Subject: categories: re: fundamental theorem of algebra

From: Michael Barr <[email-masked]>
Date: Sat, 1 Apr 2006 08:01:53 -0500 (EST)

Let me reiterate this: There can in principle be no purely algebraic proof
of the FToA because the reals have no purely algebraic definition.
(Unless you define them as a real closed field of transcendence degree c,
but that leaves the FToA as a trivial consequence and cannot be what is
wanted.)  The proof I outlined, which someone showed me 45 years ago, uses
only the fact that R is a complete ordered field.  Given that that is the
analytic definition of R, it is impossible to avoid.  That fact is, of
course, at the heart of the fact that the circle is not contractible in a
punctured plane.

Incidentally, even constructivists (well even Errett Bishop, anyway) agree
that odd order real polynomials have a real root and that positive numbers
have square roots, since there are obvious constructions for these things.
Their real line is not complete (it is countable, but the missing numbers
are not constructible), but these roots are there anyway.

The argument I outlined is elementary, even if not especially easy.  First
you have to construct the reals, the least elementary part of the
argument.  Then comes the theorem on symmetric functions.  It is not a
deep result; it needs a careful proof, but a student can follow it without
knowing anything sophisticated.  The construction of a splitting field
(without getting into UFDs) is a bit tricky.  To adjoin a root to an
irreducible polynomial p of degree n, you start with a vector space whose
basis is called 1, u, u^2,..., u^{n-1} and define a multiplication, by
having p(u) = 0.  This is analogous to how you get from R to C.  Of
course, you use the division algorithm to show you get a field.

Michael