The Fundamental Theorem of Algebra (Undergraduate Texts in Mathematics)

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Any such polynomial is the characteristic polynomial of some matrix one can use the so called companion matrix. An eigenvector corresponds to a fixed point under this action. I should probably have pointed that the conclusion follows from the Lefschetz fixed point theorem , which was the sledgehammer that I was alluding to. 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 , , doi: 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. 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. I have collected 14 proofs with different approaches, including all the proofs suggested here so far. They are available at. 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:.

The integral on the left vanishes in the limit, and we have a contradiction. A simple complex analysis and an advanced calculus proof of the fundamental theorem of algebra , Amer. Monthly 67— In Hatcher's book http: Gauss's first proof goes more or less like this. I think this proof really needs the Jordan curve theorem to be fully justified, which is a bit of an anachronism. In particular, the article by Tikhomirov and Uspenski pdf, Russian in that volume contains 10 proofs:.

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. Take your favorite one, and the proof is complete. We'll show directly that any such polynomial admits a factorization into linear factors. In other words, this map sends a set of roots to the polynomial which has precisely those roots. It is a fact that any map of positive degree between compact connected complex manifolds of the same dimension is surjective.

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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.

Also The Fundamental Theorem of Algebra: A Visual Approach by Velleman. This is not a serious answer, but one can "prove" the fundamental theorem of algebra by applying the spectral theorem to the matrix. 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. 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. 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.

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

The fundamental theorem of algebra - Ghent University Library

This polynomial is a product of distinct monic irreducibles. Let's prove that these points all contribute the same sign to the degree. 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.

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. On Gauss's first proof of the fundamental theorem of algebra, Proc. On the other hand, if the spectrum were empty, then the resolvent were holomorphic and by Cauchy's theorem, the integral is zero. Perhaps it could be of interest for you to know that there exists purely geometric proofs of this result.

Of course, this produces a contradiction since the sphere is not flat. Proof of the F. 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. One version of this proof appeared recently:. Yet another application of the Gauss-Bonnet Theorem for the sphere J.

The Fundamental Theorem of Algebra

Simon Stevin Volume 14, Number 2 , ; projecteuclid. In this paper the authors use the version of Gauss-Bonnet that relates the Gaussian curvature to the Euler characteristic. Since degree 2 polynomials have roots the quadratic formula! 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.

There's a linear algebra proof by Harm Derksen: You can also find the article posted here: What I present below is not a literal translation as if anyone on this site cares Concepts of Modern Mathematics Ian Stewart. Math Hacks Richard Cochrane.

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