Lemma 2

Let $q$ be the cardinal of K. For any integer $m\geqslant 0$, set $S_m=$ $\sum_{x\in K}x^m$. We have $S_m=-1$ if $m$ is a nonzero multiple of $q-1$ and $S_m= 0$ in all other cases.

Recall that $0^0= 1$ (I, §2, No. 1, p. 14). Suppose that $m$ is a multiple of $q-1$. Since the abelian group $K^*$ has order $q-1$, we have $x^m= 1$ for every $x\in K^*$ and $S_m= 0^m+ (q-1)\cdot 1$, which gives the assertion in this case.

Suppose that $m$ is not a multiple of $q-1$. Set $\chi (x) =x^m$ for $x\in K^*$. Since the multiplicative group $K^*$ is cyclic of order $q-1$ (V, §12, No. 1, p. 93, Proposition 1), there exists an element $a$ of $K^*$ such that $\chi(a) \neq 1$. By Lemma 1 applied to $G = K^*$, we have

this gives Lemma 2.

Let us now prove Theorem 1. Let $\mathbf{x} = (x_1, \ldots, x_n)$ be an element of $K^n$. We have $1 - f_i(\mathbf{x})^{q-1} = 0$ if $f_i(\mathbf{x}) \neq 0$ and $1 - f_i(\mathbf{x})^{q-1} = 1$ if $f_i(\mathbf{x}) = 0$. Set $P = \prod_{i=1}^r (1 - f_i^{q-1})$. We have

Let us expand the polynomial $P$ as $\sum_{\alpha \in \mathbf{N}^n} c_\alpha X^\alpha$; by assumption, it has degree $< (q-1)n$. Let $\alpha$ be an element of $\mathbf{N}^n$ such that $c_\alpha$ is nonzero. Since we have $\alpha_1 + \cdots + \alpha_n < (q-1)n$, there exists an integer $\ell$ such that $1 \leq \ell \leq n$ and $0 \leq \alpha_\ell < q-1$. By Lemma 2, we then have $\sum_{x \in K} x^{\alpha_\ell} = 0$, and therefore We consequently have Now, by formula (1), we have $\sum_{\mathbf{x} \in K^n} P(\mathbf{x}) = \operatorname{Card}(\mathbf{Z}) \cdot 1$, and therefore $\operatorname{Card}(\mathbf{Z}) \cdot 1 = 0$, which means that $\operatorname{Card}(\mathbf{Z})$ is divisible by $p$.