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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 $x^{p^n}-x$?

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2 Answers 2

up vote 15 down vote accepted

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.

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+1, this is a simpler way of proving the statement than having to deal with roots (and is an important fact, regardless). –  Zev Chonoles Aug 9 '11 at 1:37
    
Great, I completed the details: Let the set of polynomials having $\alpha$ as a zero as $ev_\alpha^{-1}(0)$. Then $f \in ev_\alpha^{-1}(0)$, so that $(f) \subseteq ev_\alpha^{-1}(0)$. Since $\mathbb{Z}_p[x]$ is a PID $(f)$ is maximal, and $ev_\alpha^{-1}(0)$ is proper since it does not contain $1\in\mathbb{Z}_p[x]$. So $(f)=ev_\alpha^{-1}(0)$. –  Gobi Aug 9 '11 at 4:09
    
@Gobi Yes! That's a very nice way to see it. –  Dylan Moreland Aug 9 '11 at 16:47

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 $E$.

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Can you give me a hint to prove that $\alpha^p$ is a root of $f$? –  Gobi Aug 9 '11 at 4:13
    
Show that, for any $f\in\mathbb{F}_p[x]$, $$\left(f(x)\right)^p=f(x^p).$$ Then the fact that $f(\alpha^p)=0$ follows from the fact that $\left(f(\alpha)\right)^p=0^p=0$. –  Zev Chonoles Aug 9 '11 at 4:15
    
Okay, but one more: How can I show that the roots $\alpha, \alpha^p, \cdots, \alpha^{p^{n-1}}$ are distinct? –  Gobi Aug 9 '11 at 4:47
    
Though I realize that this is two years later...I'll answer it anyway. –  Souparna Purohit Aug 2 '13 at 13:18
    
Suppose that $ \alpha^{p^k} = \alpha^{p^l} ,\ l<k< n.$ Then taking both sides to the power of $p^{n-k},$ we see that $ \alpha^{p^n} = \alpha^{p^{n+l-k}}.$ Note that $n+l-k < n.$ Letting $n+l-k = s,$ we see that $\alpha$ is a root of $ x^{p^s} - x,$ where $s<n.$ But this means that all the elements of the field of the $p^n$ elements that $\alpha$ is in are also roots of this above polynomial (this is because every element in that field can be expressed as a linear combination of powers of $\alpha$). –  Souparna Purohit Aug 2 '13 at 13:29

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