# Define polynomials over a finite field

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?

• What is "a prime power element"? And why do you think elements in a field are polynomials?? Feb 24 '17 at 0:07
• prime power element means that the number of elements inside the field is a prime power like 8 (2 to the power of 3) Feb 24 '17 at 0:29
• Oooh! Well, I couldn't have ever guessed that. Anyway, if you want a standard notation for that you can use $\;\Bbb F_{p^n}=\;$ the field (the unique one up to isomorphism) with $\;p^n\;$ elements, $\;p\;$ a prime. Feb 24 '17 at 0:31
• A better phrasing is "a prime power number of elements". The way you've written it sounds like you are asserting that each element of a finite field is a prime power.
– user14972
Feb 24 '17 at 0:39
• Incidentally, there's nothing wrong with polynomials having coefficients named by polynomials. Or even coefficients that are actually drawn from a polynomial ring. In fact, it is fairly useful to view the bivariate polynomial ring $R[x,y]$ as being the ring of univariate polynomials $S[y]$ over the ring $S$, where $S = R[x]$ is the ring of univariate polynomials over $R$.
– user14972
Feb 24 '17 at 0:40

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.
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.
• Absolutely. Let me add there's an isomorphism between them in characteristic $0$. Feb 25 '17 at 21:21