Two-dimensional algebras over complex numbers I have read that there are two 2-dimensional algebras over the complex numbers, but I wanted to see what they are and how they are formed. I understand how to show that there are three 2-dimensional algebras over the reals and am trying to alter that but I don't know if what I'm doing is right!
I'm quite new to this area of Maths so haven't covered Lie algebras or anything advanced.
Thanks!
 A: Let $\mathcal{u}$ be the unity of A. I construct a base { $\mathcal{u}, \mathcal{v}$} of A over $\mathbb{C}$ . Then $\mathcal{v}\cdot\mathcal{v}=a\cdot \mathcal{u}+b\cdot\mathcal{v}$. If  $a+\frac{b^{2}}{4}=0$ it is easy to find $\mathcal{w}\in A$ such that $w\cdot w=0$ and $w\notin \mathbb{C}u$($w$ has the form $\alpha\cdot u+\beta\cdot v, \beta\neq 0$). In this way I obtain a new base {$u, w$}. Now, I consider the correspondence $u\rightarrow \hat{1}$, $w\rightarrow \hat{x}$. Therefore, A is isomorphic to $\frac{\mathbb{C}[X]}{(X^2)}$ as $\mathbb{C}$-algebras. If $a+\frac{b^{2}}{4}\neq 0$ it is easy to find $\mathcal{w}\in A$ such that $w\cdot w=-1$ and $w\notin \mathbb{C}u$. In this way I obtain a new base {$u, w$}. Now, I consider the correspondence $u\rightarrow (1,0)$, $w\rightarrow (0,1)$. Therefore, A is isomorphic to $\mathbb{C} $x$\mathbb{C}$ as $\mathbb{C}$-algebras.
A: The commutative algebras have been classified up to dimension $n\le 6$ over any algebraically closed field of arbitrary characteristic, see the article of Björn Poonen "Isomorphism types of commutative algebras of finite rank over an algebraically closed field". This includes the unital associative commutative algebras.
Complex associative algebras (not necessarily commutative) have been classified up to dimension $4$, see for example here. In dimension $2$, there are four different complex associative algebras, see the table in section $2.1$.
