# How can the intersection of nonzero ideals be zero?

In Artin, exercise 6.8 of chapter 11, in part c he mentions ideals $$I,J$$ such that $$I\cap J=0$$. But if $$I$$ and $$J$$ are nonzero, then if $$i\in I,j\in J$$, then $$ij\in I$$ and $$ij\in J$$, right? So don't all ideals share nonzero elements?

On second thought, I guess $$ij$$ could be zero for all $$i$$ and $$j$$. Is this the only exception?

• You are correct. Good catch. You need to search non-integral-domains for examples. – Randall Apr 10 at 13:08
• For example, in the ring of integers modulo six, consider the ideal generated by two, and the ideal generated by three. – Gerry Myerson Apr 10 at 13:08
• $\newcommand{\Z}{\Bbb{Z}}$Possibly of interest: en.wikipedia.org/wiki/Irreducible_ring. And an example from there: "The direct product of two nonzero rings is never directly irreducible, and hence is never meet-irreducible or subdirectly irreducible. For example, in $\Z \times \Z$ the intersection of the non-zero ideals $\{0\} \times \Z$ and $\Z \times\{0\}$ is equal to the zero ideal $\{0\} \times \{0\}$." – Minus One-Twelfth Apr 10 at 13:10
• Direct products are usually great counterexamples – Randall Apr 10 at 13:14
• Consider $\,R/(I\cap J)\ \$ – Bill Dubuque Apr 10 at 14:45

It is common to require that ideals be non-degenerate in some sense, e.g. "$$s$$-unital": for every $$x\in I$$, there is $$u\in I$$ such that $$xu=x$$. In case that both $$I$$ and $$J$$ are $$s$$-unital, we have $$I\cap J$$ the set of products $$ij$$ where $$(i,j)\in I\times J$$. This property (or approximated versions) are true in several examples, e.g. closed ideals of C*-algebras, and these are usually not domains.
Here is a nice example: if $$R$$ is any domain and $$X$$ is any set (nonempty, not a singleton), consider the ring $$A=\oplus_X R$$ of finitely supported functions from $$X$$ to $$R$$. The $$s$$-unital ideals of $$A$$ are precisely those of the form $$A_Y=\left\{(r_x)_{x\in X}:r_x=0\text{ whenever }x\not\in Y\right\}$$ for subsets $$Y\subseteq X$$ (this is a discrete version of the Gelfan-Kolmogorov Theorem). Neither $$A$$ nor any of these ideals $$A_Y$$ are domains, but they are all $$s$$-unital. In this case $$A_Y\cap A_Z=A_{Y\cap Z}$$.