# Polynomials over a finite field vs polynomial functions on a finite field

Polynomials are formal sums. So in particular, $\mathbb{K}[x]$ is infinite-dimensional over $\mathbb{K}$, even if e.g. $\mathbb{K} = \mathbb{F}_2$, the field with two elements. This is true even though e.g. $x^2 – x$ is the zero function on $\mathbb{F}_2$, as you can check by substituting $0$ and $1$ for $x$.

Polynomial functions are functions e.g. on the field itself (or, more generally, on some object that can be locally parameterized by the field). But let’s simply consider $\mathrm{Poly}(\mathbb{K})$, the set of polynomial functions from the field to itself. $\mathrm{Poly}(\mathbb{K})$ is a ring with addition and multiplication given point-wise on function values. Any polynomial can be considered a polynomial function in the obvious manner, and this defines a surjection of rings:
$$\pi: \mathbb{K}[x] \to \mathrm{Poly}(\mathbb{K}).$$
In the case where $\mathbb{K}$ is a finite field, $\mathrm{Poly}(\mathbb{K})$ is a proper quotient of $\mathbb{K}[x]$ i.e. $\ker \pi \ne 0$. To see this, consider the multiplicative group $\mathbb{K}^\times$. Then $|\mathbb{K}^\times| = q – 1$, where $q = |\mathbb{K}|$. Therefore $a^{q-1} = 1$ for any $a \in \mathbb{K}^\times$, and so $a^q – a = 0$ for any $a \in \mathbb{K} = \mathbb{K}^\times \sqcup { 0 }$. We’ve shown therefore that the non-zero polynomial $x^q – x$ is in $\ker \pi$, since it is maps to the zero function on $\mathbb{K}$. This is just the generalization of our observation for $\mathbb{F}_2$, above.

In fact, we’ve found a generator for the kernel, i.e. $\ker \pi = \left\langle x^q – x \right\rangle$. One way to see this is to check that the polynomial functions defined by first $q-1$ monomials are linearly independent as functions on $\mathbb{K}$, which can be done using the Vandermonde matrix and the $q-1$ distinct non-zero field elements at your disposal. Another way to see this, since we are working over a finite field, is to simply count the elements of the quotient $\mathbb{K}[x] / {\left\langle x^q – x \right\rangle}$ and of $\mathrm{Poly}(\mathbb{K})$. There are clearly $q^q$ elements in the quotient, but what about $\mathrm{Poly}(\mathbb{K})$? It turns out that $\mathrm{Poly}(\mathbb{K})$ consists of all functions $\mathbb{K} \to \mathbb{K}$. To see this, given any function on $\mathbb{K}$, just use Lagrange interpolation to build yourself a polynomial of degree $\lt q$ that matches the function at all points. There are $q^q$ functions $\mathbb{K} \to \mathbb{K}$, and so that’s the size of $\mathrm{Poly}(\mathbb{K})$.
Thus the quotient $\mathbb{K}[x] / {\left\langle x^q – x \right\rangle}$ has the same number of elements as $\mathrm{Poly}(\mathbb{K})$. But we have a chain of surjections
$$\mathbb{K}[x] / {\left\langle x^q – x \right\rangle} \to \mathbb{K}[x] / {\ker \pi} \to \mathrm{Poly}(\mathbb{K}),$$
so $\ker \pi = \left\langle x^q – x \right\rangle$.

On the other hand, if the field $\mathbb{K}$ is infinite, then $\ker \pi$ is zero, i.e. polynomials over $\mathbb{K}$ and polynomial functions on $\mathbb{K}$ are isomorphic. To see this just show that the set of all monomials is linearly independent, using the Vandermonde matrix and the endless supply of distinct non-zero elements of $\mathbb{K}$.