What is the relationship between subfields KL and K in finite field extensions?

  • Thread starter Thread starter Mathmos6
  • Start date Start date
  • Tags Tags
    Field Product
Click For Summary

Homework Help Overview

The discussion revolves around the relationship between the subfields K and L within a finite field extension M. The original poster attempts to demonstrate that KL is a subfield of M and to establish the inequality [KL : K] ≤ [L : K ∩ L].

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the use of the tower law and consider the implications of injections. There is discussion about L being a spanning set for KL over K and the potential to find a basis for KL. Questions arise regarding the exclusion of elements from the intersection K ∩ L in the basis expansion.

Discussion Status

Participants are actively engaging with the problem, offering suggestions and exploring various approaches. Some guidance has been provided regarding the linear independence of elements in L as a basis for KL, and there is recognition of the need to clarify certain concepts related to the intersection of fields.

Contextual Notes

Participants note the challenge of understanding the implications of the intersection K ∩ L and the complexities introduced by self-taught knowledge in Galois theory.

Mathmos6
Messages
76
Reaction score
0
"Product Field"

Homework Statement


Let K and L be subfields of a field M such that M/K (the field extension M of K) is finite. Denote by KL the set of all finite sums ∑xiyi with xi ∈ K and yi ∈ L. Show that KL is a subfield of M, and that

[KL : K] ≤ [L : K ∩ L].


Homework Equations


[KL : K] etc. is the standard notation for the dimension of KL as a vector space over K (includes infinity).


The Attempt at a Solution



I've done the first part about the subfield fine, it's the inequality I'm struggling with. I've tried using the tower law ([K:M]=[K:L][L:M] where M is a subfield of L is a subfield of K. However, I can't seem to appropriately 'link up' the right subfields to relate the left-hand and right-hand side of the inequality; despite this, since it's an inequality I'm not even convinced the tower law is the right way to go: perhaps there's a nicer way to solve the problem, like finding an isomorphism between KL/K and a subfield of L/(K ∩ L) or something similar - but I have a habit of overthinking things, so maybe I'm overdoing it!

Any suggestions would be muchly appreciated - thanks in advance :)
 
Physics news on Phys.org


The tower law probably isn't the way to go.

Looking at injections is a good idea. It might help to start with a small step: Can you see why L is a spanning set of vectors for KL over the field K? From there, think about how you could pare that down to a basis, and in particular why the elements of L∩K for the most part will not be included
 


Office_Shredder said:
The tower law probably isn't the way to go.

Looking at injections is a good idea. It might help to start with a small step: Can you see why L is a spanning set of vectors for KL over the field K? From there, think about how you could pare that down to a basis, and in particular why the elements of L∩K for the most part will not be included

I've just noticed also this looks a lot like the second isomorphism theorem - perhaps another way to go about the problem?

Anyway, i can see why L is a spanning set of vectors for KL over K, and I guess you could use AoC to pick a basis by selecting an element of L at random, then another not in its spanning set over K, then another not in the spanning set of the first 2 elements picked over K, and so on. I'm not completely sure why the intersection won't be included - I can see vaguely that anything in the intersection will have an inverse in K, so can be 'included' in the K-term (i.e. k*1, for some k in K) in any basis expansion, but I don't think I really understand properly. Could you elaborate please? Thanks very much.
 


Mathmos6 said:
I've just noticed also this looks a lot like the second isomorphism theorem - perhaps another way to go about the problem?

Anyway, i can see why L is a spanning set of vectors for KL over K, and I guess you could use AoC to pick a basis by selecting an element of L at random, then another not in its spanning set over K, then another not in the spanning set of the first 2 elements picked over K, and so on. I'm not completely sure why the intersection won't be included - I can see vaguely that anything in the intersection will have an inverse in K, so can be 'included' in the K-term (i.e. k*1, for some k in K) in any basis expansion, but I don't think I really understand properly. Could you elaborate please? Thanks very much.

You can find a basis of KL over the field K containing only elements in L.

What do those elements do when you just look at them as elements of L over the field L∩K?
 


Office_Shredder said:
You can find a basis of KL over the field K containing only elements in L.

What do those elements do when you just look at them as elements of L over the field L∩K?

I'm still not completely sure sorry, I'm obviously not getting this :( When you're looking at L over L∩K, the span of your elements will have to be smaller than or equal to the span of those elements over K, but I can't really see how to formulate this idea properly - sorry to keep asking! I'm very new to Galois theory and all I know about it is currently self taught, so unfortunately it's taking me a while from time to time to get my head around things - the help is greatly appreciated.
 


We're not really interested in the span. We have a basis for KL in terms of elements of only L. I claim those elements are linearly independent vectors in L as a vector space over the subfield L∩K.

Do you see why that would wrap up the proof?
 


Office_Shredder said:
We're not really interested in the span. We have a basis for KL in terms of elements of only L. I claim those elements are linearly independent vectors in L as a vector space over the subfield L∩K.

Do you see why that would wrap up the proof?

Ah of course, it makes perfect sense when you put it like that :) The argument is fairly simple once you spot it, I was definitely overcomplicating things - thankyou for being so patient!
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 14 ·
Replies
14
Views
3K
Replies
1
Views
3K
  • · Replies 46 ·
2
Replies
46
Views
9K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K