$E/F$ finite and separable, $\alpha \in E$, then the irreducible polynomial of $\alpha$ over $F$ is separable. With Lang's definition.

Question

$E$ and $F$ have an arbitrary characteristic.

I'm stuck in this problem, i had some ideas, but they led me to nothing.

If anyone could give me some hint, some direction on how to do it, i would appreciate a lot.

The Definition of Separable extension is the Lang's definition:

• $E/F$ is separable if the number of possible extensions of an injective homomorphism $\sigma :F \rightarrow A$ (where's $A$ is an algebraically closed field) to an injective homomorphism of $E$ into $A$ is equal to $[E:F]$

Answer 1

As every subextension of a finite separable extension is separable(!), we may assume that $E=F[\alpha]$. Let $f$ be the irreducible polynomial of $\alpha$. Any extension of $\sigma\colon F\to A$ to $E$ is uniquely determined by where we send $\alpha$. As we must send $\alpha$ to a root of $f$ in $A$ and $[E:F]=\deg f$, it follows that $f$ has $\deg f$ roots in $A$. Hence $f$ is separable.