MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question

46 Answers 46

When I was a freshman, I was asked to prove the fundamental theorem of algebra on the final exam for multivariable calculus (I'm completely serious: I think the problem just stated the FTA and asked us to give a proof.)

I didn't succeed, but what I was supposed to do (I think) was apply the Gauss-Bonnet Theorem. One version of this proof appeared recently:

Yet another application of the Gauss-Bonnet Theorem for the sphere J. M. Almira and A. Romero Source: Bull. Belg. Math. Soc. Simon Stevin Volume 14, Number 2 (2007), 341-342.

In this paper the authors use the version of Gauss-Bonnet that relates the Gaussian curvature to the Euler characteristic.

I guess there's another version of this in which one instead uses the version of Gauss-Bonnet saying that the Euler characteristic is the same as the sum of the indices of any vector field (sometimes this theorem is attributed to Poincaré).

The vector field to consider is just $z \mapsto 1/p(z)$, which is well-defined for non-constant polynomials $p(z) = z^n + a_{n-1} z^{n-1} + \cdots + a_0$ without roots, because it vanishes at infinity. The index at infinity for this vector field is the degree of $p$. So if $p$ is a non-constant polynomial without roots, we'd need to have deg$(p) = \chi(S^2) = 2$. Since degree 2 polynomials have roots (the quadratic formula!), this completes the proof.

share|cite|improve this answer
It's been 11 and a half years since I failed to come up with that on my final; I hope I've got it right now! – Dan Ramras Sep 20 '10 at 1:47
Was it Hubbard that asked that question? – Ryan Budney Sep 20 '10 at 1:49
No, it was Guckenheimer. Hubbard went to France in the spring semester, and we strayed a bit from Hubbard's book. – Dan Ramras Sep 20 '10 at 1:52

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, 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 $z = z_0$.

share|cite|improve this answer
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:… – David Roberts Jul 14 '15 at 15:17

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 $w=f(z)$.

share|cite|improve this answer
This is the same as the top-voted answer…, 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

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.

share|cite|improve this answer
This looks more or less exactly the same as the top-voted answer:… – 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

There seems to be a new elementary proof using only bolzano-weierstraß and an inequality:

share|cite|improve this answer
A working link: – Emil Jeřábek Jul 31 '12 at 17:52

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.

share|cite|improve this answer
Here is the proof, as an "unconventional type of induction": – Goldstern May 31 '15 at 22:47

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.

share|cite|improve this answer

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.

share|cite|improve this answer
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. – Michael Blackmon 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-Alvarez 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

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.

share|cite|improve this answer

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.

share|cite|improve this answer

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.

share|cite|improve this answer

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 $S^4$.

share|cite|improve this answer

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 $D$. $\Box$

share|cite|improve this answer

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.

share|cite|improve this answer
Nice! Using the order 2 example makes it much clearer! – Vincent May 5 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 at 14:44

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]

share|cite|improve this answer
This is mentioned in Amy Pang's answer. – Qiaochu Yuan Jul 30 '12 at 20:23

There is an elementary proof which considers the polynomial $ p(z) $, $z \in \mathbb{C}$ as a function of $ (r,\theta) $ where $ z= r e^{-i \theta} $, $r\ge 0$ . The only assumption is that $z^{n}=a $, $a \in \mathbb{R}$ has a solution for all integer $n$. The only elementary theory used are as follows-
1. Properties of continuous functions on two dimensional plane. One important point is that continuous functions with real values divide the plane into two regions, positive and negative. The zeros separating these regions form a continuous curve or region.
2. Real and complex part of a continuous complex function are continuous.
3. The value of a polynomial $p(x) , x \in \mathbb{R}$ with real coefficients is dominated by its higher power for large absolute values of $x$.
4. For a continuous function with real values on the complex plane, intermediate value theorem holds on any continuous curve on the complex plane. This can be proved by parametrization of the curve.
5. $Re(z) < |z|, \forall z \in \mathbb{C}$.

The idea is to track the real and imaginary parts of a polynomial $p(r,\theta)$ separately. First showing that the real part $ Re(p(r,\theta))$ has atleast one zero on a big circle $r=R$ on the complex plane, we can further show that the curve containing this zero is open. Further analysis shows that the values of imaginary part of $p(r,\theta), Im(p(r,\theta))$ is positive on some point of this curve and negative at another for large $r$. Then $Im(p(r,\theta))=0$ somewhere on the particular curve $Re(p(r,\theta))=0$ we were tracking. This gives a solution for $p(r,\theta)=0$.

The proof can be visualized in the following way-

Writing $ z= r e^{-i \theta} $ we can find a big enough $r=R$ such that $ Re(p(R,\theta)) $ is positive at $\theta=\frac {2\pi k} {n} + \frac {\phi} {n} $ and negative at $\theta=\frac {2\pi k} {n} + \frac {\pi} {n} + \frac {\phi} {n} $. This is because $Re(p(R,\theta)) $ is dominated by $r^{n}cos(n\theta + \phi)$. $\phi$ is the angle of the complex coefficient of $z^{n}$. So $Re(p(R,\theta))=0$ at some point in the interval $\theta= \left( \frac {2\pi k} {n} + \frac {\phi} {n} , \frac {2\pi k} {n} + \frac {\pi} {n} + \frac {\phi} {n} \right) \forall k \in \mathbb{Z}$. So $ Re(p(r,\theta))=0 $ is nonempty. As a consequence of this and the continuity of $ Re(p(r,\theta)) $, $ Re(p(r,\theta))=0 $ containing this point forms a continuous curve on the complex plane.

The arc $r=R$ between $\theta=\frac {2\pi k} {n} + \frac {\phi} {n}$ and $\theta=\frac {2\pi k} {n} + \frac {\pi} {n} + \frac {\phi} {n} $ is called a even sector for even $k$ and is called a odd sector for odd $k$. On these sectors, the leading coefficient of $Im(p(r,\theta))$ tends towards $\infty$ in the even sectors and towards $-\infty$ in the odd sectors. So if any of the continuous curves $Re(p(r,\theta))=0$ cross the arc $r=R$ in an even sector and an odd sector, we must have a root on this curve.

If we increase $r \ge R$ continuously on any sector, we get a zero for $Re(p(r,\theta))=0$ for some $\theta$. Continuity of $p(r,\theta)$ implies that we get a continuous curve for $ Re(p(r,\theta))=0 $ in any sector for $r\ge R$. Now, zeros of a continuous function divide the plane into two disconnected parts. One of the parts is finite in area if the curve is closed. Otherwise both areas are unbounded. Here, since $Re(p(r,\theta) \ne 0$ $ \forall r \ge R$ and $\theta=\frac {2\pi k} {n} + \frac {\phi} {n} $ or $\theta=\frac {2\pi k} {n} + \frac {\pi} {n} + \frac {\phi} {n} $, $Re(p(r,\theta))=0$ cannot be closed.

Now, there is some curve $Re(p(r,\theta))=0$ which crosses the arc $r=R$ odd number of times on any sector. If we start from an even sector, it must cross $r=R$ on another sector which is odd. This can be demonstrated using exhaustion since there are only finite numbers of arcs (a sequence of $2m$ consecutive integers cannot be paired in only disjoint or nested (even,even) and (odd,odd) pairs). Let us call this curve $f(r,\theta)$.

$Im(p(R,\theta)) $ is dominated by $r^{n}sin(n\theta + \phi)$. So, on $f(r,\theta)$, $Im(p(r,\theta)) $ is positive somewhere in the even sector and negative somewhere in the odd sector for big enough $r$. So $Im(p(z)=0) $ somewhere on the continuous curve $f(r,\theta)$. This gives a solution for $p(r,\theta)=0$.

Additionally, $f(r,\theta)$ is not asymptotic to $\theta=\frac {2\pi k} {n} + \frac {\phi} {n}$ or $\theta=\frac {2\pi k} {n} + \frac {\pi} {n} + \frac {\phi} {n}$. This can be proven by using the fact that $Re(z) < |z|, \forall z \in \mathbb{C}$. If this was not true, domination of $Im(p(r,\theta))$ on $f(r,\theta)$ by $r^{n}$ would be problematic since $sin(n \theta + \phi)$ might tend towards zero for increasing $r$.

Finally, for every root found, we can do division algorithm to get another polynomial of degree $n-1$. We know that quartic and lower order polynomials with complex roots are always solvable using radicals. So induction shows that all polynomials with real coefficients are solvable on the complex plane.

share|cite|improve this answer
"odd ones always have a rreal root by intermediate value theorem." What about $z-i = 0$? – David Roberts Aug 3 '15 at 15:20
This proof only considers real coefficients. The motivation is the original question from which complex numbers were developed, which is - what is the smallest field containing real numbers as subset in which all polynomials with real coefficients have roots?. This is similar to the question- Does $x^{n}=q$ where $q$ is a positive rational have a root?. Real numbers were developed to solve this kind of equations. Later on, the answers to these questions led to much more mathematical structures than these simple questions. – sobasu Aug 3 '15 at 20:04
Thanks David Roberts for your input. I edited the proof to consider complex coefficients. However, I still think for a beginner it is more important to answer the questions for real coefficients and can be later expanded to consider complex coefficients. – sobasu Aug 3 '15 at 22:12

protected by quid Jan 18 at 19:19

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.