Is there any proof of the fundamental theorem of algebra that can be introduced to undergraduates who have just completed Calc III?

I am aware of three proofs of the fundamental theorem of algebra, using:

1. Liouville's theorem
2. The fundamental group of the punctured plane, or
3. Multiplicativity of field extensions together with the intermediate value theorem

If I had a small group of undergraduates who had taken Calc III, is there any of these proofs or another proof that I could show them over a period of 3 1-hour sessions or less?

• Now that you have at least one good answer, will you share your goal in showing these students such a proof? – user173 Apr 13 '14 at 1:59
• The real reason is that I wanted to understand a proof of it at that point of my education, and I wish someone had shown me one. – Brian Rushton Apr 13 '14 at 2:14
• That theorem never cried out to me for a proof. But when I was 9, I wanted a proof of V=Bh/3 for pyramids, which I finally got in calculus. – user173 Apr 13 '14 at 3:35
• Artin gave an incredible proof that uses algebra. See Dummit and Foote p. 616-17. Although this is also a proof that may transcend what your normal Calc III student can do. (i.e. I learned it in Math 672). – Vladhagen Apr 23 '14 at 0:35
• @MattF. My favorite argument for that (not really a proof) is the following: first realize that we must have $V = kBh$ for some constant $k$ by scaling arguments: namely we can partition the base into small squares, and so reduce the problem to square base pyramids. Formula is true for these with a universal constant by "shearing" (aka cavaleri's principle). So all we need is to determine the constant. But six pyradmids with square $1\times1$ base and height $\frac{1}{2}$ fit in a cube, so we must have $k = \frac{1}{3}$. No need to know how to integrate $x^2$. – Steven Gubkin Aug 29 '14 at 0:16

I think that the proof based on the fundamental group of the punctured plane can be re-wrigged, just by omitting references to homotopies and the fundamental group, to be convincing to students who don't even necessarily know calculus, as long as they have sufficient comfort with the complex plane to be able to think about how a polynomial maps loops about the origin. From an aggressively technical point of view this introduces some handwaving, but IMHO the underlying topological theorems are sufficiently intuitively reasonable that this more or less counts as a proof anyway.

More specifically:

(1) Do you think the students could (be made to) see that for $t$ very small, $f(te^{i\theta})$ (for $\theta\in[0,2\pi]$) traces a very small loop of some kind about the constant term of $f$, that therefore gets nowhere near the origin?

(2) Do you think they could see that for $t$ large, $f(te^{i\theta})$ traces some kind of incredibly huge $n$-fold loop that encloses the origin?

If the answers are both yes, then I bet they will also see that if $t$ varies continuously from very small to very large, the image of $f(te^{i\theta})$ is going to vary continuously and therefore have to pass through the origin.

You can find a very elementary and direct proof here.

Personally, while I think that any of the more sophisticated proofs is very cool the direct proof is valuable since it really shows the fundamental theorem of algebra is not a theorem of algebra but rather of the complex numbers, and that it really is fundamental in the sense that it is very elementary (and important). It shows the proof is really not that complicated and amounts to a relatively simple analysis exercise. Using other methods (in my humble opinion) is using a sledgehammer where it really is not needed.

• I disagree. The fact that it is possible to give such a proof is not an argument in favor that it represents causality. The Liouville-corollary argument is, to me (yes, I do acknowledge the subjectivity), the only sane, understandable, memorable proof. All other arguments are labyrinthine in comparison, immemorable to say the least, in my opinion. – paul garrett Apr 13 '14 at 0:04
• I disagree that the Liouville-corollary proof is the unique simplest / most memorable. Here are two others that are very short and elegant and both I find extremely memorable. (But, they all use higher tech theorems than the OP wants.) – benblumsmith Apr 13 '14 at 2:02
• (1) A polynomial extends to a continuous map $S^2\rightarrow S^2$ via stereographic projection. By sliding all non-leading coefficients to zero, the map is homotopic to $z\mapsto z^n$ and therefore has Brouwer degree $n$. If $n\neq 0$, it is therefore surjective. (2) $f(te^{i\theta})$ is a homotopy from a constant map ($t=0$) to an $n$-fold loop about the origin (large $t$). If $n\neq 0$, these represent different elements of $\pi_1(\mathbb{C}\setminus \{0\})$, so the image of $f$ can't be contained in $\mathbb{C}\setminus\{0\}$. – benblumsmith Apr 13 '14 at 2:04
• Another memorable/(not-quite-as) short one: let $F$ be an algebraic extension of $C$ that is Galois over $R$. The fixed field of a Sylow 2-subgroup of $G=Gal(F/R)$ has odd degree over $R$ thus is generated by an odd-degree irreducible polynomial, which must be linear because odd-degree real polynomials always have real roots. Thus $G$ is a 2-group, thus so is $H=Gal(F/C)$. If $F>C$, any maximal subgroup of $H$ is index 2 normal thus its fixed field is a quadratic extension of $C$. But $C$ can't have a nontrivial quadratic extension since every complex number has a square root, so $F=C$. – benblumsmith Apr 13 '14 at 2:17
• My point is, what counts as "sane / memorable" depends on what the student knows. Which proof you go for depends what you have the best working knowledge of: Brouwer degree? Fundamental group? Liouville's theorem? Sylow theorems and the fundamental theorem of Galois theory? Etc. – benblumsmith Apr 13 '14 at 2:24

Your students might find it useful to see this "visual approach" to proving the FTA:

Velleman, D. J. (2007). The Fundamental Theorem of Algebra: A Visual Approach. Link.

For a more rigorous approach by the same author, see:

Velleman, D. J. (1997). Another proof of the fundamental theorem of algebra. Mathematics Magazine, 216-217. Link.

(The latter proof uses "the fact that entire functions can always be represented by power series.")

Finally, an entire list of proofs can be found here. I have not read closely through these, so I can neither make a more specific recommendation nor vouch for all of them; in particular:

Remark. In the list above, #60 "A topological proof of the fundamental theorem of algebra" (Arnold, 1949) is known to have errors. This is why he published a correction paper a couple years later (#58); the main idea, though, of using the Brouwer Fixed Point Theorem to prove the FTA has been carried out (though perhaps this is a result your post-Calc III students have not yet seen). In case this is useful, see:

Fort, M. K. (1952). Some properties of continuous functions. American Mathematical Monthly, 372-375. Link.

(The last two sources I became aware of while writing up an answer here.)

Unsurprisingly, there is a list of ways to prove the FTA on MathOverflow.

This is likely to include most any suggestion that could be posted here (on MESE) but the reverse inclusion is sure not to hold: The majority of these proofs are not written for students who just completed Calc III. On the other hand, three lectures with the FTA as a sole goal could cover a fair bit of material...

You can find a proof that goes back to Gauss, which is based only on multivariable calculus (double integrals and partial derivatives) at http://www.math.uconn.edu/~kconrad/blurbs/fundthmalg/fundthmalgcalculus.pdf. I think the proof will come across as rather mysterious even if it can be followed line by line. The last paragraph of the file gives an indication of what proof it is related to in complex analysis.

You can short-cut the Liouville proof as follows. Assuming that $f$ is a monic polynomial of degree $d>0$ with no roots, define $$h(r) = \int_{\theta=0}^{2\pi} f(r\,e^{i\theta})^{-1}\,d\theta.$$ If you are willing to differentiate under the integral sign, then it is not hard to show that $h'(r)=0$, so $h$ is constant. When $r$ is large enough the $z^d$ term in $f(r\,e^{i\theta})$ will dominate and we see that $|h(r)|$ is approximately $2\pi r^{-d}$ at most. It follows that $h(r)\to 0$ as $r\to\infty$, but $h$ is constant, so $h(0)=0$, which is impossible as $h(0)=2\pi/f(0)$.

Various things need to be justified to turn this into a complete proof, but I think that they are all quite plausible and intuitive.

In this video, David Eisenbud (a math prof at Berkeley) explains a proof of the fundamental theorem of algebra that is very enlightening / beautiful, and almost makes the result seem obvious.

Following this answer, students can experiment with Velleman, D. J. (2007). The Fundamental Theorem of Algebra: A Visual Approach using interactive versions of the paper's plots. (Disclosure: I am a dev for the site.)

• I would have commented, but didn't have the rep. – gballan Apr 13 '14 at 23:40
• Added opening reference to Benjamin's answer. – gballan Apr 14 '14 at 1:04

Here is an elementaryproof which considers $p(z)$ as a function of $(x,y)$ where $z=x+iy$.

The only assumptions are that $z^{n}=r$ has a solution for all integer $n$ and real $r$ and that intermediate value theorem holds for continuous real valued functions along any continuous curve on the complex plane. The second assumption can be proved by parametrizing the curve and then using the least upper bound principle for real numbers.

Writing $z= r e^{-i \theta}$ we can find a big enough $r=R$ such that $Re(p(R,\theta))$ is positive at $\theta=\frac {2\pi k} {n} + \frac {\phi} {n}$ and negative at $\theta=\frac {2\pi k} {n} + \frac {\pi} {n} + \frac {\phi} {n}$. This is because $Re(p(R,\theta))$ is dominated by $r^{n}cos(n\theta)$. $\phi$ is the angle of the complex coefficient of $z^{n}$.

Now , $p(z)$ being continous, $Re(p(z))$ is also continuous. So the zeros of $Re(p(z))$ form a continuous curves on the complex plane. The point is to show that on one such continuous curve there is a point where $Im(p(z))>0$ and another point where $Im(p(z))<0$. Then on the particular curve $Re(p(z))=0$ there must be a point where $Im(p(z))=0$. However, this is the solution of $p(z)=0$.

The proof can be visualized in the following way-

The arc $r=R$ between $\theta=\frac {2\pi k} {n} + \frac {\phi} {n}$ and $\theta=\frac {2\pi k} {n} + \frac {\pi} {n} + \frac {\phi} {n}$ is called a even sector for even $k$ and is called a odd sector for odd $k$.

Let us start from the even arc with $k=0$. The values of $Re(p(R,\theta))$ at the ends of this arc are of opposite sign, so it much be zero somewhere on the curve. The same thing holds if we increase $r \ge R$ continuously. So we get a continuous curve for $Re(p(z))=0$ in the sector $\theta \in$ { $0$, $\frac {\pi} {n}$} for $r\ge R$. Let us call this curve $f(z)$ Now, zeros of a continuous function divide the plane into two disconnected parts. One of the parts is finite in area if the curve is closed. Otherwise both areas are unbounded. Here, since we cannot have $Re(p(z))=0$ for $\theta =0$ and $\theta=\frac {\pi} {n}$ for $r\ge R$, the curve $f(z)$ does not intersect the rays $\theta =0$ and $\theta=\frac {\pi} {n}$ for $r \ge R$. This means $f(z)$ cannot be closed.

Now, $f(z)$ must cross one of the odd arcs. This can be constructed using exhaustion (there are only finite numbers of arcs). If not, there must be another curve satisfying $Re(p(z))=0$ and the desired conditions.

$Im(p(R,\theta))$ is dominated by $r^{n}sin(n\theta)$. So, for big enough $r$, $Im(p(r,\theta))$ is positive somewhere in the even sector and negative somewhere in the odd sector. So $Im(p(z)=0)$ somewhere on the continuous curve $f(z)$. This gives a solution for $p(z)=0$.

Additionally, $f(z)$ is not asymptotic to $\theta=\frac {2\pi k} {n} + \frac {\phi} {n}$ or $\theta=\frac {2\pi k} {n} + \frac {\pi} {n} + \frac {\phi} {n}$. This can be proven by using the fact that $Re(z) < |z|, \forall z \in \mathbb{C}$. If this was not true, domination of $Im(p(z))$ on $f(z)$ by $r^{n}$ would be problematic since $sin(n \theta)$ might tend towards zero.

Finally, for real coefficients, if $z$ a solution, $\overline z$ is also a solution for $p(z)=0$, since $\overline p(z) = p(\overline z)=0$. So for every root found, we can do division algorithm to get another polynomial of degree $n-2$. We know that quartic polynomials are always solvable using radicals. So induction shows that all polynomials with real coefficients are solvable on the complex plane.

I admit I'm a bit late to the party, but for completeness let me add a link to an interactive version of one of the visual proofs:

The fundamental theorem of algebra - a visual proof

(Disclaimer: I'm the author of that page.)

$\def\RR{\mathbb{R}}\def\CC{\mathbb{C}}$ Here is a proof via ODE. I imagine it's been found before, but I hadn't seen it before. The essential idea is that, if $z: \RR \to \CC$ is an inverse function to $f$ then we should have $z'(t) = 1/f'(z(t))$ (formula for the derivative of the inverse function, in an exotic setting). Therefore, we can solve this ODE and get an inverse function, and we can evaluate the inverse function at $0$ to find a root of $f$.

Let $f(z) = f_n z^n + \cdots + f_1 z + f_0$ be the polynomial which we wish to prove has a root. We first make a simplifying assumption:

Simplifying Assumption: We may assume that there is no point $w$ where $f(w) \in \mathbb{R}$ and $f'(w)=0$.

Proof: Let $w_1$, $w_2$, ..., $w_k$ be the zeroes of $f'(z)$, there are finitely many of them by the easy part of the FTA. If $f(w_i)=0$ for any $i$, we are done. If not, replace $f(z)$ by $e^{- i \alpha} f(z)$ for some $\alpha$ not equal to the arguments of any $f(w_i)$. $\square$

With this assumption, we will show

Theorem: There is a smooth curve $t \mapsto x(t) + i y(t)$, parametrized by $\RR$, so that $f(x(t)+i y(t))=t$.

In particular, the point $x(0)+i y(0)$ is a zero of $f$.

Lemma 1: There is a constant $c>0$ so that $|f'(z)| > c$ for $z \in f^{-1}(\RR)$.

Proof: Choose $R$ large enough that $n |f_n| S^{n-1} > \sum_{k=0}^{n-1} k |f_k| S^{k-1} + 1$ for $S>R$. Then $|f'(z)| > 1$ for $|z|>R$. The set $f^{-1}(\RR) \cap \{ |z| \leq R \}$ is closed and bounded, hence compact, so $|f'(z)|$ has a minimal value $b$ on that set, and by the simplifying assumption $b>0$. Take $c = \min(b,1)$. $\square$.

In order to set up an ODE, we also need an initial value:

Lemma 2: There is a point $z_0 \in \CC$ where $f(z_0) \in \RR$.

Proof: Choose $R$ large enough that $|f_n| R^n > 2 \sum_{k=0}^{n-1} |f_k| R^k$. Let $f_n = |f_n| e^{i \alpha}$. Then $\mathrm{Im} f(R e^{i (-\alpha-\pi/2)/n})<0$ and $\mathrm{Im} f(R e^{i (-\alpha+\pi/2)/n})>0$. So, by the intermediate value theorem, there is some $\theta$ with $f(R e^{i \theta}) \in \RR$. $\square$

Write $t_0$ for $f(z_0)$.

Consider the ODE $z'(t) = 1/f'(z(t))$ on $\RR \times \CC$, with initial value $z(t_0) = z_0$. Then this ODE is solvable for all $t \in \RR$, and $f(z(t)) = t$ for all $t$.

We first check that any solution to the ODE obeys $f(z(t))=t$. The proof is simply to differentiate the left hand side: $z'(t) f'(z(t)) = \frac{1}{f'(z(t))} f'(z(t)) = 1$ and $f(z(t_0)) = t_0$, so $f(z(t)) = t$. However, this assumes that students are comfortable differentiating complex valued functions of real arguments by the usual rules. So I'd first do the quick but questionable way and then I'd write it out: Put $f(x(t)+iy(t)) = \sum (g_k + i h_k) (x(t)+iy(t))^k$ and carefully take the derivative with respect to $t$, seeing that you get $z'(t) f'(z(t))$ at the end.

Now, we must check that solutions to the ODE extend to all $\RR$. Suppose, for the sake of contradiction, that there is a solution on $[t_0, q)$ but not beyond $q$; a similar argument works to show the ODE is solvable for all negative time.

By the previous computation, $f(z(t)) = t$ for $t \in [t_0,q)$, so in particular $z(t) \in f^{-1}(\RR)$. By Lemma 1, the right hand side of the ODE is bounded above by $1/c$ for $t \in (p,q)$, so $\lim_{t \to q^{-}} z(t)$ exists; call this limit $w$. By continuity of $f$, we have $f(w) = f \left( \lim_{t \to q^{-}} z(t) \right) = \lim_{t \to q^{-}} f(z(t)) = \lim_{t \to q^{-}} t = q$. In particular, by our Simplifying Assumption, $f'(w) \neq 0$. So we can solve the ODE again with initial condition $z(q) = w$. This gives a solution which agrees with the previous $z$ on a neighborhood to the left of $q$ (by uniqueness of solutions to ODEs), and is defined to the right of $q$, a contradiction.