Lemma 4

Let K be a commutative field, V a finite dimensional vector space over K, G a finite group of automorphisms of V whose order q is invertible in K, S the symmetric algebra of V, and R the subalgebra of S consisting of the elements invariant under G. A prime ideal B of height 1 of S is ramified over $p = B \cap R$ (Commutative Algebra) if and only if there exist a non-zero element a of V and a non-zero element f of $V^*$ such that $B = Sa$ and the pseudo-reflection $s_{a,f}$ belongs to G. The decomposition group $G^Z(B)$ is then the subgroup of elements of G leaving Ka stable, and the inertia group $G^T(B)$ is the cyclic subgroup $H_a$ of G consisting of the pseudo-reflections of G with vector a. The residue field S(B) of S at B is separable over the residue field $R(p)$ of R at p, and the ramification index $e(B/p)$, equal to the coefficient of B, augmented by 1, in the divisor $\operatorname{div}(D_{S/R})$ of the different, is equal to $\operatorname{Card}(H_a)$.

To say that B is ramified over R means that its inertia group $G^T(B)$ does not reduce to the identity, in other words that there exists $g \neq 1$ in G such that $g(z) \equiv z \pmod{B}$ for all $z \in S$. Since S is a factorial ring, B is a principal ideal Sa, and a must divide all the elements $g(z) - z$ ($z \in S$); now, for $z \in V$, these elements are homogeneous of degree 1 and are all non-zero (since $g \neq 1$); thus, a must be homogeneous of degree 1, in other words $a \in V$. Then there exists a linear form f on V such that $g = s_{a,f}$. Conversely, if g is a pseudo-reflection $s_{a,f}$ different from 1, then $g(z) \equiv z \pmod{Sa}$ for all $z \in S$, so g belongs to the inertia group of the prime ideal $B = Sa$. This proves the first assertion of the lemma and the characterizations of $G^Z(B)$ and $G^T(B)$. Since q is coprime to the characteristic p of K (which is also that of S(B)), the extension S(B) of $R(p)$ is separable (Commutative Algebra, Chap. V, § 2, no. 2, Cor. of Prop. 5). The equality $e(B/p) = \operatorname{Card}(H_a)$ follows from this (Commutative Algebra). Since $e(B/p)$ is coprime to p, the coefficient of B in $\operatorname{div}(D_{S/R})$ is $e(B/p) - 1$ (Commutative Algebra). This proves the lemma.