Proposition 1   If $D$ is a gcd domain, and $x$ is an irreducible element, then $I=(x)$ is an irreducible ideal.

Proof. If $x$ is a unit, then $I=D$ and we are done. So we assume that $x$ is not a unit for the remainder of the proof.

Let $I = J\cap K$ and suppose $a\in J-I$ and $b\in K-I$ Then $ab=x^n$ for some $n\in \mathbb{N}$ Let $c$ be a gcd of $a$ and $x$ So $$cd=x$$ for some $d\in D$ Since $x$ is irreducible, either $c$ is a unit or $d$ is. The proof now breaks down into two cases:

• $c$ is a unit. Let $t$ be a lcm of $a$ and $x$ Then $tc$ is an associate of $ax$ But $c$ is a unit, $t$ and $ax$ are associates, so that $ax$ is a lcm of $a$ and $x$ As $ab=x^n$ both $a\mid ab$ and $x\mid ab$ hold, which imply that $ax\mid ab$ Write $axr=ab$ where $r\in D$ Then $b=xr\in I$ which is impossible by assumption.
• $d$ is a unit. So $c$ is an associate of $x$ Because $c$ divides $a$ we get that $x\mid a$ as well, or $a\in I$ which is again impossible by assumption.

Therefore, the assumption that $J-I\ne \varnothing$ and $K-I\ne \varnothing$ is false, which is the same as saying $J\subseteq I$ or $K\subseteq I$ But $I\subseteq J$ and $I\subseteq K$ either $I=J$ or $I=K$ or $I$ is irreducible.