Every finite extension of a finite field is separable

Solution 1:

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

Solution 2:

$E=F(\alpha)$ 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).