Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
add comment

2 Answers

up vote 4 down vote accepted

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 $\,p^nx^{p^n-1}-1=-1\neq 0\pmod p\,$

share|improve this answer
    
why the multiplicative group has $p^n -1$ elements? which element is excluded in $F$? Thank you for your answer :) –  user42912 Dec 17 '12 at 3:29
1  
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. –  DonAntonio Dec 17 '12 at 3:31
    
yes, of course, gracias :) –  user42912 Dec 17 '12 at 3:54
add comment

$E=F(\alpha)$ is a 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).

share|improve this answer
    
$x^{p^n} -x$ is irreducible over $F$? Thank you for your answer :) –  user42912 Dec 17 '12 at 3:54
    
It can't be irreducible as $$x^{p^n}-x=x\left(x^{p^{n-1}}-1\right)$$ –  DonAntonio Dec 17 '12 at 3:56
    
@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? –  user42912 Dec 17 '12 at 3:59
1  
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. –  DonAntonio Dec 17 '12 at 4:17
2  
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. –  JSchlather Dec 17 '12 at 4:27
show 3 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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