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

$$E$$ 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 $$[E:F]$$
• What is your definition of "separable field extension" if it doesn't make the claim trivial? – Hagen von Eitzen Jun 16 at 13:22
• 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 $[E:F]$. – user8785084 Jun 16 at 13:30

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.