Proposition 7

Assume that $k$ is infinite and that condition (AC) is satisfied. Let $\widetilde{S}$ be the set of $s\in S$ such that $V^0(s) = V^0(S)$. If $s\in S$, let $P(s)$ be the determinant of the endomorphism of $V/V^0(S)$ defined by $r(s)$ (no. 1, Cor. 2 (i) of Th. 1).

(i) The function $s \rightarrow P(s)$ is polynomial on S. We have $\widetilde{S} =\{s\in S|P(s)\not= 0\}$; this is an open subset of S in the Zariski topology (App. 1).

(ii) $\widetilde{S}$ is non-empty, and $V^+(s) = V^+(S)$ for all $s\in \widetilde{S}$.

The fact that $s \rightarrow P(s)$ is polynomial follows from the linearity of $r$. If $s\in S, V^0(s)\supset V^0$(S), with equality if and only if $r(s)$ defines an automorphism of $V/V^0$(S), hence (i).

Now let $k'$ be an algebraic closure of $k$, and introduce $V',S', r'$ as in Prop. 6. We remark that $S'$ satisfies condition (AC) by continuation of the polynomial identity (ad $r(s_1)$)$^{2 dim V-1}r(s_2) = 0$ valid for $s_1, s_2\in S$ (no. 1, Remark ). Applying Th.1, we deduce a decomposition

with $\lambda_i\not= 0$ for $1\leq i\leq m$. For $1\leq i\leq m$, there exists a polynomial function $P_i$ non-zero on $S'$ and an integer $q_i$ such that $\lambda^{q_i}_i= P_i$ (Prop. 5). Since $k$ is infinite, there exists $s\in S$ such that $(P_1. . .P_m)(s)\not= 0$, cf. Algebra, Chap. IV, §2, no. 3, Cor. 2 of Prop. 9. Then $\lambda_i(s)\not= 0$ for all $i$, so ${V'}^0(S'{) = V'}^0(s)$ and consequently $V^0(S) = V^0(s)$ (Prop. 6), which shows that $\widetilde{S}\not=\emptyset$. If $s\in \widetilde{S}$, the fact that $V^+(S)$ is stable under $r(s)$ and is a complement of $V^0(s)$ in V implies that $V^+(S) = V^+(s)$ (Cor. 2 of Th. 1).