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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?


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Add a unit and classify all unital three dimensional :) – Norbert Jan 27 '12 at 17:17
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
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
up vote 2 down vote accepted

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

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The Classification of Two-dimensional Algebras

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