What are the $2$-dimensional algebras over any arbitrary field?

Question

As a follow-up of this question, I would like to ask, what are the $2$-dimensional algebras over $\mathbb R$, $\mathbb Q$, or any arbitrary field? Can we classify them?

Answer 1

Suppose $\{\mathbb{1},x\}$ is a basis for $A$, and the span of $\{\mathbb{1}\}$ is the copy of $\mathbb{K}$, with $k\mathbb{1}\in A$ identified with $k\in\mathbb{K}$. So $\mathbb{1}\cdot\mathbb{1}=\mathbb{1}$.

The algebra is determined by how basis vectors multiply. $$\begin{align} \mathbb{1}\cdot x&= c_{1x}\mathbb{1}+d_{1x}x\\ x\cdot\mathbb{1}&= c_{x1}\mathbb{1}+d_{x1}x\\ x\cdot x&= c_{xx}\mathbb{1}+d_{xx}x\\ \end{align}$$

Note that I am not assuming things like $\mathbb{1}\cdot x=x$ or even $\mathbb{1}\cdot x=x\cdot\mathbb{1}$, even though of course $1\cdot x=x$. I am guessing that you intend to assume associativity, so to move forward, expand equations like $(\mathbb{1}x)x=\mathbb{1}(xx)$ and $(\mathbb{1}x)\mathbb{1}=\mathbb{1}(x\mathbb{1})$ to get relations on the six unaccounted for structure constants. For instance, the relation $(\mathbb{1}x)x=\mathbb{1}(xx)$ implies that either $c_{1x}=0$ or $d_{1x}=0$ once you expand and examine coefficients of $x$. And the relation $(xx)x=x(xx)$ implies that either the algebra is commutative, or $c_{xx}=0$.

In this manner you can break things down into a bunch of cases, each having narrow restrictions on the structure constants. Once you have done this, you can examine whether any of the classes are redundant due to existence of an isomorphism between them.