# Unique Factorization Domains (UFDs)

Definition: Let $(R, +, \cdot)$ be an integral domain. Then $R$ is a Unique Factorization Domain if the following properties are satisfied:1) Every element $a \in R$ that is nonzero and that is not a unit can be expressed as a product of irreducible elements in $R$.2) If $a = p_1p_2…p_m$ and $a = q_1q_2…q_n$ where $p_1, p_2, …, p_m, q_1, q_2, …, q_n$ are irreducible elements in $R$ then $m = n$ and the factors of both expressions can be arranged so that $p_i \sim q_i$ for each $i \in \{ 1, 2, …, m \}$. |

We will soon be able to show that every principal ideal domain is a unique factorization domain. We first need to prove the following lemma.

Lemma 1: Let $(R, +, \cdot)$ be a principal ideal domain and let $I_1, I_2, …$ be ideals of $R$ such that $I_1 \subseteq I_2 \subseteq … \subseteq I_n \subseteq …$. Then there exists a positive integer $m$ such that $I_n = I_m$ for all $n \geq m$. |

**Proof:**Let:

(1)

\begin{align} \quad I = \bigcup_{i=1}^{\infty} I_i \end{align}

- Then $I$ is an ideal of $R$. Since $R$ is a principal ideal domain we have that $I = aR$ for some generator $a \in R$. Since $a \in I$ there exists at least one set in the union $I$ containing $a$. Let $m$ be the smallest positive integer such that $a \in I_m$. Then:

(2)

\begin{align} \quad I = aR \subseteq I_m \end{align}

- So $I = I_m$. Hence $I_n = I_m$ for all $n \geq m$. $\blacksquare$

Theorem 2: Every principal ideal domain is a unique factorization domain. |