Special project: factoring over $\mathbb{C}$

In the case where $F$ is the complex numbers, let use $z=x+iy$ as our variable name.

The most important result about a polynomial $p$ over the complex numbers is that it will have exactly deg$(p)$ roots. In other words, we have the following result.

Theorem 2.10.1 (Fundamental Theorem of Algebra)   If $p(z)$ is a polynomial of degree $n$ in $\mathbb{C}[z]$ then there are $n$ complex roots $r_1,...,r_n$ (counted according to multiplicity) such that $p(z)=a(z-r_1)...(z-r_n)$, where $a$ is the leading coefficient of $p(z)$.

C. F. Gauss was the first to give a complete proof of this fact. It's simplest proof (as far as we know of) uses complex-analytic techniques which go beyond the scope of this book. A topological argument is given in [Ar], ch 13, §9.

Corollary 2.10.2   If $p(z)$ is an irreducible polynomial in $\mathbb{C}[x]$ then $p(z)=az+b$, for some $a,b\in \mathbb{C}$.

To factor a polynomial $p(z)$ over the complex numbers, one may use the following procedure.

• Eliminate any multiple factors by dividing $p(z)$ by $gcd(p(z),p'(z))$, as in Lemma 2.4.6.
• Find a root of the resulting polynomial.
• Use the division algorithm to factor $p(x)$ into a linear factor times a polynomial of lower degree.
• Repeat this process until $p(z)$ is completely factored.

Example 2.10.3   If $p(z)=z^n-1$ then all $n$ roots of $p(z)$ lie equally distributed on the unit circle in the complex plane. In fact, and complex number $z=x+iy$ may be written in its polar decomposition $z=re^{i\theta}$, where $r=\sqrt{x^2+y^2}$ and $\theta=arg(z)$ is the ``argument'' of $z$. If $z=re^{i\theta}$ is a root of $z^n=1$ then $r^ne^{n\theta}=1$, then $r=1$ (which implies that such a root is on the unit circle) and $n\theta$ is a multiple of $2\pi$. The $n$ distinct numbers $z_1=1,\ z_2=e^{2\pi i/n},\ z_3=e^{4\pi i/n},\ ..., z_n=e^{2\pi i(n-1)/n}$ satisfy $z_i^n=1$. By the fundamental theorem of algebra, they must be all the roots of $p(z)$.

In case $n=2$, this implies that the roots are $1,-1$.

In case $n=3$, this implies that the roots are $1, {-1+\sqrt{3}i\over 2}, {-1-\sqrt{3}i\over 2}$.

In case $n=4$, this implies that the roots are $1,-1,i,-i$.