# Ways to prove the fundamental theorem of algebra

This seems to be a favorite question everywhere, including Princeton quals. How many ways are there?

Please give a new way in each answer, and if possible give reference. I start by giving two:

1. Ahlfors, Complex Analysis, using Liouville's theorem.

2. Courant and Robbins, What is Mathematics?, using elementary topological considerations.

I won't be choosing a best answer, because that is not the point.

There's a linear algebra proof by Harm Derksen: http://www.jstor.org/pss/3647746. You can also find the article posted here: http://math.berkeley.edu/~ribet/110/f03/derksen.pdf

This is one of the proofs currently on the nLab. My goal in writing it was to see how elementary I could make it, that if you squint a little it might have been a proof from the late $18^{th}$ century. I think it's pretty close; it uses the Bolzano-Weierstrass theorem, proven by Bolzano in 1817. (It's also similar in form to the answer by Timothy Gowers https://mathoverflow.net/a/16671/2926, but with more detail, suitable perhaps for a presentation to an advanced calculus class.)

Let $f\colon \mathbb{C} \to \mathbb{C}$ be a nonconstant polynomial mapping, and suppose $f$ has no zero.

1. Let $s$ be the infimum of values ${|f(z)|}$; choose a sequence $z_1, z_2, z_3, \ldots$ such that ${|f(z_n)|} \to s$. Since $\lim_{z \to \infty} f(z) = \infty$, the sequence $z_n$ must be bounded; by the Bolzano-Weierstrass theorem it has a subsequence $z_{n_k}$ that converges to some point $z_0$. Then ${|f(z_{n_k})|}$ converges to ${|f(z_0)|}$ by continuity, and converges to $s$ as well, so ${|f(z)|}$ attains its absolute minimum $s$ at $z = z_0$. By supposition, $f(z_0) \neq 0$.

2. The polynomial $f$ may be uniquely written in the form $$f(z) = f(z_0) + g(z)(z - z_0)^n$$ where $g$ is polynomial and $g(z_0) \neq 0$. Put $$F(z) = f(z_0) + g(z_0)(z - z_0)^n$$ and choose $\delta \gt 0$ small so that $${|z - z_0|} = \delta \Rightarrow {|g(z) - g(z_0)|} \lt {|g(z_0)|}.$$

3. $F$ maps the circle $C = \{z : {|z - z_0|} = \delta\}$ onto a circle of radius $r = {|g(z_0)|}\delta^n$ centered at $f(z_0)$. (This uses the fact that any complex number has an $n^{th}$ root; an $18^{th}$ century mathematician might have invoked Euler's formula or De Moivre's formula.) Choose $z' \in C$ so that $F(z')$ is on the line segment between the origin and $f(z_0)$ (we can always choose $\delta$ so that also $r \lt {|f(z_0)|}$). Then $${|F(z')|} = {|f(z_0)|} - r.$$ We also have $${|f(z') - F(z')|} = {|g(z') - g(z_0)|} {|z' - z_0|^n} \lt {|g(z_0)|} \delta^n = r$$ according to how we chose $\delta$ in 2. We conclude by observing the strict inequality $${|f(z')|} \leq {|F(z')|} + {|f(z') - F(z')|} \lt {|f(z_0)|} - r + r = {|f(z_0)|},$$ which contradicts the fact that ${|f(z)|}$ attains an absolute minimum at $z = z_0$.

• I wonder if the reverse mathematical strength of FTA is the same as Bolzano-Weierstrass (in the sense that BW is necessary to prove FTA). This would make even stronger Friedman's point that BW over the rationals (a variant on 'every bounded sequence has a Cauchy subsequence') is equivalent to Con(PA), over some base theory. – David Roberts Jun 29 '15 at 2:58
• @DavidRoberts I really don't know. I'm having some difficulty finding the reverse mathematics of FTA through a Google search. Interesting question, though. – Todd Trimble Jun 29 '15 at 12:15
• I would ask on the fom mailing list, but I got off it. So, maybe a new question here? – David Roberts Jun 30 '15 at 0:08
• For completeness, here's the question: mathoverflow.net/questions/210432/… – David Roberts Jul 14 '15 at 15:17

I came across this article when I was pondering teaching the FTA to my multi-variable calculus class. In the end, I didn't have time to include it. It is nice in that it relies only on Green's theorem which we get through in the first semester. On the other hand, it is clear that they are largely being clever at avoiding introducing new definitions or standard theorems (for instance they use Cauchy-Riemann equations for polynomials without really saying so). I think it's nice nevertheless.

http://www.math.binghamton.edu/paul/papers/LoyaFundThmAlg.pdf

• The referee assigned to review that Monthly article didn't carry out proper due diligence: that proof goes back to Gauss! See, for instance, the discussion of the third proof of Gauss in math.huji.ac.il/~ehud/MH/Gauss-HarelCain.pdf. I saw this proof for the first time in Volume II of Fikhtengoltz's "Course of Differential and Integral Calculus" and translated it into English in case I want to use it myself. It's posted on math.uconn.edu/~kconrad/blurbs/analysis/fundthmalgcalculus.pdf and at the end the hidden connection with the argument principle is indicated. – KConrad Jan 17 '10 at 19:30
• ok thanks. I wasn't sure if the place it was posted was for original papers anyways? I thought it was just an exposition. Anyways, it's good to know, thanks! – David Jordan Jan 18 '10 at 21:55
• The place where that article appeared, the American Math. Monthly, is not intended specifically for original papers. But the author really found essentially the proof of Gauss and neither he nor the referee realized it. It should have been presented in the article as a proof that goes back to Gauss but is not so well-known (except if you translate it into complex variables it becomes one of the known complex-analytic proofs). I don't have a problem with the proof appearing in the Monthly, but the lack of awareness that it is a known proof is kind of unfortunate. – KConrad Feb 7 '10 at 1:52

Two very short proofs, mostly topological, that a nonconstant polynomial map $f:{\bf C} \to \bf C$ is surjective (joint work with Robert Palais):

(1) Complex analysis shows that $f$ is an open map (images of open sets are open). A standard estimate, $|f(z)|\to\infty$ as $|z|\to\infty$, implies$f$ is also a closed map (images of closed sets are closed). Thus $f(\bf C)$ is an open, closed, nonempty subset of the connected space $\bf C$, therefore $f(\bf C)=\bf C$.

The openness of $f$ is nontrivial, but it can be replaced by elementary algebra and topology:

(2) The set $K$ of roots of $f'$ is finite. The inverse function theorem shows that the set $A:=f({\bf C})\setminus f(K)$ is open, with finite boundary $A'=f(K) \setminus A$ because $f$ is closed. Thus $A$ has closure $\bar A = f({\bf C})$. Since a finite set cannot disconnect the plane, $\bar A = \bf C$.

A nice feature of these proofs is that they have straightforward (but not trivial) generalizations to higher dimensions:

Theorem: Every nonconstant, closed, holomorphic map between connected, complex n-dimensional manifolds, is surjective.

• This looks more or less exactly the same as the top-voted answer: mathoverflow.net/questions/10535/… – Ryan Reich Aug 8 '12 at 18:43
• By the way, if this is the same as that answer, you should go answer the question in the comments there as to where the proof came from, since it seems that may be you. – Ryan Reich Aug 9 '12 at 2:17

Re Gauss's first proof, Smale pointed out the following (discussed by Weber in the collected papers of Gauss): In order to show that the zero sets of the real and imaginary parts of a complex polynomial intersect, Gauss states (paraphrasing): "If a [polynomial] curve C in the plane enters a region, it must leave it. No one to whom I have explained the meaning of this result doubts it. I will give a proof in a later paper." This seems to mean, for example, that no point has an open neighborhood in the plane meeting C in a half open interval, or in a set homeomorphic to 9. In modern terms: (A) For every p in C the number of branches containing p is even. This is true-- but to use it Gauss would have to prove it without using FTA! (A) follows from a vast generalization due independently to D. Sullivan and Deligne: (A') Let p be a point in an analytic variety X over C or R, and S the boundary of a sufficiently small ball centered at p. Then the Euler characteristic of the intersection of S and X is 0 in the complex case, and even in the real case. So Gauss's celebrated proof has an enormous gap.

This proof uses the open mapping theorem, which I found in Narasimhan's book.

Theorem: Let $f\in C^\omega(\mathbb{C})$, and suppose that $|f(z)|\to\infty$ as $|z|\to\infty$. Then $f(\mathbb{C})=\mathbb{C}$.

Proof: Obviously $f$ is not constant, and so the open mapping theorem implies that $f(\mathbb{C})$ is open. Let us show that $f(\mathbb{C})$ is also closed. Suppose that $f(z_k)\to w\in\mathbb{C}$ as $k\to\infty$. Then $\{z_k\}$ is bounded, so taking a subsequence if necessary, there is $z\in\mathbb{C}$ such that $z_k\to z$. By continuity $f(z_k)\to f(z)$, concluding that $w=f(z)$.

• This is the same as the top-voted answer mathoverflow.net/questions/10535/…, with a different justification for why $f$ is open. – Ryan Reich Aug 8 '12 at 18:50
• @Ryan: That's great. So perhaps this answer should be voted up too :) – timur Aug 8 '12 at 20:18

Another probabilistic proof:

Pascu, Mihai N. A probabilistic proof of the fundamental theorem of algebra. (English summary) Proc. Amer. Math. Soc. 133 (2005), no. 6, 1707–1711 (electronic).

Summary: "We use Lévy's theorem on invariance of planar Brownian motion under conformal maps and the support theorem for Brownian motion to show that the range of a non-constant polynomial of a complex variable consists of the whole complex plane. In particular, we obtain a probabilistic proof of the fundamental theorem of algebra.''

It is different from the probabilistic proof already listed among answers to this question, which uses martingale convergence theorem.

I don't have it on hand, but Ronald Solomon's Abstract Algebra has an interesting proof using symmetric polynomials and induction on the 2-adic valuation of the degree.

• Here is the proof, as an "unconventional type of induction": mathoverflow.net/a/166014/14915 – Goldstern May 31 '15 at 22:47
• This proof is an elementary variant of Artin's proof: it constructs directly the “dévissage” of a 2-group (which is necessarily nilpotent) via its translation action on the set of unordered pairs. – ACL Aug 20 '19 at 20:20

In this post the OP wonders how one can prove the FTA using the hairy ball theorem, after having read a book of Ian Stewart containing an allusion to such a proof. The answer I gave could be added to the current list of proofs as well. There is no doubt that this proof is known. However, it does not seem to be mentioned in the other items of this list.

The main idea of proof is already present when one wants to show that a polynomial $f\in\mathbf C[z]$ of degree $2$ has a zero using the hairy ball theorem. Suppose that it does not have any zeros. Then the holomorphic vector field $$f\tfrac{\partial}{\partial z}$$ is a nonvanishing vector field on the Riemann sphere $S^2$. This is because $f$ has a pole of order $2$ at infinity, while $\partial/\partial z$ has a zero of order $2$ at infinity. The contradiction comes from the hairy ball theorem that states that such vector fields cannot exist. The general case of a polynomial $f$ of even degree $2d$ can be proved by arguing that the vector field $$\sqrt[d]{f\cdot\left(\tfrac{\partial}{\partial z}\right)^{\otimes d}}$$ is well defined, and nonvanishing on $S^2$ if $f$ does not have any zeros in $\mathbf C$. See here for more details.

• Nice! Using the order 2 example makes it much clearer! – Vincent May 5 '16 at 15:06
• I just came across a nice paper of Milnor's containing an elementary and surprising proof of the hairy ball theorem. I join it to make this proof of the FTA complete: Milnor, John: Analytic proofs of the "hairy ball theorem'' and the Brouwer fixed-point theorem. Amer. Math. Monthly 85 (1978), no. 7, 521–524 – Johannes Huisman Jun 20 '16 at 14:44

Not quite an answer, but relevant:

Eilenberg and Niven proved that every "polynomial" in the quaternions has a root (provided it has only one term of highest degree). The trick is familiar: they show that such a polynomial is homotopic to $q\mapsto q^n$, which induces a map of degree $n$ on the one-point compactification of $\mathbb{H}$, namely $S^4$.

• What do you mean by "provided it has only one highest degree"? do you mean the highest term would not be in the form $aq^n+q^nb$? What can be said about such kind polynomials?Do they have always roots? – Ali Taghavi May 28 '18 at 7:27
• @AliTaghavi Because $\mathbb{H}$ is not commutative, you can't always collect all the terms of the form $a_0q a_1 q a_2\cdots a_{n-1}q a_n$ (with $a\in \mathbb{H}$) into a single term of the form $a q^n$. – Jeff Strom May 28 '18 at 23:57

There seems to be a new elementary proof using only bolzano-weierstraß and an inequality: http://de.arxiv.org/PS_cache/arxiv/pdf/1109/1109.1459v1.pdf

Gauss's first proof contained an enormous gap, since he presumed facts equivalent to the Jordan curve theorem to be true. Jordan curve theorem was proven a century later.

There is a modification on Gauss's first proof that uses only basic real analysis concepts (continuity and least upper bound principle for real numbers) on the real and complex parts of a complex polynomial (which are bivariate polynomials in either $(r,\theta)$ or $(x,y)$: On Gauss's first proof of the fundamental theorem of algebra

Sorry for the self promotion, I am an author in the proof.

There is an alternative proof for FTA using "Fredholm operators on Hilbert spaces":

Assume that $P(z)=z^n+a_{n-1}z^{n-1}+\ldots+a_1 z+a_0$ has no root in $\mathbb{C}$. Then for every $\epsilon$ the polynomial $Q(z)=\epsilon^nP(z/\epsilon)=z^n+\epsilon a_{n-1} z^{n-1} +\ldots+\epsilon^{n-1}a_1z+\epsilon^n a_0$ has no root in $\mathbb{C}$, too.

The space of entire Holomorphic functions $Hol(\mathbb{C)}$ is densely embedded in $\ell^2$ via $f(z)=\sum_{n=0}^{\infty} a_nz^n \mapsto (a_0,a_1,\ldots)$. We substitute "$z$" in $Q(z)$ by the shift operator on $\ell^2$. Then it turns out that every polynomial $Q(z)$ defines a bounded linear operator $Q$ on $\ell^2$ which restricts to "multiplicative operator" by polynomial $Q(z)$ on $Hol(\mathbb{C})$ so the operator $Q$ keeps $Hol(\mathbb{C})$ invariant. Moreover a non vanishing polynomial $Q(z)$ determines an operator on $\ell^2$ which restricts to a surjective operator on $Hol(\mathbb{C})$.

Note that the operator $Q\in B(\ell^2)$ described above is a perturbation of the $n$-shift operator so $Q$ is a fredholm operator of index $-n$. On the other hand it is a perturbation of the $n$-shift operator which is an isometry. So $Q$ satisfies $|Q(v)|>k|v|$ for all $v \in \ell^2$ and for some constant $k$. Then $Q$ is an injective operator so it can not be a surjective operator on $\ell^2$ since its index( $-n$ )is nonzero. On the other hand, since the restriction of $Q$ to a dense subspace of $\ell^2$ is surjective and $Q$ satisfies $|Q(v)>k|v|$, then $Q$ must be a surjective operator on $\ell^2$, a contradiction.

Added: This proof is based on the following note in arxiv however this arxived paper was not written well.(Full of typos, grammar problem, mistakes in english writing).

https://arxiv.org/abs/math/0509113v1

At the risk of being highly downvoted, I can't resist reposting my comment to Andrew L's answer (or rather, question) below:

is there a purely algebraic proof that for any non constant $P$ in $\mathbb{Q}[i][X]$ and $\epsilon>0$ in $\mathbb{Q}$, there is $q$ in $\mathbb{Q}[i]$ s.t. $|P(q)|<\epsilon$?

I think the statement above is purely algebraic, but I have to admit I'm a bit uncertain as to where the boundary between algebra and analysis falls.

• It is not so clear to me whether to consider "<" a purely algebraic concept or not. – Qfwfq Oct 17 '10 at 16:30
• formally its just a relation, a symbol in the language of ordered spaces with no intrinsic significance, or assumptions, beyond those associated with it in the definition. – Not Mike Feb 3 '11 at 0:27
• @Michael: that is probably the comment with least mathematical content I have ever read one MO :D – Mariano Suárez-Álvarez Apr 15 '12 at 4:20
• Rule of thumb: there is an epsilon of difference between algebra and analysis :) – Ryan Reich Aug 8 '12 at 18:59
• @Qfwfq: please see Matt F.'s comments here: mathoverflow.net/a/348665/2480 – Yaakov Baruch Dec 23 '19 at 13:42

Maybe I should have posted this as a comment to Gian Maria Dall'Ara very nice proof, because is a mere variation.

He uses the lemma : Any open proper map to a locally compact space surjects the connected components it reaches. Now any non-constant polynomial corestricted to its regular value locus is as in the lemma, so it is surjective.

Here is a "constructive proof" of the D'Alembert-Gauss theorem. Fix a degree $n > 0.$

Consider the $M_n$ the affine space of monic polynomials of degree $n$ and the proper map $\mathbb C^n \to M_n$, mapping a tuple $(z_1, \ldots , z_n)$ to $\Pi (X-z_i).$ Its critical values locus is non-disconnecting because it is a complex (singular) hypersurface : it is a polynomial image of the arrangement of hyperplanes { $z_i = z_j$}.

The map is therefore surjective.

Let $p$ be a complex polynomial of degree $n$.

• By analytic continuation arguments applied to any branch of $p^{-1}$, it can be shown that any level curve $\Lambda=\{z:|p(z)|=\epsilon\}$ of $p$ which does not contain a zero of $p'$ is a Jordan curve.

• Since $p'$ has at most $n-1$ zeros, we can find some point $z$ such that the level curve $\Lambda_z$ of $p$ which contains $z$ does not contain a critical point.

• Let $D$ denote the bounded face of $\Lambda_z$. By the maximum modulus theorem applied to $f$ on $D$, $f$ has a zero in $D$. $\Box$

There is a proof using clutching functions over the sphere and the first Chern class. It is quite similar to the fundamental group proof of FTA. The trick is a polynomial without zeroes allows one to construct an isomorphism between a vector bundle with first Chern class $\deg d$ and a vector bundle with first Chern class $0$ using the polynomial restricted to circles concentric with the origin as clutching functions.

The details: From a polynomial $p: \mathbb{C} \to \mathbb{C}$, $p(z) = \sum_{j=0}^d a_j z^j$ we can construct a continuous family $p_t: \mathbb{C} \times[0,1] \to \mathbb{C}$ of polynomials such that $p_1(z)(z) = a_d z^d$ and $p_0(z) = a_0$. If $p$ has no zeroes, one can construct $p_t$ in such a way that $p_t$ has no zeroes on the unit circle.

This means that for a fixed $t \in [0,1]$ we can use $p_t$ restricted to the circle as a clutching function for $S^2$. Since $p_t$ is continuous family, this gives a vector bundle $E$ over $S^2 \times [0,1]$. It is a standard fact in the theory of vector bundles that $E$ restricted to $S^2 \times \{0\}$ isomorphic to $S^2 \times\{0\}$. But our construction allows us to read off that the former has first chern class $d$, while the latter has first chern class $0$. Hence $d = 0$ and $p$ must be constant.

I'm surprised that no-one's mentioned the proof using Roueche's theorem:

Given $f,g$ holomorphic and $C$ a closed contour if $|g(z)|< |f(z)|$ on $C$ then $f$ and $f+g$ have the same number of zeros (counting multiplicity) in the interior of $C$. There's an easy proof of this using the Cauchy integral formula.

If Let $g(z) = a_{n-1} z^{n-1} + \cdots + a_0$, and $f(z) = z^n$. If $R$ is sufficiently big then $|g(z)|<|f(z)|$ on the circle of radius $R$ with the center at 0. Thus $p(z) := z^n + g(z)$ has $n$ zeros inside that circle.

[As a side note, when I was taught this by Lipman Bers, he picturesquely referred to it as the "dog on the leash theorem" -- it's essentially a winding number argument]

• This is mentioned in Amy Pang's answer. – Qiaochu Yuan Jul 30 '12 at 20:23