Proposition 5

Assume that condition (AC) is satisfied, and let $\lambda : S\rightarrow k$ be such that $V^{\lambda}(S)\not= 0$. If $k$ is of characteristic 0, the map $\lambda$ is linear. If $k$ is of characteristic $p\not= 0$, there exists a power $q$ of $p$ dividing dim $V^{\lambda}(S)$, and a homogeneous polynomial function $P : S\rightarrow k$ of degree $q$, such that $\lambda (s)^q= P(s)$ for all $s\in S$.

Since $V^{\lambda}(S)$ is stable under $r(S)$ (Lemma 1 and formula (1) of no. 1), we can assume that $V = V^{\lambda}(S)$. Let $n=$ dim V. Thus, for $s\in S$,

det(X $-r(s)$) $= (X-\lambda (s))^n$.

On the other hand, the expansion of the determinant shows that

det(X $-r(s)$) $= X^n+a_1(s)X^{n-1}+\cdots +a_i(s)X^{n-i}+\cdots$

where $a_i: S\rightarrow k$ is a homogeneous polynomial function of degree $i$. Write $n=qm$ where $q$ is a power of the characteristic exponent of $k$ and $(q, m) = 1$. Then $(X-\lambda (s))^n= (X^q-\lambda (s)^q)^m$; hence $-m\lambda (s)^q=a_q(s)$, which implies the result.