Proving K/F is Separable Extension

  • Thread starter Palindrom
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In summary, the student is trying to find F automorphisms of K, but is having trouble doing so because E/F is separable.
  • #1
Palindrom
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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 [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. :smile:
 
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  • #2
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.
 
  • #3
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...
 
  • #4
O.K., reached a new block. :biggrin:

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?
 

1. What is a separable extension in the context of field theory?

A separable extension in field theory is a type of field extension where the minimal polynomial of an element in the extension field has no repeated roots. In other words, the minimal polynomial can be factored into distinct linear factors in the base field.

2. Why is it important to prove that a field extension is separable?

Proving that a field extension is separable is important because it guarantees that the extension is a normal extension, which means that every irreducible polynomial in the base field has all of its roots in the extension field. This is a fundamental property of field extensions that allows for the use of Galois theory and other important techniques in abstract algebra.

3. How do you prove that a field extension is separable?

To prove that a field extension K/F is separable, we need to show that every irreducible polynomial in F[x] has all of its roots in K. This can be done by using the fact that a polynomial is separable if and only if its derivative is relatively prime to the polynomial itself. Therefore, we can check if the derivative of each irreducible polynomial is relatively prime to the polynomial itself, and if so, we can conclude that the extension is separable.

4. Can a field extension be both separable and inseparable?

No, a field extension cannot be both separable and inseparable. This is because these two properties are mutually exclusive - a field extension is either separable or inseparable. If a field extension is separable, then it cannot be inseparable, and vice versa.

5. Are all finite field extensions separable?

Yes, all finite field extensions are separable. This is a result of the fact that every finite extension is algebraic, and every algebraic extension is separable. Therefore, any finite field extension can be proven to be separable by showing that every irreducible polynomial in the base field has all of its roots in the extension field.

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