Product of two ideals doesn't equal the intersection

But in your example $IJ = I = I \cap J$. How about taking $I = J = 2 \mathbb{Z}$. Then $I \cap J = 2\mathbb{Z}$ while $IJ = 4\mathbb{Z}$. In general $IJ \subset I \cap J$, and the equality holds if $I+J = R$ where $R$ is the ring you are working with.

In a PID, we have $(a) \cap (b) = (\mathrm{lcm}(m,n))$, whereas we have $(a) \cdot (b) = (a \cdot b)$. So this ideal-theoretic question becomes a number-theoretic one and we see: $(a) \cap (b) = (a \cdot b)$ iff $a \cdot b$ is a lcm of $a$ and $b$ iff $a,b$ are coprime.

Pick any ring with a nonzero ideal $I$ such that $I^2=0$. Then $I=I\cap I\neq I*I=\{0\}$.

A concrete example of this is $I=2\mathbb{Z}/4\mathbb{Z}$ in the ring $R=\mathbb{Z}/4\mathbb{Z}$.

Maybe it is helpful for you to realise what really happens for ideals in the integers. You probably know that any ideal in $\mathbb Z$ is of the form $(a)$ for $a\in \mathbb Z$, i.e. is generated by one element. The elements in $(a)$ are all integers which are divisible by $a$.

If we are given two ideals $(a)$ and $(b)$, their intersection consists of those numbers which are divisible by $a$ and divisible by $b$. Their product consists of all numbers which are divisible by the product $ab$.

If $a$ and $b$ are coprime they are the same. E.g. all numbers which are divisible by $2$ and $3$ are also divisible by $6$, and vice versa. If they are not coprime the situation changes. If a number is divisible by $4$ and $2$, then it is not necessarily divisible by $8$.

Another way of saying that two integers $a$, $b$ are coprime is that there exist $x,y$, such that $xa+by=1$ (cf. Euclidean algorithm). In the language of ideals this translates to $(a)+(b)=\mathbb Z$ and the circle closes.