Subject: categories: Fundamental theorem

From: Michael Barr <[email-masked]>
Date: Thu, 30 Mar 2006 19:43:55 -0500 (EST)

The proof that Vaughan outlined is found in almost exactly the same form
in Courant & Robbins, published around 60 years ago.

Here is a proof I learned in grad school, based on three facts, one
analytic and two algebraic.  The analytic fact, which is an irreducible
minimum, given that the reals cannot be defined algebraically (except as a
real closed field of continuum transcendence degree, which misses the
point) is the intermediate value theorem.  In other words, order
completeness.  From this follows the fact that every odd order polynomial
has a (real) root and that every real number--and with a bit of
manipulation, every complex number--has a complex square root.  The
algebraic facts are the existence of a splitting field and the theorem on
elementary symmetric functions, neither of which is quite trivial, nor
very deep.

The way you do it is by proving that every real polynomial of degree n =
2^k*m with m odd has a complex root, by induction on k.  The case k = 0 is
quite trivial, of course.  So suppose that f is a real polynomial of even
degree n and, in some splitting field has roots r_1,...,r_n.  For each
integer s form the polynomial f_s = \prod{i<j}(x - r_i - r_j - sr_ir_j)
which has degree n(n-1)/2, which is less 2 divisible than n.  The theorem
on symmetric functions implies it is real and hence for some i and j
dependent on s, r_i + r_j + sr_ir_j is in C.  Since there are only
finitely many pairs i and j and infinitely many integers, there are
distinct s and t for which both r_i + r_j + sr_ir_j and r_i + r_j +
tr_ir_j belong to C.  Given that C has square roots, one easily discovers
that r_i and r_j are both complex numbers.  To go from real to complex
polynomials, just multiply a complex polynomial by its conjugate; for a
root r of the product, one of r and its conjugate is a root of the
original.  And the process of factoring completely is well-known.

Incidentally, it is the case that if K is an algebraic extension of L and
every polynomial with coefficients in L has at least one root in K, then K
is an algebraic closure of L.  The argument above pretty much does the
characteristic 0; the prime case is trickier.

This is not entirely elementary, but then neither is the winding number
argument.  Comparing them is difficult because the algebraic argument I
have given could all be taught in a lower division course, while the
winding number argument, intuitively appealing, really requires some
sophisticated stuff to deal with rigorously.

Which raises an interesting question.  If we agree, as we seem to, that
proof is the essence of what we mean by mathematics, then what does it
mean to give an intuitively appealing non-proof and call it a proof?

Michael