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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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48 Answers 48

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Here is the proof of the equivalent statement "Every complex non-constant polynomial $p$ is surjective".
1) Let $C$ be the finite set of critical points , i.e. $p'(z)=0$ for all $z\in C$. $C$ is finite by elementary algebra.

2) Remove $p(C)$ from the codomain and call the resulting open set $B$ and remove from the domain its inverse image $p^{-1}\left( p (C) \right)$, and call the resulting open set $A$. Note that the inverse image is again finite.

3) Now you get an open map from $A$ to $B$, which is also closed, because any polynomial is proper (inverse images of compact sets are compact). But $B$ is connected and so $p$ is surjective.

I like this proof because you can try it for real polynomials and it breaks down at step 3) because if you remove a single point from the line you disconnect it, while you can remove a finite set from a plane leaving it connected.

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  • $\begingroup$ This is my favorite proof yet! $\endgroup$ – Pietro KC May 16 '10 at 23:56
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    $\begingroup$ Hi Gian, do you know who proved this first ? I came up with precisely this proof. It was published in the American Mathematical Monthly, November 2000 issue. $\endgroup$ – Cosmonut May 18 '10 at 5:04
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    $\begingroup$ Alternatively presented: Consider the map from the Riemann sphere to the Riemann sphere induced by a polynomial. This is a continuous map from a compact space to a Hausdorff space, and thus its image is closed. It is also trivially holomorphic, and thus, if non-constant, its image is open. But the domain is inhabited and the codomain is connected, so this map must be surjective. $\endgroup$ – Sridhar Ramesh Oct 14 '11 at 20:20
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    $\begingroup$ Perhaps it is worth noting that this answer led to another MO question: mathoverflow.net/questions/132036/… $\endgroup$ – Benjamin Dickman Nov 7 '13 at 10:14
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    $\begingroup$ @GianMariaDall'Ara citing the Open Mapping Theorem of Complex Analysis isn't the best, since the proof would then fail earlier for real polynomials. But you could instead say that the map is a local homeomorphism on A (from the inverse function theorem and A only having regular points), and is therefore open on A. $\endgroup$ – Fred Akalin Aug 28 '18 at 0:53
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Here is a standard algebraic proof. It suffices to show that if $L/\mathbb{C}$ is a finite extension, then $L=\mathbb{C}$. By passing to a normal closure we assume that $L/\mathbb{R}$ is Galois with Galois group $G$. Let $H$ be the Sylow-2 subgroup of $G$ and $M=L^H$.

By the Fundamental Theorem of Galois Theory, $M/\mathbb{R}$ has odd degree. Let $\alpha\in M$ and $f(x)$ be its minimal polynomial. Then $f(x)$ has odd degree and by the Intermediate Value Theorem, a real root. As $f(x)$ is irreducible, it must have degree one. Then $\alpha\in\mathbb{R}$ and $M=\mathbb{R}$. So $G=H$ is a 2-group. Then $G_1:=Gal(L/\mathbb{C})$ is a 2-group as well.

Assuming that $G_1$ is not trivial, there must exist a degree 2 subextension $K$ of $\mathbb{C}$. But every quadratic complex polynomial has a root (by the quadratic formula), so we have a contradiction.

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    $\begingroup$ I believe this proof is due to E. Artin. (So Dummit & Foote attribute it, anyway.) $\endgroup$ – Charles Rezk Jan 4 '10 at 0:09
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    $\begingroup$ This proof can be extended to real closed fields, that is ordered fields in which every polynomial of odd degree has at least one root and every positive element is a square (and so has a square root). $\endgroup$ – lhf Jan 4 '10 at 0:10
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    $\begingroup$ This prove can also be extended to prove that the "field" that you obtain by adjoining a square root of -1 to the "field" of surreal numbers is algebraically closed. ("field" is not actually precise, because these sets are proper classes) $\endgroup$ – Andrea Ferretti Feb 22 '10 at 14:18
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    $\begingroup$ I learned this proof from Serge Lang in 1965 when I was a freshman in college. He attributed it to Emil Artin (and he was the keeper of the Artin flame, so I'd definitely believe it). $\endgroup$ – Victor Miller Apr 15 '12 at 2:39
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    $\begingroup$ I learned this proof in college, from Tate no doubt (another Artin disciple), and I guess I've been misremembering all these years: I thought he said it went back to Gauss. $\endgroup$ – Tom Goodwillie Jun 28 '15 at 15:15
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You can prove it using only basic facts about continuity/compactness and the same estimate which makes the winding number/fundamental group of $S^1$ proofs work: first check that if $p(z)=z^n + a_{n-1}z^{n-1} + \ldots + a_0$ is a polynomial then $|p(z)|$ tends to infinity as $|z|$ tends to infinity (this is the "leading term dominates" estimate for large $|z|$). It follows easily that $|p|$ attains a minimum value, since outside a large disc centered at $0$ the value of $|p|$ is really big, and the disc is compact, so $|p|$ attains a minimum on it. We want this minimum value to be zero, so suppose for the sake of contradiction it isn't, then we can change coordinates if necessary so that minimum is attained at $0$, and rescale $p$ so the minimum is $1$.

Then you just have to show that if $p(z) = 1+b_kz^k + \ldots + b_n z^n$ (where $k \geq 1$) then you can make $|p|$ smaller than $1$ for some nonzero $z$. But this is just the same kind of estimate as before: the term $b_kz^k$ dominates the other terms for $z$ small, and we can easily arrange for $b_kz^k$ to be a negative real giving the required contradiction.

The proof has the advantage that it makes the theorem "obvious" once you have some notion of compactness in the plane, so you could use this proof pretty early in a course that talks about functions on $\mathbb R^n$ or $\mathbb C$. As I said before though, what makes the proof tick is the same as what makes the $\pi_1(S^1)$ proofs work, it just uses simpler techniques to get a contradiction. Unfortunately I have no idea who it's due to (which is why I explained it rather than giving a reference...)

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    $\begingroup$ You can fund this proof here: jstor.org/stable/2315823 $\endgroup$ – Gian Maria Dall'Ara Jan 4 '10 at 8:51
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    $\begingroup$ It's also in Rudin, Principles of Mathematical Analysis, Chapter 8. $\endgroup$ – lhf Jan 4 '10 at 9:58
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    $\begingroup$ This proof did not come from that jstor reference (to an article of Fefferman). It is basically the proof by d'Alambert from 1746 and it is essentially the first correct proof, except that properties of compactness and its consequences (continuous real-valued functions assume a minimum value) were not rigorously established back then. $\endgroup$ – KConrad Feb 12 '10 at 23:07
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    $\begingroup$ The result that the minimum of $|p|$, if it exists, must be zero is sometimes called d'Alembert's lemma. But his proof is very hard to follow, because he attempts to solve the equation $p(z)=w$ as a fractional power series in $w$. A much easier proof was given by Argand in 1806, using simple algebra and the geometric interpretation of complex numbers. $\endgroup$ – John Stillwell May 17 '10 at 1:07
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    $\begingroup$ To make the estimate in the second paragraph very precise, you may make another change of variable: substituting $(-1/b_k)^{1/k}\cdot z$ for $z$ makes the $k$'th coefficient $-1$ and you get $p(z)=1-z^k+z^{k+1}q(z)$ for some new polynomial $q$ (note that this is the step that fails for $\mathbb{R}$). Now, using the fact that $zq(z)\to 0$ at $0$, you may find $0<z<1$ such that $|zq(z)|<1/2$ and get $|p(z)|\leq |1-z^k|+|p(z)-(1-z^k)|=(1-z^k)+z^k|zq(z)|<(1-z^k)+1/2z^k=1-1/2z^k<1$, giving you your desired contradiction. $\endgroup$ – Uri Bader Oct 3 '17 at 7:57
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Here's an extract of my post to sci.math.research from 2001. The proof definitely uses a "sledgehammer method", but perhaps it has some pedagogical value. I have no doubt that other people may have come up with similar arguments.


Sketch: It suffices to check that any complex monic polynomial has a root. Any such polynomial is the characteristic polynomial of some matrix (one can use the so called companion matrix). Thus one is reduced to showing that an $n\times n$ complex matrix $A$ has an eigenvalue or equivalently an eigenvector. $A$ may be assumed to be invertible, since otherwise $0$ is an eigenvalue. Then $A$ acts on complex projective space $P = \mathbb{C}\mathbb{P}^{n-1}$ by sending the span of $v$ to the span of $Av$. An eigenvector corresponds to a fixed point under this action. Since the general linear group of $\mathbb{C}$ is connected, $A$ can be connected by a path to the identity I (this can be done explicitly by writing $A$ as a product of elementary matrices and deforming these to $I$ in the obvious way). It follows that the $A$ is homotopic to $I$, and therefore its Lefschetz number on $P$ coincides with the Euler characteristic of $P$ which is nonzero. Therefore, $A$ has a fixed point on $P$.

Added: I should probably have pointed that the conclusion follows from the Lefschetz fixed point theorem, which was the sledgehammer that I was alluding to.

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  • $\begingroup$ I like it! $ $ $\endgroup$ – Victor Protsak May 18 '10 at 6:50
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    $\begingroup$ Late to the party, but: my high-school self would have been jealous to read this proof. He figured out the first three sentences but had no idea what to do next :) $\endgroup$ – Ryan Reich Apr 3 '11 at 14:44
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    $\begingroup$ Nice! I think this is the first time I've seen someone prove that every endomorphism of a finite-dimensional complex vector space has an eigenvector without using the fundamental theorem of algebra. Maybe we should have another big-list for proofs of that fact... $\endgroup$ – Vectornaut Apr 12 '12 at 16:24
  • $\begingroup$ Why does this argument fail over $\mathbb{R}?$ $\endgroup$ – Igor Rivin Nov 18 '12 at 19:19
  • $\begingroup$ I had almost forgotten about this... but, Igor, to answer your question $\mathbb{R}\mathbb{P}^{n-1}$ is a very different animal. When $n=1$, it is circle, which need not be any fixed points. $\endgroup$ – Donu Arapura Nov 18 '12 at 19:52
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Here is a nice proof that was posted on the ALGTOP list recently. See also the ensuing discussion.

An algebraic extension of $\mathbb{C}$ is a unital division algebra over $\mathbb{C}$, say of dimension $n$, so induces $$\mathbb{CP}^{n-1} \times \mathbb{CP}^{n-1} \to \mathbb{CP}^{n-1}$$ satisfying $$ y \mapsto 1 \otimes y + y \otimes 1 $$ in second integral cohomology. Since $y^n = 0$, we must have $n=1$.

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  • $\begingroup$ If you put something else in place of C in there, it doesn't work. So what property is used? $\endgroup$ – Gerald Edgar Jan 4 '10 at 1:10
  • $\begingroup$ @Gerald: you need to know that $\mathbb{CP}^n$ has a specific integral cohomology ring. $\endgroup$ – Mariano Suárez-Álvarez Jan 4 '10 at 5:48
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    $\begingroup$ The proof of the ring structure of $H^* \mathbb CP^n$ is the core connectivity/analytical argument comparable to the intermediate value theorem or $\pi_1 S^1 \simeq \mathbb Z$ (used in most of the other proofs). $\endgroup$ – Ryan Budney Jan 4 '10 at 7:08
  • $\begingroup$ @gerald: if you look through the discussion bit following the original post by Bob to the mailing list you will find some of this discussion, but i suspect that Ryan's answer is the more relevant one. @Kevin:this is my favorite proof, Bob Bruner happens to be my advisor so i am a tidge biased. In fact, i think you should know that he was very pleased to see this posted here with the link. $\endgroup$ – Sean Tilson Jun 17 '10 at 18:06
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A recent and very important contribution to the literature on the fundamental theorem of algebra is Joe Shipman's article "Improving the Fundamental Theorem of Algebra," Math. Intelligencer 29 (2007), 9-14, doi:10.1007/BF02986170. Here is one of his results: A field with the property that every polynomial whose degree is a prime number has a root is algebraically closed. This result is sharp in the sense that if any prime is omitted then the conclusion is false.

Shipman's paper should go a long way towards addressing Andrew L's question of whether there is a "purely algebraic proof" of the FTA. The above result of Shipman's shows that we can limit the topology/analysis to proving that every polynomial over $\mathbb{C}$ of prime degree has a root; the rest is pure algebra. If you wanted to try to limit the use of topology or analysis even further, then this part of the proof is where you should focus your attention.

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    $\begingroup$ Wow,I was completely unaware of this article. I'll definitely give it a long look.Thanks! $\endgroup$ – The Mathemagician Sep 19 '10 at 16:14
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In Hatcher's book http://www.math.cornell.edu/~hatcher/AT/ATpage.html Theorem 1.8 he deduces the fundamental theorem as a corollary of the fact that the fundamental group of the circle is isomorphic to the integers.

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    $\begingroup$ I have always thought that the algebraic topology proofs of the fundamental theorem of algebra were far simpler than the complex analytic proofs. $\endgroup$ – Daniel Miller Sep 8 '10 at 12:34
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I have collected 15 proofs with different approaches, including all the proofs suggested here so far. They are available at

https://github.com/andreaferretti/math-notes/blob/master/TFA.pdf

Unfortunately they are in italian. I hope someone is able to read them.

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    $\begingroup$ This is a really good article. Thanks for posting. The Italian is not hard. Also the really good thing is the detailed bibliography in the end. $\endgroup$ – Anweshi Jan 4 '10 at 16:56
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    $\begingroup$ Proof XIV is fantastic. It proves the non existence of field structures on R^n (n>2) using only a pair of fundamental group computations! Grazie! $\endgroup$ – Gian Maria Dall'Ara Jan 5 '10 at 9:38
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    $\begingroup$ I was going to write, "The Italian is not hard, and Proof XIV is fantastic," but then I noticed both of those things had been written already. $\endgroup$ – Vectornaut Apr 12 '12 at 16:20
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    $\begingroup$ The link is a 404. $\endgroup$ – Julien Puydt Aug 8 '12 at 19:46
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    $\begingroup$ The entire base URI is down for me. $\endgroup$ – Toby Bartels Nov 7 '13 at 5:33
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Two (or three) more complex analysis approaches, the first is "essentially the same" as the proof in Alfors I think, but the second is different (I'm afraid I don't have a reference, but I can type up the full proofs if you want):

Let p be a polynomial and n be its degree.

  • apply the residue theorem to $1/[zp(z)]$. If $p$ has no roots, then this function is analytic except at $z=0$, where it has a simple pole of non-zero residue. But $1/[zp(z)]$ is bounded, so integrating it along a circular contour centred at the origin gives something inversely proportional to the length of this contour ie this integral can be made as small as possible, which contradicts the residue at $z=0$ being non-zero.

  • a variant of the above: If p has no roots, then Cauchy's integral formula implies $$\int_{|z|=r}\frac{dz}{zp(z)}=\frac{2\pi i}{p(0)}\ne0.$$ Now let $r\to\infty$: The integral on the left vanishes in the limit, and we have a contradiction. See Anton R. Schep: A simple complex analysis and an advanced calculus proof of the fundamental theorem of algebra, Amer. Math. Monthly 116 (2009) 67–68.

  • by Rouche's theorem, $p(z)$ and $z^n$ have the same number of roots inside a large circular contour centred at the origin, and this number is n.

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    $\begingroup$ The second one is quite interesting, and new for me. Thanks. The first one is not exactly the same as in Ahlfors book; but some arguments are common to both. $\endgroup$ – Anweshi Jan 3 '10 at 19:18
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    $\begingroup$ Rouche's theorem was always my favorite method =) $\endgroup$ – Sean Rostami Jan 5 '10 at 0:15
  • $\begingroup$ I added a variant (due to Schep) of the first one. Now I notice that this messes up Anweshi's reference to the “second one”, which is now the third one. Sorry about that. $\endgroup$ – Harald Hanche-Olsen Jan 5 '10 at 4:36
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    $\begingroup$ One can rephrase the second argument as follows: if P has no roots, then 1/P is entire, and decays to zero at infinity, hence is zero by Liouville's theorem, a contradiction. Though this proof is perhaps somewhat unsatisfying since Liouville's theorem is generally regarded as being less intuitive than the fundamental theorem of algebra. $\endgroup$ – Terry Tao Sep 19 '10 at 22:43
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Gauss's first proof goes more or less like this. Let $p(z)$ be a polynomial of degree $n$ and complex coefficients. Write $p(x+iy) = a(x,y) + ib(x,y)$, where $a,b$ have real coefficients. The crucial observation is that the branches of $a=0$ and $b=0$ as real curves interlace at infinity (as can be seen from the degree $n$ terms). Also, real algebraic plane curves don't just stop somewhere in the affine plane, so a branch of $a=0$ must be connected to another branch of $a=0$. Now, in between them, there is a branch of $b=0$ which connects to another branch of $b=0$. If the connections alternate, they have to meet and we get a common zero of $a$ and $b$, which gives a complex zero of $p$. If they don't alternate, find a branch of $a=0$ in between the two connecting branches of $b=0$ and repeat.

I think this proof really needs the Jordan curve theorem to be fully justified, which is a bit of an anachronism.

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Pukhlikov has a proof using only real numbers (the page in English, full text in Russian) of the fact that indecomposable elements in $\mathbb R[x]$ have degree 1 or 2 as is part of the 1997 issue of Russian journal "Математическое Просвещение" which dedicated 50 pages of that issue to proving the main theorem of algebra by different methods.

In particular, the article by Tikhomirov and Uspenski (pdf, Russian) in that volume contains 10 proofs:

  1. topological, by considering a circle of big radius;
  2. algebro-geometric, using $\mathbb C\mathbb P^1$;
  3. algebro-geometric, using general facts about Riemann surfaces;
  4. complex-analytic, by referring to number of zeroes = number of poles or Stokes theorem;
  5. using topology to prove that holomorphic maps between compact surfaces are constant or surjectivet;
  6. similarly, but using inverse theorem from calculus;
  7. finding a minimum of $|p(z)|$;
  8. applying the Liouville's theorem to $1/p(z)$;
  9. applying Lefschetz theorem to $\mathbb C\mathbb P^n$ to prove that every linear operator over $\mathbb C$ has eigenvector;
  10. using Galois theory + simple facts about $\mathbb C$ to show that $\mathbb C$ has no algebraic extensions;

as well as extensive historical notes.

Though I don't know whether a translation exists, I think that this collection of articles deserves it and I'm sure the authors will be happy to give their permission to republish. A translation could be done as a project for an undergraduate student with knowledge of the language.

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  • $\begingroup$ It had been much more fun to read short proofs in this page itself. Figuring out Russian is a headache if you are a non-Russian. It is hard enough even if you are a Russophile, but not Russophile enough to learn the language. Of course, all this is unless you were at Princeton, and you had to pass an exam in that course! $\endgroup$ – Anweshi Jan 4 '10 at 16:46
  • $\begingroup$ And I had meant this question mainly for non-Princeton folks, since they are the ones who didn't have to look up various proofs of this for purposes of clearing the quals! :) $\endgroup$ – Anweshi Jan 4 '10 at 16:52
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    $\begingroup$ Илья, хотя я профессор вместо студента, я был бы рад перевести эту статью. Это очень легко и интересно, конечно. Я уже перевел до Леммы 1 (на четвертой странице), а даже нашел математические опечатки. После того, как я перевел статью, я хотел бы, что вы прочитал бы мой перевод. Хорошо? Я не знаю ваш емейл адрес. Можно найти мой адрес с Гугла, используя мое название с этого сайта. Пошлите мне, пожалуйста, емейл. Гмм, что вы думаете, люди здесь ненавидят иностранные языки? :) $\endgroup$ – KConrad Jan 17 '10 at 1:21
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    $\begingroup$ Hey, no fair. But +1 for Konrad's русский язык comment for coolness :) $\endgroup$ – David Roberts May 17 '10 at 11:58
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    $\begingroup$ It's no problem, this is why God gave us Google Translate. $\endgroup$ – Toby Bartels Nov 7 '13 at 5:40
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Here is another short proof, showing that no finite extension of $\mathbb{C}$ exists. If $A$ were such an extension of dimension $d$, then the projective space $\mathbb{P} A \cong \mathbb{CP}^{d-1}$ were a compact commutative Lie group, hence a torus. There are plenty of ways showing that $\mathbb{CP}^{m}$ is not a torus (if $m>0$). Take your favorite one, and the proof is complete.

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  • $\begingroup$ Showing that every compact connected Lie group is a torus (without using that $\mathbb{C}$ is algebraically closed) is not too hard, but not entirely trivial either. You don't need to do that here. Considering the exponential map $A\to A^\times$, $a\mapsto \sum a^n/n!$ you get a covering homomorphism $A/\mathbb{C} \to \mathbb{P}^{d-1}(\mathbb{C})$. Then it is easy to see that the latter is a torus (which implies $d=1$). $\endgroup$ – Uri Bader Sep 29 '17 at 10:20
  • $\begingroup$ hmm... reading further down the list I realized that my previous comment aligns with the answer mathoverflow.net/a/10685/89334. Then I saw the comment by Benjamin Steinberg to that answer, which made me post a new answer mathoverflow.net/a/282333/89334. $\endgroup$ – Uri Bader Sep 29 '17 at 13:13
  • $\begingroup$ I am slow here... Why does the projective space has to be a Lie group? $\endgroup$ – Michael Sep 5 at 18:14
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As in one of the previous posts, consider the projective space $CP^n$ of nonzero all polynomials

$$ c_nT^n + c_{n-1} T^{n-1} + \cdots + c_1 T + c_0 $$ considered up to nonzero scalar multiple. We'll show directly that any such polynomial admits a factorization into linear factors.

Consider the map $\phi: CP^1 \times \cdots \times CP^1 \to CP^n$ given by

$$ ([\alpha_1:\beta_1],\dots,[\alpha_n:\beta_n]) \mapsto \prod_{i=1}^n (\alpha_i T - \beta_i) $$

In other words, this map sends a set of roots to the polynomial which has precisely those roots. It suffices for us to show that $\phi$ is surjective.

If the points $[\alpha_i:\beta_i]$ are distinct, it is easy to check that the differential $d\phi$ is nonzero. Hence the polynomial $T^n - 1$ (for example) is a regular value of $\phi$ with exactly $n!$ preimages (here we've used the fact that polynomials factor uniquely into irreducibles). Thus the map $\phi$ has positive degree.

It is a fact that any map of positive degree between compact connected complex manifolds of the same dimension is surjective. (Proof: Since any such map is orientation preserving, the number of preimages of any regular value must be exactly equal to the degree - not just up to multiplicity. Hence the image contains the set of regular values, which is dense by Sard's theorem. But the image is also closed since it is the image of a compact set, hence the map is surjective.)

We conclude that $\phi$ is surjective. In other words, every polynomial of degree $n$ has a factorization into linear factors.

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I add another proof separately since it is rather different.

It is based on the observation that an irreducible polynomial of degree $n$ on a field $F$ gives rise to a field structure on $F^n$, compatible with the vector addition (just quotient out polynomials with coefficients in $F$ by the maximal ideal generated by the irreducible). Then the FTA is a consequence of the non-existence of real commutative division algebras in dimension greater than $2$.

This last assertion follows from the observation that in such a $R^n$ you can define an $exp$ which gives an epimorphism $R^n\rightarrow R^n$ \ $0$, and hence a homeo $S^j \times R^k\cong R^n$ \ $0$, which is forbidden by a fundamental group computation (this was in Andrea Ferretti's notes).

In fact you can use this argument to prove that any real irreducible polynomial has degree $1$ or $2$.

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  • $\begingroup$ This is Kevin Lin's answer, isn't it? $\endgroup$ – Ryan Budney Jan 4 '10 at 15:44
  • $\begingroup$ Yes, I didn't understand it before, and thought was different. $\endgroup$ – Gian Maria Dall'Ara Jan 4 '10 at 18:43
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    $\begingroup$ I don't think they are the same, you can say that they are related, just like nearly all complex analytic proofs are related. At any rate, this one is much clearer to me (a Lie theorist, not an algebraic topologist) - I can actually digest it at a glance, and it avoids unnecessary cohomological machinery. $\endgroup$ – Victor Protsak May 18 '10 at 6:46
  • $\begingroup$ I agree. When I wrote the comment I hadn't supplied the easier proof of the non existence of field structures on real vector spaces. $\endgroup$ – Gian Maria Dall'Ara May 18 '10 at 9:42
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    $\begingroup$ An alternative version of the proof is to note that the map $x\mapsto x^2/\|x^2\|$ induces a continuous injection from $\mathbb P^{n-1}\to S^{n-1}$ for $n>2$ which must be an open mapping by invariance of domain. Thus this map is a homeomorphism, contradicting the fundamental groups being different. $\endgroup$ – Benjamin Steinberg May 14 '12 at 21:27
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Also The Fundamental Theorem of Algebra: A Visual Approach by Velleman.

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    $\begingroup$ Pretty! For some reason, looking at the figures finally made me understand how to read those "rainbow plots" of functions $\mathbb{C} \to \mathbb{C}$. $\endgroup$ – Vectornaut Apr 12 '12 at 16:42
  • $\begingroup$ I don't think that I've every seen such rainbow plots, but Figures 1–3 made them very clear. Awesome! $\endgroup$ – Toby Bartels Nov 7 '13 at 5:46
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See the book The fundamental theorem of algebra by Fine and Rosenberger.

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Volume I of Diffusions, Markov processes and martingales by Rogers and Williams has a probabilistic proof of the fundamental theorem (Prop. 19.5, page 41).

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  • $\begingroup$ Fantastic! As soon as I have time, I will add it to my notes. :-) $\endgroup$ – Andrea Ferretti Feb 22 '10 at 14:23
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    $\begingroup$ Somebody once told me "Every area of mathematics has its own proof of the fundamental theorem of algebra... except maybe probability theory". There goes that caveat! $\endgroup$ – Paul Siegel May 17 '10 at 20:26
  • $\begingroup$ There is another probabilistic proof, by Mihai Pascu; I put it as a separate answer. $\endgroup$ – Margaret Friedland Apr 12 '12 at 15:36
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Here is a translation into English of a second "real" proof from the journal Ilya mentioned in his answer. This proof is due to Petya Pushkar', it is found at

http://www.mathnet.ru/php/archive.phtml?wshow=paper&jrnid=mp&paperid=7&option_lang=eng

and it is based on the notion of the degree of a map. Recall that for a smooth proper mapping of oriented manifolds, its degree is defined by picking a regular value and adding up the signs of the determinants of the differential of the mapping at the points in the inverse image. That the degree is well-defined is rather complicated to prove, but it explains the following topological fact.

Fact: Let $M^n$ and $N^n$ be smooth connected oriented manifolds and $f \colon M^n \rightarrow N^n$ be a smooth proper mapping of degree not equal to zero. Then $f$ is surjective.

To prove the Fundamental Theorem of Algebra in a "real" version, we will focus on polynomials of even degree (any of odd degree have a real root). We will show that any real polynomial of degree $2n$ can be factored into a product of $n$ polynomials of the second degree.

We identify each monic polynomial $x^d + a_{d-1}x^{d-1} + \cdots + a_1x + a_0$ with the point $(a_{d-1},\dots,a_0)$ in $\mathbf R^d$. We will be particularly interested in the space of monic quadratic polynomials $x^2 + ax + b$, which are identified with the plane $\mathbf R^2$. Consider the multiplication mapping $$ u \colon (\mathbf R^2)^n \rightarrow \mathbf R^{2n} \ \ \text{ where} \ \ \ (f_1,f_2,\dots,f_n) \mapsto f_1f_2\cdots f_n. $$ Proving the Fundamental Theorem of Algebra amounts to showing that $u$ is surjective.

First we show $u$ is proper.

For any $d \geq 1$, identify the nonzero polynomials of degree at most $d$, considered up to scaling by nonzero real numbers, with $\mathbf P^d(\mathbf R)$ by $[a_dx^d + a_{d-1}x^{d-1} +\cdots + a_0] \mapsto [a_d,a_{d-1},\dots,a_0]$. (The polynomials of exact degree $d$, after being scaled to be monic, are a copy of $\mathbf R^d$ in $\mathbf P^d(\mathbf R)$.) Consider the multiplication mapping $$ \widehat{u} \colon (\mathbf P^2(\mathbf R))^n \rightarrow \mathbf P^{2n}(\mathbf R) \ \ \text{ where} \ \ \ ([f_1],[f_2],\dots,[f_n]) \mapsto [f_1f_2\cdots f_n]. $$ The mapping $\widehat{u}$ is proper since it is defined on a compact manifold and is continuous.

The mapping $\widehat{u}$ is a natural "compactification" of the mapping $u$. The space $(\mathbf P^2(\mathbf R))^n$ can be written as the union of $(\mathbf R^{2})^n$ and an "infinitely distant part" $B_1$ ($n$-tuples of polynomials of degree at most 2 where at least one polynomial has degree less than 2), while $\mathbf P^{2n}(\mathbf R)$ can be written as the union of $\mathbf R^{2n}$ and an "infinitely distant part" $B_2$ (polynomials of degree less than $2n$). From this point of view, $\widehat{u}$ on $(\mathbf R^2)^n$ agrees with $u$ and, clearly, $\widehat{u}^{-1}(B_2) = B_1$. Therefore the map $u$ is proper.

Next we show the degree of $u$ is equal to $n!$. Orient the space of monic polynomials of degree 2 (we denote this space as $\mathbf R^2$) arbitrarily and give $(\mathbf R^2)^n$ the product orientation (as a product of oriented manifolds). As an exercise, show the polynomial $p(x) = \prod_{i=1}^n (x^2+i)$ is a regular value of the mapping $u$. (Hint: This polynomial is a product of distinct monic irreducibles. Now use the description of the regular values of the multiplication mappings $\mu_k$ in Pukhlikov's proof of the Fundamental Theorem of Algebra, which is written in a separate answer on this page.)

The polynomial $p(x)$ has $n!$ inverse images under $u$: all ordered $n$-tuples with coordinates $x^2+i$ for $i = 1,\dots,n$. Let's prove that these points all contribute the same sign to the degree.

The mapping $u$ is invariant under permutations of its arguments, and any such permutation preserves orientation (exercise). Therefore the sign of the determinant of the differential at all the inverse images is the same, which shows $u$ has degree $n!$. By the topological fact at the start, $u$ is surjective, so all monic real polynomials of degree $2n$ are a product of monic quadratic real polynomials.

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  • $\begingroup$ Thank you! I was preparing to post this proof myself today, but fortunately you did it! I want to mention, that this proof was based on the following idea. I had this idea while studying at school (but I could not realize it at this time), I think it is almost first what can come in mind: polynomial is decomposable if and only if its coefficients satisfy to some system of equations on coefficients of factors. $\endgroup$ – Petya Feb 28 '10 at 15:03
  • $\begingroup$ So Fundamental Theorem of Algebra is equivalent to a statement that some system of equations is solvable. If we restrict to monic polynomials then this system of equations is squared system (the number of variables is equal to the number of equations) and the theory of degree of a map can help to prove that it is solvable! The problem is to show that the theory of degree is applicable (since we are in non-compact situation)... $\endgroup$ – Petya Feb 28 '10 at 15:03
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This is not a serious answer, but one can "prove" the fundamental theorem of algebra by applying the spectral theorem to the matrix

$$ A := \begin{pmatrix} 0 & 1 & 0 & \ldots & 0 \\\ 0 & 0 & 1 & \ldots & 0 \\\ \vdots & \vdots & \vdots & \ddots & \vdots \\\ -a_0 & -a_1 & -a_2 & \ldots & -a_{n-1} \end{pmatrix} $$

to locate an eigenvalue $\lambda$, which is then a root of the monic polynomial $z^n + a_{n-1} z^{n-1} + \ldots + a_0$.

Of course, this argument is usually circular, because most of the standard proofs of the spectral theorem for matrices requires the fundamental theorem of algebra (either by explicitly citing that theorem, or implicitly, by borrowing one of the proofs given here, e.g. by applying Liouville's theorem to the resolvent $(A-zI)^{-1}$) in the first place...

However, one could imagine a weird proof of the spectral theorem that somehow avoids the fundamental theorem and would thus give a non-circular proof of that theorem. I thought about proceeding by showing that the set of diagonalisable matrices is a generic subset of the set of all matrices, but I realised that in order to have enough algebraic geometry to talk about "generic", I need to know the ambient field is algebraically closed, which of course is precisely the fundamental theorem of algebra. Deducing the spectral theorem for matrices from the spectral theory of more general objects, such as elements of a C^* algebra, doesn't work either, for much the same reason.

Perhaps it is best to view the above arguments not as proofs of the fundamental theorem of algebra, but rather as "consistency checks" that show that this result is compatible with the basic theory of other mathematical subjects, such as linear algebra and algebraic geometry.

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  • $\begingroup$ It seems like this has been done here jstor.org/stable/3647746 (I can't say for sure because I haven't read the whole thing). $\endgroup$ – Peter Samuelson Sep 19 '10 at 23:48
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    $\begingroup$ Terry, this can be done for matrices. In fact, Harm Derksen wrote a nice proof of it in the Monthly (2003). The structure of his inductive argument actually resembles very much Gauss' approach in one of his proofs. $\endgroup$ – Andrés E. Caicedo Sep 20 '10 at 0:50
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    $\begingroup$ Can't one prove the spectral theorem by applying the method of Lagrange multipliers to find the extremal points of some function? I seem to remember some proof along these lines... $\endgroup$ – Mark Apr 16 '11 at 12:22
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    $\begingroup$ Donu Arapura's answer (mathoverflow.net/questions/10535/…) seems to give such a proof of the spectral theorem in finite dimensions, and Ilya Nikokoshev's answer (mathoverflow.net/questions/10535/…) mentions what appears to be the same proof. Are those not good for some reason? $\endgroup$ – Vectornaut Apr 12 '12 at 16:34
  • $\begingroup$ @Mark Schwarzmann: Is this the proof you're thinking of? (ncatlab.org/nlab/show/…) I like it! $\endgroup$ – Vectornaut Apr 12 '12 at 16:35
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A proof using dynamical systems:

If $p$ is a non-constant polynomial without roots then $f = \text{Re}(\frac{1}{p})$ is a bounded harmonic function which goes to zero at infinity. Consider the gradient flow for $f$. This flow is area preserving because $f$ is harmonic. Also, the value of $f$ is strictly increasing along the orbit of any non-singular point.

Consider the bounded set where $f > \epsilon > 0$. This set is invariant for the flow and by Poincarè's recurrence theorem almost all orbits in it are recurrent. However because of the monotonicity of the values of $f$ along orbits this is impossible unless all these orbits are singular. Hence the derivative of $f$ on the set where it is positive is zero. The same argument applied to $-f$ shows that the derivative of $f$ is zero on the set where $f$ is negative. Hence $f$ is constant.

The same argument applied to the imaginary part of the reciprocal of $p$ implies that $p$ is constant.

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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; projecteuclid.

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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  • $\begingroup$ 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! $\endgroup$ – Dan Ramras Sep 20 '10 at 1:47
  • $\begingroup$ Was it Hubbard that asked that question? $\endgroup$ – Ryan Budney Sep 20 '10 at 1:49
  • $\begingroup$ No, it was Guckenheimer. Hubbard went to France in the spring semester, and we strayed a bit from Hubbard's book. $\endgroup$ – Dan Ramras Sep 20 '10 at 1:52
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I'm partial to Milnor's proof in Topology from the Differentiable Viewpoint, a slightly simpler variant of the "every complex non-constant polynomial $p$ is surjective" proof given above, published somewhat earlier (1965). In brief:

Definition: Let $f: M \to N$, $M,N \subset \mathbb{R}^n$, $M$ compact, then, for each regular value $y \in N$, $$\#f^{-1}(y) = \text{number of points in the inverse image of $y$}.$$

Lemma: $\#f^{-1}$ is locally constant on the set of regular values.

Proof of lemma: Since $f$ is a diffeomorphism in a neighborhood of each $x_i \in f^{-1}(y)$, we can choose pairwise disjoint neighborhoods $U_i$ for the $x_i$, let $V_i = f(U_i)$, and then $$\#f^{-1}(V_1 \cap \cdots \cap V_k - f(M - U_1 - \cdots - U_k)) = \{\#f^{-1}(y)\}.$$

Proof of the F.T.A.: Using stereographic projection, we can consider the polynomial as a smooth map $f: S^2 \to S^2$. Since $f$ has only a finite number of critical points, the set of regular values is connected; the locally constant $\#f^{-1}$ is therefore constant on this set. As $\#f^{-1}$ cannot be zero for all regular values of $f$, it must be zero for none. Thus $f$ is surjective, and the polynomial has a root.

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  • $\begingroup$ Why is the stereographic projection essential to this proof? Can we obtain the result considering the polynomial as a map from the complex plane to the complex plane, instead of involving the spheres? $\endgroup$ – Pait Feb 9 '17 at 21:25
  • $\begingroup$ I would be grateful for an explanation! $\endgroup$ – Pait Feb 14 '17 at 15:17
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    $\begingroup$ @Pait: You need to apply the lemma to a smooth map of compact manifolds, hence compactify the polynomial map $f\colon\mathbf C\to\mathbf C$ to a smooth map $f\colon \mathbf S_2\to\mathbf S_2$. $\endgroup$ – ACL Nov 7 '17 at 18:43
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Here's another complex analysis proof that I heard about for the first time under a week ago (because it was set as a question on a course I am teaching for). Pick a circle large enough for the modulus of p(z) to be greater than |p(0)| everywhere in that circle. Inside that circle take a point w where the modulus of p is minimal (which obviously you can do by compactness). There are many ways of proving that p(w)=0. One can use the minimum modulus theorem (that any point of minimum modulus not on the boundary must be a zero), the open mapping theorem, the local mapping theorem, or an elementary bare-hands argument.

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Gersten and Stallings gave a proof using free groups:

On Gauss's first proof of the fundamental theorem of algebra, Proc. Amer. Math. Soc. 103 (1988), 331-332

of course giving all of the credit to Gauss.

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There is one short proof on wikipedia, that shows the statement that any endomorphism $A$ of a finite-dimensional vector space $V$ of positive dimension has an eigenvalue. Look at the resolvent function $R_A:z \mapsto (A-z)^{-1}$. Outside the disc of radius $\|A\|$, it can be developped into a geometric series. Use this geometric series to compute the integral of $R_A$ around a big circle; the result is $2 \pi i \cdot id_V$. On the other hand, if the spectrum were empty, then the resolvent were holomorphic and by Cauchy's theorem, the integral is zero. This is a contradiction if $dim (V) >0$.

To derive the FTA from the existence of eigenvalues (without using determinants), let $p(z)$ be a normed polynomial of degree $n$ and look at the vector space $V:= \mathbb{C}[z]/(p(z))$ and let $A:V \to V$ be the multiplication by $z$. $dim (V)=n$ is easy to see (division with remainder) and clearly $p(A)=0$. Take a nonzero $v \in V$ with $Av=\lambda v$. Then $0=p(A) v = p(\lambda) v$ shows that $\lambda$ is a zero of $p$.

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Perhaps it could be of interest for you to know that there exists purely geometric proofs of this result. Concretely, it can be shown that, if the FTA fails then there exists a plane Riemannian metric over the Sphere S^2. Of course, this produces a contradiction since the sphere is not flat. This proof can be located at the very recent paper by J M Almira and A Romero: "Some Riemannian Geometric proofs of the fundamental theorem of algebra", which is available at: http://www.mathem.pub.ro/dgds/v14/D14-al.pdf

J. M. Almira

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Thanks to Tim Chow for citing me. Technically, you don't need to show every polynomial of prime degree in F[x] has a root, you just need to show that there is a field G such that every polynomial of odd prime degree in G[x] has a root and every element or its additive inverse has a square root; then G[i] will be algebraically closed. Even more interesting, to show that all polynomials of degree d have a root, all you need is that all polynomials of degree p have a root for those p which divide d, plus the existence of any sufficiently large degree d' such that all polynomials of degree d' have a root (an explicit algorithm for how large d' must be is easily derivable from my proof).

Of course, this is not a proof of the Fundamental Theorem of Algebra, what I did was identify the pure algebraic core of the requirement that a field be algebraically closed. To show that the complex numbers are algebraically closed, you still need some way of showing that real polynomials of odd prime degree have roots, which depends on the Intermediate Value Theorem or some other analytical or topological argument in all the proofs I know.

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    $\begingroup$ If $-1$ has no square root in $G$, then the condition “every polynomial of odd prime degree has a root” trivially implies “every polynomial of odd degree has a root”, and this has nothing to with with primes (any infinite set of odd natural numbers will do): given an odd degree polynomial $f$, consider $(x^2+1)^df$, where $d$ is chosen so that $\deg(f)+2d$ is prime. $\endgroup$ – Emil Jeřábek Jul 31 '12 at 12:34
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I believe the following proof had not appeared before, though it is motivated by this previous answer and a comment by Benjamin Steinberg to this answer.


Consider a field extension $F$ of $\mathbb{R}$ of finite degree $d>1$. We'll see that $d=2$, which forces $F\simeq \mathbb{C}$.

Note that the map $\alpha(x)=x^2$ is an endomorphism of the multiplicative group $F^\times$. The differential of $\alpha$ at $x$ is the multiplication by $2x$ operator, which is invertible, so the image of $\alpha$ is open in $F^\times$. Since $F^\times$ is connected (it is homeomorphic to $\mathbb{R}^d-\{0\}$ and $d>1$), and a connected topological group has no proper open subgroup (open subgroups are closed), we conclude that $\alpha$ is onto.

Since $\mathbb{R}^\times=\alpha^{-1}(\mathbb{R}^\times_+)$ we get that $\alpha$ induces a continuous isomorphism $F^\times/\mathbb{R}^\times\simeq F^\times/\mathbb{R}^\times_+$. This yields a homeomorphism $\mathbb{P}^{d-1}(\mathbb{R}) \simeq S^{d-1}$. There are several ways to see that there is no such a homeomorphism unless $d=2$ (the most standard one is by using the simply-connectedness of the sphere, but using an ad-hoc argument would make the above proof available for younger students), thus I leave the end game to the reader.

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  • $\begingroup$ How do see that $\alpha$ is onto? I guess it doesn't matter, since $F^\times/R^\times$ is compact, and that is what we really care about. Alternately, you could say that squaring on $F^\times/R^\times_+$ gives a connected 2-fold covering space over $S^{d-1}$, forcing $d=2$ since higher spheres are simply connected. $\endgroup$ – Charles Rezk Sep 29 '17 at 13:48
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    $\begingroup$ @Charles, the image of $\alpha$ is an open subgroup of a connected topological group. Now, every open subgroup in a topological group is closed, as its complement is a union of open cosets, and every clopen subset of a connected space is empty or full. It follows that every open subgroup of a connected group is full. Thus $\alpha$ is onto. $\endgroup$ – Uri Bader Sep 29 '17 at 13:58
  • $\begingroup$ Ah right, thanks. This is a nifty proof. $\endgroup$ – Charles Rezk Sep 29 '17 at 14:27
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When you consider how polynomial $f$ of degree $n$ acts on a big circle $R$, it gives rise to a map $S^1 \to S^1$ of degree $n$. Such a map cannot be continued to a map of a disk $D \to D-\{0\}$, thus $f(D)$ contains point 0.

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    $\begingroup$ A variant on this is to think of the polynomial as a self-map of the Riemann sphere $f : S^2 \to S^2$, and the degree of the polynomial is the degree of the map (same argument). $\endgroup$ – Ryan Budney Jan 4 '10 at 7:03
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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

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protected by user9072 Jan 18 '16 at 19:19

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