Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $A$ be a two dimensional commutative associative algebra over field $K$ of reals or complex numbers.

Assume that $A$ has units $e$. Let $u \notin Ke$. Then $\{e,u\}$ is basis of $A$. In order to determine that algebra it suffices to know $u\cdot u$. Let $u=pu+qe$. Let's consider polynom $f(x)=x^2-px+q \in K[x]$. Three cases may occur: $f$ has two, one or none roots. In the first case putting $v=(y_2-y_1)^{-1}(u-y_1)$, where $y_1,y_2$ are roots of $f$, we have $v^2=v$ and $\{e,v\}$ is the basis of $A$. In the second putting $v=u-y_1e$, where $y_1$ is a root of $f$, we have $v^2=0$ and ${e,v}$ is the basis of $A$. In the third case $A$ is a field.

How to determine all two dimensional commutative associative algebras without units?

Thanks.

share|improve this question
1  
Add a unit and classify all unital three dimensional :) –  Norbert Jan 27 '12 at 17:17
1  
I tried to analyze the three dimensional faithful module, but the number of cases got pretty unwieldy. k[x,y] has lots of quotients isomorphic to quotients of k[x] and it seemed hard to distinguish. –  Jack Schmidt Jan 27 '12 at 17:33
2  
I suggest looking at two main cases: e^2 is a multiple of e, or not. Each breaks into sub cases that are basically easy to handle, but the details are getting too long, and I think there is a nice Jacobson theory to eliminate most of the calculations. –  Jack Schmidt Jan 27 '12 at 17:41
add comment

1 Answer

Quick googling gives the following article on two dimensional algebras over arbitrary field:

1) 2-dimensional algebras

The more general classification for nonassociative algebras over arbitrary field is given here

2) The Classification of Two-dimensional Nonassociative Algebras

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.