Generalization of the Gelfand-Mazur theorem: A $\mathbb C$-algebra that is also a division ring

Question

The Gelfand-Mazur theorem states that if $\mathscr A$ is a $\mathbb C$-Banach algebra that is also a division ring, then $\mathscr A=\mathbb C$.

Proof. If $a$ is an element, then the spectrum of $a$ is nonempty. If $\lambda$ is in its spectrum, then $a-\lambda$ has no inverse. But by definition of division ring $a-\lambda=0$.

This proof clearly utilizes the norm since it is an application of the spectral theory.

However, what if $\mathscr A$ is just a $\mathbb C$-algebra that is also a division ring? Do we have this result or some other classification?

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My attempt: If $\mathscr A$ is a $\mathbb C$ algebra, then $\mathscr A$ is also a $\mathbb R$ algebra. By Corollary IX.6.8(Frobenius) of Hunderford's Algebra, an (thanks to the comments, finite dimensional) division algebra over the field $\mathbb R$ of real numbers is isomorphic to one of $\mathbb R$, $\mathbb C$ and the division algebra of real quaternions. But $\mathbb R$ and $\mathbb H$ don't seem to be $\mathbb C$-algebras.

Answer 1

To summarize the comments:

1. Frobenius' Theorem only applies to finite dimensional algebras. If $\mathscr A$ is a finite dimensional division algebra over $\mathbb C$, $\mathscr A\cong \mathbb C$ by arguments in the question.

2. For infinite dimensional algebras, this no longer hold. In fact, $\mathbb C(x)$ is a division algebra over $\mathbb C$ that is not isomorphic to $\mathbb C$.

3. This survey discusses some notable generalizations of the Gelfand-Mazur theorem. In particular, the following hold:

Theorem 1 (Real Version). Every real normed division algebra is isomorphic to the set of all real numbers $\mathbb R$, the complex numbers $\mathbb C$ or the quaternions $\mathbb H$.

Theorem 2 (Complex Version). Every complex normed division algebra is isometrically isomorphic to $\mathbb C$.

Here a normed algebra is just a algebra with a norm (not necessarily complete) such that $\|a\cdot b\|\le\|a\|\|b\|$.