What $R$-modules $M$ and ideals $I$ satisfy $IM=M$?

Question

Let $R$ be a unital associative ring and $I\subseteq R$ be an ideal. Let $M$ be a (left) $R$-module. What unital associative rings $R$, $R$-modules $M$ and ideals $I\subseteq R$ satisfy $IM=M$? Do they have a name?

I am looking for a list of interesting examples. Of course $M=0$ works for any choice of $R$ and $I$. Similarly, the equality always holds for $R=I$. It never holds for $M=R$ (considered as a left regular $R$-module) and $I\neq R$.

A less trivial example I can think of is $M=\mathbb{Q}$, $R=\mathbb{Z}$, $I=n\mathbb{Z}$ with $n\in \mathbb{N}$. More generally, let $F$ be a field and $R\subseteq F$ be a subring. Let $0\neq I\subset R$ be an ideal of $R$. Then the $R$-module $F$ (restriction of scalars) satisfies $F/IF$.

Answer 1

What unital associative rings $R$, $R$-modules $M$ and ideals $I\subseteq R$ satisfy $IM=M$? Do they have a name?

This seems like an overly general question, but it's possible someone has focused on special subcases. I have never encountered a name for this. If one exists, perhaps it is "divisiblity" themed, since it seems akin to how divisible modules work.$^\ast$

I am looking for a list of interesting examples.

To go a step further with your initial observations, if you want $M\neq 0$ and $M$ finitely generated, then it is necessary that $I$ is not contained in the Jacobson radical (because of the conclusion fo Nakayama's lemma).

One obvious large class that would get you a lot of examples is just to find a ring $R$ with an idempotent ideal $I$ and let $I=M$. These are plentiful. For example, in a direct product of matrix rings over fields (or even division rings) any ideal will do.

Another distinct example would be $R=k[x^{1/2},x^{1/4},x^{1/8},...]/(x)$ for a field $k$, in which the Jacobson radical $(x^{1/2},x^{1/4},x^{1/8},...)=I=M$ itself is nonzero and idempotent. (This does not contradict what we said before because in this case $M$ is not finitely generated.)

Along the lines of $M=\mathbb Q$ and $I=n\mathbb Z$, you could generalize the idea to this: let $R$ be a nonfield domain with nonzero ideal $I$ and $M$ be a divisible $R$ module. Then $IM=M$. Again, it is not hard to cook up specific examples. You can take any nonzero $R$ module and let $M$ be its injective hull.

$^\ast$ Actually I did find something like this here: http://profs.sci.univr.it/~angeleri/angarc.pdf