The Fundamental Theorem of Algebra tells us that any polynomial with real coefficients can be written as a product of linear factors over $\mathbb{C}$. If we don't want to use $\mathbb{C}$, the best we can say is:

If $p(x)$ is a polynomial with real coefficients, $\deg p > 2$, then there exists some quadratic polynomial $q(x)$ such that $q(x)$ is a factor of $p(x)$. (There are then two cases: either $q(x)$ is irreducible over $\mathbb{R}$, or it can be further factored as a product of linear factors.)

This is easily proven as a corollary of the FTA: If we work over $\mathbb{C}$, $p(x)$ can be written as a product of linear factors, and since a complex number $z$ is a root of $p(x)$ if and only if its conjugate $\bar{z}$ is, we can pair up linear factors so as to get a real quadratic.

(In fact, the theorem above -- which seems at first glance like a weaker form of the FTA -- is equivalent to it, since every real quadratic can be factored over $\mathbb{C}$.)

Is there a way to prove the above theorem without invoking complex numbers? It seems to have a certain value on its own as a property of the reals. For example, the fact that any even-degree real polynomial of $\deg 2n$ can be factored into $n$ (real) quadratics seems like something one should be able to prove without needing to change fields.

A version of this question was asked at Factorize real polynomials to quadratic factors. Proof without fundamental theorem of algebra., but the accepted (and only) answer there just concluded that such a proof would be equivalent to the FTA. I already know that; I'm wondering how one could write such a proof without passing to the algebraic closure.

Or, to put it another way: The statement of the theorem would make sense (and would be true) even if complex numbers had never been invented (or discovered, if you are a Platonist). So it seems like it ought to be provable without using complex numbers. Is it?

  • $\begingroup$ Have you considered the problem through the lens of Kronecker field extensions? Because these technically preceded the FTA. But you would ultimately reconstruct the complex plane. $\endgroup$ Feb 29, 2016 at 17:06
  • $\begingroup$ I think you may be able to imitate the fundamental group proof of FTA by considering the map $\mathbb{R}^2\to\mathbb{R}^2$ that sends $(a,b)$ to $(c,d)$ where $cx+d$ is the remainder when you divide your polynomial by $x^2+ax+b$. I haven't worked out the details (in particular, I haven't checked that the relevant winding number really is nonzero). $\endgroup$ Apr 20, 2017 at 20:03

2 Answers 2


Here is a topological proof. It suffices to show that every irreducible polynomial over $\mathbb{R}$ has degree $\leq 2$, or equivalently that every finite field extension of $\mathbb{R}$ has degree $\leq 2$. So suppose $K$ is a finite extension of $\mathbb{R}$ of degree $d>2$. Note that then $K$ has a natural topology homeomorphic to $\mathbb{R}^d$ and the field operations are continuous, and so $K^\times\cong\mathbb{R}^d\setminus\{0\}$ is a topological abelian group.

We can now invoke some heavy machinery from topology to conclude this is impossible for $d>2$. For instance, any topological abelian group $X$ is weak homotopy equivalent to a product $\prod_n K(\pi_n(X),n)$ of Eilenberg-Mac Lane spaces (one for each of its homotopy groups). But $K(\pi_{d-1}(\mathbb{R}^d\setminus\{0\}),d-1)=K(\mathbb{Z},d-1)$ has nontrivial cohomology in infinitely many dimensions if $d>2$, and so $\mathbb{R}^d\setminus\{0\}$ cannot be weak equivalent to a product with $K(\mathbb{Z},d-1)$ as a factor.

Alternatively, you can observe that the topological abelian group structure on $\mathbb{R}^d\setminus\{0\}$ is actually smooth, so it makes $\mathbb{R}^d\setminus\{0\}$ an abelian Lie group. Any connected abelian Lie group is a product of copies of $\mathbb{R}$ and $S^1$, so this is impossible for $d>2$.


The existence of such factorization can indeed be shown by only real analysis, without invoking FTA. The way is to show that the two remainders obtained during division of any polynomial (coefficients of the remaining linear polynomial) with a real quadratic $x^2-ax-b$ attain zero value simultaneously for some $a$ and $b$.

Showing the existence of such $a$ and $b$ comes from topological arguments on interlacing and continuity on the $a-b$ plane (Similar to Gauss's first proof of FTA). Showing the interlacing is the part which takes a bit of work.

For details, here is a link to my horrible write-up on this (It got rejected straight away from a journal due to the bad writing). Some day I will find time to write a neat publication on this.

Please feel free to ask for any clarifications. It will help me organize the write-up better.


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