This post assumes familiarity with some basic concepts in algebraic topology, specifically what a group is and the definition of the fundamental group of a topological space.
The fundamental theorem of algebra has quite a few number of proofs (enough to fill a book!). In fact, it seems a new tool in mathematics can prove its worth by being able to prove the fundamental theorem in a different way. This series of proofs of the fundamental theorem also highlights how in mathematics there are many many ways to prove a single theorem, and in re-proving an established theorem we introduce new concepts and strategies. And perhaps most of all, we accentuate the unifying beauty of mathematics.
Problem: Let be a non-constant polynomial with coefficients in
. Prove
has a root in
.
Solution: We may assume without loss of generality that is monic. So let
.
Supposing has no roots in
, we will show
is constant. First, consider for a fixed
the loop
Indeed, by assumption the denominators are never zero, so this function is continuous for all . Further, each value
is on the unit circle in
by virtue of the scaling denominator (
for all
). Finally,
and
yields the same value, so this is a closed path based at 1.
We note this function is continuous in both and
(indeed, they are simply rational functions defined for all
), so that
is a homotopy of loops as
varies. If
then the function is constant for all
, and so for any fixed
the loop
is homotopic to the constant loop.
Now fix a value of which is larger than both
and
. For
, we have
And hence . It follows that the polynomial
has no roots when both
and
. Fixing this
, and replacing
with
in the formula for
, we have a homotopy from
(when
, nothing is changed) to the loop which winds around the unit circle
times, where
is the degree of the polynomial. Indeed, plug in
to get
, which is the loop
.
In other words, we have shown that the homotopy classes of and
are equal, but
is homotopic to the constant map. Translating this into fundamental groups, as
, we note that
, but if
then it must be the case that
, as
is the free group generated by
. Hence, the degree of
to begin with must have been 0, and so
must be constant.
Thank you! The details you provided really helped me to understand the proof.
LikeLike