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There have been questions before on Math SE about whether there is a purely algebraic proof of the FTA. But as far as I know, nobody has proven rigorously that there is no purely algebraic proof. Has anyone rigorously defined what a purely algebraic proof is, and proven that there is no purely algebraic proof in some book or paper?

user107952
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    How can you define "purely algebraic"? Anyway, the real (and those complex) numbers are coming from analysis, so it's hard to believe you can avoid using at least a bit of analysis in this theorem. – Mark Jul 29 '21 at 11:47
  • To better understand the comment by @Mark you may find it instructive to locate the non-algebraic part(s) of the following proof (or incomplete part(s) that usually require additional reasoning supplied by analysis): Appendix (pp. 577-581) in Advanced Course in Algebra by Webster Wells (1904). – Dave L. Renfro Jul 29 '21 at 12:48
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    Lagrange and others gave proofs of FTA that assumed (in modern terminology) the existence of an algebraic closure of $\Bbb{R}$ and then showed algebraically that that algebraic closure must be $\Bbb{R}[i]$ (see https://www.jstor.org/stable/27642032 or the book Numbers by Ebbinghaus et al.). These proofs were disparaged by Gauss, but are now vindicated since we now know that every field has an algebraic closure. However, the proof of that fact involves some form of the axiom of choice, and so is not really algebraic in the sense of only involving finitary operations. – Rob Arthan Jul 29 '21 at 22:22

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If $R$ is a real closed field, then $R[i]$ is algebraically closed. This theorem and its proof using Galois theory seem "purely algebraic" to me (if we define "real closed field" to mean "ordered field in which every positive element has a square root and every odd degree polynomial has a root").

The "non-algebraic" part comes in verifying that the field of real numbers $\mathbb{R}$ is real closed (from which it follows by the theorem above that $\mathbb{C}$ is algebraically closed). We need to use the intermediate value theorem to verify the axioms.

But now I'll second Mark's comment: to me, the very definitions of $\mathbb{R}$ and $\mathbb{C}$ are analytic, not "purely algebraic". So on the face of it, it is impossible to give a "purely algebraic" proof of any statement about these fields.

The proof of the fundamental theorem of algebra via real closed fields is appealing to me because it "factors out" the analysis from the algebra, and it seems to minimize the analytic part of the proof as much as possible, down to some easy applications of the intermediate value theorem.

Alex Kruckman
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