Algebraic Extensions and Separability 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?
 A: 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.
