Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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!

share|improve this question
    
You presumably meant that the transcendence degree of $L^G$ over $K$ is $n$, not $|G|$. –  Andreas Blass Jun 29 '11 at 0:41
    
Yes this what I meant, I just changed it –  Hugo Chapdelaine Jun 29 '11 at 1:01
    
Do you mean to assume transitivity of $G$? For example, if $n = 4$, a Klein 4-group $V$ can be embedded in $S_4$ in two quite different ways. One as the unique normal subgroup of order $4$ of $S_4$ which is regular (and, in particular transitive). In this case $V$ is eembedded as $\langle (1), (12)(34),(13)(24),(14)23) \langle$. Alternatively, $V$ may be embedded as $\langle (1),(12),(34),(12)(34) \rangle$, which is an intransitive subgroup of $S_4$. –  Geoff Robinson Jul 2 '11 at 16:01
add comment

2 Answers

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 $U'/G$.

share|improve this answer
    
Do you have a reference for this result? –  Hugo Chapdelaine Jun 29 '11 at 21:19
    
Thanks for the simple and elegant proof. –  Hugo Chapdelaine Jun 30 '11 at 12:16
add comment

If I were going to think about this I would do the following. Take some situation where I know Noether's problem has a negative answer, like the generalized quaternion group of order 16 in its regular permutation representation on 16 variables. (I didn't know that off the top of my head, and maybe it's not the easiest example -- it came up on Google, ascribed to Serre.)

Now this group has lots of embeddings in S_16, I suppose. There are so many cases where Noether's theorem is known to have a positive answer that you might be able to show that one of these has a rational field of invariants.

share|improve this answer
    
I'm pretty sure the extension is purely transcendental whenever $G$ is nilpotent. –  Steve D Jun 29 '11 at 4:01
1  
Isn't Noether's problem over $Q$ already negative for $C_8$ (Lenstra 1974)? math.leidenuniv.nl/~hwl/PUBLICATIONS/1974c/art.pdf –  Junkie Jun 29 '11 at 4:16
    
@Junkie: Yes, you're right, sorry, I was thinking of the positive characteristic case. –  Steve D Jun 29 '11 at 4:23
    
In fact the $\mathbb Q$ case goes back to Swan (though for $n=47$). –  Torsten Ekedahl Jun 29 '11 at 6:02
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.