# A PDE-analytic proof of the fundamental theorem of algebra

The fundamental theorem of algebra is one of the most important theorems in mathematics, being core to algebraic geometry and complex analysis. Unraveling the definitions, it says:

Fundamental theorem of algebra. Let $f$ be a polynomial over $\mathbf C$ of degree $d$. Then the equation $f(z) = 0$ has $d$ solutions $z$, counting multiplicity.

Famously, most proofs of the fundamental theorem of algebra are complex-analytic in nature. Indeed, complex analysis is the natural arena for such a theorem to be proven. One has to use the fact that $\mathbf R$ is a real closed field, but since there are lots of real closed fields, one usually defines $\mathbf R$ in a fundamentally analytic way and then proves the intermediate value theorem, which shows that $\mathbf R$ is a real closed field. One can then proceed by tricky algebraic arguments (using, e.g. Galois or Sylow theory), or appeal to a high-powered theorem of complex analysis. Since the fundamental theorem is really a theorem about algebraic geometry, and complex analysis sits somewhere between algebraic geometry and PDE analysis in the landscape of mathematics (and we need some kind of analysis to get the job done; purely algebro-geometric methods will not be able to distinguish $\mathbf R$ from another field $K$ such that $-1$ does not have a square root in $K$) it makes a lot of sense to use complex analysis.

But, since complex analysis sits between algebraic geometry and PDE analysis, why not abandon all pretense of respectability (that is to say, algebra — analysis is not a field worthy of the respect of a refined mathematician) and give a PDE-analytic proof? Of course, this proof will end up “looking like” multiple complex-analytic proofs, and indeed it is basically the proof by Liouville’s theorem dressed up in a trenchcoat (and in fact, gives Liouville’s theorem, and probably some other complex-analytic results, as a byproduct). In a certain sense — effectiveness — this proof is strictly inferior to the proof by the argument principle, and in another certain sense — respectability — this proof is strictly inferior to algebraic proofs. However, it does have the advantage of being easy to teach to people working in very applied fields, since it entirely only uses the machinery of PDE analysis, rather than fancy results such as Liouville’s theorem or the Galois correspondence.

The proof
By induction, it suffices to prove that if $f$ is a polynomial with no zeroes, then $f$ is constant. So suppose that $f$ has no zeroes, and introduce $g(z) = 1/f(z)$. As usual, we want to show that $g$ is constant.

Since $f$ is a polynomial, it does not decay at infinity, so $g(\infty)$ is finite. Therefore $g$ can instead be viewed as a function on the sphere, $g: S^2 \to \mathbf C$, by stereographic projection. Also by stereographic projection, one can cover the sphere by two copies of $\mathbf R^2$, one centered at the south pole that misses only the north pole, and one centered at the north pole that only misses the south pole. Thus one can define the Laplacian, $\Delta = \partial_x^2 + \partial_y^2$, in each of these coordinates; it remains well-defined on the overlaps of the charts, so $\Delta$ is well-defined on all of $S^2$. (In fancy terminology, which may help people who already know ten different proofs of the fundamental theorem of algebra but will not enlighten anyone else, we view $S^2$ as a Riemannian manifold under the pushforward metric obtained by stereographic projection, and consider the Laplace-Beltrami operator of $S^2$.)

Recall that a function $u$ is called harmonic provided that $\Delta u = 0$. We claim that $g$ is harmonic. The easiest way to see this is to factor $\Delta = 4\partial\overline \partial$ where $2\partial = \partial_x - i\partial_y$. Then $\overline \partial u = 0$ exactly if $u$ has a complex derivative, by the Cauchy-Riemann equations. There are other ways to see this, too, such as using the mean-value property of harmonic functions and computing the antiderivative of $g$. In any case, the proof is just calculus.

So $g$ is a harmonic function on the compact connected manifold $S^2$; by the extreme value theorem, $g$ has (or more precisely, its real and imaginary parts have) a maximum. By the maximum principle of harmonic functions (which is really just the second derivative test — being harmonic generalizes the notion of having zero second derivative), it follows that $g$ is equal to its maximum, so is constant. (In fancy terminology, we view $g$ as the canonical representative of the zeroth de Rham cohomology class of $S^2$ using the Hodge theorem.)