- #1

Palindrom

- 263

- 0

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 [tex]\[

\theta \in K

\]

[/tex], I know that [tex]\[

\mu _\theta \left( x \right)

\]

[/tex] (the minimal polynomial over E) is separable. What I need to prove is that [tex]\[

\hat \mu _\theta \left( x \right)

\]

[/tex], which is the minimal polynomial over F, is separable.

I know that [tex]\[

\mu '_\theta \left( x \right) \ne 0

\]

[/tex], since [tex]\[

\mu _\theta \left( x \right)

\]

[/tex]is separable and therefore [tex]\[

\left( {\mu ,\mu '} \right) = 1

\]

[/tex]. I now need to prove that [tex]\[

\hat \mu '_\theta \left( x \right) \ne 0

\]

[/tex].

...

Tried all kinds of things, didn't get me far though... Any hints would be appreciated.