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. $\square$

Want to respond? Send me an email or find me elsewhere on the internet.