# Separable extension..

• Sep 2nd 2009, 09:35 PM
ynj
Separable extension..
Let $\displaystyle F$ be a field with characteristic 0.$\displaystyle E$is a finite field extension of $\displaystyle F$. Prove that $\displaystyle E$is a separable extension...
I know that for an $\displaystyle \alpha$, if the minimal polynomial $\displaystyle f(x)$ splits on $\displaystyle E$, then $\displaystyle f(x)$is separable on $\displaystyle E$. But why $\displaystyle f(x)$splits?
• Sep 3rd 2009, 11:57 AM
ThePerfectHacker
Quote:

Originally Posted by ynj
Let $\displaystyle F$ be a field with characteristic 0.$\displaystyle E$is a finite field extension of $\displaystyle F$. Prove that $\displaystyle E$is a separable extension...
I know that for an $\displaystyle \alpha$, if the minimal polynomial $\displaystyle f(x)$ splits on $\displaystyle E$, then $\displaystyle f(x)$is separable on $\displaystyle E$. But why $\displaystyle f(x)$splits?

You are misunderstanding what "seperable" means. An irreducible polynomial $\displaystyle f(x)\in F[x]$ is "seperable" over $\displaystyle F$ iff $\displaystyle f(x)$ has no repeated roots in its splitting field. Now $\displaystyle \alpha \in E$ is seperable iff the minimal polynomial for $\displaystyle \alpha$ is seperable over $\displaystyle F$. This does not mean that the miniminal polynomial must split over $\displaystyle E$.