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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?

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    What is "a prime power element"? And why do you think elements in a field are polynomials?? – 2017-02-24
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    prime power element means that the number of elements inside the field is a prime power like 8 (2 to the power of 3) – 2017-02-24
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    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. – 2017-02-24
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    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. – 2017-02-24
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    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 $R$. – 2017-02-24

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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.

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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.

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    In short: with polynomials it's all about coefficients, with polynomial functions it's all about values. – 2017-02-25
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    Absolutely. Let me add there's an isomorphism between them in characteristic $0$. – 2017-02-25
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    I'm not sure if I understand how this answers the question. – 2017-02-27
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    @Morgan Rodgers: I'm not sure either, as the O.P. question was rather confusing. I answered what seemed most likely to me, waiting for some reaction to adjust what I could explain – 2017-02-27
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There is not really a problem here with having polynomials as coefficients.

We would form, for example $\mathbb{F}_{4}$ as $\mathbb{F}_{2}[x]/\langle x^{2}+x+1\rangle$, to get a new field. It makes sense to define something like $\alpha$ to be the equivalence class of $x$ in this new field, so $\alpha$ satisfies $\alpha^{2}+\alpha+1 = 0$. So the elements of your field $\mathbb{F}_{4}$ are polynomials in $\alpha$, so $\mathbb{F}_{4} = \{0,1,\alpha, \alpha+1\}$.

If we want to think of polynomials over $\mathbb{F}_{4}$, we have an indeterminate $x$ and the elements $\{ 0, 1, \alpha, \alpha+1 \}$ of $\mathbb{F}_{4}$ are our set of possible coefficients.