(Fraleigh Theorem 33.12)
Let $p$ be a prime. If $E$ and $E'$ are fields of order $p^n$, then $E \simeq E'$.
Proof in the text:
$E$ and $E'$ both have $\mathbb{Z}_p$ as prime field, up to isomorphism. Then $E$ is a simple extension of degree $n$ and thus isomorphic to $\mathbb{Z}_p[x]/\langle f(x) \rangle$. Since elements of $E$ are zeros of $\smash{x^{p^n} – x}$, $f(x)$ is a factor of $\smash{x^{p^n} – x}$. Because $E'$ also contains zeros of $\smash{x^{p^n} – x}$, $E'$ 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 $\mathbb{Z}_p[x]/\langle f(x) \rangle$.
I don't know how to get to "$E'$ is isomorphic to $\mathbb{Z}_p[x]/\langle f(x) \rangle$" from the facts that $E'$ contains the zeros of $f(x)$ and has $p^n$ elements.
Let $E' = \mathbb{Z}_p(\alpha')$, with $\alpha' \in E'$ algebraic over $\mathbb{Z}_p$. I thought I needed to show that $f(\alpha') = 0$. But I only know there is a zero of $f(x)$ in $E'$, not necessarily $\alpha'$, how do I proceed?
Best Answer
From the first part of the proof we know three things:
Now, let $\alpha \in E'$ be a root of $f$. Since $f$ is irreducible, $f$ is (perhaps up to a non-zero constant) the minimal polynomial of $\alpha$ over $\mathbb{Z}_p$. Since $f$ has degree $n$, the subfield $\mathbb{Z}_p(\alpha) \subseteq E'$ generated by $\alpha$ has degree $n$ over $\mathbb{Z}_p$. Therefore, $\mathbb{Z}_p(\alpha)$ must have exactly $p^n$ elements. But $E'$ also has $p^n$ elements, and contains $\mathbb{Z}_p(\alpha)$. Therefore we get $E' = \mathbb{Z}_p(\alpha) = \mathbb{Z}_p[X]/\langle f\rangle$.
I hope this helps!