Re: fundamental theorem of algebra

["John Baez" <baez@math.ucr.edu>]

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