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 $x^{p^n}-x$?
Best Answer
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.