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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.

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41 Answers 41

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 $w=f(z)$.

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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.

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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
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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

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.

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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.

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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

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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.

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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.

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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.

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It is not so clear to me whether to consider "<" a purely algebraic concept or not. –  Qfwfq Oct 17 '10 at 16:30
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Rule of thumb: there is an epsilon of difference between algebra and analysis :) –  Ryan Reich Aug 8 '12 at 18:59

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 $S^4$.

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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]

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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.

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