Finite reflection invariant generators are algebraically independent

A field has only ideals $0$ and itself: an ideal containing nonzero $a$ contains $a^{-1}a=1$. They are generated by the empty set and by $1$. Thus $\mathbb{C}$ is Noetherian by F3, and finitely many applications of F2 make $S$ Noetherian. F3 supplies finite generators for $I$. Each is a finite sum of multiples of positive-degree invariants by F1; replace each invariant by its finitely many homogeneous components. The resulting finite family of homogeneous positive-degree invariants still generates $I$. Successively discard a redundant member until none remains, producing the required minimal family. If an invariant homogeneous $f$ has degree $d>0$, write $f={\sum }_j{s}_j{f}_j$ with $s_j$ homogeneous of degree $d-\mathrm{deg}{f}_j<d$, discarding negative-degree terms. Reynolds gives $f={\sum }_j\mathcal{R}(s_j)f_j$. Induction on $d$, starting with constants at degree zero, proves that all invariants are polynomials in the $f_j$.

We prove a relation fact. Suppose homogeneous invariants $F_1,\dots ,F_m$ satisfy $F_1\mathrm{\notin }(F_2,\dots ,F_m)S$, and homogeneous coefficients obey ${\sum }_j{g}_j{F}_j=0$. Then $g_1\in I$. Induct on $D=\mathrm{deg}{g}_1$ when $g_1\mathrm{\ne }0$; the zero case is immediate. At $D=0$, a nonzero scalar $g_1$ would solve for $F_1$ in the other ideal, impossible. For $D>0$ and a generating reflection $\sigma$, choose a nonzero linear equation $\alpha$ for its fixed hyperplane. Each $\sigma g_j-g_j$ vanishes there, hence equals $\alpha h_j$: take $\alpha$ as a coordinate and set that coordinate to zero in the polynomial, so the constant term in it is zero. Since $\sigma F_j=F_j$, subtracting the original relation from its transform and canceling the nonzero factor $\alpha$ gives ${\sum }_j{h}_j{F}_j=0$. Cancellation is valid since a polynomial ring over a field is a domain, as its leading monomial products have nonzero leading coefficient. Now $\mathrm{deg}{h}_1=D-1$ if it is nonzero, so induction yields $h_1\in I$, hence $\sigma g_1-g_1\in I$. The ideal $I$ is $G$-stable by F1. Writing any $g\in G$ as a product of generating reflections and telescoping therefore gives $g{g}_1-g_1\in I$. Averaging gives $\mathcal{R}(g_1)-g_1\in I$, while $\mathcal{R}(g_1)$ is either zero or a positive-degree invariant, hence in $I$. Therefore $g_1\in I$.

Suppose a nonzero polynomial relation $H(f_1,\dots ,f_r)=0$ exists. Give its variables weights $d_j=\mathrm{deg}{f}_j>0$. Separating weighted homogeneous parts and choosing least positive weighted degree gives such an $H$ of minimal degree $D$. Some partial derivative $H_j=\mathrm{\partial }H\mathrm{/}\mathrm{\partial }{y}_j$ is nonzero in characteristic zero. Its weighted degree is $D-d_j<D$, so $H_j(f)\mathrm{\ne }0$ by minimality (a nonzero constant also cannot evaluate to zero). From the finite family $H_j(f)$ choose a minimal subfamily generating their $S$-ideal and relabel it $H_1(f),\dots ,H_m(f)$, with $m\ge 1$. For $i>m$ write $H_i(f)={\sum }_{j=1}^m{a}_{ij}{H}_j(f)$, taking homogeneous coefficients of degrees $d_j-d_i$ and zero coefficients when this number is negative.

Differentiate $H(f)=0$ with respect to each coordinate $x_k$. The chain rule, verified on monomials, gives ${\sum }_j{H}_j(f){\mathrm{\partial }}_k{f}_j=0$. Substituting 2.1 yields ${\sum }_{j=1}^m{H}_j(f)({\mathrm{\partial }}_k{f}_j+{\sum }_{i>m}{a}_{ij}{\mathrm{\partial }}_k{f}_i)=0$. Minimality of the derivative generating subfamily gives $H_1(f)\mathrm{\notin }(H_2(f),\dots ,H_m(f))S$. Apply 1.2 to obtain ${\mathrm{\partial }}_k{f}_1+{\sum }_{i>m}{a}_{i1}{\mathrm{\partial }}_k{f}_i\in I$. It is homogeneous of degree $d_1-1$, so by 1.1 write it ${\sum }_j{q}_{jk}{f}_j$ with $\mathrm{deg}{q}_{jk}=d_1-d_j-1$, again zero in negative degrees. In particular $q_{1k}=0$.

Multiply these identities by $x_k$ and sum over $k$. The monomial identity ${\sum }_k{x}_k{\mathrm{\partial }}_k{f}_j=d_j{f}_j$ gives $d_1{f}_1+{\sum }_{i>m}{a}_{i1}{d}_i{f}_i={\sum }_{j\mathrm{\ne }1}({\sum }_k{x}_k{q}_{jk})f_j$. Since $d_1$ is a nonzero integer, this puts $f_1$ in the ideal of the other basic generators, contrary to their minimality. Thus no polynomial relation exists, proving algebraic independence. The precise exclusion of the self-coefficient is $\mathrm{deg}{q}_{1k}=-1$, not a positivity claim about arbitrary invariant forms. If $I=0$, step 1.1 gives $R=\mathbb{C}$ and the empty generating set is algebraically independent. The argument allows degree-one invariants and the trivial group; for $V=0$ it gives that empty set. All selections are finite or least positive degrees. F2 uses only the finite-generation and ascending-chain consequences in its proof, and F3 is invoked only in its explicitly choice-free branch; no AC is needed.