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For a small project of mine, I am tasked with finding proofs of the fundamental theorem of algebra that require as little analysis as possible.

I also know quite well that a truly algebraic proof is not feasible, since $\mathbb R$ is by definition complete: a fundamentally analytic concept.

I am aware of the galois-theoretic proof that uses only these 3 facts about $\mathbb R$:

  1. $\mathbb R$ is an ordered field
  2. Every non-negative real number has a square root
  3. Every odd degree real polynomial has a root

Property (1) is, in my opinion, an algebraic notion, since the rational numbers are also an ordered field, with every field of characteristic 0 having its prime subfield isomorphic to $\mathbb Q$. Property (2) can be proven directly from the completeness axiom, which is the best we can do.

However, (3) is a consequence of the intermediate value theorem, which requires the non-algebraic notion of continuity. I feel like we can do a little better here...

Is it possible to dispense of the notion of continuity to prove the fundamental theorem of algebra? Can (3) be deduced with only the completeness axiom and algebraic notions? Or perhaps is there a different, 'less analytic' condition than (3) that would suffice?

I also kindly request to have references in answers, as I will need them for citation purposes.

Mafematician
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  • I’d be interested in a link to this Galois-theoretic proof. I suspect a ‘purely algebraic’ proof does not exist, since we know that we can’t really express the roots of arbitrary polynomials - though they exist - from the basic field operations. – FShrike Nov 20 '22 at 14:39
  • For a reference to the galois-theoretic proof, see Ian Stewart's Galois Theory (4th or 5th edition) Chapter 23 – Mafematician Nov 21 '22 at 10:24
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    The notion of an ordered field is algebraic. Artin and Schreier noticed a key property of such fields : $\sum_{i=1}^n a_i^2=0\implies a_i=0,\forall i$ and used this property to define formally real fields and proved that such fields can be ordered. The central idea in this theory is the notion of real closed fields which have properties very similar to $\mathbb{R} $. In particular intermediate value theorem and mean value theorem holds for polynomials in real closed fields. – Paramanand Singh Nov 22 '22 at 00:49
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    The three properties of $\mathbb{R} $ listed in your question hold for any real closed field. I think this is the best one can mimic $\mathbb{R} $ using algebraic concepts. – Paramanand Singh Nov 22 '22 at 00:54
  • @ParamanandSingh Thank you for your comment. I am currently investigating real closed fields to see if I could prove that $\mathbb R$ is real closed without resorting to continuity. – Mafematician Nov 24 '22 at 17:48
  • @FShrike Locally there is this thread. Not very satisfactory as I gloss over most of the details. Anyway, that matches with Mafematician's description. Both Paramanand Singh and your truly looked at Jacobson's Basic Algebra I. – Jyrki Lahtonen Nov 27 '22 at 11:08
  • Anyway, my .02 cents suggest that we need the existence of roots of odd degree polynomials for this approach to work. Either from the intermediate value theorem or as an axiom of real closed fields. The construction of the field of complex numbers has this unavoidable topological component because the field of real numbers is based on it (via completeness). Constructing algebraically closed fields by other means (when we need Zorn's lemma and such at a critical step) is possible, but relating the resulting field with the complex numbers is then a huge obstacle. – Jyrki Lahtonen Nov 27 '22 at 11:13

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