Exercise 12.3.5

Proof. Suppose that $n=1$, and $g\in F[t_1],g\ne 0$. Then $g$ has at most $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(g)\le N_1$ roots. The cardinality of $\{0,1,\dots ,N_1\}$ is $N_1+1$, therefore some integer $a_1\in \{0,1,\dots ,N_1\}$ is not a root of $g$. The property is so established if $n=1$.

Reasoning by induction, suppose that the property is true for $n-1$ variables $t_1,\cdots ,t_{n-1}$, and let $g\in F[t_1,\dots ,t_n]$ a nonzero polynomial. Write

where $d$ is the partial degree of $g$ relative to the variable $t_n$.

If $d=0$, then $g=c_0\ne 0$, and the induction hypothesis gives $(a_1,\dots ,a_{n-1})$, with $0\le a_i\le N_i$ for each $i$, $0\le i\le n-1$, such that $c_0(a_1,\dots ,a_{n-1})\ne 0$. If we take $a_n=0$, then $(a_1,\dots ,a_n)\in A$ is such that $g(a_1,\dots ,a_n)\ne 0$.

If $d>0$, the induction hypothesis gives $(a_1,\dots ,a_{n-1})$, with $0\le a_i\le N_i$ for each $i$, $0\le i\le n-1$, such that $c_d(a_1,\dots ,a_{n-1})\ne 0$. Then $h(t_n)=g(a_1,\dots ,a_{n-1},t_n)$ is a polynomial in $t_n$ with degree $d\le N_n$, so with the same argumentation as in the case $n=1$, there exists some $a_n$, $0\le a_n\le N_n$ such that $h(a_n)\ne 0$. Therefore $(a_1,\dots a_n)\in A$ and $g(a_1,\dots ,a_n)\ne 0$. The induction is done. (b) Now suppose that $F$ has characteristic $p$ and is infinite. A nonzero polynomial $p$ in $F[x]$ has at most $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(p)$ roots. The same induction gives an element $a_n$ in the infinite field which is not a root of the polynomial, so the property is true in any infinite field. (c) If $F={𝔽}_p$, and $g=t_1^p-t_1$, then $g\ne 0$ but all elements $a_1$ in ${𝔽}_p$ satisfy $g(a_1)=0$ (Fermat’s little Theorem).

Another such counterexample with $n=2$ is the nonzero polynomial $g=t_1^p{t}_2-t_1{t}_2^p$ in ${𝔽}_p[t_1,t_2]$, such that $g(a_1,a_2)=0$ for all $a_1,a_2\in {𝔽}_p$. □