Kernel of "polynomial evaluation map" for finite fields For a field $k$ and $n$ a natural number (at least $1$) define the 'evaluation at a point' $k$-homomorphism
$$
\varphi_{k, n} : k[X_1, \ldots, X_n] \to k^{k^n} : f \mapsto (f(a))_{a \in k^n}.
$$
One easily shows by induction on $n$ that for $f \in k[X_1, \ldots, X_n]$ with $\deg_{X_i}(f) < \lvert k \rvert$ for all $i = 1, \ldots, n$ either $f = 0$ or $\varphi_{k, n}(f) \neq 0$. In particular it follows that $\varphi$ is injective if $k$ is infinite. If $k$ is the finite field with $q$ elements, it follows that the kernel of $\varphi_{k, 1}$ is the ideal generated by $X^q - X$.
Can a description of the kernel of $\varphi_{k, n}$ be given for finite fields $k$ and $n > 1$? Clearly it contains $X_i^q - X_i$ for all $i$, but is it the ideal generated by those polynomials?
 A: The kernel for $k=\Bbb F_q$ is the ideal $I$ generated by $X_i^q-X_i$ for all $i$.
Modulo $I$ each polynomial is congruent to one having degree $\le q-1$
in all variables. These polynomials form a vector space $V$ of dimension
$q^n$. Every map from $k^n$ to $k$ is represented by an element of $V$
and as $|V|=q^{q^n}$ then the only element of $V$ representing the
zero map is the zero polynomial. So the kernel of the "universal evaluation" is indeed $I$.
A: Here's another proof. Observe that
$$ \mathbf{F}_q[X]/(X^q - X) \cong \prod_{a \in \mathbf{F}_q}\mathbf{F}_q$$
where the $a$-th component is "evaluation at $a$". Now, for finite $I$,
$$ \begin{align}\left( \prod_{i \in I} \mathbf{F}_q \right) [X]/(X^q - X)
&\cong \prod_{i \in I} \mathbf{F}_q [X]/(X^q - X)
\\&\cong \prod_{i \in I} \prod_{a \in \mathbf{F}_q}\mathbf{F}_q
\\ &\cong \prod_{i \in I \times \mathbf{F}_q} \mathbf{F}_q
\end{align}
$$
Repeating $n$ times gives
$$\mathbf{F}_q[X_1, \ldots, X_n]/(X_1^q - X_1, \ldots, X_n^q - X_n) 
\cong \prod_{v \in \mathbf{F}_q^n} \mathbf{F}_q$$
