Noetherian Domain is Factorization Domain

Theorem
Let $R$ be a noetherian integral domain.

Then $R$ is atomic.

Proof
Let $\mathcal F$ be the set of ideals of $R$ of the form $xR$, with $x$ not a unit and such that $x$ cannot be decomposed in the form:
 * $x = u p_1 \cdots p_r$

with $u$ a unit and $p_1,\ldots,p_r$ irreducible.

We show by contradiction that $\mathcal F = \emptyset$.

Suppose $\mathcal F \neq \emptyset$.

Since $R$ is noetherian, we can choose a maximal element $a R \in \mathcal F$.

By construction, $a$ is not irreducible, so we can write $a = bc$ with $b,c$ non-units and not associates.

By the definition of associates in a commutative ring this means that $bR \subsetneq aR$ and $aR \subsetneq bR$.

Since $aR$ is assumed maximal, this means that $bR$ and $cR$ do not belong to $\mathcal F$.

Therefore there exist units $u,v$ and irreducible elements $p_1,\ldots,p_r,q_1,\ldots,q_s$ such that:
 * $b = u p_1 \cdots p_r,\quad \text{ and } c = v q_1 \cdots q_s$

But this implies that:
 * $a = bc = \left({ uv }\right) p_1 \cdots p_r \cdot q_1 \cdots q_s$

a contradiction, since we assumed that $a$ could not be written in this form.