Unique Factorization Domains (UFDs)

Table of Contents

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

(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}

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

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