Define polynomials over a finite field

Question

I have a very basic question in number theory. How could we define a polynomial over a finite field which has "prime power" elements? As far as I know, if a field has prime power elements, it will include polynomials as field elements. But if we try to define a polynomial over this field, we will get a polynomial for which the coefficients are also polynomials (field elements).

I'm confused. Could someone give me a simple example?

Answer 1

Not sure of what you mean by prime power elements, but I'm afraid you're confusing polynomials and polynomial functions.

For instance, let's consider the prime field $\mathbf F_p=\mathbf Z/p\mathbf Z$, and the polynomial ring $\mathbf F_p[X]$. In this ring, the non-zero polynomial $X^p-X$ induces the polynomial function \begin{align}f\colon\mathbf F_p&\longrightarrow \mathbf F_p\\x&\longmapsto x^p-x, \end{align} which is $0$ by Little Fermat.

Answer 2

It may help to understand that there is more than one way of thinking about a polynomial. One way that you may be more familiar with is as a function that transforms the elements of a field. Consider for example the polynomial $p(x) = x^2 + 1$ over the field $\mathbb{F}_2$ (we'll consider prime powers later). As a function, $p(0) = 1$ and $p(1) = 0$. We could also consider the polynomial $q(x) = x + 1$ over $\mathbb{F}_2$. As a function, $q$ corresponds to the same mapping as $p$. However, as polynomials, we cannot say that $p(x) = q(x)$ because they have different coefficients.

In a similar fashion, a polynomial over a field of prime power order is defined by its coefficients. To see this, it may help to use a different indeterminate to write out the coefficients if you must write them explicitly. For an arbitrary example, you might write the polynomial where the coefficient of $x^2$ is $y + 2$ and the constant term is $3$ over $\mathbb{F}_{25}$ as $(y + 2) x^2 + 3$. In this context, this would be a polynomial of a single variable because $y + 2$ is a coefficient and not a second indeterminate. You cannot "simplify" a polynomial by mixing coefficients with indeterminates. Conceptually, we could also define the polynomial as the vector $(y + 2, 0, 3)$ to make the distinction even clearer.