Factorization in Noetherian Domains

Definition/Theorem (Noetherian Rings).
A commutative ring $R$ is said to be Noetherian if it satisfies one of the following equivalent conditions:

  1. (Ascending Chain Condition) Every ascending chain of ideals

    \[I_1 \subseteq I_2 \subseteq I_3 \subseteq \cdots\]

    stabilizes. That is, there exists $n_0 \in \mathbb{Z}^+$ such that $I_n = I_{n_0}$ for all $n \geq n_0$.

  2. (Finitely Generated Ideals) Every ideal of $R$ is finitely generated.

Proof.
$1\implies 2$: Let $I \subset R$ be an ideal. Suppose $I$ is not finitely generated. Then, we can construct a strictly increasing chain:

\[\langle a_1 \rangle \subsetneq \langle a_1, a_2 \rangle \subsetneq \langle a_1, a_2, a_3 \rangle \subsetneq \cdots\]

where each $a_i \in I$ is chosen so that $a_i \notin \langle a_1, \dots, a_{i-1} \rangle$. Such $a_i$ exists since $I$ is not finitely generated, and thus $\langle a_1, \dots, a_{i-1} \rangle\subsetneq I$. However, this contradicts the ascending chain condition.

$2\implies 1$: Let $I_1 \subseteq I_2 \subseteq \cdots$ be an ascending chain of ideals. Then, the union

\[I = \bigcup_{n=1}^{\infty} I_n\]

is also an ideal. Since $R$ is Noetherian, $I$ is finitely generated: $I = \langle a_1, \dots, a_k \rangle$. Then all $a_i$ lie in some $I_{n_0}$, hence $I = I_{n_0}$, and the chain stabilizes at $I_{n_0}$.

Theorem (Existence of Factorizations in Noetherian Integral Domains).
Let $D$ be a Noetherian integral domain. Then every nonzero, non-unit element $a \in D$ can be written as a product of irreducible elements.

Proof.
Let

\[\mathcal{G} = \left\{ \langle a \rangle : a \in D \setminus (\{0\} \cup D^\times),\ a \text{ is not a product of irreducibles} \right\}\]

Suppose for the contrary that $\mathcal{G} \ne \emptyset$. Since $D$ is Noetherian, every chain in $\mathcal{G}$ has an upper bound, and by Zorn’s Lemma, $\mathcal{G}$ has a maximal element $\langle m \rangle$.

If $m$ is irreducible, then $m$ is trivially a product of irreducibles—contradiction.
If $m$ is reducible, write $m = ab$ with $a, b \notin D^\times$. Then $\langle a \rangle, \langle b \rangle \supsetneq \langle m \rangle$, and by maximality, both $a$ and $b$ are products of irreducibles. Hence $m = ab$ is also a product of irreducibles—contradiction.

Thus, $\mathcal{G} = \emptyset$ and the theorem holds.

Theorem (Irreducible Is Prime $\Leftrightarrow$ Uniqueness of Factorization).
Let $D$ be an integral domain in which every nonzero, non-unit element can be written as a product of irreducibles. Then:

\[D \text{ is a UFD} \iff \text{every irreducible in } D \text{ is prime}\]

Proof.
Suppose $D$ is a UFD and let $p \in D$ be irreducible. If $p \mid ab$, then write $ab = pc$ for some $c \in D$. Factor $a$, $b$, and $c$ into irreducibles:

\[a = q_1 \cdots q_m,\quad b = r_1 \cdots r_k,\quad c = s_1 \cdots s_l\]

then:

\[ab = q_1 \cdots q_m r_1 \cdots r_k = p s_1 \cdots s_l\]

By uniqueness of factorization, $p$ is associate to one of the $q_i$ or $r_j$, hence divides $a$ or $b$.

Conversely, we suppose that every irreducible is prime. To show uniqueness of factorization, we consider:

\[p_1 \cdots p_n = q_1 \cdots q_m\]

where $p_i, q_j$ are irreducibles. Since $p_1$ is prime, it divides one of the $q_j$. By possible reordering, we suppose that $p_1 \mid q_1$. Then, we have $q_1 = p_1u_1$ for some $u_1\in D$. Since $q_1$ is irreducible, and $p_1$ is not a unit, we know that $u_1$ is a unit, and thus $p_1\sim q_1$.

Then, we cancel $p_1$ on both sides, and proceed by induction on $n$.

Corollary.
Let $D$ be a Noetherian integral domain. Then $D$ is a UFD if and only if every irreducible element in $D$ is also prime.

Corollary.
Every Principal Ideal Domain (PID) is a UFD.

Remark.
The implication “irreducible ⇒ prime” is essential for uniqueness. This condition holds in any UFD and in all PIDs. However, in general integral domains, irreducibles need not be prime, and factorization may fail to be unique.