Algebraic and normal field extensions.

[Artusartos]

Homework Statement

Let K be an algebraic field extension of a field F, and let L be a subfield of K such that F \subseteq L and L is normal in F. Show that if \sigma is an automorphism of K over F, then \sigma(L)=L.

Homework Equations

The Attempt at a Solution

I've been thinking about this for a while, but I couldn't really prove anything. So I'm just looking for a hint that will help me get started...

 

[micromass]

What is your definition of a normal extension?

 

[Artusartos]

What is your definition of a normal extension?

If f \in F[x] has one root in L, then it splits over L.

 

[micromass]

Take an element $a\in L$. Use that $K$ is algebraic to find a polynomial $f\in F[x]$ such that $f(a)=0$. What can you tell about $f(\sigma(a))$?

 

[Artusartos]

Take an element $a\in L$. Use that $K$ is algebraic to find a polynomial $f\in F[x]$ such that $f(a)=0$. What can you tell about $f(\sigma(a))$?

I'm guessing that I can assume that $\sigma$ fixes $F$ since it says "$\sigma$ is an automorphism of $K$ over $F$". If that's true, then

$f(a)=b_0 + b_1a + b_2a^2 + ... + b_na^n = 0$

So,

$f(\sigma(a))= b_0 + b_1\sigma(a) + ... + b_n\sigma(a) = \sigma(b_0+b_1a+...+b_na^n)= \sigma(0)=0.$

So $\sigma(a)$ must be a root of $f$. But since #L# is normal, it must contain all the other roots of $f$. So, basically, $\sigma$ just permutes the roots of a given polynomial over $F$ with roots in $L$, meaning that $\sigma(L)=L$. Is that correct?

 

[micromass]

I'm guessing that I can assume that $\sigma$ fixes $F$ since it says "$\sigma$ is an automorphism of $K$ over $F$". If that's true, then

#f(a)=b_0 + b_1a + b_2a^2 + ... + b_na^n = 0#

So,

#f(\sigma(a))= b_0 + b_1\sigma(a) + ... + b_n\sigma(a) = \sigma(b_0+b_1a+...+b_na^n)= \sigma(0)=0.#

So #\sigma(a)# must be a root of #f#. But since #L# is normal, it must contain all the other roots of #f#. So, basically, #\sigma# just permutes the roots of a given polynomial over #F# with roots in #L#, meaning that #\sigma(L)=L#. Is that correct?

Yes, that's what I had in mind.