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Consider the product of complex linear monic polynomials times polynomials of degree less than $n$, that is $\big( (z-\lambda), p(z)\big)\mapsto (z-\lambda)p(z)$. If we represent a polynomial by its coefficients, this produces a map $$h:\mathbb{C}\times\mathbb{C}^n\to \mathbb{C}^{n+1},$$ \begin{align} & \begin{bmatrix} \lambda \\ p_{0} \\ p_{1} \\ \vdots \\ p_{n-2} \\ p_{n-1} \end{bmatrix} \mapsto \begin{bmatrix} \phantom{p_0}-\lambda p_0, \\ p_0-\lambda p_1 \\ p_1-\lambda p_2\\ \vdots \\ p_{n-2}-\lambda p_{n-1}\\ p_{n-1} \phantom{-\lambda p_{n-1}} \end{bmatrix} \end{align} which is, by the Fundamental Theorem of Algebra, surjective onto the space of polynomials of positive degree at most $n$; in fact, a.e. with fiber of cardinality $n$. So the Gaussian measure of the image of $h$ with multiplicity (i.e., the integral of the multiplicity function $\# h^{-1}$) w.r.to Gaussian probability measure) is exactly $$n=\frac{1}{\pi^{n+1}} \int_{\mathbb{C}^{n+1}} \# h^{-1}(q)e^{-\|q\|^2} dq.$$ On the other hand, if we apply the Change of Variable formula, we find the following multiple integral with respect to the $2(n+1)$-dimensional Lebesgue measure on $\mathbb{C}\times\mathbb{C}^n\sim \mathbb{R}^{2(n+1)}$: $$\frac{1}{\pi^{n+1}}\int_{\mathbb{C}\times\mathbb{C}^n} \Big|\sum_{k=0}^{n-1}c_k\lambda^k\Big|^2e^{{-\sum_{k=0}^{n}| c_{k-1}-\lambda c_k|^2}} \mathrm d(\lambda,c_0,\dots,c_{n-1}),$$ where for the sake of notation we put $c_{-1}=c_n=0$ in the sum. If we compute directly this integral and find the value $n$, and use the obvious bound $\# h^{-1}\le n$ (meaning that a polynomial can't have more than $n$ roots), we have that the image of $h$ has full measure, so it is dense. Due to the elementary a priori bounds on the roots of a polynomial, we conclude that $h$ is indeed surjective onto the polynomials of positive degree, thus proving the Fundamental theorem of Algebra by means of multiple integrals.

The computation of the above integral turns out to be an elementary, yet not completely trivial exercise of iterated integrals and linear algebra (I will link the complete computation later). I wonder if there is a more clever way to perform the computation (of course, without using the FTA, nor any proof of it). Also, I am pretty sure -quia nihil sub sole novi, that somebody should already have done this computation, and I'd be glad to learn who.

edit Here is the elementary computation, for this question, and to whom is interested.

Pietro Majer
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    @FedorPetrov I uploaded it few hours ago, it is still on hold. Does this work? https://www.dropbox.com/s/v0ra4duxh85iloq/Multiple%20Integrals%20and%20TFDA.pdf?dl=0 – Pietro Majer Jan 28 '21 at 20:18
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    Doesn't this just prove that almost every polynomial has zeroes? (I.e. the set of those degree $n$ polynomials without zeroes has measure 0) – Achim Krause Jan 30 '21 at 20:19
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    @AchimKrause yes, but a polynomial close enough to a polynomial without roots also does not have roots (it is bounded from below). – Fedor Petrov Jan 30 '21 at 20:49
  • @FedorPetrov, what is "it" in "it is bounded from below"? – LSpice Jan 30 '21 at 20:51
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    @LSpice it=a close polynomial. If $p(x)$ of degree $n$ does not have roots, then we have $|p(x)|>c>0$ for all $x\in \mathbb{C}$. Then $|q(x)|>c/2$ whenever $q$ of degree $n$ has coefficients close enough to those of $p$ (consider separately a large disc and its complement). – Fedor Petrov Jan 30 '21 at 22:13
  • Or also, the set of monic polynomials of degree $n$ with a zero is closed. Indeed if for a sequence of monic polynomials $p_j\to p$ and $p_j(\lambda_j)=0$, then the sequence ${\lambda_j}$ is bounded. So up to a subsequence converges, and the limit is a zero of $p$. – Pietro Majer Jan 31 '21 at 08:56
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    This reminds me of a proof of FTA by P. Pushkar formulated entirely in terms of real numbers: each real polynomial of even degree at least $2$ can be factored into a product of real quadratic polynomials. Your use of the fiber of $h$ is analogous to Pushkar's use of the degree of a smooth proper map. His paper is at http://www.mathnet.ru/php/archive.phtml?wshow=paper&jrnid=mp&paperid=7&option_lang=eng and I wrote up a translation of it into English in the last section of https://kconrad.math.uconn.edu/blurbs/fundthmalg/propermaps.pdf. Properness of the mapping is proved by "compactification". – KConrad Jan 31 '21 at 14:31
  • I also wrote up Pushkar's proof on this site back in 2010 when there was a big-list question about different proofs of FTA: https://mathoverflow.net/questions/10535/ways-to-prove-the-fundamental-theorem-of-algebra. – KConrad Jan 31 '21 at 15:15
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    Thank you. I am aware of similar proofs. E.g. one can study the map "zeros $\mapsto$ polynomial"as a covering projection with singularities; or by continuation arguments, or topological degree.The existence result in what I proposed above is somehow different, more of measure theory's nature -(maybe less natural, for this theorem). – Pietro Majer Jan 31 '21 at 16:51
  • Surjectivity of the map $(\lambda, p(z))\mapsto (\lambda-z)p(z)$ is just an obvious reformulation of the statement of the FTA, so it is not a real element of similarity, otherwise we should say all proofs are similar because they all have the same thesis). The aim of the above proof is testing the power of integral calculus in $\mathbb R^n$ by proving the FTA, with no or little use of topology. – Pietro Majer Nov 30 '22 at 14:44
  • Actually, the topological degree relies on deep facts of topology of $\mathbb R^n$ (that one can prove in several ways: by differential topology, differential geometry, algebraic topology, integral calculus etc: but always there is a consistent set of proofs behind). So, if you put together the topological degree and polynomials, whatever you do, you end up with a proof of the FTA; we may say by a pun, the difficult problem is how not to prove the FTA by the topological degree. – Pietro Majer Nov 30 '22 at 14:47

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