Gauss proof of fundamental theorem of algebra
Gauss actually appends a footnote to this statement "if a branch of an algebraic curve enters a limited space, it necessarily has to leave it again" (Latin original follows below), in which he argues that:
It seems to be well demonstrated that an algebraic curve neither ends abruptly (as it happens in the transcendental curve $y = 1/\log x$), nor lose itself after an infinite number of windings in a point (like a logarithmic spiral). As far as I know nobody has ever doubted this, but if anybody requires it, I take it on me to present, on another occasion, an indubitable proof.
As explained by Harel Cain (see also Steve Smale), this outline of the proof shows that Gauss’s geometric proof of the FTA is based on assumptions about the branches of algebraic curves, which might appear plausible to geometric intuition, but are left without any rigorous proof by Gauss. It took until 1920 for Alexander Ostrowski to show that all assumptions made by Gauss can be fully justified.
Alexander Ostrowski. Über den ersten und vierten Gauss’schen Beweis des Fundamental satzes der Algebra. (Nachrichten der Gesellschaft der Wissenschaften Göttingen, 1920).
Here is the Latin original and an English translation of
Carl Friedrich Gauss. Demonstratio nova theorematis omnem functionem algebraicam rationalem integram unius variabilis in factores reales primi vel secundi gradus resolvi posse (PhD thesis, Universitat Helmstedt, 1799); paragraph 21 and footnote 10.
Iam ex geometria sublimori constat, quamuis curvam algebraicam, (sive singulas cuiusuis curvae algebraicae partes, si forte e pluribus composita sit) aut in se redientem aut utrimque in infinitum excurrentem esse, adeoque si ramus aliquis curvae algebraicae in spatium definitum intret, eundem necessario ex hoc spatio rursus alicubi exire debere. [*]
[*] Satis bene certe demonstratum esse videtur, curvam algebraicam neque alicubi subito abrumpi posse (uti e.g. evenit in curva transscendente, cuius aequatio $y=1/\log x$), neque post spiras infinitas in aliquo puncto se quasi perdere (ut spiralis logarithmica), quantumque scio nemo dubium contra rem movit. Attamen si quis postulat, demonstrationem nullis dubiis obnoxiam alia occasione tradere suscipiam.
[English translation]:
But according to higher mathematics, any algebraic curve (or the individual parts of such an algebraic curve if it perhaps consists of several parts) either turns back into itself or extends to infinity. Consequently, a branch of any algebraic curve which enters a limited space, must necessarily exit from this space somewhere. [*]
[*] It seems to have been proved with sufficient certainty that an algebraic curve can neither be broken off suddenly anywhere (as happens e.g. with the transcendental curve whose equation is $y = 1/\log x$ ) nor lose itself, so to say, in some point after infinitely many coils (like the logarithmic spiral). As far as I know, nobody has raised any doubts about this. However, should someone demand it then I will undertake to give a proof that is not subject to any doubt, on some other occasion.
A "modern" version of the proof will soon appear in American Mathematical Monthly (Authors: Daniel J. Velleman and Soham Basu). Here is the arxiv link. The outline is as follows:
- Get an auxiliary real polynomial for any complex polynomial $f(z)=\sum_{n=0}^{N}c_{n}z^{n}$ as follows $\tilde f(z)=(\sum_{n=0}^{N}c_{n}x^{n})(\sum_{n=0}^{N}\bar{c_{n}}x^{n})$
- Gauss already showed that the zeros of the real and complex parts of $\tilde f(z)$ interlace for any $\lvert z \rvert=R$ for large enough $R>0$. Certainly the interlacing condition fails for $R=0$.
- There is a subset of $R$ satisfying the interlacing which is half open interval open towards $+\infty$. Theory of real numbers says that this set has a least upper bound $R_{0}$. Further analysis (specially continuity considerations of the zeros of real and complex part) shows that the only possibility that interlacing fails at $R_{0}$ is when the real and complex parts of the polynomial both have a common zero at $R_{0}$.
There is also an alternative proof in the upcoming paper using straight line contours instead of circles. I hope the information is sufficient to build up a "modern" proof. I will be happy to elaborate if any of the points need further clarification.
$\def\RR{\mathbb{R}}\def\CC{\mathbb{C}}$Not sure this is useful, but a few years ago I put some thought into what it would take to rigorously give this proof in a Honors Multivariable calculus course (the sort where everything is proved). Here is what I came up with. Let $f:\CC \to \CC$ be a polynomial of degree $n$. It turns out to be convenient to make the following simplifying assumption -- letting $w_1$, $w_2$, ..., $w_k$ be the zeroes of $f'$, none of $f(w_i)$ are either purely real or purely imaginary. This is possible because, if $f(w_i)=0$ we are done, and otherwise we can replace $f$ by $e^{i \theta} f$ for some generic $\theta$.
Let $R = f^{-1}(\RR)$ and $I = f^{-1}(i \RR)$. The argument which we are trying to make rigorous is "near infinity, $R$ and $I$ look like $4n$ interleaved spokes, so they must cross somewhere in the interior." The implicit function theorem shows that $I$ and $R$ are closed one dimensional submanifolds of $\CC$. (This is why we required that the zeroes of $f'$ be disjoint from $I \cup R$.)
From this perspective, we can see that it would be bad if one of the components of $R$, for example, stopped at a point, or spiralled infinitely into a point -- we need them to disconnect $\mathbb{C}$.
I have heard people say that fixing this proof comes down to proving the Jordan curve theorem in the form "If $\phi: \RR \to \CC$ and $\psi: \RR \to \CC$ are smooth maps which are interleaved at infinity, then they $\phi(\RR)$ and $\psi(\RR)$ cross. In fact, I claim the hard thing is to show that the unbounded components of $R$ may be parametrized by $\RR$ in the first place, and that each component has two ends!
First, let's see why things are easy if we assume such a parametrization exists. Let $\Gamma$ be a connected component of $R$ touching one of the unbounded spokes. Suppose we could show there is a parametrization $\phi: \RR \to \Gamma$. Note that the composite $f \circ \phi: \RR \to \RR$ has nowhere vanishing derivative, so it is monotone and without loss of generality we can assume it is increasing. If we know that $\Gamma$ is unbounded in both directions (which is what is implicitly assumed when you draw a picture of $R$ as a bunch of strands connecting the spokes at infinity), then there is no need to use the Jordan curve theorem -- just apply the implicit function theorem to $f \circ \phi$! We have $\lim_{t \to \pm \infty} f(\phi(t))=\pm \infty$, so somewhere in the middle $f(\phi(t))=0$ and we win.
So the real challenge is to show that $\Gamma$, a connected unbounded $1$-dimensional submanifold of $\CC$, can be parametrized by $\RR$ and goes to $\infty$ in both directions. I looked up various proofs of the classification of $1$-dimensional manifolds, but they all seemed a little messier than I wanted to do in class.
Then I came up with the idea of just trying to invert the map $f: \Gamma \to \RR$, which is how I came up with this argument. I must admit, though, that the geometric origins are no longer visible.