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The fundamental theorem of algebra is the statement that a complex polynomial of positive degree has at least one root. I do not know complex analysis but I searched for proofs of the statement and came across proofs using complex analysis which seemed rather short and elegant. This is to say: I am aware that there exist very easy proofs using tools of complex analysis. Since I do not know complex analysis yet I started to wonder if there are any known proofs that are short and easy that use only tools that a first year undergrad knows? Like real analysis and linear algebra?

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    $\begingroup$ Spivak's book Calculus has an accessible proof. $\endgroup$ – littleO Aug 12 '13 at 8:46
  • $\begingroup$ For a first yearer it is going to be a little hard and not precisely because of the mathematical level itself but because of the abstraction level required. There's a paper (in italian) with 14 differnet proofs of the FTA, but all of them, AFAIK, are of higher level than first underg. year. Even the shortest proof I know, that of Loya's, requires some complex analysis... $\endgroup$ – DonAntonio Aug 12 '13 at 8:50
  • $\begingroup$ @DonAntonio What is the shortest proof? If you are allowed the assume things, the shortest proof has to be that every polynomial map lifts to an endomorphism of the Riemann sphere which, being a compact Riemann surface, is either constant or surjective. $\endgroup$ – Alex Youcis Aug 12 '13 at 9:03
  • $\begingroup$ @AlexYoucis, read this paper of 1.2 pages: math.binghamton.edu/loya/papers/LoyaFTA.pdf It doesn't require topology nor complex analysis, though some complex numbers stuff is included (but, for example, it is at high school level of complex numbers). By "shortest" I also meant, of course, very elementary. $\endgroup$ – DonAntonio Aug 12 '13 at 9:15
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    $\begingroup$ The usual proofs are not very hard. You may have a look at this blog post where two proofs are presented and the second one uses a bare minimum of algebra and analysis and is sort of a gem. $\endgroup$ – Paramanand Singh Feb 16 '18 at 2:23
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Here are three accessible proofs, via Keith Conrad:

http://www.math.uconn.edu/~kconrad/blurbs/fundthmalg/fundthmalgcalculus.pdf

http://www.math.uconn.edu/~kconrad/blurbs/fundthmalg/fundthmalglinear.pdf

http://www.math.uconn.edu/~kconrad/blurbs/fundthmalg/propermaps.pdf

(the last one requires a bit more sophistication, but isn't too bad)

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    $\begingroup$ Perhaps the second proof is the closest to a regular first years undergraduate level, and even that is doubtable. The other two use stuff which is usually covered after first year. $\endgroup$ – DonAntonio Aug 12 '13 at 8:53
  • $\begingroup$ @DonAntonio I disagree. The OP said "like linear algebra and analysis". Anyone with that background should be able to understand the first two with a bit of elbow grease. Admittedly, the last proof does seem a little to hard in these circumstances. $\endgroup$ – Alex Youcis Aug 12 '13 at 8:58
  • $\begingroup$ linear algebra is a wide subject and in my experience inner product, eigenvalues and stuff many times isn't covered in first year, and even if it is Hermitian and complex linear algebra, as required in Conrad's paper (not much, but still) already seems above the OP's requirements. $\endgroup$ – DonAntonio Aug 12 '13 at 9:18
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There is a proof using linear algebra due to Derksen :

H.Derksen, The fundamental Theorem of Algebra and Linear Algebra, Amer. Math. Monthly, 110, (2003), 620-623. http://www.math.lsa.umich.edu/~hderksen/preprint.html

A somewhat expanded version of it is also available (due to S. Kumaresan) :

http://main.mtts.org.in/expository-articles (See #15 under "Analysis")

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  • $\begingroup$ Note that the second link Alex Youcis gave is a modification of Derksen's proof. $\endgroup$ – littleO Aug 12 '13 at 8:59
  • $\begingroup$ I think your first link is down. $\endgroup$ – Viktor Glombik Apr 22 '19 at 6:05
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Following proof is an ad hoc proof designed for FTA so that every undergraduate can follow it without studying general path integrals in the Real or Complex plane.

If we suppose that our polynomial $P(z)$ doesn't vanish in the Complex plane, then $1/P(z)$ is a continuous function. Since the components of the polynomial $P(x,y)=u(x,y)+iv(x,y)$ satisfy the Cauchy-Riemann conditions, we can infer that the components of $1/P(x,y)=u/(u^2(x,y) + v^2(x,y))+i(-v(x,y))/(u^2(x,y)+v^2(x,y))$ also satisfy the Cauchy-Riemann conditions. Therefore, by the simple special case of Green's theorem in a disk, we have the Cauchy Integral Theorem in a disk for the function $1/P(z)$ and in turn the Gauss Meanvalue Theorem for the function $1/P(z)$.

We know that $1/P(0)$ is not zero, but due to the Gauss Meanvalue Theorem $1/P(0) \rightarrow 0$ as the radius of the disk goes to infinity - a desired contradiction.

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