The fundamental theorem of algebra (polynomials have roots in the complex numbers) is a classic result which one has very little idea of how to do at school, and at the end of an undergraduate course has about four different ways to prove, maybe using Liouville’s theorem, facts about maps between Riemann spheres (compactness implies surjective), maybe using Rouche’s theorem and facts about winding numbers, which amounts to the intuitive topological proof involving thinking about small circles and big circles and how many times they orbit the origin.

While reading around, I stumbled across the first almost purely algebraic proof I’ve ever seen in Lang’s epic “Algebra” book. I’d say purely algebraic, but we are going to need the fact that every odd degree polynomial has a solution in the reals, which is an analytic fact, but a much more obvious one than Liouville’s theorem. However, from this we’ll be able to get to the fundamental theorem of algebra just using basic Galois theory and group theory, and the easily verifiable fact that every quadratic equation with complex coefficients has a complex root. Indeed, once one knows that these are the ingredients, the proof is just “throw them all into a pan.”

Here we go. Consider an arbitrary finite extension $E/\mathbb{C}$. Since the reals have characteristic zero, E is separable over $\mathbb{R}$ and we may therefore take $F/\mathbb{R}$ Galois and containing $E$. Consider the Sylow 2-group of the Galois group, and note by Galois theory that it fixes a field of odd degree over $\mathbb{R}$. Since there are no nontrivial odd algebraic extensions of $\mathbb{R}$ (there are no odd degree irreducible polynomials except the linear ones), this must in fact be the entire Galois group. Consider the subgroup corresponding to $\text{Gal}(F/\mathbb{R}(i))$. If nontrivial, this is a 2-group, so has a subgroup of index 2, and hence by Galois theory $\mathbb{R}(i)$ has a degree 2 extension. But this contradicts the fact that all quadratic equations with complex coefficients have complex roots, so $\text{Gal}(F/\mathbb{R}(i))$ is trivial, hence $E =\mathbb{C}$, so the complex numbers have no nontrivial algebraic extensions.