[FOM] Only one proof

Vaughan Pratt pratt at cs.stanford.edu
Wed Sep 16 16:03:16 EDT 2009

Rob Arthan wrote:
> On Saturday 12 Sep 2009 4:18 am, Vaughan Pratt wrote:
>> Certainly one can, by purely algebraic means, form from the field C
>> of complex numbers and a polynomial p an extension C' of C containing
>> a root of p.  But to prove the Fundamental Theorem of Algebra that
>> way you need the extra step of showing how to collapse C' to C.
>> How do you do that without appealing to the completeness of C, or
>> some topological counterpart thereof?
> Once you have that every real polynomial of odd degree has a root, no topology 
> is required. The elegant algebraic proof is due to Laplace. See, for example, 
> Ebbinghaus et al., Numbers, Springer GTM no 123, appendix to chap. 4.

We should agree to agree.  As the book says a little earlier (p.109), 
"All proofs require, in the final analysis, the aid of non-algebraic 
(analytic, transcendental) methods and concepts."  In the case of 
Laplace's method it enters with roots of odd degree polynomials.

Laplace's proof is reasonable, but in my view is not as easy to see 
intuitively as the following considerable simplification of Gauss's 
third proof, which oddly does not appear in Ebbinghaus et al (Remmert, 
the author of that chapter, seems to have been unaware of it even though 
it predates the book), and which unlike all other proofs I know delivers 
all n roots in one stage.

The monomial z^m maps any circle of radius r centered on zero to a loop 
that traverses a circle of radius r^m m times.  Hence any polynomial a_0 
+ a_1 z + ... + a_n z^n, as a finite linear combination of such 
monomials for various m <= n, maps such a circle continuously to a loop 
of indefinite but continuous shape, even if the a_i's are complex.  As r 
tends to zero this loop tends to a circle of radius |a_1| r about a_0. 
  If a_0 = 0 then z = 0 is a root, if not then for sufficiently small r 
the loop does not wind around zero, that is, its winding number about 
zero is zero.  As r tends to infinity this loop tends to an n-fold 
circle of radius |a_n| r^n about zero, whence its winding number about 
zero is n.  Hence for some real r > 0 the winding number must change 
discretely, at which instant the loop must pass through zero, giving a root.

In fact we can associate a root with each increment by one of the 
winding number, so this argument delivers all n roots at once.  If the 
winding number changes by more than one at any instant, remove this 
degeneracy by a suitable perturbation of the coefficients and then let 
the coefficients return continuously to their original values to see 
that there must have been repeated roots there.  "Suitable" can be any 
weighted sum of the original coefficients with the coefficients of a 
polynomial known to have all roots distinct; shift the weight 
continuously from one polynomial to the other.

Vaughan Pratt

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