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I'm trying to prove that every finite extension of a finite field is separable. I found a solution on internet which says:

Let $F$ be a finite field and $E$ be an extension of $F$ having $p^n$ elements. Then $E=F(\alpha)$, where $\alpha \in E$ and so $\alpha^{p^n} -\alpha=0$. This implies $\alpha$ is a separable element, and hence $F(\alpha)$ is a separable extension of F.

I don't understand why $\alpha^{p^n} -\alpha=0$ and why $\alpha$ is a separable element, I need help.

Thanks

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Since any finite subgroup of the multiplicative group of any field is cyclic, if a field has $\,p^n\,$ elements, then its multiplicative group has $\,p^n-1\,$ elements, and thus for any non-zero element $\,\alpha\,$ in the field,

$$\alpha^{p^n-1}=1\Longrightarrow \alpha^{p^n}=\alpha\Longleftrightarrow \alpha^{p^n}-\alpha=0$$

Note that the above equality is true also for the zero element in the field.

Thus, any element in a field with $\,p^n\,$ elements is a root of $\,x^{p^n}-x\,$ , and this pol. is separable since its derivative is $\,p^nx^{p^n-1}-1=-1\neq 0\pmod p\,$

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  • $\begingroup$ why the multiplicative group has $p^n -1$ elements? which element is excluded in $F$? Thank you for your answer :) $\endgroup$ – user42912 Dec 17 '12 at 3:29
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    $\begingroup$ El cero, of course. You can do this with any field (reals, rationals, complex, etc.), but only in case you have a finite subgroup of it you can be sure it will be cyclic. $\endgroup$ – DonAntonio Dec 17 '12 at 3:31
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    $\begingroup$ yes, of course, gracias :) $\endgroup$ – user42912 Dec 17 '12 at 3:54
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$E=F(\alpha)$ is a vector space over the field of $p$ elements, where $p$ is the characteristic of $F$, so $E$ has $p^n$ elements for some positive integer $n$. That means that that multiplicative group of $E$ has $p^n-1$ elements, so $\alpha^{p^n-1}=1$, and $\alpha^{p^n}-\alpha=0$.

This extension is separable because the minimal polynomial of $\alpha$ has no multiple roots: the minimal polynomial of $\alpha$ divides $X^{p^n}-X$, and $X^{p^n}-X$ has no multiple roots because it does not have any roots in common with its derivative, $p^nX^{p^n-1}-1=-1$ (which has no roots).

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  • $\begingroup$ $x^{p^n} -x$ is irreducible over $F$? Thank you for your answer :) $\endgroup$ – user42912 Dec 17 '12 at 3:54
  • $\begingroup$ It can't be irreducible as $$x^{p^n}-x=x\left(x^{p^{n-1}}-1\right)$$ $\endgroup$ – DonAntonio Dec 17 '12 at 3:56
  • $\begingroup$ @DonAntonio but it has a root in $F$, $1\in F$ is a root of this polynomial, then it can't be irreducible. Am I wrong? $\endgroup$ – user42912 Dec 17 '12 at 3:59
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    $\begingroup$ No, and that's precisely what I said, Rafael! Actually, ALL the elements of $\,F\,$ are that polynomial's roots: in fact, this is one fancy possible way to define $\,F\,$, so yes: the polynomial is not irreducible(i.e., it is reducible), just as I wrote in the comment above. $\endgroup$ – DonAntonio Dec 17 '12 at 4:17
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    $\begingroup$ It's worth noting that $X^{p^n}-X$ has degree $p^n$ so it has at most $p^n$ roots. Since each element of $\mathbb F_{p^n}$ is a root it follows that it has no repeated roots. $\endgroup$ – JSchlather Dec 17 '12 at 4:27

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