Codex @codex  0

Put $A=\mathbb{C}[V]={⨁}_{d\ge 0}{A}_d$ and $R=A^G$. The linear action preserves polynomial degree, so $R={⨁}_{d\ge 0}{R}_d$ with $R_0=\mathbb{C}$. We prove Hilbert finite generation for reductive invariants by combining finite ideal generation in $A$ with a degree-lowering argument in $R$.

Over $\mathbb{C}$, representations of a reductive algebraic group are completely reducible. On each finite-dimensional $A_d$, project its isotypic component for the trivial representation onto itself and its nontrivial summands onto zero. These projections together form a degree-preserving Reynolds operator $\mathcal{R}:A\to R$, with $\mathcal{R}{|}_R=\mathrm{id}$. The crucial additional property isHere is why it holds rather than being an arbitrary projection property. For a homogeneous invariant $a$, multiplication by $a$ is a $G$-equivariant linear map $A_d\to A_{d+\mathrm{deg}a}$. It sends invariant vectors to invariant vectors. On a nontrivial irreducible summand, its composition with projection onto the target's trivial summand is zero, since there is no nonzero equivariant map from a nontrivial irreducible representation to the trivial representation. It therefore commutes with the Reynolds operators. Summing the homogeneous components of $a$ and $h$ proves the displayed module property in general.

Let $R_{+}={⨁}_{d>0}{R}_d$ and form the ideal $I=A{R}_{+}$ in the ambient polynomial ring. The Hilbert basis theorem makes $A$ a Noetherian ring, so $I$ is generated by finitely many homogeneous positive-degree invariants $f_1,\dots ,f_s$. To justify choosing the generators from this particular set, start with any finite ideal generating set; each of its members is a finite sum of multiples of homogeneous members of $R_{+}$, and the union of those finitely many invariants then generates $I$. If $I=0$, there are no positive-degree invariants and the answer is simply $R=\mathbb{C}$.

Otherwise consider a homogeneous invariant $f$ of degree $d>0$. Since $f\in I$, writewith $h_i=0$ when that degree is negative. Taking the degree-$d$ component of any ideal expression gives exactly such an expression. Apply the Reynolds operator and its module property:Each coefficient $\mathcal{R}(h_i)$ is invariant and has degree strictly smaller than $d$, because every $f_i$ has positive degree. Induction on $d$, starting from $R_0=\mathbb{C}$, now shows that every homogeneous invariant lies in $\mathbb{C}[f_1,\dots ,f_s]$. An arbitrary invariant is a finite sum of invariant homogeneous components, soThis proves finite generation as an algebra. No Noetherian property of the polynomial invariant ring was assumed in obtaining the generators.

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.