Theorem 1

Let K be a commutative field, V a finite dimensional vector space over K, S = S(V) the symmetric algebra of V, G a finite group of automorphisms of V, and R the graded subalgebra of S consisting of the elements invariant under G. Assume that G is generated by pseudo-reflections (§ 2, no. 1) and that $q = \mathrm{Card}(G)$ is coprime to the characteristic of K. Then the R-module S has a basis consisting of q homogeneous elements.

a) Since every submodule of $S/(R_+S)$ is free over $R_0 = K$, it is enough to show (in view of Algebra, Chap. II, §11, no. 4, Prop. 7) that the canonical homomorphism from $R_+ \otimes_R S$ to S is injective. For any R-module E, denote by T(E) the R-module $\mathrm{Ker}(R_+ \otimes_R E \to E)$ (\*in other words, $T(E) = \mathrm{Tor}_1^R(R/R_+, E)$\*). If E, $E'$ are two R-modules and $u$ is a homomorphism from E to $E'$, the homomorphism $1 \otimes u$ from $R_+ \otimes E$ to $R_+ \otimes E'$ defines by restriction to T(E) a homomorphism from T(E) to T($E'$) that we denote by $T(u)$. If $u'$ is a homomorphism from $E'$ to an R-module $E''$, we have $T(u' \circ u) = T(u') \circ T(u)$. Thus, if G acts R-linearly on E, then G acts on T(E).

b) The group G acts R-linearly on S, and hence also on T(S). Moreover, T(S) has a natural structure of graded S-module. We show first that, if $g \in G$, then $g$ transforms every element $x$ of T(S) into an element congruent to $x$ modulo $S_1 T(S)$. It is enough to do this when $g$ is a pseudo-reflection. Then there exists a non-zero vector $v$ in V such that $g(x) - x \in K v$ for all $x \in V$. Since V generates S, it follows that $g_S$ acts trivially on $S/Sv$. Thus, for any $y \in S$, there exists an element $h(y)$ in S such that

Since S is integral and $v$ is non-zero, this element is determined uniquely by $y$; it is immediate that $h$ is an endomorphism of degree – 1 of the R-module S. Thus, $g_S - 1_S = m_v \circ h$, where $m_v$ denotes the homothety with ratio $v$ in S. Hence,

the image of which is contained in $vT(S)$, proving our assertion.

c) We show that any element of T(S) invariant under G is zero. Indeed, let Q be the endomorphism of the R-module S defined by

for all $y \in S$. Then $Q(S) = R$. We can thus write $Q = i \circ Q'$, where $Q'$ is a homomorphism from the R-module S to the R-module R and $i$ denotes the canonical injection of R into S. Thus $T(Q) = T(i) \circ T(Q')$ and $T(Q') = 0$ since $T(R) = \mathrm{Ker}(R_+ \otimes R \to R) = 0$. Thus

But $q^{-1} \sum_{g \in G} T(g_S)$ leaves fixed the elements of T(S) invariant under G. These elements are therefore zero.

d) Assume that $T(S) \neq 0$. There exists in T(S) a homogeneous element $u \neq 0$ of minimum degree. By b), $u$ is invariant under G. By c), $u = 0$. This is a contradiction, so $T(S) = 0$. Q.E.D.