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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 birational equivalence forms a semigroup, denoted by $T(\pi)$. We will give a complete proof of the following theorem due to Endo and Miyata \cite{EM4}. 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 one of 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 $m$).

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.

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 $\bm{C}$. It is known that, if the unramified Brauer group, denoted by ${\rm Br}_u(\bm{C}(M)^G)$, is non-trivial, then the fixed field $\bm{C}(M)^G$ is not rational (= purely transcendental) 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 $4$. There is a direct decomposition ${\rm Br}_u(\bm{C}(M)^G)= B_0(G) \oplus H^2_u(G,M)$ where $H^2_u(G,M)$ is some subgroup of $H^2(G,M)$. The first summand $B_0(G)$, which is related to the faithful linear representations of $G$, has been investigated by many authors. But the second summand $H^2_u(G,M)$ doesn't receive much attention except when the rank is $\le 3$. 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 ${\rm Br}_u(\bm{C}(M)^G)\neq 0$. The {\rm GAP IDs} (actually the {\rm CARAT IDs}) of the corresponding groups $G$ may be determined explicitly. A similar result for lattices of rank $6$ is found also. Motivated by these results, we construct $G$-lattices $M$ of rank $2n+2, 4n, p(p-1)$ ($n$ is any positive integer and $p$ is any odd prime number) satisfying that $B_0(G)=0$ and $H^2_u(G,M)\neq 0$; and therefore $\bm{C}(M)^G$ are not rational over $\bm{C}$.

Let $k$ be a field and $k(x_0,\ldots,x_{p-1})$ be the rational function field of $p$ variables over $k$ where $p$ is a prime number. 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 $P_0=\{p\in P:\bm{Q}(\zeta_{p-1})$ is of class number one$\}$ where $\zeta_n$ a primitive $n$-th root of unity in $\bm{C}$ for a positive integer $n$; $P_0$ is a finite set by \cite{MM}. Theorem. Let $k$ be an algebraic number field and $P_k=\{p\in P: p$ is ramified in $k\}$. Then $k(x_0,\ldots,x_{p-1})^G$ is not stably rational over $k$ for all $p\in P\backslash (P_0\cup P_k)$.