The title says it all: I want to show that an arbitrary cubic polynomial can be factored as a product of linear terms without appealing to the fundamental theorem of algebra and (preferably) without appealing to the general formula for solving a cubic equation. This is exercise $2.11$ from Ian Stewart's Galois Theory text. Explicitly, it says

Prove, without using the winding number, that every cubic polynomial over $\Bbb C$ can be expressed as a product of linear factors over $\Bbb C$.

In the exercise before, he asks us to show that a cubic polynomial over $\Bbb R$ can be factored as a product of linear terms over $\Bbb C$. This is pretty much immediate since it necessarily tends to $-\infty$ in one direction and $\infty$ in the other direction, hence by the IVT it has a zero. The remainder theorem says that we can factor this linear term out, and then we just use the quadratic formula and we're done. I've been thinking about how we can show something similar for a polynomial over $\Bbb C$ and I haven't figured anything out. Is there a method besides appealing to the formula for cubic polynomials?

The proof of the FTA in this book uses the winding number and presents the general formula for a cubic, so I'm not inclined the solution for this problem relies on either of these results.

  • $\begingroup$ I think that the answer to this question depends on the way the fundamental theorem of algebra is proved in Stewart's book - or if it is proved at all. Even though this feels like cheating, the proof of the fundamental theorem that I thought was the standard proof does something similar to the (real) case you describe and invokes Liouville's theorem. This may well be what you're looking for (in the case of a degree-three polynomial, if you want to). I'm curious now about the intended solution. Could you post it, when you know it? $\endgroup$ – HSN Jan 17 '13 at 20:34
  • $\begingroup$ In the book, Stewart uses something about winding numbers to prove the fundamental theorem of algebra. I've read it pretty closely, but I don't understand winding numbers fluently enough to think that I've completely grasped the proof. I've been thinking about this, though. I'll post it when I know. $\endgroup$ – Clayton Jan 18 '13 at 1:03
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    $\begingroup$ You can prove the fundamental theorem of algebra without winding numbers, but it seems rather unnatural to me to prove the result for cubics without showing it for all polynomials at once, i.e. without proving the whole fundamental theorem of algebra. $\endgroup$ – Ewan Delanoy Jan 20 '13 at 17:53
  • $\begingroup$ @EwanDelanoy: The problem is starred (indicating it is slightly more difficult than the average problem), but it doesn't seem like the author would leave it as an exercise to prove the FTA in a different manner. It would seem there is some straightforward approach as in $\Bbb R$, but without the use of IVT. $\endgroup$ – Clayton Jan 20 '13 at 18:01
  • $\begingroup$ There's a related question $\endgroup$ – Hurkyl Jan 25 '13 at 19:28

For what it's worth here's a trick method for cubics specifically.


Define complex exponential. Can use $e^{(x+iy)}=e^x(\cos(y)+i\sin(y))$ or the power series. From this first definition it's clear that this is onto as map from $\mathbb{C}$ to $\mathbb{C}\setminus{0}$. Note that $e^{(z_1+z_2)}=e^{z_1}e^{z_2}$.

Now define complex sine by $\sin z= \frac{1}{2i} (e^{iz}-e^{-iz})$. We can check that this is onto by solving $\sin z=w$- namely, $u=e^{iz}$ gives $u-\frac{1}{u}=2iw$ which is quadratic, and we solve for (non-zero) $u$ and $iz$ in turn.

Finally, note that $\sin(3z)=3\sin z -4 \sin^3 z$ either because this is true for real $z$ and hence always, or by checking explicitly from definitions (using the fact $(u-\frac{1}{u})^3=u^3-\frac{1}{u^3}-3(u-\frac{1}{u})$ this is easy).

Main part:

If $z^3+az^2+cz+d=0$ we can sub $w=z-a/3$ to get rid of the quadratic term, getting $w^3+kw+l=0$. Now sub $w=k^{1/2}u$ and divide out by $k^{3/2}$ to get $u^3+u+p=0$. Finally, let $u=(-4/3)^{1/2}t$, which gives $-4t^3+3t=q$. We are done: there exists $z$ with $\sin 3z =q$, and $t=\sin z$ is a solution.

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    $\begingroup$ The case $k=0$ is left as exercise for the reader :) $\endgroup$ – Max Jan 25 '13 at 17:09
  • $\begingroup$ That was a typo. I edited it out. Hope it makes sense now - just plug in and simplify. $\endgroup$ – Max Jan 25 '13 at 18:00
  • $\begingroup$ Yes, it makes perfect sense now. Thanks! $\endgroup$ – Clayton Jan 25 '13 at 18:15
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    $\begingroup$ To add to this, if you wanted to recover the formula for the roots in terms of the coefficients of your original polynomial, this is possible from this approach, because finding $\sin z$ from $\sin 3z$ reduces to computing cubic and square roots, using the complex exponential representation. $\endgroup$ – Andrés E. Caicedo Jan 25 '13 at 19:18

This is probably not what you want, but for the record here is a proof that does not use the winding number. (as you can see, the problem is that it can be obviously modified to show the whole fundamental theorem of algebra).

Let $f(z)=a_3z^3+a_2z^2+a_1z+a_0$ be a cubic polynomial : $a_3\neq 0$. Since you already have the quadratic formula, it suffices to show that $f$ has a root.

The closed ball with center $0$ and radius $M$ (let us call it $K$) is compact, so $|f|$ attains a minimum $\rho$ on $K$. There is a $z_0\in K$ such that $|f(z_0)|=\rho$. Replacing $f$ with $g(z)=f(z_0+z)$, we may assume that $z_0=0$. Let $k$ be the smallest value in $\lbrace 1,2,3 \rbrace$ such that $a_k\neq 0$.

Then we can write $f(z)=a_0+a_kz^k+z^{k+1}h(z)$, where $h$ is another polynomial. There is a constant $C$ such that $|h(z)| \leq C$ whenever $|z| \leq M$. Then we have

$$ |f(z)| \leq |a_0+a_kz^k| + C |z|^{k+1} \tag{1} $$

Suppose by contradiction that $\rho=|a_0|\neq 0$. Write $a_0=\rho e^{i\theta}$ with $\theta\in {\mathbb R}$, and $a_k=\rho_ke^{i\theta_k}$ with $\rho_k>0$ and $\theta_k \in {\mathbb R}$. For small $\varepsilon >0$, put

$$ z_{\varepsilon}=\big(\frac{\varepsilon}{r_k}\big)^{\frac{1}{k}} e^{i\frac{\pi+\theta-\theta_k}{k}} \tag{2} $$

Then $a_0+a_kz^k=(\rho-\varepsilon)e^{i\theta}$ and $|z|=\big(\frac{\varepsilon}{r_k}\big)^{\frac{1}{k}} $. So by (1), we have

$$ |f(z_{\varepsilon})| \leq \rho-\varepsilon + C \big(\frac{\varepsilon}{r_k}\big)^{\frac{k+1}{k}} $$

This will be $\lt \rho$ for small enough $\varepsilon$, contradicting the minimality of $\rho$. So $\rho=0$, $f$ has a root and we are done.

  • $\begingroup$ I like the answer, but it does seem perhaps too difficult, if that is possible. The text seems to be aimed very low for undergraduates, so I don't think it would involve some of the more difficult aspects here... If at the end of the week, I haven't received any better answers, I'll be sure to accept your solution! $\endgroup$ – Clayton Jan 21 '13 at 17:29
  • $\begingroup$ I tend to think that if such an elementary proof existed, everybody would know about it. Perhaps the location of this exercice in Stewart’s book is an error. $\endgroup$ – Ewan Delanoy Jan 21 '13 at 17:35
  • $\begingroup$ By the way, I seem to remember that a “completely explicit”, computer-checkable proof of the FTA is an active research project in computer science today. $\endgroup$ – Ewan Delanoy Jan 21 '13 at 17:40
  • $\begingroup$ I can't say I would find that too hard to believe. Well, we'll see, I suppose. It's not that I dislike the answer, please don't misunderstand me. It just seems much more difficult than even the other starred problems. $\endgroup$ – Clayton Jan 21 '13 at 17:40

If using a complex-analytical proof of the fundamental theorem of algebra specifically for cubic polynomials is allowed (which I pointed out earlier I don't expect to be the case), the following might be a bit easier.

We will assume that $f(z)$ is a complex polynomial without zeroes in $\mathbb{C}$. This has two relevant consequences:

(1) There exists a real $R>0$ such that $|f(z)|\geq R$ for all $z\in\mathbb{C}$. (This may take a bit of effort to prove.)

(2) The function $\phi:=\frac{1}{f(z)}$ is entire, i.e. holomorphic on all of $\mathbb{C}$.

Due to observation (1), we know that $|\phi(z)|\leq \frac{1}{R}$, hence the function $\phi$ is bounded. It therefore is bounded and entire, so by Liouville's theorem (link), it is constant. But, if $\phi$ is constant and nowhere zero, then $\frac{1}{\phi}=f$ is also constant. So if $f$ where a complex polynomial without zeroes, it most certainly wouldn't be a cubic polynomial. It follows that every cubic polynomial has a zero.


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