When is a product of two ideals strictly included in their intersection?

Let $I,J$ two ideals in a ring $R$. The product of ideals $IJ$ is included in $I \cap J$. For example we have equality in $\mathbb{Z}$ if generators have no common nontrival factors, in a ring $R$ when $I+J=(1)$, or when $R/IJ$ has no nonzero nilpotent elements. My question is not about equality, instead it is about strict inclusion.

Under what conditions $IJ \subsetneq I \cap J$ ?

If the question appears a little too general, then my primary aim is to see what happens under the hypothesis that $R$ is a Dedekind domain.

Solution 1:

Hint: in a Dedekind (or Prüfer) domain $\:(I+J)\: (I\cap J)\: =\: IJ\ \ $ (gcd $*$ lcm law)

Solution 2:

There is actually a complete answer, when we assume that the ring $R$ is commutative (which you do when you state the condition $I + J = R$). I stole it from David Speyer's answer here, where he gives a short proof too. Here is the statement:

Let $I,J \leq R$ for a commutative ring $R$. Then $IJ = I \cap J$ if and only if $Tor^R_1(R/I,R/J) = 0$.

So the inclusion $IJ \subsetneq I \cap J$ is strict if and only if $Tor^R_1(R/I,R/J) \neq 0$.