Algebraic and normal field extensions.

  • Thread starter Thread starter Artusartos
  • Start date Start date
  • Tags Tags
    Field Normal
Click For Summary

Homework Help Overview

The discussion revolves around the properties of algebraic field extensions, specifically focusing on the relationship between a field extension K, a subfield L, and an automorphism σ of K over a field F. The original poster seeks to understand why σ(L) equals L given that L is a normal extension of F.

Discussion Character

  • Conceptual clarification, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the definition of a normal extension and its implications. They discuss the relationship between elements of L and their corresponding polynomials in F[x], questioning how the properties of these polynomials relate to the action of the automorphism σ.

Discussion Status

Some participants have provided insights into the relationship between the roots of polynomials and the normality of the extension. There is an ongoing exploration of how σ interacts with elements of L, with some participants affirming the reasoning presented by others.

Contextual Notes

Participants are working under the assumption that σ fixes F, as stated in the problem. There is also a focus on the implications of L being a normal extension, which is central to the discussion.

Artusartos
Messages
236
Reaction score
0

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...
 
Physics news on Phys.org
What is your definition of a normal extension?
 
micromass said:
What is your definition of a normal extension?

If f \in F[x] has one root in L, then it splits over L.
 
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))##?
 
micromass said:
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?
 
Artusartos said:
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.
 
  • Like
Likes   Reactions: 1 person

Similar threads

Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K