$) and of exponent $2^e$ such that (i) $e\ge n-2$ and (ii) $\zeta_{2^{e-1}} \in k$, then $k(G)$ is $k$-rational. Theorem 2. Let $G$ be a group of order $4n$ where $n$ is any positive integer (it is unnecessary to assume that $n$ is a power of 2). Assume that {\rm (i)} $\fn{char}k \ne 2$, $\zeta_n \in k$, and {\rm (ii)} $G$ contains an element of order $n$. Then $k(G)$ is rational over $k$, except for the case $n=2m$ and $G \simeq C_m \rtimes C_8$ where $m$ is an odd integer and the center of $G$ is of even order (note that $C_m$ is normal in $C_m \rtimes C_8$) ; for the exceptional case, $k(G)$ is rational over $k$ if and only if at least one of $-1, 2, -2$ belongs to $(k^{\times})^2$., Comment: arXiv admin note: incorporates virtually all of arXiv:1009.2299

Let $G$ be a finite group and $k$ be a field. Let $G$ act on the rational function field $k(x_g:g\in G)$ by $k$-automorphisms defined by $g\cdot x_h=x_{gh}$ for any $g,h\in G$. Noether's problem asks whether the fixed field $k(G)=k(x_g:g\in G)^G$ is rational (i.e. purely transcendental) over $k$. We will prove that, if $G$ is a group of order $2^n$ ($n\ge 4$) and of exponent $2^e$ such that (i) $e\ge n-2$ and (ii) $\zeta_{2^{e-1}} \in k$, then $k(G)$ is $k$-rational.13A50,14E08,14M20,12F12, Comment: The content of this paper became part of a new paper "Noether's problem for the groups with a cyclic subgroup of index 4" which has been posted in arXiv

Let $k$ be a field, $G$ be a finite group and $k(x_g:g\in G)$ be the rational function field over $k$, on which $G$ acts by $k$-automorphisms defined by $h\cdot x_g=x_{hg}$ for any $g,h\in G$. Noether's problem asks whether the fixed subfield $k(G):=k(x_g:g\in G)^G$ is $k$-rational, i.e.\ purely transcendental over $k$. If $\widehat{S_n}$ is the double cover of the symmetric group $S_n$, in which the liftings of transpositions and products of disjoint transpositions are of order $4$, Serre shows that $\bm{Q}(\widehat{S_4})$ and $\bm{Q}(\widehat{S_5})$ are not $\bm{Q}$-rational. We will prove that, if $k$ is a field such that $\fn{char} k \neq 2, 3$, and $k(\zeta_8)$ is a cyclic extension of $k$, then $k(\widehat{S_4})$ is $k$-rational. If it is assumed furthermore that $\fn{char}k=0$, then $k(\widehat{S_5})$ is also $k$-rational.

Let $G$ be a finite subgroup of $GL_4(\bm{Q})$. The group $G$ induces an action on $\bm{Q}(x_1,x_2,x_3,x_4)$, the rational function field of four variables over $\bm{Q}$. Theorem. The fixed subfield $\bm{Q}(x_1,x_2,x_3,x_4)^G:=\{f\in\bm{Q}(x_1,x_2,x_3,x_4):\sigma \cdot f=f$ for any $\sigma\in G\}$ is rational (i.e.\ purely transcendental) over $\bm{Q}$, except for two groups which are images of faithful representations of $C_8$ and $C_3\rtimes C_8$ into $GL_4(\bm{Q})$ (both fixed fields for these two exceptional cases are not rational over $\bm{Q}$). There are precisely 227 such groups in $GL_4(\bm{Q})$ up to conjugation; the answers to the rationality problem for most of them were proved by Kitayama and Yamasaki \cite{KY} except for four cases. We solve these four cases left unsettled by Kitayama and Yamasaki; thus the whole problem is solved completely.

Let $k$ be an infinite field. The notion of retract $k$-rationality was introduced by Saltman in the study of Noether's problem and other rationality problems. We will investigate the retract rationality of a field in this paper. Theorem 1. Let $k\subset K\subset L$ be fields. If $K$ is retract $k$-rational and $L$ is retract $K$-rational, then $L$ is retract $k$-rational. Theorem 2. For any finite group $G$ containing an abelian normal subgroup $H$ such that $G/H$ is a cyclic group, for any complex representation $G \to GL(V)$, the fixed field $\bm{C}(V)^G$ is retract $\bm{C}$-rational. Theorem 3. If $G$ is a finite group, then all the Sylow subgroups of $G$ are cyclic if and only if $\bm{C}_{\alpha}(M)^G$ is retract $\bm{C}$-rational for all $G$-lattices $M$, for all short exact sequences $\alpha : 0 \to \bm{C}^{\times} \to M_{\alpha} \to M \to 0$. Because the unramified Brauer group of a retract $\bm{C}$-rational field is trivial, Theorem 2 and Theorem 3 generalize previous results of Bogomolov and Barge respectively (see Theorem \ref{t5.9} and Theorem \ref{t6.1})., Comment: Several typos in the previous version were corrected

Let $K$ be any field and $G$ be a finite group. Let $G$ act on the rational function field $K(x_g:g\in G)$ by $K$-automorphisms defined by $g\cdot x_h=x_{gh}$ for any $g,h\in G$. Noether's problem asks whether the fixed field $K(G)=K(x_g:g\in G)^G$ is rational (=purely transcendental) over $K$. We will prove that if $G$ is a non-abelian $p$-group of order $p^n$ ($n\ge 3$) containing a cyclic subgroup of index $p^2$ and $K$ is any field containing a primitive $p^{n-2}$-th root of unity, then $K(G)$ is rational over $K$. As a corollary, if $G$ is a non-abelian $p$-group of order $p^3$ and $K$ is a field containing a primitive $p$-th root of unity, then $K(G)$ is rational.

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.

We will give several reduction theorems for Noether's problem.

We will show the raitonality of some twisted symmetric group actions.