The Fundamental Theorem of Algebra (with the Fundamental Group)

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 p(z) be a non-constant polynomial with coefficients in \mathbb{C}. Prove p(z) has a root in \mathbb{C}.

Solution: We may assume without loss of generality that p(z) is monic. So let

\displaystyle p(z) = a_0 + a_1z + \dots + a_{n-1}z^{n-1} + z^n.

Supposing p(z) has no roots in \mathbb{C}, we will show p is constant. First, consider for a fixed r \in \mathbb{C} the loop

\displaystyle f_r(s) = \frac{p(re^{2 \pi is})/p(r)}{\left | p(re^{2 \pi is})/p(r) \right |}

Indeed, by assumption the denominators are never zero, so this function is continuous for all s \in [0,1]. Further, each value f_r(s) is on the unit circle in \mathbb{C} by virtue of the scaling denominator (|f_r(s)| = 1 for all s,r). Finally, f_r(0) = (p(r)/p(r)) / |p(r)/p(r)| = 1, and f_r(1) yields the same value, so this is a closed path based at 1.

We note this function is continuous in both s and r (indeed, they are simply rational functions defined for all s,r), so that f_r(s) is a homotopy of loops as r varies. If r=0, then the function is constant for all s, and so for any fixed r, the loop f_r(s) is homotopic to the constant loop.

Now fix a value of r which is larger than both |a_0| + \dots + |a_{n-1}| and 1. For |z| = r, we have

\displaystyle |z^n| = r \cdot r^{n-1} > (|a_0| + \dots + |a_{n-1}|)|z^{n-1}|

And hence |z^n| > |a_0 + a_1z + \dots + a_{n-1}z^{n-1}|. It follows that the polynomial p_t(z) = z^n + t(a_{n-1}z^{n-1} + \dots + a_0) has no roots when both |z| = r and 0 \leq t \leq 1. Fixing this r, and replacing p with p_t(z) in the formula for f_r(s), we have a homotopy from f_r(s) (when t=1, nothing is changed) to the loop which winds around the unit circle n times, where n is the degree of the polynomial. Indeed, plug in t=0 to get f_r(s) = (r^ne^{2 \pi ins}/r^n)/|r^ne^{2 \pi ins}/r^n|, which is the loop \omega_n(s) = e^{2 \pi ins}.

In other words, we have shown that the homotopy classes of f_r and \omega_n are equal, but f_r is homotopic to the constant map. Translating this into fundamental groups, as \pi_1(S^1,1) = \mathbb{Z}, we note that [\omega_n] = [f_r] = 0, but if \omega_n = 0 then it must be the case that n = 0, as \mathbb{Z} is the free group generated by \omega_1. Hence, the degree of p to begin with must have been 0, and so p must be constant. \square

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