# Algebraic and normal field extensions.

1. Jul 29, 2013

### Artusartos

1. The problem statement, all variables and given/known data

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 $\sigma(L)=L$.

2. Relevant equations

3. 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...

2. Jul 29, 2013

### micromass

Staff Emeritus
What is your definition of a normal extension?

3. Jul 29, 2013

### Artusartos

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

4. Jul 29, 2013

### micromass

Staff Emeritus
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 $f(\sigma(a))$?

5. Jul 29, 2013

### Artusartos

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?

6. Jul 29, 2013

### micromass

Staff Emeritus
Yes, that's what I had in mind.