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


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.

  • $\begingroup$ +1, this is a simpler way of proving the statement than having to deal with roots (and is an important fact, regardless). $\endgroup$ – Zev Chonoles Aug 9 '11 at 1:37
  • $\begingroup$ 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 $(f)=ev_\alpha^{-1}(0)$. $\endgroup$ – Gobi Aug 9 '11 at 4:09
  • $\begingroup$ @Gobi Yes! That's a very nice way to see it. $\endgroup$ – 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 $E$.

  • $\begingroup$ Can you give me a hint to prove that $\alpha^p$ is a root of $f$? $\endgroup$ – Gobi Aug 9 '11 at 4:13
  • $\begingroup$ 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 $\left(f(\alpha)\right)^p=0^p=0$. $\endgroup$ – Zev Chonoles Aug 9 '11 at 4:15
  • $\begingroup$ Okay, but one more: How can I show that the roots $\alpha, \alpha^p, \cdots, \alpha^{p^{n-1}}$ are distinct? $\endgroup$ – Gobi Aug 9 '11 at 4:47
  • $\begingroup$ Though I realize that this is two years later...I'll answer it anyway. $\endgroup$ – user24503 Aug 2 '13 at 13:18
  • $\begingroup$ 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 $\alpha$). $\endgroup$ – user24503 Aug 2 '13 at 13:29

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.