Theorem 4

Let K be a commutative field, V a finite dimensional vector space over K, S the symmetric algebra of V, G a finite group of automorphisms of V, and R the subalgebra of S consisting of the elements invariant under G. Assume that q = Card G is invertible in K. The following conditions are equivalent:

(i) G is generated by pseudo-reflections; (ii) S is a free graded R-module; (iii) R is a graded polynomial algebra over K.

The equivalence of (ii) and (iii) follows from no. 2 and Lemma 5. The implication (i) $\Longrightarrow$ (ii) follows from Th. 1.

We show that (iii) $\Longrightarrow$ (i). Let G' be the subgroup of G generated by the pseudo-reflections belonging to G, and let R' be the subalgebra of S consisting of the elements invariant under G'. We have R $\subset$ R' $\subset$ S. By Lemma 4, div(D_{S/R}) = div(D_{S/R'}), so div(D_{R'/R}) = 0. Assume then that R is a graded polynomial algebra. Since this is also the case for R' (since G' is generated by pseudo-reflections), Lemma 5 shows that the R-module R' admits a homogeneous basis (Q_1, ..., Q_m); let q_i = deg(Q_i). Put

The fact that div(D_{R'/R}) is zero shows that div(d) = 0 (Commutative Algebra), which means that d belongs to K*. On the other hand Tr_{R'/R}(Q_i Q_j) is a homogeneous element of degree q_i + q_j, and d is homogeneous of degree $2 \sum_i q_i$. Thus, $\sum_i q_i = 0$, i.e. $q_i = 0$ for all i, which means that R' = R, and so G' = G by Galois theory. This proves that G is generated by pseudo-reflections. Q.E.D.