Simple comm. algebra is a field

[losiu99]

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.

 

[Fredrik]

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.

 

[Citan Uzuki]

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:

Spoiler

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.

 

[losiu99]

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

 

[NobodySpecial]

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

 

[losiu99]

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

 

[NobodySpecial]

You can't say donkey on the internet !