# Can't follow a proof involving Prime-Power Fields

Theorem: Let $p$ be a prime and let $n\in\mathbb{Z}^{+}$. If $E$ and $E'$ are fields of order $p^{n}$, then $E\cong E'$.

Proof: Both $E$ And $E'$ have $\mathbb{Z}_{p}$ as prime fields (up to isomorphism). By Corollary 33.6, $E$ 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 $E\cong \mathbb{Z}_{p}[x] / \langle f(x)\rangle$.

I'm hoping someone can get me unstuck! Thanks!

Corollary 33.6 that is referenced states: A finite extension $E$ of a finite field $F$ is a simple extension of $F$.

Edit: The original question below has been answered by DonAntonio. I understand how Corollary 33.6 is being applied, but I don't see where the polynomial comes from. This looks like the construction of $\mathbb{Z}_{p}(\alpha)$, if $\alpha$ were a root of the polynomial $f(x)$, so I'm guessing that is related. But what if $E = \mathbb{Z}_{p}(\alpha)$ for some transcendental $\alpha$ over $\mathbb{Z}_{p}$? This isn't outlawed by the Corollary.

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## 1 Answer

It can't be $\,E=\Bbb F_p(\alpha)=\Bbb Z_p(\alpha)\,\,,\,\,\alpha$ transcendental over $\,\Bbb F_p\,$ , as then $\,E\,$ wouldn't be a finite extension of $\,\Bbb F_p\,$ and thus not even a finite field, of course.

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That sounds true. I'll take it as an exercise to verify afterwards for the moment and try to finish the proof. Thanks for the help. –  Kyle Schlitt Aug 4 '12 at 1:19
OK. This was actually theorem in a previous chapter: A finite extension field $E$ of $F$ is an algebraic extension of $F$. Thanks again. –  Kyle Schlitt Aug 4 '12 at 1:22
Anytime @Kyle , though I'd rather word that theorem as "a finite field extension is algebraic", just to make crystal clear that the other way around is pretty false: there are algebraic extensions that are infinite. –  DonAntonio Aug 4 '12 at 1:45