TITLE: On the field of invariants of a finite group
QUESTION [9 upvotes]: So let $G$ be a finite group and let $\iota:G\rightarrow S_n$ be an embedding of $G$ in a symmetric group of degree $n$ for some fixed integer $n$. Let $K$ be a fixed field of characteristic $0$. The group $S_n$ permutes the variables $\{x_1,\ldots,x_n\}$ and therefore
acts on the field 
$$
L:=K(x_1,\ldots,x_n).
$$
One may look at the invariant subfield $L^{G}\subseteq L$. From Galois theory one has that
$L/L^G$ is a Galois extension with Galois group $G$. In particular, the transcendence degree
of $L^G$ over $K$ is equal to $n$. In general, the field $L^G$ is not purely transcendental
so the following question makes sense:
Q: Does the isomorphism class of $L^G$ depend on the embedding $\iota$ ?
Intuitively I would say no, but this is really just a guess!

REPLY [6 votes]: It seems likely (and may be known) that it does depend on the permutation representation. However, it is a subtle question as the stable isomorphism class (of any faithful permutation, or in fact arbitrary, representation) does not. Here two extensions $K$ and $K'$ of a field $k$ are stably isomorphic if for some $m$ and $n$ $K(x_1,\ldots,x_m)$ and $K'(x'_1,\ldots,x'_n)$ are isomorphic as $k$-extensions.
Addendum: The result is well-known but I cannot at the moment come up with a reference so instead I give the proof: Let $V$ be a faithful $G$-representation and $U$ the non-empty Zariski open subset where $G$ acts freely. Then $k(V)^G$ is the fraction field of $U/G$. If now, $V'$ is another faithful representation with open subset $U'$ we have that $U\times V'$ has a free $G$-action with a linear action on the second factor. Hence $U\times V'/G$ is a vector bundle over $U/G$ (by descent theory) and in particular its fraction field is stably isomorphic to that of $U/G$. However, $U\times V'/G$ is birational to $U\times U'/G$ which in turn is birational to $V\times U'/G$. The fraction field of the latter is for the same reason stably isomorphic to that of $U'/G$.