I read in the book "Concepts of modern mathematics" by Ian Stewart that it was possible to proof the fundamental theorem of algebra using the hairy ball theorem (complete reference to the page is in the following quora question : https://www.quora.com/How-can-the-hairy-ball-theorem-be-used-to-prove-the-fundamental-theorem-of-algebra). I am not aware of such a proof. It is not obvious to me (compactification does not seem to lead to a satisfactory answer) and I would be delighted if someone could help me with that.

  • $\begingroup$ I see the idea, but I am not sure that it works that easy. It is not obvious to me that we can extend a polynomial into a continuous function on its one-point compactification. For instance P(z) = z tends to +\infty when $z$ real tends to $+\infty$ or $-\infty$ if $z$ tends to $-\infty$, and thus the function on the compactified space would not be continuous, isn't it ? Also, take a constant polynomial, and the approach seems to be contradictory. maybe this is the right approach, but there are some technical difficulties that I do not manage to overcome. $\endgroup$ May 2, 2016 at 10:22
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    $\begingroup$ non-constant Polynomials are proper hence extend to the one point compactification $\endgroup$
    – Thomas Rot
    May 2, 2016 at 11:49
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    $\begingroup$ Right, I didn't read your comment properly. The 'point' of the one-point compactification is that all infinities ($+\infty, -\infty, i \infty$ etc) are the same point (often just denoted $\infty$) so that the extension to the sphere IS continuous (as Thomas also says). But this extension takes values in the sphere rather than $\mathbb{C}$ as, for instance, $f(\infty) = \infty$ $\endgroup$
    – Vincent
    May 2, 2016 at 11:53
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    $\begingroup$ Both theorems amount to $\pi_1(S^1)=\mathbb{Z}$, so in this sense they are equivalent. $\endgroup$ May 2, 2016 at 11:58
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    $\begingroup$ You probably want the slightly more sophisticated version of the Hairy Ball theorem which says that the sum of the indices of the zeros equals 2. $\endgroup$
    – HJRW
    May 2, 2016 at 17:50

2 Answers 2


Maybe this is an idea. It suffices to prove that any polynomial $f\in \mathbf C[z]$ of even degree $2d\geq2$ has a zero in $\mathbf C$. Assume that it has no zero. Then the rational section $$ s=f\cdot\left(\tfrac{\partial}{\partial z}\right)^{\otimes d} $$ of the $d$-th tensor power $T^{\otimes d}$ of the holomorphic tangent bundle on the complex projective line $\mathbf P^1(\mathbf C)$ is a nowhere vanishing global section. This is so since the holomorphic vector field $\partial/\partial z$ does not vanish on $\mathbf C$, has a zero of order $2$ at $\infty$, the polynomial $f$ does not vanish on $\mathbf C$, and has a pole of order $2d$ at $\infty$. Consider the section $s$ as a topological surface in the complex line bundle $T^{\otimes d}$ over $\mathbf P^1(\mathbf C)$. As image of the sphere $S^2$, it is homeomorphic to $S^2$ and does not have any nontrivial topological coverings. Its inverse image under the power-$d$-map $$ T\rightarrow T^{\otimes d} $$ from the complex line bundle $T$ into the complex line bundle $T^{\otimes d}$ is, therefore, a disjoint union of $d$ disjoint copies of the sphere $S^2$ in the complement of the zero section of the tangent bundle of $S^2$. Each would define a nowhere vanishing vector field on $S^2$. Contradiction by the hairy ball theorem.


This is a (very) partial answer: Suppose the polynomial is real and of odd degree. Then this polynomial is the characteristic polynomial of some matrix acting in an odd number of dimensions. Restricting this matrix to the sphere and then projecting onto the tangent space of the sphere defines a vector field on an even dimensional sphere, hence by the Hairy Ball Theorem it vanishes somewhere. The place where it vanishes is an eigenvector of the matrix, hence there is a corresponding eigenvalue, hence the polynomial has a root.

I'm not sure if something similar can be done for arbitrary polynomials though.

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    $\begingroup$ Note that in the real odd degree case the polynomial takes different signs for large positive and negative arguments, so you can get by with a topologically simpler connectedness and continuity argument. $\endgroup$
    – Noah Stein
    May 2, 2016 at 18:20
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    $\begingroup$ @NoahStein Yes of course, that's the standard proof, but it gives no connection of the result with the hairy ball theorem which is what the question is about. $\endgroup$ May 2, 2016 at 22:09

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