Proposition 5

(i) Let D be the element $\prod_{g \in H} e_g$ of S. The elements of S which are anti-invariant under G are the elements of RD.

(ii) Identify S with the polynomial algebra $K[X_1, \ldots, X_l]$ by choosing a basis $(X_1, \ldots, X_l)$ of V, and let $(P_1, \ldots, P_l)$ be algebraically independent homogeneous elements of S generating the algebra R (Th. 3). Then the jacobian $J = \det \left( \frac{\partial P_i}{\partial X_j} \right)$ is of the form $\lambda D$, where $\lambda \in K^*$.

a) With the notation in (ii), we have

so, for all $g \in G$,

hence J is anti-invariant under G. On the other hand, the field of fractions N of S is a Galois extension of the field of fractions E of R (no. 2); if $\Delta$ is a derivation of E with values in an extension field $\Omega$ of N, $\Delta$ extends to a derivation of N with values in $\Omega$ (Algebra, Chap. V, § 16, no. 4, Th. 3); since the $P_i$ are algebraically independent, it follows that $dP_1 \wedge \ldots \wedge dP_l \neq 0$, hence $J \neq 0$.

b) Let z be an element of S anti-invariant under G. We show that z is divisible by D in S. Let a be a non-zero vector in V. The elements of G that leave Ka stable leave stable a complementary hyperplane L (Appendix, Prop. 2); an element of G leaving Ka stable is 1 or a pseudo-reflection with vector a if and only if it induces 1 on L; the pseudo-reflections with vector a belonging to G thus constitute, together with 1, a cyclic subgroup $G'$ of G; let t be its order. There exists a basis $(X_1, \ldots, X_l)$ of V such that $a = X_1$, $X_2 \in L, \ldots, X_l \in L$, and z can be identified with a polynomial $P(X_1, \ldots, X_l)$ with coefficients in K. From $g(z) = (\det g)^{-1} z$ for $g \in G'$, we see that $X_1$ only appears in $P(X_1, \ldots, X_l)$ with exponents congruent to -1 modulo t. Thus, $P(X_1, \ldots, X_l)$ is divisible by $X_1^{t-1} = a^{t-1}$. Now D is, up to a scalar factor, the product of the $a^{t-1}$ for those $a \in V$ such that $t > 1$, and these elements of S are mutually coprime. Since S is factorial, z is divisible by D.

c) By a) and b), J is divisible by D in S. Now

(Prop. 3), so $\deg J = \deg D$, hence $J = \lambda D$ with $\lambda \in K$. Since $J \neq 0, \lambda \in K^*$. This proves (ii).

d) Parts a) and c) of the proof show that D is anti-invariant under G. Next, if $y \in R$, it is clear that $yD$ is anti-invariant under G. Finally, if $z \in S$ is anti-invariant under G, we have seen in b) that there exists $y \in S$ such that $z = yD$. Since S is integral, $y \in R$. This proves (i).