I know there are already a couple of questions on this on the site, but I haven't seen an answer to this particular form...


We know, from the Fundamental Theorem of Algebra, that the complex algebraic numbers contain a unique maximal ordered subfield, namely the real algebraic numbers, and the complex algebraic numbers are obtained by adjoining a square root of $-1$ to the real algebraic numbers.


This is a purely algebraic statement (I think!) and one could reasonably ask for a purely algebraic proof. Is there such a proof? Or is there a barrier to such a proof (such as a model of ZF where the statement is false)?


I'm aware of the standard (purely algebraic, or at least I'm happy to call it so) Zorn's Lemma proof that the complex algebraic numbers exist, since any field $K$ has an algebraic closure and this is unique up to isomorphism. However the 'up to isomorphism' here includes a great many isomorphisms of the complex algebraic numbers that fix the rationals. I think a more or less equivalent question to what I am asking is: from this construction, is there a good algebraic way to pick out the privileged automorphism $i\to-i$ and put an order on the subfield this automorphism fixes?


There do exist proofs of the usual Fundamental Theorem of Algebra which use a `minimal' amount of analysis, such as <https://arxiv.org/abs/1504.05609> (due to Piotr Błaszczyk) which doesn't really use more analysis than required to define the reals. However even this is more than (I think!) I am asking for.



Fundamental Theorem of Algebra, via algebra
Despite the extensive discussion in the comments, I'm not sure I completely understand the question, but I think the following fact may settle the question in the negative.


Work of Läuchli implies that there is a model of ZF in which $\mathbb Q$ has an algebraic closure with no real-closed subfield. For more details, see for example Wilfrid Hodges, "Läuchli's algebraic closure of $Q$," *Math. Proc. Cambridge Philos. Soc.* **79** (1976), 289–297.


I think your desired conclusion requires every algebraic closure of $\mathbb Q$ to have a real-closed subfield. If you agree, then the above result shows that you at least need some weak form of AC, which arguably is "not algebraic." But this depends on what you mean by "algebraic"; maybe a weak form of choice doesn't automatically disqualify a proof from being "algebraic"?


