$\Bbb R$-algebras of rank $2$

Question

Say I want to find all associative, unital but not-necessarily commutative $\Bbb R$-algebras of rank $2$.

Then, these all arise as quotients of $\Bbb R\langle x\rangle$ by an argument given here.

Let $A$ be an associative and unital $\Bbb R$-algebra of rank $2$. Then $A$ is a (say) left $\Bbb R$-module, and since $\Bbb R$ is a field, $A$ must be free, and isomorphic as an $\Bbb R$-module to $\Bbb R^2$ (but not as an algebra).

Since $A$ is a quotient ring $\Bbb R\langle x\rangle / I$, $I$ must be a two-sided ideal of $\Bbb{R}\langle x\rangle=\Bbb R[x]$ (since there is nothing for $x$ to not commute with). Since $\Bbb R$ is a field $\Bbb R[x]$ is a PID and thus $I=(f)$ for some $f\in \Bbb R[x]$. Since $R[x]/(f)$ is an algebra of rank $2$, $f$ must be a polynomial of degree $2$ and this encodes how $x\cdot x$ is defined. In particular $(f)=(a^{-1}f)$ for a unit $a$, so we can without loss of generality take $f$ to be monic, say $f=x^2+bx+c$, which defines $x\cdot x = -bx-c$. I can see this captures $\Bbb C = \Bbb{R}[x]/(x^2+1)$ for example.

How do I finish this classification?

Answer 1

Hint: consider these cases:

• $f$ is irreducible, thus has no real roots, and $\mathbb R[x]/f$ is a field
• $f$ is reducible, thus has two real roots (counting multiplicity). In particular:
  • $f$ has distinct roots, that is $f = (x-a)(x-b)$
  • $f$ has a multiple root, that is $f = (x-a)^2$