Noether's problem for the groups with a cyclic subgroup of index 4

Details Noether's problem for the groups with a cyclic subgroup of index 4 Noether's problem for the groups with a cyclic subgroup of index 4

2011-08-17

arxiv.org/abs/1108.3379v1

Mathematics

Commutative Algebra

arXiv admin note: incorporates virtually all of arXiv:1009.2299

Scientific paper

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 $(k^{\times})^2$.

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