Simple comm. algebra is a field

  • Thread starter Thread starter losiu99
  • Start date Start date
  • Tags Tags
    Algebra Field
Click For Summary

Homework Help Overview

The discussion revolves around the properties of a simple, commutative, associative algebra over a field, specifically focusing on proving that such an algebra is a field itself and that the base field is isomorphic to a subfield of the algebra.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the existence of a multiplicative identity and inverses in the algebra. Questions arise regarding the implications of the algebra being simple and commutative, and how these properties relate to the proof.

Discussion Status

Some participants have offered hints regarding the mapping from the field to the algebra and the significance of the ideal generated by nonzero elements. There is acknowledgment of the complexity involved in proving the algebra is a field, indicating a productive exploration of the topic.

Contextual Notes

There are references to assumptions about the algebra's structure and the nature of the original poster's inquiry, as well as a light-hearted exchange about a misunderstanding related to terminology.

losiu99
Messages
132
Reaction score
0
Simple comm. ***. algebra is a field

Homework Statement
Let A be a simple, commutative, associative algebra over a field k. A is an extension field for k.

I appreciate any hints.
 
Physics news on Phys.org


You're trying to prove that k is isomorphic to a subfield of A, right?

Does your algebra have an identity element? If so, I would try to use the map k\ni c\mapsto c1\in A.

I haven't actually done this problem. That's just where I would start. I don't know where "simple and commutative" enters the picture. Hm, I do know that the only simple and commutative Banach algebra with identity is the complex numbers. Not sure if that implies anything useful about simple and commutative algebras.
 


The isomorphism Fredrik describes is the easy part. The difficult part is to prove that A is a field in the first place -- given the assumptions on A, this amounts to proving the existence of a multiplicative identity and inverses. The trick is to use the fact that the ideal generated by any nonzero element is the entire algebra. You should be able to do the problem from here, but if you're still stuck, here's one more hint to get you started:

Start by picking a nonzero element x. Since the ideal generated by x is A, there exists y such that xy = x. Try to show that this implies that y is a multiplicative identity for A.
 


That's just soo embarrassing... Thank you so much for you help!
 


I have no idea about the maths - but I am intrigued as to what the apparently censored word is?
I can think of many obscene 3letter words - algebra related or otherwise
 


Oh, sorry for that, it was supposed to be a shorthand for associative... Guess I should have expected that.
 


You can't say donkey on the internet !
 

Similar threads

  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K