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?
 A: This is more of a long comment.
It is interesting to consider the connections between intersection and product of ideals; see this MO topic.
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 $A_Y\cap A_Z=A_{Y\cap Z}$.
