What are the three non-isomorphic 2-dimensional algebras over $\mathbb{R}$?

What are the three non-isomorphic $2$-dimensional algebras over $\mathbb{R}$?

Am I right in thinking they are
$\lbrace x+iy: x,y\in\mathbb{R},\ i^{2}=1\rbrace$,
$\lbrace x+jy: x,y\in\mathbb{R},\ j^{2}=-1\rbrace$ and
$\lbrace x+\varepsilon y: x,y\in\mathbb{R},\ \varepsilon^{2}=0\rbrace$?

Solution 1:

You're right, these are in fact the three isomorphism classes of two-dimensional commutative real algebras.

Let $A$ be a real two-dimensional algebra.

Let $x \in A \setminus \mathbb{R}$. Then, the set $1,x,x^2$ must be linearly dependent, so we have a non-trivial equation of the form $\lambda_1x^2+\lambda_2x+\lambda_3=0$ Note that $\lambda_1 \neq 0$, as $1,x$ is linearly independent, as the subspace generated by it properly contains $\mathbb{R}$, so must be two-dimensional, i.e. $A$. So after divison by $\lambda_1$, we may assume that $\lambda_1 = 1$. Now we complete the square and get $0=x^2+\lambda_2x+\lambda_3 = (x+\frac{\lambda_2}{2})^2+\lambda_3-\frac{\lambda_2^2}{4}$ Now take $v = x + \frac{\lambda_2}{2}$, then the last equation shows that $v^2 \in \mathbb{R}$ . Now $A$ is spanned by $1,v$ (as the subspace generated by $1,v$ properly contains $\mathbb{R}$, so it is two-dimensional). If $v^2=0$, then clearly we have $A \cong \lbrace x+\varepsilon y: x,y\in\mathbb{R}, \varepsilon^{2}=0\rbrace$ if $v^2 \neq 0$, normalize $v$ by setting $u= \frac{v}{\sqrt{|v^2|}}$, now $u^2= \pm 1$ and we have $A \cong \lbrace x+iy: x,y\in\mathbb{R}, i^{2}=1\rbrace$, if $u^2=1$ and $A \cong \lbrace x+jy: x,y\in\mathbb{R}, j^{2}=-1\rbrace$ if $u^2=-1$

This shows that any two-dimensional real algebra is isomorphic to one of these three. To show that there are no isomorphisms among them, note that $\lbrace x+jy: x,y\in\mathbb{R}, j^{2}=-1\rbrace \cong \mathbb C$ is a field, so it contains neither non-trivial idempotents nor non-trivial nilpotents. And note that $ \lbrace x+\varepsilon y: x,y\in\mathbb{R}, \varepsilon^{2}=0\rbrace \cong \mathbb{R}[x]/(x^2)$ is a local ring with maximal ideal $(\varepsilon)$, so it does not contain non-trivial idempotents either, whereas $\lbrace x+iy: x,y\in\mathbb{R}, i^{2}=1\rbrace \cong \mathbb{R}[x]/(x^2-1) \cong \mathbb{R}[x]/(x-1) \times \mathbb{R}[x]/(x+1) \cong \mathbb{R} \times \mathbb{R}$ does contain non-trivial idempotents.