This post will take a familiar theorem in a few less familiar directions.

The Fundamental Theorem of Algebra (FTA) says that an *n*th degree polynomial over the complex numbers has *n* roots. The theorem is commonly presented in high school algebra, but it’s not proved in high school and it’s not proved using algebra! A math major might see a proof midway through college.

You’re most likely to see a proof of the Fundamental Theorem of *Algebra* in a course in complex *analysis*. That is because the FTA depends on analytical properties of the complex numbers, not just their algebraic properties. It is an existence theorem that depends on the topological completeness of the complex numbers, and so it cannot be proved from the algebraic properties of the complex numbers alone.

(The dividing lines between areas of math, such as between algebra and analysis, are not always objective or even useful. And for some odd reason, some people get touchy about what belongs where. But the categorization of math subjects implicit in the creation of course catalogs puts the proof of the FTA on the syllabus of complex analysis courses, sometimes topology courses, but not algebra courses.)

Gauss offered a proof of the FTA in 1799, but his proof was flawed because he implicitly assumed the Jordan curve theorem, a seemingly obvious theorem that wasn’t rigorously proven until 1911. In keeping with our the theme above, the hole was due to overlooking a topological subtlety, not due to an algebraic mistake.

The proof of the FTA using complex analysis is a quick corollary of Liouville’s theorem: If a function is analytic everywhere in the complex plane and bounded, then it must be constant. Assume a non-constant polynomial *p*(*z*) is nowhere zero. Then 1/*p*(*z*) is analytic everywhere in the complex plane. And it must be bounded because *p*(*z*) → ∞ as |*z*| → ∞. We have a contradiction, so *p*(*z*) must be zero for some *z*.

The usual statement of the FTA says that an *n*th degree polynomial has *n* roots, but the proof above only says it must have one root. But you can bootstrap your way to the full result from there. The analyst tells the algebraist “I’ve done the hard part for you. You can take it from there.”

In some sense the rest of the theorem properly belongs to algebra, because the bookkeeping of how to count zeros is more of an algebraic matter. In order for every *n*th degree polynomial to have exactly *n* roots, some roots must be counted more than once. For example, (*z* − 42)² has a double root at 42. From a strictly analytical view point, there is one point where (*z* − 42)² is zero, namely at *z* = 42, but it is useful, even to analysts, to count this as a double root.

It may seem pedantic, or even Orwellian, so say that some roots count more than others. But a great deal of theorems are simpler to state if we agree to count roots according to multiplicity. This idea is taken further in algebraic geometry. For example, Bézout’s theorem says that if *X* is an *m*th degree curve and *Y* is an *n*th degree curve, then *X* and *Y* intersect in *mn* points. It takes a lot of prep work for this theorem to have such a simple statement. You have to count intersections with multiplicity, and you have to add points at infinity.

Liouville’s theorem’s is useful for more than proving the FTA. It is also useful in studying elliptic functions. (I believe Liouville proved his theorem for this purpose. Someone into math history can check me on that.) If an analytic function is periodic in two distinct directions in the complex plane (more on that here), it must have a singularity because otherwise it would be bounded.

The FTA may seem less profound than it is. You might object that of course every *n*th degree polynomial has *n* roots: you’ve cheated by adding in the roots you need. But that’s not true. The complex numbers are created by adding in one root to one particular polynomial, namely *z*^{2} = −1. The impressive part is that adding *one* root of *one* particular quadratic polynomial is enough to force *all* quadratic polynomials and even *all higher degree* polynomials also to also have roots.

This does not happen in finite fields. A finite field cannot be algebraically complete. If you extend a finite field so that every quadratic equation has a root, then only quadratic equations have roots. It does not follow that cubic polynomials, for example, have roots. You could extend it further so that all cubic equations have roots, but then 4th degree polynomials will not all have roots. If you stop the extension process after any finite number of steps, there will be some polynomials without roots.

What kind of high school does complex analysis? :-)

Btw, there are recent proofs of FTA that are elementary in the sense that they don’t require complex analysis. As I recall, they use linear algebra. I suppose that the same high school that taught complex analysis taught linear algebra too :).

You can avoid using complex analysis, but you have to use the topological completeness of the complex numbers somewhere. Some proofs, for example, use the intermediate value theorem.

If you wanted to prove the FTA as quickly as possible, starting with high school algebra, you wouldn’t take Liouville’s route. But if you want to prove FTA with minimal detour from exploring widely applicable math, Liouville’s proof is hard to beat.

Actually the first place I saw a proof of the fundamental theorem of algebra was in an algebra course. The proof basically made use of a lot of Galois theory to reduce the problem down, then used the intermediate value theorem. It felt like it was making use of rather deep algebraic theory just to avoid using analysis as much as possible. The intermediate value theorem is one of the simplest results from analysis. On the other hand, the simplicity of the complex analysis version of the proof belies the complexity needed to achieve Liouville’s theorem. I’m not sure the one proof is actually simpler, rather it just hides more of its steps behind a lemma.