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 have been wrestling with the following problem for the past few hours, but I have made no progress whatsoever, namely:

Let $K/k$ be an algebraic extension with characteristic $p>0$ and let $\alpha \in K$. Show that $\alpha$ is separable over $k$ if and only if $k(\alpha^p)=k(\alpha)$. Conclude that every field extensions between finite fields is separable.

I wish I had progress to post, but, as of yet, I have made no progress beyond setting up the problem.

Could someone give a hint to push me in the right direction?

share|improve this question
2  
Please see the following link: math.stackexchange.com/questions/126242/… –  user38268 May 9 '12 at 14:25
    
To those who wish to vote to close: Please do not close this question as the OP has an extra bit at the end that was different from mine. I just realised this but it is a bit too late to retract my own close vote. –  user38268 May 9 '12 at 14:55
add comment

1 Answer 1

up vote 4 down vote accepted

Ok let us see how to prove one direction, which I did not post in my question in that link. We would like to prove that if $\alpha$ is not separable over $k$, in other words if the minimal polynomial of $\alpha$ over $k$ is not separable then $k(\alpha^p) \subsetneqq k(\alpha)$. Now this is tantamount to proving that $\Big[k(a) : k(a^p)\Big] > 1$. Now

$$\Big[k(a) : k(a^p)\Big] = \frac{\Big[k(a):k\Big]}{\Big[k(a^p):k \Big]}. $$

The numerator on the right hand side is the degree of $f$ while the denominator is the degree of $g$, the minimal polynomial of $a^p$ over $F$. However since $f$ is not separable it is a polynomial in $x^p$. This is also saying that $f$ has $a^p$ as a root so that $g$ must divide $f$. It follows from here that $\deg g > \deg f$ so that

$$\Big[k[a]:k[a^p]\Big] > 1.$$

Can you prove the other direction?

Now for the last part of the problem in proving why every finite extension of a finite field is separable, we want that given any $\alpha$ in our extension $K$, the minimal polynomial $f$ of $\alpha$ over $K$ is separable. By your problem above, if we can show that

$$k(\alpha) = k(\alpha^p)$$

then we are done. Now one subset inclusion is already clear. For the other inclusion, we want to show that $k(\alpha) \subseteq k(\alpha^p)$. Now suppose that $[K:\Bbb{Z}/p\Bbb{Z}] = n$. This has to be finite in the first place because $K/k$ is a finite extension and $k/\Big(\Bbb{Z}/p\Bbb{Z}\Big)$ is a finite extension too. Then $K$ has $p^n$ elements so that its multiplicative group has $p^n-1$ elements. It follows by Lagrange's Theorem that

$$\alpha^{p^n - 1} = 1$$

so that for all $\alpha \in K$, we have that $\alpha^{p^n} = \alpha$. But then the left hand side can be written as $(\alpha^p)^{p^{n-1}}$ showing that $\alpha \in k(\alpha^p)$. Hence $k(\alpha) = k(\alpha^p)$ and the result follows.

share|improve this answer
add comment

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.