Lemma 1

— Let L be a field, V a vector space over L and u an endomorphism of V such that $u^p = u$. For each $i \in \mathbf{F}_p$ let $V_i$ be the kernel of $u - i$. Then we have

For each $i \in \mathbf{F}_p$ denote by $P_i(X)$ the polynomial $- \prod_{j \neq i} (X - j)$. We have

by Formula (2) ($V$, p. 94). On deriving the quoted formula (2), we find

The formulae (8) and (9) show that the endomorphism $P_i(u)$ of $V$ maps $V$ into $V_i$ and that $\sum_{i \in \mathbf{F}_p} P_i(u) = 1$, whence $V = \sum_{i \in \mathbf{F}_p} V_i$. It remains to show that the sum is direct. For each $i \in \mathbf{F}_p$ let $v_i \in V_i$; it is clear that $P_i(u)$ annihilates $v_j$ for all $j \neq i$ and we have $P(u)v_i = av_i$ with $a = -\prod_{n \in \mathbf{F}_p^*} n \neq 0$. The relation $\sum_{i \in \mathbf{F}_p} v_i = 0$ thus implies $v_i = 0$ for all $i \in \mathbf{F}_p$ and so the sum is direct.