Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 23 / 2 / ii / Solution

Let $R=\mathbb{C}[V]=\mathrm{Sym}(V^{\ast })$, with its degree grading, and let $A=R^G$ be the polynomial invariant ring. For a compact group average polynomials over $G$; for a complex reductive group average over its Zariski-dense compact form from part (i). Each degree piece is finite dimensional, so this defines a polynomial and gives the Reynolds operator on a polynomial invariant ringIt preserves degree. For the reductive case compact invariance is full group invariance by Zariski density.

Let $A_{+}$ be the positive-degree invariants and form the ideal $I=R{A}_{+}$. The Hilbert basis theorem makes $R$ a Noetherian ring, so finitely many homogeneous invariants $f_1,\dots ,f_s\in A_{+}$ generate $I$ as an $R$-ideal. Indeed start with a finite ideal-generating set and collect the finitely many invariants occurring in expressions for those generators; then replace them by homogeneous pieces.

We claim that these same $f_i$ generate $A$ as an algebra. For a homogeneous invariant $f$ of degree $d>0$, write $f={\sum }_i{a}_i{f}_i$ with homogeneous coefficients of degree $d-\mathrm{deg}{f}_i$; terms of negative degree are absent. Apply the Reynolds operator on a polynomial invariant ring to get $f={\sum }_i\mathcal{R}(a_i)f_i$. The averaged coefficients are invariant and have smaller degree than $d$. Starting from the constants, induction on degree puts them and then $f$ in $\mathbb{C}[f_1,\dots ,f_s]$. Taking homogeneous pieces covers every invariant. This proves Hilbert finite generation for reductive invariants:Noetherianity alone would not suffice for a subring; the Reynolds module property is the step that makes the induction work.