Injective but not divisible module

Question

I want to find an injective but not divisible $R$-module.

If $R$ is integral domain, every injective is divisible so it should be $R$ is not an integral domain. Is there any example?

Answer 1

Consider the ring $R=F_2[Z]/(Z^2)=M$ where $F_2$ is the field of two elements. This is a self-injective ring, so $M$ is an injective $R$-module.

But now consider $x=1$ and $r=Z$, where I abuse notation for the images of $1$ and $Z$ in this ring. Saying that there exists $y\in R$ such that $yZ=1$ implies that $Z$ is a unit, but it is clearly not since it is a nilpotent element. So this module is not divisible in your sense.

What user says is good information: while this definition of "divisible" seems like the simplest extension, it turns out to be undesirable. T.Y. Lam's definition cited by user is the nicest one I know.

Notice that in the example above, $ann(Z)=(Z)\not\subseteq ann(1)=\{0\}$, so this is no longer a counterexample for Lam's definition. Indeed, all injective modules are divisible by that definition.

For domains this is not an issue since $ann(r)=\{0\}$ for every nonzero $r$, and therefore the condition that $ann(r)\subseteq ann(x)$ is met automatically. That's how this defintion reduces to the domain definition.