Two finite fields with the same number of elements are isomorphic 
Fraleigh(7ed) Theorem33.12. Let $p$ be a prime and let $n\in\mathbb{Z}^+$. If $E$ and $E'$ are fields of order $p^n$, then $E \simeq E'$.

Proof in the text: Both $E$ and $E'$ have $\mathbb{Z}_p$ as prime field, up to isomorphism. By Corollary 33.6(A finite extension $E$ of a finite field $F$ is a simple extension of $F$), $E$ is a simple extension of $\mathbb{Z}_p$ of degree $n$, so there exists an irreducible polynomial $f(x)$ of degree $n$ in $\mathbb{Z}_p[x]$ such that $E\simeq \mathbb{Z}_p[x]/ \langle f(x) \rangle$. Because the elements of $E$ are zeros of $x^{p^n}-x$, *we see that $f(x)$ is a factor of $x^{p^n}-x$ in $\mathbb{Z}_p[x]$*. Because $E'$ also consists of zeros of  $x^{p^n}-x$, we see that $E'$ also contains zeros of irreducible $f(x)$ in $\mathbb{Z}_p[x]$. Thus, because $E'$ also contains exactly $p^n$ elements, $E'$ is also isomorphic to $E\simeq \mathbb{Z}_p[x]/\langle f(x) \rangle$.
I don't know why $f(x)$ divides $x^{p^n}-x$. $E$ has a zero $\alpha$ of $f(x)$, but it doesn't need to have all the zeros of $f(x)$. So $f(x)=(x-\alpha)g(x)$ in $E[x]$ and $g(x)$ need not be splitted into linear factors. How can $f(x)$ divide $x^{p^n}-x$?
 A: If $\alpha$ is a root of $f\in\mathbb{F}_p[x]$, i.e.
$$f(\alpha)=0\in\mathbb{F}_p,$$
show that $\alpha$ is also a root of $f(x^p)$, so that $\alpha^p$ is also a root of $f$. If $f$ is irreducible of degree $n$, this implies that in fact, all of the other roots of $f$ are $\alpha^p,\alpha^{p^2},\ldots,\alpha^{p^{n-1}}$ (the key is to show that these elements are distinct). So all the roots of $f$ are in $E$.
A: You can show that $f$ divides any polynomial in $\mathbf Z_p[x]$ having $\alpha$ as a zero: the set of such polynomials is an ideal in the principal ring $\mathbf Z_p[x]$, and since $f$ is irreducible it follows that $f$ must generate this ideal.
A: Because $f(x)$ has a root $\alpha$ in $E$, we know that ${\rm gcd}(f(x),x^{p^{n}}-x) \neq 1$ in $\mathbb{Z}_{p}[x]$. For otherwise, we could write $a(x)f(x) + b(x)(x^{p^{n}}-x) = 1$ for polynomials $a(x),b(x) \in \mathbb{Z}_{p}[x]$. But evaluating this expression at $\alpha$ gives a contradiction, as the left hand side takes value $0$ at $\alpha$.
Since $f(x)$ is irreducible in $\mathbb{Z}_{p}[x]$, we must have ${\rm gcd}(f(x),x^{p^{n}}-x) = f(x)$ in $\mathbb{Z}_{p}[x]$ ( up to a constant multiple, where the constant is a non-zero element of $\mathbb{Z}_{p}$). But then $f(x)$ divides $x^{p^{n}}-x$ in $\mathbb{Z}_{p}[x]$ as required.
