Polynomial divisors

Ivan has defined the terminology excellently, except to add that a unique factorization domain cannot have zero divisors . There cannot be nonzero elements $a,b$ whose product $ab$ equals $0$ . To throw another bit of terminology about, a commutative ring without zero divisors is called a integral domain , and UFDs are always integral domains.

The point is that, outside a UFD, you cannot determine the number of factors of a number simply by finding a factorization into irreducible elements.

For example, consider the ring $\mathbb{Z}[2i]$ consisting of Gaussian integers of the form $a + 2bi$ where $a,b \in \mathbb{Z}$ . If we consider the number $8i$ , then the identity $8i \; = \; 2 \times 2 \times 2i \;,$ is a factorization of $8i$ into irreducibles in $\mathbb{Z}[2i]$ . If $\mathbb{Z}[2i]$ were a UFD, then all factors of $8i$ would be equal to plus or minus (the only units here are $1$ and $-1$ ) products of these factors, namely $\begin{array}{rclcrclcrcl}\pm 1 &&&\qquad& \pm2 &&&\qquad& \pm2^2 &=& \pm4\\ \pm 2i &&&& \pm2\times2i &=& \pm4i && \pm 2^2\times2i &=& \pm8i \end{array}$ However, we also have $8i = (2 + 2i)^2$ , and we have found a totally different irreducible factor $2+2i$ of $8i$ . Note that neither $2$ nor $2i$ divides $2+2i$ in $\mathbb{Z}[2i]$ , in that no element $z \in \mathbb{Z}[2i]$ can be found with either $2+2i = 2z$ or $2 + 2i = 2iz$ .

$\mathbb{Z}[x]$ is the ring of polynomials with integer coefficients. For example, $x, 1, 2x^2 + 3x - 4$ are elements of $\mathbb{Z}[x]$ .

A unit of a ring is an element that has a multiplicative inverse. In the natural numbers, the only unit is 1, because 2 has "inverse" $\frac{1}{2}$ that is not a natural number. In the integers, -1 also becomes a unit. Surprisingly, going to $\mathbb{Z}[x]$ , there are no new units; it's still 1 and -1. ( $x$ is not a unit because its inverse would be $x^{-1}$ , not a polynomial.) Two elements of a ring are associated if you can multiply a unit to one element to obtain the other.

A unique factorization domain is a (commutative) ring where every nonzero element can be factorized uniquely into irreducible elements (or prime elements), except probably multiplying the factors by units, and changing their orders. A common example is the ring of positive integers or the ring of integers (due to the fundamental theorem of arithmetic). $12 = 2^2 \cdot 3$ is the unique factorization (you can also have $12 = (-1) \cdot (-2) \cdot 2 \cdot 3$ , but they are just multiplying the factors by units, so it's considered the same). It turns out that $\mathbb{Z}[x]$ is also a unique factorization domain.

The importance of that is that we can factorize any element $e$ of $\mathbb{Z}[x]$ , take all those factors (with multiplicities) and throw in $-1$ (because it's a unit), and now all factors of $e$ can be generated by taking some subset of them and multiplying them all together. For example, $12 = 2^2 \cdot 3$ , so take $2, 2, 3$ , throw in $-1$ as a unit, and now all factors of 12 can be generated by taking some subset of $2, 2, 3, -1$ and multiplying them together. For example, taking $2, 2, -1$ gives the factor -4 of 12. The same also applies to $x^4 - 1$ in the problem; this time, it factorizes to $(x-1)(x+1)(x^2+1)$ , so all factors can be generated by taking some subset of $x-1, x+1, x^2+1, -1$ and multiplying them all together.