irreducible ideal

Let $R$ be a ring. An ideal $I$ in $R$ is said to be if, whenever $I$ is an intersection of two ideals: $I=J\cap K$, then either $I=J$ or $I=K$.

Irreducible ideals are closely related to the notions of irreducible elements in a ring. In fact, the following holds:

Proposition 1.

If $D$ is a gcd domain, and $x$ is an irreducible element, then $I\mathrm{=}\mathrm{(}x\mathrm{)}$ 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 $J-I\ne \mathrm{\varnothing }$ and $K-I\ne \mathrm{\varnothing }$ 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.∎

Remark. In a commutative Noetherian ring, the notion of an irreducible ideal can be used to prove the Lasker-Noether theorem: every ideal (in a Noetherian ring) has a primary decomposition.