Kernel of "polynomial evaluation map" for finite fields

Question

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?

Answer 1

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 $I$.

Answer 2

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

$$\mathbf{F}_q[X_1, \ldots, X_n]/(X_1^q - X_1, \ldots, X_n^q - X_n) \cong \prod_{v \in \mathbf{F}_q^n} \mathbf{F}_q$$