Conditions for the equality $IJ = I$ for proper ideals of a domain

Question

$I,J$ are proper ideals in $R$, an integral domain with unity, and $I$ or $J$ is finitely generated. Can we say $IJ\subsetneq I?$ I would also like to know if we could more generally say something like this for modules over $R$.

The only ideals I could find where the equality $IJ=I$ was achieved were in non-domanins or when both $I$ and $J$ are infinitely generated, specifically in the rings $\mathbb{Z}/6\mathbb{Z}$ and $\mathbb{Z} [\{x^{1/2^k}|k\in \mathbb{N} \cup {0}\}]$ respectively.

I can see if $I$ were principal, then my question would be true, as if $I=(a)$ and $IJ=I$ then $a\in aJ$, but we can cancel $a$ as $R$ is a domain, which would give us $1\in J$, a contradiction. However I am unsure of how I can extend this to even when $I$ is finitely generated.

Answer 1

In the case that $I\neq 0$ is finitely generated one always has $IJ\neq I$: the equation $IJ=I$ by Nakayamas lemma implies the existence of some $x\not\in J$ such that $xI=0$, which is impossible in a domain.

Answer 2

Consider the subring $\mathbb{Z} + X\mathbb{Q}[X]$ of the polynomial ring $\mathbb{Q}[X]$. This is the ring of rational polynomials where the constant term has to be an integer. Let $I$ be the ideal of all elements with constant term $0$ and $J=(2)$. Both are proper ideals, but $IJ = I$ because everything in $I$ can be halved and still be in $I$. This shows $J$ need not be infinitely generated at least.