Exercise 6, § 2

Let $K$ be a finite field having $q$ elements. (a) In the ring $K[X_1, X_2, ..., X_n]$, let $a$ be the ideal generated by the $n$ polynomials $X_i^q - X_i$ (1 \leq i \leq n). Now prove that if $f \in a$, one has $f(x_1, x_2, ..., x_n) = 0$ for every $(x_i) \in K^n$ (remark that the multiplicative group $K^*$ is of order $q - 1$).

(b) If $f$ is any polynomial of $K[X_1, X_2, ..., X_n]$, now prove that there exists one and only one polynomial $\overline{f}$ such that $\deg_i \overline{f} \leq q - 1$ for $1 \leq i \leq n$, and such that $f \equiv \overline{f}$ (mod. $a$). We therefore have $\deg \overline{f} \leq \deg f$; if $f$ is such that $f(x_1, x_2, ..., x_n) = 0$ for every $(x_i) \in K^n$, $f$ belongs to the ideal $a$, which is thus the inverse image of 0 by the homomorphism $f \mapsto \overline{f}$ (use exerc. 1).

(c) Let $f_1, f_2, ..., f_m$ be non-zero polynomials of $K[X_1, X_2, ..., X_n]$, such that $f_i(0, 0, ..., 0) = 0$ ($1 \leq i \leq m$), and such that the sum of the total degrees of the $f_i$ is $< n$. Now prove that there exists an element $(x_1, x_2, ..., x_n) \in K^n$, distinct from $(0, 0, ..., 0)$, and such that

(Remark that, if this were not so, the polynomial $\prod_{i=1}^m (1 - f_i^{q^{-1}})$ would be congruent modulo $a$ to the polynomial $\prod_{i=1}^n (1 - X_i^{q^{-1}})$, and use b).)