Proving K/F is Separable Extension

[Palindrom]

Hi all,

Suppose E/F and K/E are finite separable extensions. Prove: K/F is a separable extension.

I tried, but I'm stuck again. (Have you noticed I have one like this every week? And it seems I'm the only one that's even trying to submit these H.W. weekly...).

Anyway, that was my direction: Given $$\[<br /> \theta \in K<br /> \]$$, I know that $$\[<br /> \mu _\theta \left( x \right)<br /> \]$$ (the minimal polynomial over E) is separable. What I need to prove is that $$\[<br /> \hat \mu _\theta \left( x \right)<br /> \]$$, which is the minimal polynomial over F, is separable.
I know that $$\[<br /> \mu '_\theta \left( x \right) \ne 0<br /> \]$$, since $$\[<br /> \mu _\theta \left( x \right)<br /> \]$$is separable and therefore $$\[<br /> \left( {\mu ,\mu '} \right) = 1<br /> \]$$. I now need to prove that $$\[<br /> \hat \mu '_\theta \left( x \right) \ne 0<br /> \]$$.
...
Tried all kinds of things, didn't get me far though... Any hints would be appreciated.

 

[mathwonk]

didnt i already answer this in some detail a day or 2 ago? i.e. use the connection between separability, degree, and number of automorphisms.

 

[Palindrom]

It was something else back then (the extension E(a) when a is separable over E is separable. Here a is separable over a separable extension of E). But I think your hint might have just solved my question anyway.
Thanks a lot! (And sorry for all the trouble I'm causing you with this class. But I love it! It really is a fascinating class. I'm actually the only one, almost, that tries to solve the questions he publishes).
I'm starting to get the hang of this separable thing though. You might have noticed that's what causing me the most trouble...

 

[Palindrom]

O.K., reached a new block.

Given an E automorphism of K, I want to find [E:F] F automorphisms of K, right?
Now I already have [E:F] F automorphisms of E, since E/F is separable- but that doesn't allow me to expand the definition of my E automorphism of K. The natural way to expand it wouldn't give me even a homomorphism. (I've just checked).
So where's the catch? How can I find [K:F] F automorphisms of K?