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?

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    $\begingroup$ You are correct. Good catch. You need to search non-integral-domains for examples. $\endgroup$ – Randall Apr 10 at 13:08
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    $\begingroup$ For example, in the ring of integers modulo six, consider the ideal generated by two, and the ideal generated by three. $\endgroup$ – Gerry Myerson Apr 10 at 13:08
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    $\begingroup$ $\newcommand{\Z}{\Bbb{Z}}$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 $\{0\} \times \{0\}$." $\endgroup$ – Minus One-Twelfth Apr 10 at 13:10
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    $\begingroup$ Direct products are usually great counterexamples $\endgroup$ – Randall Apr 10 at 13:14
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    $\begingroup$ Consider $\,R/(I\cap J)\ \ $ $\endgroup$ – Bill Dubuque Apr 10 at 14:45

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 $A_Y\cap A_Z=A_{Y\cap Z}$.


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