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This is homework, but I am pretty stuck and I feel I am lacking intuition on this so I ask.

The question is as follows : let $k$ be a field and $R = k[x_1, \dots, x_n]$ (this notation means a polynomial ring in $n$ variables with coefficients in $k$ but there are possibly relations between the variables, i.e. $R \cong k[X_1, \dots, X_n] / I$ where $I$ is an ideal of the ring $k[X_1, \dots, X_n]$ where the elements $X_i$ are algebraically independent indeterminates). Consider $G \le \mathrm{Aut}(R)$ a finite subgroup of automorphisms of the ring and let $R^G$ denote the fixed subring of $R$ by the action of $G$, i.e. $$ R^G = \{ r \in R \, | \, \forall g \in G, \quad g \cdot r = r\}. $$ Assume $k \subseteq R^G$. We were asked to show that the extension $R^G \to R$ is an integral extension and that $R$ is a finitely generated $R^G$-module (which is okay). Writing $$ R = \langle r_1, \dots, r_m \rangle_{R^G}, $$ for each generator $r_i$ we know that there exists an integrality relation of the form $$ r_i^{n_i} + s_{i1} r^{n_i - 1} + \dots + s_{in_i} = 0, \quad s_{ij} \in R^G. $$ Fix such a relation for each generator. Define $$ S = k[ s_{11},\dots,s_{1n_1},s_{21},\dots,s_{2n_2},\dots,s_{mn_m}] \subseteq R^G $$ to be the $k$-subalgebra generated by the coefficients of these polynomials in $R^G[X]$. We are asked to show that $R$ and $R^G$ are finitely generated $S$-modules and that $R^G$ is a finitely generated $k$-algebra (i.e. is a polynomial ring over some elements in $R^G$).

All I have managed to show is that if $R^G$ is a finitely generated $S$-module, then $R$ also is, because since $R$ is a finitely generated $R^G$-module, I can use the fact that $$ R = \sum_{i=1}^m R^G r_i = \sum_{i=1}^m \left( \sum_{j=1}^p S q_j \right) r_i = \sum_{i=1}^m \sum_{j=1}^p r_i q_j. $$ I don't need a full proof, just a hint towards a solution would be appreciated ; right now I just have no idea where to start.

Edit: I asked my teacher and he confirms that all he really wanted was to give steps to show that $R^G$ was an affine $k$-algebra, so we can use whatever generators we want (which wasn't explicit in the question). I'll be fine for the rest, but thank you all for noticing I was doing an impossible question (or uselessly too hard).

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closed as too localized by Alex Becker Jun 10 '13 at 3:38

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Comments or ideas from someone who does not know a full solution are also okay, the discussion might generate the idea. – Patrick Da Silva Feb 19 '13 at 18:09
I would try to go the other way. First show that $R$ is a finitely generated $S$-module, and then that $R^G$ is. All this assuming that $r_i$ are really the $x_i$:s, so that we know $R$ to be generated by the $r_i$:s as a $k$-algebra (and hence also as an $S$-algebra). – Jyrki Lahtonen Feb 19 '13 at 18:09
I have the same question as Jyrki - do you have the condition that $r_i$ generate $R$ as a $k$-algebra? It seems to be a natural requirement here by your setting. (which also leads to a proof of your question) – user27126 Feb 19 '13 at 18:22
@Sanchez : I guess that in the example I used for my proof the $r_i$'s do generate $R$ as a $k$-algebra because the $x_i$'s are $r_i$'s (but I need more than just the $x_i$'s to get the $R^G$-module). But I don't see how that helps me. And the question doesn't seem to assume the condition you mentioned. – Patrick Da Silva Feb 19 '13 at 23:30
@Jyrki Lahtonen : I tried both directions, i.e. I tried showing that either $R$ or $R^G$ was finitely generated... all I know is that if I get $R^G$ first then I don't have to work out the second.. but all I know for $R$ is that $S[r_1, \dots, r_n]$ is integral over $S$... I guess that if the $x_i$'s are among the $r_i$'s, that gives me a proof that $R = S[r_1, \dots, r_n]$ is an f.g. $S$-module? – Patrick Da Silva Feb 19 '13 at 23:32