Show a Cubic Polynomial over $\Bbb C$ can be factored as a product of linear terms

Question

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.

Answer 1

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

Preliminaries:

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.

Answer 2

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.

Answer 3

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.