Injective modules under change of rings Let $R$ be a  ring with identity, $I$ an  ideal and $M$ a left injective module with $IM= 0$. How can I show that $M$ is an injective $\frac RI$ module? 
 A: I'm sure there's a one-liner abstract nonsense answer, but I think the point is that you can let an $R/I$ module be an $R$ module by letting $I$ annihilate the module, and the arrow that completes the universal property for $R$ is also an $R/I$-module morphism completing the same triangle.  
A: The "pullback" module structure is a way by which every $R/I$ module can be viewed as an $R$ module via the action $rx:=\overline{r}x$. Check that any $\overline{R}$ homomorphism remains an $R$ homomorphism when the modules are given the pullback structure, and conversely an $R$ homomorphism of the pulled-back modules is an $\overline{R}$ homomorphism.
Starting with $_RM$, the condition that $I\subseteq ann(M)$ is necessary to view $M$ as an $\overline{R}=R/I$ module via the action $\overline{r}x:=rx$, reversing the pullback construction.
Let $A\to B$ be a monomorphism of left $R/I$ modules, and $f:A\to M$ be an $R/I$ homomorphism. By pulling these back up into the category of $R$ modules, we can use the injective property of $M$ to complete the diagram, and then pass the completing homomorphism down to an $R/I$ homomorphism.
This seems to show that an injective $R$ module annihilated by $I$ is also an injective $R/I$ module, even without the $I^2=I$ hypothesis.
A: In general, "If M is an injective left R-module, then ann_(I)(M) is an injective left R/I-module".
The submodule annI(M) = { m in M : im = 0 for all i in I } is a left submodule of the left R-module M, and is the largest submodule of M that is an R/I-module.
