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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?

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.

fantasie
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    You can apply Frobenius' theorem only for finite-dimensional $\mathbb{R}$-algebras. – richrow Apr 30 '22 at 10:04
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    If $\mathscr{A}$ is a finite dimensional space over $\Bbb{C}$ then the classical results force it to be equal to $\Bbb{C}$. Also, in a finite dimensional complex space every linear operator has an eigenvalue, so you can reuse the argument in your question. If we allow infinite dimensional algebras, then the result is false. For example the field of rational functions $\Bbb{C}(x)$ is an infinite dimensional extension field of $\Bbb{C}$, and hence a division ring. – Jyrki Lahtonen Apr 30 '22 at 10:19
  • I would expect this to have been explained on our site somewhere, but I don't have the time to search for one. Hence the terse outline. – Jyrki Lahtonen Apr 30 '22 at 10:20
  • @JyrkiLahtonen Thanks for your response (as well as @richrow's). Hungerford didn't stress finite dimensional in that Corollary, but it was clear from the proof the dimension is finite. Can you please write an answer? I know a bit of Galois theory, and I think your example is very clear. – fantasie Apr 30 '22 at 10:37
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    You may consult https://www.google.com/url?sa=t&source=web&rct=j&url=https://iitpkd.ac.in/sites/default/files/2020-02/gelmazsurvey.pdf&ved=2ahUKEwiGgLCs1Lv3AhWCvYsKHYu7AD8QFnoECAYQAQ&usg=AOvVaw18z5B_gxdxp1sRGCNCm5Th Perhaps this helps. – Ryszard Szwarc Apr 30 '22 at 11:20

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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 $\|a\cdot b\|\le\|a\|\|b\|$.

fantasie
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