Unique Factorisation Domains

Definition We say a nonzero non-unit element in a integral domain is irreducible if it cannot be written as the product of two non-units.

Definition We say a nonzero element, \( p \), in a commutative ring \( R \), is prime if \( \forall a, b \in R \):

\[ p | ab \implies p | a \text{ or } p | b \]

Theorem All prime elements in an integral domain are irreducible.

Proof By contradiction, suppose a prime element \( p\in R \) is not irreducible, ie there exist non-units \( x, y \in R \ s.t. \ p= xy \). By the definition of prime elements, we must have:

\[ p | x \text{ or } p | y \]

WLOG suppose \( x = p*r \) for some \( r \in R \). So then:

\begin{align*} &\implies pry = p &\\ &\implies p(ry - 1) = 0 &\\ &\implies ry = 1 & \text{ Since p is nonzero and R is an integral domain}\\ &\implies y \text{ is invertible } \end{align*}

A contradiction, hence \( p \) is irreducible.

The converse however is not true in general, for example in \( Z[\sqrt{-5}] \), 3 is irreducible but \( 9 = (2 + \sqrt{-5})(2 - \sqrt{-5}) \), and 3 does not divide either of these elements. The converse is true if \( R \) is a UFD (or more generally a GCD domain).

Definition We say an integral domain, \( R \), is a unique factorisation domain if:

1. Every nonzero non-unit is a product of irreducibles.
2. The decomposition from 1. is unique up to order and multiplication by units.

Equivalently, since uniqueness is hard to verify, an ID is a UFD if every nonzero element can be written as a product of a unit and prime elements of \( R \).

Intuitively, an integral domain is a UFD if a statement analagous to the fundamental theorem of arithmetic holds.

Theorem If \( R \) is a PID, then \( R \) is a UFD.

Proof