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Choose finitely many homogeneous positive-degree invariants generating the ideal generated by all positive-degree invariants in the ambient polynomial ring. Such a finite set exists because a polynomial ring is Noetherian. Apply the Reynolds operator on a polynomial invariant ring to an expression for any homogeneous invariant in that ideal. Its coefficients become invariants of smaller degree, so induction proves that the chosen set generates the entire invariant ring as an algebra.

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.