Consider the following question where the ring is assumed to be commutative.
For ideals $I, J$ of a ring $R$, their product $I J$ is defined as the ideal of $R$ generated by the elements of the form $x y$ where $x \in I$ and $y \in J$.
- Prove that, if a prime ideal $P$ of $R$ contains $I J$, then $P$ contains either $I$ or $J$.
- Give an example of $R, I$ and $J$ such that the two ideals $I,J$ and $I \cap J$ are different from each other.
The first part is easy enough. For the second part I came up with the example $I=J=3\mathbb{Z}$ as then $IJ = 9\mathbb{Z}$ and $I \cap J = 3\mathbb{Z}$. However, I am worried that the question may be asking for distinct $I$ and $J$. The only prime ideals on top of my head are ones of the form $n\mathbb{Z}$ in the context of the integers and I am convinced that finding an example in the integers with $I \neq J$ is not possible. I am unsure were to look for one with $I \neq J$ . Could someone guide me in the right direction?
An easy thing to do is to find a ring with a nonzero ideal $I$ such that $I^2=\{0\}$, containing a nonzero ideal $J$.
You could use, for example, $I=4\mathbb Z/16\mathbb Z$ and $J=8\mathbb Z/16\mathbb Z$ in $R=\mathbb Z/16\mathbb Z$.
There you have it: $\{0\}=IJ\neq J\cap I=J$