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Because the complex numbers are constructed from the real numbers, they are inherently an 'analytic object'. It is thus perhaps not surprising that "there is no" proof that $\mathbb C$ is algebraically closed "without using analysis". Cf. Is there a purely algebraic proof of the Fundamental Theorem of Algebra?

A friend once imagined that, if we take a purely algebraic characterization of $\mathbb C$ as its definition, we may be able to prove that it is algebraically closed "without using analysis".

Specifically, he proposed the following characterization due to Pontryagin: $\mathbb C$ and $\mathbb F_2$ are the only locally compact Hausdorff topological fields with connected group of units. (See Are $\Bbb R$ and $\Bbb C$ the only connected, locally compact fields?)

With $\mathbb C$ defined as the unique topological field different from $\mathbb F_2$ satisfying Pontryagin's characterization, can we give a direct proof that it is algebraically closed?

Ideas: if we assume $F/\mathbb C$ is a finite extension, then the product topology on $F$ does not depend on the choice of a $\mathbb C$-basis and makes $F$ a topological field. With this topology, $F$ satisfies Pontryagin's characterization and so it is isomorphic to $\mathbb C$. If it were true that a field cannot contain itself as a finite-index copy, we would be done. Unfortunately, that is not true (it is not hard to construct a counterexample from a transcendence basis).

Bart Michels
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  • Using topology instead of analysis has already been done. This is not to say this proof wouldn't be new, but the goal would also be to eliminate topology as well, since topology has already been used to replace analysis. – Matt Samuel Mar 03 '19 at 22:14
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    I would hardly consider using the topology of $\mathbb{C}$ to be "without analysis"! – Eric Wofsey Mar 03 '19 at 22:20
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    Why would you want to do this? Any time you actually need to invoke the fundamental theorem, you're working with $\mathbb{C}$ in the usual sense, not in this sense. So at some point you still need to prove that $\mathbb{C}$ is locally compact Hausdorff etc. and at this point analysis is unavoidable again. – Qiaochu Yuan Mar 03 '19 at 22:56
  • I agree that this isn't going to tell us anything new about $\mathbb C$ nor give a self-contained proof that $\mathbb C$ is algebraically closed. Rather, I am interested in this as an experiment, if you want, and I'm curious which techniques show up at the interplay of topology and field extensions. – Bart Michels Mar 05 '19 at 12:45

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