Vaughan writes:
However I've been reflecting on just what is behind the very uniform insistence on the distinction between an algebraic proof and an analytic one. Since algebra is descended from analysis, it seems unkind for algebra to deny its parentage in this way.
I thought people knew how to add before they knew how to take limits. :-)
But I see now that this denial is logically necessary. For consider the algebraic plane, the least algebraically closed subfield of the complex plane, consisting of the algebraic numbers. The FTAlg is by definition true there, so it ought to be provable there.
Hmm. How do you propose to show there *exists* an algebraically closed subfield of the complex numbers? I would do it using the fundamental theorem of algebra - the usual one, for the complex numbers. Unless you have some other way, I don't understand how you hope to circumvent the use of analysis by introducing such an entity. Indeed, the usual proof that the real numbers contains a square root of 2 uses the completeness of the real numbers, which also counts as "analysis".
It is ironic that a theorem of algebra about an algebraic domain that itself has no element of analysis to it, being just the algebraic closure of the rationals, a small and totally disconnected space, should require analysis, the parent of algebra, for its proof.
That the rational numbers has an algebraic closure is a purely algebraic result, with no mention of topology in either the statement or proof. That the complex numbers is algebraically closed is not an algebraic result: it has topology built into the statement, and also the proof(s). That the algebraic closure of the rationals embeds in the complex numbers has topology in the statement - and I bet also in every proof. Best, jb