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Is there already in the literature a proof of the fundamental theorem of algebra as a consequence of Brouwer's fixed point theorem?

N.B. The original post contained superfluous information, but it did generate one answer with a source that claims such a proof is impossible, and another answer with a source that claims to carry out precisely such a proof. Clearly these cannot both be correct.

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The way you've written this question is "thinking out loud", which doesn't really lend itself to an answer. In a sense, you've given a partial answer yourself. Is it possible for you to summarize, in an organized way, the stuff you know, and then collect the actual question into a concise block at the end? Because it's going to get closed as written. –  Ryan Reich May 27 '13 at 18:33
Also, you can use TeX math in approximately the way you'd normally expect to. Please try it; it's more pleasant. –  Ryan Reich May 27 '13 at 18:47
I played around once with proving FTA from the Banach fixed point theorem but I couldn't get it to work. You can prove FTA from the Lefschetz fixed point theorem, though. Does that still qualify as "fixed point theory"? –  Qiaochu Yuan May 27 '13 at 18:48
Too bad "tl;dr" is too short for a comment. –  Andrej Bauer May 28 '13 at 7:11
After Todd's clean-up I have cast a vote to re-open –  Yemon Choi Jun 5 '13 at 16:51
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2 Answers 2

I believe the answer to this question is actually yes, and would be quite pleased if someone could communicate a copy of the last reference I have listed here. (My about-page has an email address.)

With regard to the answer already provided:

The Arnold proof is well known to be erroneous, but a correct (as far as I know) version is cited in an earlier MO post here. In particular, it is a proof of the FTA via the Brouwer Fixed Point Theorem.

The latter source is:

Some Properties of Continuous Functions. M. K. Fort, Jr. The American Mathematical Monthly, Vol. 59, No. 6 (Jun. - Jul., 1952), pp. 372-375. http://www.jstor.org/stable/2306806.

[Edit: Todd Trimble has kindly provided a link to the Fort paper that does not require jstor access.]

Separately, I see the following quotation:

"Recently, there have been very interesting proofs of the Brouwer theorem. Kulpa deduced a generalization of the Brouwer theorem from the Fubini theorem and the Weierstrass approximation theorem, and applied it to give a simple proof of the fundamental theorem of algebra."

The source of this excerpt is:

Park, S. (1999). Ninety years of the Brouwer fixed point theorem. Vietnam Journal of Mathematics, 27(3), 187-222. http://www.math.ac.vn/publications/vjm/vjm_27/No.3/187-222_Park.PDF

And the reference under discussion is:

W. Kulpa, An integral criterion for coincidence property, Radovi Mat.6 (1990) 313-321.

I gathered this information at the request of D. Goroff some time ago, at which point my search for the Kulpa paper was, unfortunately, fruitless. If anyone can find an accessible copy of Kulpa's paper, I would be most interested in it (and I know he would be as well).

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For those who do not have J-Stor access, here is another web-link to Fort's article: oldweb.cecm.sfu.ca/personal/jborwein/Expbook/Manuscript/… –  Todd Trimble Jun 5 '13 at 13:44
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Suppose $K/\mathbb C$ is a finite extension of fields. From the multiplication of $K$ we can turn the projectivization $P(K)$ of $K$ as a vector space into a group, which is then a compact connected Lie group. If $g\in P(K)$ is any point, then the map $L_g$ given by left multiplication by $g$ on $P(K)$ is homotopic to the identity map. Its Lefschetz number $\Lambda(L_g)$ is therefore equal to the Euler characteristic of $P(K)$ which is non-zero (because we know that $P(K)$ is a complex projective space!) It follows from Lefschetz's fixed point theorem, the map $L_g$ has a fixed point on $P(K)$. This is only possible if $g$ is the identity element of $P(K)$, and therefore $P(K)$ has exactly one point.

This means that the extension $K/\mathbb C$ is trivial, and that $\mathbb C$ is algebraically closed.

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Somewhere on MO there is a sketch of a proof of the FTA which as above turns the projective space $P(K)$ into a Lie group, observes it is commutative, invokes the classification of compact abelian Lie groups and using, say, the computation of the cohomology of tori and projective spaces, notes that no positive-dimensional torus is a projective space. My argument above is a minor variation, really. –  Mariano Suárez-Alvarez Dec 18 '13 at 0:12
This is also a variation of the proof linked to by David in the comments to the question, of course. –  Mariano Suárez-Alvarez Dec 18 '13 at 0:17
Mariano: the page you are probably thinking of is mathoverflow.net/questions/10535/…. Two answers there mention a proof by the Lefschetz fixed point theorem that is essentially the proof you write above. –  KConrad Dec 18 '13 at 0:48
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