polynomial ring over finite field Can someone provide a proof for the following? I wrote a one page proof. I must be doing something wrong. There has to be a quicker way to prove this. Thank you ahead of time. I will highly rate for your help.


I'm curious what approach you took in your proof, as this would help in saying what you're doing wrong—though I've often seen students write very long proofs which are basically correct and elegant, but have unnecessary steps, or prove the same thing several times in different forms, et cetera.
Anyway, in principle this should be easy.  First, let's give this map a name: $\psi : \mathbb{F}_p [x] \to F$ is defined by $\psi(x) = a$.
Surjectivity is easy: each nonzero element of $F$ is some power $a^k$, and $\psi(x^k) = a^k$.  $\psi(0) = 0$, so $\psi$ is surjective.
By the theorem that's usually called something like The First Isomorphism Theorem, we have $F \cong \mathbb{F}_p [x] / I$ for some ideal $I$, namely $I = \ker{\psi}$. Since $\mathbb{F}_p [x]$ is a principal ideal domain (which probably requires the Euclidean Algorithm to prove, if you don't have it already), $I = (g(x))$ for some $g$.  Let $m = \deg{g}$.
$I$ must be a prime ideal, since its quotient is an integral domain.  This forces $g(x)$ to be irreducible; otherwise $g(x) = g'(x)g''(x) \in I$, and some polynomial of degree $<m$ would lie in $I$, contradiction.
Figuring out the degree of $g$ is immediate with some basic field theory, but we can also get it from first principles.  An element of $\mathbb{F}_p [x] / (g(x))$ is determined uniquely by its remainder on division by $g(x)$.  So the size of this ring is exactly the number of polynomials of degree $<m$, which is $p^m$ (each of $m$ coefficients is arbitrarily chosen from $\mathbb{F}_p$).
Since $| F| = p^n$, we have $p^n=p^m$, so $n=m$.
Well, this is getting close to page maybe—there are a lot of tiny things to check.  Verbosity can be a good thing. :)
The final part is to show that $g$ divides $x^q-x$.  The point is that $\psi(x^q-x) = a^q-a=0$.  But earlier, we agreed that $I=(g(x))$ was the kernel, so $x^q-x \in (g(x))$, which is what we need.
Clear?
