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I am trying to make sense of the notation and certain sets in two articles by Annick Valibouze whose results I'm using for my bachelor's thesis, I hope it's relevant enough to merit an answer.

  1. In one article, $F\in k[x_1,...,x_n]$ is called an L-primitive H-invariant ($H \subset L \subset S_n$) if $stab_{L}(F)=H$, and the L-relative resolvent by F is defined as $\prod (X-\sigma_i F)$, where the $\sigma_i$ are a transversal of $L/H$. Now the assertion is that the L-relative resolvent by F lies in $K(x_1,...,x_n)^L[X]$, where $K=k(x_1,...,x_n)^{S_n}$, i.e. the resolvent lies in $k(x_1,...,x_n)^{S_n}(x_1,...,x_n)^L[X]$.

Let $M=k(x_1...,x_n)^L$ and $N=k(x_1,...,x_n)^{S_n}(x_1,...,x_n)^L$. Suppose $g\in N$. I believe we can simply look at one of the terms of g, say $e_1^{i_1}\cdots e_n^{i_n} x_1^{j_1}\cdots x_n^{j_n}$, where the $e_i$ are the elementary symmetric polynomials. Then g is L-invariant, hence $g\in M$, i.e. $N \subset M$. Seems to me also that if $h\notin N$, it cannot be in $M$ either, hence N=M. Would you say this is correct?

2a. This is from the other article but regards the same thing. Let $H \subset L$ again. The article defines a relative primitive invariant of H with respect to L as any primitive element of the extension $k(x_1,...,x_n)^H/k(x_1,...,x_n)^L$ belonging to something called $A_H$. This $A_H$ is defined as the integral closure of $k[e_1,...,e_n]$ in $k(x_1,...,x_n)^H$ (e_i as above), and $A_H = k[x_1,...,x_n] \cap k(x_1,...,x_n)^H$. So about the intersection: it can only contain polynomials, so we need only consider the intersection $k[x_1,...,x_n] \cap k[x_1,...,x_n]^H$, which must be $k[x_1,...,x_n]^H$. Is this correct? Normally I wouldn't have thought twice about it, but since the author has taken the time to introduce the notation $A_H$ I get worried.

2b. Another assertion is that $k(x_1,...,x_n)^H=Frac(A_H)$, which if I'm right about (2a) seems correct, but he then adds that $k(x_1,...,x_n)^H = A_H[1/k[e_1,...,e_n]]$. I have no idea what this particular notation means, my guess is that it means $\{f/g \mid f \in A_H, g\in k[e_1,...,e_n]\}$. So my question here is, what is $Frac(A_H)$, is it $k(x_1,...,x_n)^H$ or that last set?

Thanks in advance.

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1. The $k(x_1,...,x_n)^{S_n}(x_1,...,x_n)^L$ notation looks fishy to me -- while it does make sense formally, I am surprised that anyone is using it because (as you have correctly noticed, although not correctly proven) $k(x_1,...,x_n)^{S_n}(x_1,...,x_n)^L = k(x_1,...,x_n)^L$. (About your proof: The elements of $k(x_1,...,x_n)^{S_n}(x_1,...,x_n)^L$ are not polynomials in the $x_1,...,x_n$, but rational functions, so you cannot decompose them into monomials.) I think it would be best if you let the cat out of the sack and name the article so we can tell what's going on. –  darij grinberg Mar 31 '13 at 16:32
    
Thank you. The article for question 1 is "Galois Theory and Reducible Polynomials" by Valibouze; you'll have to google it. The other article is "Lagrange Resolvents" by Arnaudiès and Valibouze, available from mathscinet. –  Erik Vesterlund Mar 31 '13 at 16:51
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2a. It looks to me like you're correct. I am bewildered by some of his notations as well; it seems that he is putting rational function fields unnecessarily into the foreground where I'd normally think in terms of polynomial rings. But it doesn't really complicate the understanding once one gets used to it. 2b. I think both your guess and his assertion are correct. The field $k\left(x_1,x_2,...,x_n\right)^H$ is the same as $\left\lbrace f/g \mid f\in A_H,\ g\in k\left[e_1,...,e_n\right],\ g\neq 0\right\rbrace$ (don't forget $g\neq 0$), because ... –  darij grinberg Mar 31 '13 at 23:01
    
... whenever you have an $H$-invariant rational function, you can write it as a quotient of an $H$-invariant polynomial and an $\mathfrak S_n$-invariant polynomial (which $\mathfrak S_n$-invariant polynomial, as you know, is a polynomial in the $e_1$, $e_2$, ..., $e_n$). This is a particular case of a more general fact that if $G$ is a group acting on an integral domain $D$ by ring automorphisms, and $H$ is a subgroup of $G$, then $\left(Q\left(D\right)\right)^H = \left\lbrace f/g \mid f \in D^H,\ g\in D^G,\ g\neq 0\right\rbrace$ (where $Q\left(D\right)$ means the quotient field of $D$). ... –  darij grinberg Mar 31 '13 at 23:07

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