# Factorization

Let $$R$$ be a ring.

An element $$x\in R$$ is irreducible if $$x$$ is not a unit and for all $$y, z\in R$$, $$x = y z$$ implies $$y$$ or $$z$$ is a unit.

An element $$x \in R$$ is prime if $$x \ne 0$$, $$x$$ is not a unit and for all $$y,z\in R$$, we have $$x | yz \implies x|y \text { or } x|z$$. Note $$x$$ is prime if and only if $$x R$$ is a prime ideal.

In an integral domain, all primes are irreducbile. The converse is not always true. For example, take $$R =\mathbb{Z}[\sqrt{-5}]$$. Then by using the norm, it can be deduced that the units of $$R$$ are $$\pm 1$$. We have $$6 = 2\cdot 3 = (1+\sqrt{-5})(1-\sqrt{-5})$$, where all the factors are irreducible but not prime.

An integral domain $$R$$ is a unique factorization domain (UFD) if every nonzero nonunit of $$R$$ can be expressed as a product of irreducibles and furthermore the factorization is unique up to order and associates. In a UFD, all irreducibles are prime.

Example: $$\mathbb{Z}$$ is a UFD. All fields are trivially UFD’s.

Gauss' Theorem: If $$R$$ is a UFD then the polynomial ring $$R[x]$$ is a UFD.

Proof: Consult an abstract algebra textbook.

Example

1. Let $$R = F[x_1,...,x_n]$$ where $$F$$ is a field. By Gauss' Theorem $$R$$ is a UFD. If $$p \in R$$ is any irreducible polynomial over $$F$$ then it is also prime, and the principal ideal $$p R$$ is a prime ideal of $$R$$.

2. Let $$R = \mathbb{Z}$$ or $$R = F[x]$$ for some field $$F$$. By using the Euclidean algorithm, it can be seen that every ideal of $$R$$ is principal. (In $$\mathbb{Z}$$, the nonzero prime ideals are generated by a prime, while in $$F[x]$$, the nonzero prime ideals are generated by irreducible polynomials.) In these rings, it turns out that all nonzero prime ideals are also maximal, for reasons we shall see below.

A principal ideal domain (PID) is an integral domain in which all ideals are principal.

Proposition: Let $$R$$ be a PID. Then every nonzero prime ideal is maximal.

Proof: Let $$I = x R$$ be some nonzero prime ideal. Suppose $$I$$ is strictly contained in some ideal $$y R$$ of $$R$$. Then $$x = y z$$ for some $$z\in R$$. Since $$I$$ is prime, we must have $$y \in I$$ or $$z \in I$$. The former would imply $$y R$$ is contained in $$I$$, a contradiction. So we must have $$z \in I$$. In other words, $$z = x t$$ for some $$t \in R$$, hence $$x = x y t$$. implying that $$y t = 1$$. Thus $$y$$ is a unit so $$y R = R$$.

Example: Going back to the example $$R = F[x_1,...,x_n]$$ for some field $$F$$, set $$M = \{p\in R|p(0) = 0\}$$ (the set of polynomials with zero constant term). We have $$A/M \cong F$$ thus $$M$$ is maximal. Now if $$n \gt 1$$, $$M$$ is not principal, because any set of elements that generates $$M$$ must contain at least $$n$$ elements.

Ben Lynn blynn@cs.stanford.edu 💡