My question concerns the argument given by Gauss in his "geometric proof" of the fundamental theorem of Algebra. At one point he says (I am reformulating) :

A branch (a component) of any algebraic curve either comes back on itself (I suppose that means : it is a closed curve) or it goes to infinity on both sides.

I have a geometric intuition of what it means, but I am not sure where or how to get a "modern" formulation of such result, and a proof.

  • $\begingroup$ I would understand the second case as '$\mathbb{C}\setminus \gamma$ has at least two unbounded connected components'. $\endgroup$ Commented Mar 29, 2016 at 16:31
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    $\begingroup$ It could make sense to give the original formulation, too, as well as an exact reference. $\endgroup$
    – user9072
    Commented Mar 29, 2016 at 17:06
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    $\begingroup$ This answer to a question on different proofs of FTA seems relevant. $\endgroup$
    – user9072
    Commented Mar 29, 2016 at 17:16
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    $\begingroup$ @FedorPetrov: I don't think that's what He means. It sounds more like thinking of the curve as we usually draw one: there is a point where we start drawing and there is one where we end. Those are the two sides. If they meet, the curve "comes back to itself" if they don't, then you could keep continuing with drawing in either direction="both sides". By the way, this sounds like an argument over the reals. Over the complex numbers every algebraic curve "goes to infinity".... $\endgroup$ Commented Mar 29, 2016 at 17:24
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    $\begingroup$ Side comment: although Gauss is often credited as giving the "first proof" of FTA, I've heard it said that it was Argand who gave the first rigorous proof (1814). $\endgroup$
    – Todd Trimble
    Commented Mar 29, 2016 at 18:38

5 Answers 5


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] (Wayback Machine):

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.

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    $\begingroup$ Thank you for this very interesting answer. I suppose then that Gauss uses (but does not prove) the following true assertion : "For any compact $K$ of the plane, any branch of any plane real algebraic curve $\gamma$ that intersects the boundary of $K$ properly (in the sense that the curve is not included in $K$ or its complement), actually intersects it properly at least two times". Am I correct in saying that this is true and that this is proven in the reference Alexander Ostrowski ? Is that a classical result in the "modern" theory ? If so, where can I find it "easily" (in a textbok) ? $\endgroup$
    – JonP
    Commented Mar 30, 2016 at 11:18

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:

  1. Get an auxiliary real polynomial for any complex polynomial $f(z)=\sum_{n=0}^{N}c_{n}z^{n}$ as follows $\tilde f(z)=(\sum_{n=0}^{N}c_{n}x^{n})(\sum_{n=0}^{N}\bar{c_{n}}x^{n})$
  2. 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 $R=0$.
  3. 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.

  • $\begingroup$ Is that really difficult? Start with some point $y\in \Gamma$, and let $\phi_0:I_0\to \mathbb{C}$ be a natural parametrisation of a piece of $\Gamma$ near $y$. Consider all possible extensions of $\phi_0$ to natural parametrisations of sub-curves of $\Gamma$ by larger intervals. Any two such extensions must agree on their common domain, therefore, we can consider a maximal extension $\phi:(a,b)\to\mathbb{C}$. If there is a sequence $t_k\nearrow b$ such that $\phi(t_k)$ is bounded, then we can pick a subsequence such that $\phi(t_k)\to z\in\Gamma$, and a contradiction... $\endgroup$
    – Kostya_I
    Commented Jan 10, 2023 at 10:32
  • $\begingroup$ ... is obtained by looking at a neighborhood of $z$: the part of $\Gamma$ in that neighborhood looks like a graph in right coordinates, the abscisses of $\phi(t_k)$ must be monotone, so $\phi$ in fact parametrizes one-half of the graph, and the other half is untouched since we coundn't have escaped without passing through $z$. In particular, $b<\infty$, but then the parametrization can be continued beyond $b$, a contradiction. $\endgroup$
    – Kostya_I
    Commented Jan 10, 2023 at 10:38
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    $\begingroup$ @Kostya_I I think this works! Let me say it in other words. Let $\Gamma$ be a connected component of $R$. Suppose that $f$ is bounded above on $\Gamma$. Then we can choose $z_k \in \Gamma$ such that $f(z_k)$ approaches $\text{sup}_{z \in \Gamma} f(z)$. These $z_k$ lie in a bounded region of $\mathbb{C}$ (details omitted) so we can choose a convergent subsequence; let its limit be $z$. Parametrizing $\Gamma$ near $z$ gives a contradiction. Similarly, $f$ is not bounded below on $\Gamma$. So $f$ is both positive and negative on $\Gamma$, and $f$ must be zero somewhere on $\Gamma$, QED. $\endgroup$ Commented Jan 10, 2023 at 12:22

David Savitt and some REU students write about "harmonic curves" and "meanders". I have not read the estimates in details and I heard they can be optimized.

The idea is to check $\mathrm{Re}[p(x)]=0$ and $\mathrm{Im}[p(x)]=0$ and to show that they intersect at exactly $n = \mathrm{deg} \, p$ points. Then we can have a topological proof since we can split the plane into $\mathbb{C} \backslash \mathbb{D} \cup \mathbb{D}$ where on the outside we have an alternating series of lines extending out to infinity and inside we have some kind of non-crossing matching of the lines.

They come across an objection similar to what you have, is that topological picture correct? Can't the curves $I$ or $R$ be self-intersecting or worse?

Back then, there was no systematic study of shape such as topology. Savitt merely proves these curves are non-singular without explicitly ruling out these phenomena.

Then Savitt studies $\mathrm{Re}[e^{-i\theta}p(x)] = 0$ which contains $R$ and $I$ as special cases (set $\theta = 0, 90^\circ$)

  • $\begingroup$ Without being too hand-wavy "singular" could mean we cover the Euclidean plane $\mathbb{C}$ with disks and either we the disk is empty, or a line passes through the disk as a chord or two lines intersect at exactly one point. Then we can count the intersections we find and there are exactly $n = \mathrm{deg} \, p$ of them. $\endgroup$ Commented Mar 30, 2016 at 14:04
  • $\begingroup$ The three proofs in Cain's note look kind of similar. I think this corresponds to #3. A "modern" approach could be Morse Theory or Picard-Lefschetz Theory. The original work of Picard seems to have precursor to the notion of divisor $\endgroup$ Commented Mar 30, 2016 at 14:27

One simple way to fill this "gap" in the Gauss' proof is as follows.

Let $p(z)$ be a polynomial of degree $n$, and let $T=\mathfrak{Re}(p)$, $U=\mathfrak{Im}(p)$. The proof starts by observing that for a large $R$, the level lines $T=0$ intersect the circle $|z|=R$ transversally at $2n$ points, say, $\eta_1,\dots ,\eta_{2n}$; the signs of $U$ at those points alternate, and the signs of $T$ alternate on the arcs $(\eta_1,\eta_2),\dots,(\eta_{2n},\eta_1)$ This is easily done by comparison to the leading term.

Then, let us enter the dics $|z|<R$ along the level line $\gamma$ ($T\equiv 0$ on $\gamma$) at $\eta_1$, and follow this line so that we have $T>0$ on the right and $T<0$ on the left. The idea of Gauss is that we must end up somewhere, and this somewhere can only be on the boundary of the disc, i. e., at one of the points $\eta_k$. Now, since we have $T>0$ on our right, $k$ must be even, so the signs of $U$ at the beginning and at the end are opposite, which means that there was a zero along the way, i. e., a point where $T=0$ and $U=0$. Q. E. D.

It remains to justify "we must end up somewhere, and this somewhere can only be on the boundary of the disc". First, observe that by Cauchy-Riemann equations, every critical point $a$ of $T$ is a zero of $p'$ (say, of multiplicity $k$), which means that there are only finitely many of them, and locally, the set $T=0$ near $a$ consists of $2k+2$ smooth curves ensuing from $a$ at equal angles (look at the leading term of Taylor expansion at $a$). Now, we can consider the maximal initial segment of the curve $\gamma$ starting at $\eta_1$ such that all the points on that segment are not critical (this segment is a simple curve), and a sequence $z_1,z_2,\dots$ of points that escapes any smaller initial segment. If this sequence is unbounded, then this initial segment crosses $|z|=R$, and we are done. Otherwise, we may pass to a subsequence and assume that $z_k\to z_0$, which by maximality must be a critical point. But then we can continue $\gamma$ from there, say, by going straight, and ask for the new maximal segment, and in a finite number of steps the process terminates. Isn't it exactly what Gauss claimed: "algebraic curve either comes back on itself or it goes to infinity on both sides"?

Ostrowski's approach is even more "hands-on". He doesn't even use any complex analysis or the fact that the polynomial $T$ is harmonic. He comes up with a simple algebraic argument that after taking away suitable factors from $T$, there will be only finitely many common zeros of $T$ and $\partial_x T$ and finitely many common zeros of $T$ and $\partial_y T$. If $x_1,\dots ,x_m$ are abscisses of all these common zeros, then on each for the segments $(x_1,x_2),\dots $, the set $T=0$ breaks into several disjoint graphs $y(x)$ of smooth functions; in fact, since $y'=\partial_x T/\partial_y T$, these functions will even be monotone and thus have limits at the edges of the intervals. Then, at every point where these "elementary branches" meet, there is an even number of them meeting - after all, they separate domains of different sign of $T$. And then the proof is concluded as above.

All of this leaves a question as to why people keep claiming that Gaussian proof "contained a gap" in the first place, and why some of them call the gap "immense", and its justification - "subtle" and "requiring advanced topology". Certainly, any pre-Cauchy (pre-Dedekind, pre-Zermelo, you name it) proof of virtually anything is not completely rigorous from today's standpoint. But what Gauss used was intuitively obvious, and more importantly, the rigorous proof is straightforward - it hardly involves any mathematical ideas, just standard techniques to put epsilons and deltas in place and make the intuition rigorous. Nothing compared to the pre-Gauss proofs that tacitly assumed existence of splitting field, say.

  • $\begingroup$ Before Gauss's first proof, Bolzano theorem (Later known as Bolzano-Weistrass theorem) was already known. Concept of supremum and infimum was also known to Gauss. It was theoretically possible to build a proof by putting these together (reference). Of course completeness of real numbers would have to be assumed, but that is less subtle a gap than considering algebraic curves entering and leaving a closed region. $\endgroup$
    – sobasu
    Commented Nov 7, 2017 at 15:44
  • $\begingroup$ Can you clarify how a theorem proven in 1817 (and infamous for being completely ignored even after that) was already known in 1799? $\endgroup$
    – Kostya_I
    Commented Nov 9, 2017 at 8:07
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    $\begingroup$ I am completely wrong there, of course! Thanks for clarifying. I was considering his fourth proof from 1849, which is very similar to his first proof and in which he tried to unsuccessfully plug the loopholes. So in principle he could have filled in the "gap". As for ignoring Bolzano's theorem, the personal biases of mathematicians are at fault, not available mathematical knowledge. And the reference that I have included fills in the "gap" using post-Dedekind knowledge, so it indeed does not do chronological justice to Gauss's first proof. $\endgroup$
    – sobasu
    Commented Nov 9, 2017 at 9:57

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