Exercise 6, § 2

Let $K$ be a finite field with $q$ elements. a) In the ring $K[X_1, X_2, \ldots, X_n]$ let $a$ be the ideal generated by the $n$ polynomials

X_i^q - X_i \quad (1 \leq i \leq n). Show that if $f \in a$, then $f(x_1, x_2, ..., x_n) = 0$ for all $(x_i) \in K''$ (observe that the multiplicative group $K^*$ is of order $q - 1$). b) If $f$ is any polynomial of $K[X_1, X_2, ..., X_n]$, show that there exists precisely one polynomial $\overline{f}$ such that $\deg, \overline{f} \leq q - 1$ for $1 \leq i \leq n$ and such that $f \equiv \overline{f} \pmod{a}$. We thus have $\deg \overline{f} \leq \deg f$; if $f$ is such that $f(x_1, x_2, ..., x_n) = 0$ for each $(x_i) \in K''$, $f$ belongs to the ideal $a$, which is thus the inverse image of 0 under the homomorphism $f \mapsto \overline{f}$ (use Exercise 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 the sum of the total degrees of the $f_i$ is < $n$. Show that there exists an element $(x_1, x_2, ..., x_n) \in K''$ distinct from $(0, 0, ..., 0)$ such that

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