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.

 A: 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?
A: Let $\varphi: \mathbb{F}_p\lbrack x\rbrack \to \mathbb{F}_q$ ($= F$ in your notation) be the alleged homomorphism from the prompt (i.e., $\varphi: f(x) \mapsto f(a)$). Here are some notes ...


*

*Showing $\varphi$ is a homomorphism is very quick. I'll leave you to it.

*For surjectivity, for each $b \in \mathbb{F}_q$ find a polynomial $f(x) = \alpha x+\beta$ with $f(a) = b$.

*$f(x) \in \ker\varphi \iff a$ is a root of $f(x)$. This tells you all you need to know to find an irreducible factor $g(x)$ of $f(x)$ and, accordingly, a maximal ideal $I$ containing $f(x)$.

*The remaining two statements are seen, respectively, by applying the First Isomorphism Theorem and by recalling the relationship between $x^q - x$ and $\mathbb{F}_q$.

A: It is quite immediate that $f\mapsto f(a)$ is a ring homomorphism: The polynomial ring $A[x]$ is defined by a universal property, namely that a ring homomorphism $A[x]\to B$ "is the same as" a ring homomorphism $A\to B$ and an element $b\in B$ (the image of $x$). Here we take the zero homomorphism $\mathbb F_p\to F$ and $a\in F$.
Since $a$ generates $F^\times$ and the image of a ring homomorphism is closed under multiplication, at least $F^\times$ is in the image. But as also $0\mapsto 0$, the homomorphism is in fact onto.
As $\mathbb F_p[x]$ is a principle ideal domain, it is clear that the kernel is a principle ideal and can be written as $(g)$ for some $g\in\mathbb F_p[x]$.
As $1\mapsto 1$, we have $1\notin(g)$, i.e. $\deg g\ge 1$. If we have $g=g_1g_2$ with $g_1g_2\in\mathbb F_p[x]$, then $g\mapsto 0$ implies $g_q\mapsto 0$ or $g_2\mapsto 0$ ($F$ is a field, hence has no zero divisors), i.e. one of the factors is a multiple of $g$. Thus $g$ is irreducible.
By the isomorphism theoprems, $\mathbb F_p[x]/(g)\cong F$.
If $d=\deg g$, note that $1, x, \ldots, x^{d-1}$ for a basis of $F_p[x]/(g)$ so it has $p^d$ elements. By comparison, $d=n$.
