Commutative Algebra - Factorization

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 $x \neq 0$ , $x$ is not a unit and for all $y, z \in R$ , we have $x ∣ yz \Rightarrow x ∣ y or x ∣ z$ . 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 $R = ℤ [\sqrt{- 5}]$ . 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: $ℤ$ 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

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 $R$ .
Let $R = ℤ$ 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 $ℤ$ , 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 $A / M ≅ F$ thus $M$ is maximal. Now if $n > 1$ , $M$ is not principal, because any set of elements that generates $M$ must contain at least $n$ elements.