Unique Factorization Domains

Unique factorization domains generalize the fundamental theorem of arithmetic to arbitrary integral domains, identifying the precise algebraic structure needed for unique factorization into irreducibles.

Irreducibles and Primes

Definition6.1Irreducible and prime elements

In an integral domain $R$ :

A nonzero non-unit element $p$ is irreducible if $p = ab$ implies $a$ or $b$ is a unit.
A nonzero non-unit element $p$ is prime if $p \mid ab$ implies $p \mid a$ or $p \mid b$ (equivalently, $(p)$ is a prime ideal).
Two elements $a, b$ are associates if $a = ub$ for some unit $u$ .

Every prime element is irreducible. The converse holds in UFDs but fails in general.

Definition6.2Unique factorization domain

An integral domain $R$ is a unique factorization domain (UFD) if:

Every nonzero non-unit element can be written as a product of irreducibles.
This factorization is unique up to order and associates: if $p_1 \cdots p_m = q_1 \cdots q_n$ with all $p_i, q_j$ irreducible, then $m = n$ and (after reordering) $p_i$ and $q_i$ are associates for each $i$ .

Examples and Non-Examples

ExampleUFDs and failures of unique factorization

$\mathbb{Z}$ is a UFD (fundamental theorem of arithmetic).
$k[x]$ for $k$ a field is a UFD (in fact a PID).
$k[x_1, \ldots, x_n]$ is a UFD (Gauss's lemma extends factorization from $k[x]$ ).
$\mathbb{Z}[\sqrt{-5}]$ is not a UFD: $6 = 2 \cdot 3 = (1+\sqrt{-5})(1-\sqrt{-5})$ , and $2, 3, 1 \pm \sqrt{-5}$ are all irreducible but not associates.
$\mathbb{Z}[i]$ is a UFD (it is a Euclidean domain).

RemarkIrreducible vs. prime in non-UFDs

In $\mathbb{Z}[\sqrt{-5}]$ : the element $2$ is irreducible (check norms: $N(2) = 4$ , and there is no element with $N = 2$ ) but not prime ( $2 \mid (1+\sqrt{-5})(1-\sqrt{-5}) = 6$ but $2 \nmid (1 \pm \sqrt{-5})$ ). This failure of primality of irreducibles is the hallmark of non-UFD behavior.

Characterizations

Theorem6.1Characterization of UFDs

An integral domain $R$ is a UFD if and only if:

Every nonzero non-unit is a product of irreducibles (the ascending chain condition on principal ideals, or ACCP, suffices for this), and
Every irreducible element is prime.

Theorem6.2Gauss's lemma

Let $R$ be a UFD with field of fractions $K$ . If $f \in R[x]$ is a non-constant polynomial that is irreducible in $R[x]$ , then $f$ is irreducible in $K[x]$ . Consequently, $R[x]$ is a UFD.