I'm currently studying quotients of affine varieties by finite groups. I use the book "Algebraic Geometry" by J. Harris (http://userpage.fu-berlin.de/aconstant/Alg2/Bib/Harris_AlgebraicGeometry.pdf). On page 123, Harris shows that the quotient of an affine variety under the action of a finite group does exist. For that, he shows that the invariant ring $A(X)^G$ is a finitely generated $K$-algebra. However, a few steps are not very clear to me:
I understand why $K[x_1, … ,x_m]^{S_m}$ is a finitely generated $K$-algebra: it gets finitely generated by the elementary symmetric polynomials $y_i$, okay. But now my question is: How can we see that $K[x_1, … , x_m]$ is a finitely generated $K[y_1, … , y_m]$-module? That's not clear for me at all. That would mean that we could write any $P\in K[x_1, … , x_m]$ as $P = \sum_{i=1}^k s_i f_i $, where $s_i$ are symmetric polynomials, so $s_i \in K[y_1, … , y_m]$, and the $f_i \in K[x_1, … , x_m]$ are the generators. But what about polynomials of the form $P_n (x) = x_m^n + 1$? I am very sure that these polynomials cannot all be written in the from as above, because they are not symmetric, so the $s_i$ must be constant functions, but $K[x_1, … , x_m]$ is not a finitely generated $K$-module, but that would contradict the highlighted sentence. Can please someone tell me what I am doing wrong?
Then, I am also not quite sure, why it follows that also $K[x_1, … , x_m]^G$ must be a finitely generated $K[y_1, … , y_m]$ as well. Is it maybe because it is a submodule of $K[x_1, … , x_m]$, and maybe $K[x_1, … , x_m]$ is a Noetherian $K[y_1, … , y_m]$-module? I'm completely stumped.
Best Answer
Consider the polynomial $\prod_{i=1}^m (T-x_i) \in K[x_1,\dots, x_m]^{S_m}[T]$, this is monic and has $x_1, \dots, x_m$ as its roots. It follows that all the $x_i$ are integral over $K[x_1,\dots, x_m]^{S_m}$. A finitely generated integral ring extension is finite as a module. As $K[x_1,\dots, x_m]^{S_m}$ is a Noetherian ring, any finitely generated module over $K[x_1,\dots, x_m]^{S_m}$ is Noetherian, thus $K[x_1, \dots ,x_m]^G$ is finite over $K[x_1,\dots, x_m]^{S_m}$ as well.