# Injective module over a quotient ring

I want to make sure I am not totally confused about this.

Suppose $R$ is an integral domain, $I$ a (non-zero) ideal of $R$. Then if $M$ is a (non-zero) $R/I$-module is it necessarily not injective because it is also an $R$-module with the action induced by the projection to $R/I$, and obviously $M$ is not a divisible module over $R$ since its annihilator in $R$ is not zero.

Is this correct? If so, it seems that there are no finite dimensional vector spaces (of non-zero dimension) over a field $F$ which are injective over the module $F[x]$, since these are all torsion modules and so the action by $F[x]$ is equivalent to the action by $F[x]$ modulo the annihilator of $R$, which if also true surprises me.

thanks for any help/comments

• Yes, I agree with it now, though I would pick another term besides "torsion module". Perhaps you can rephrase by simply saying "since these all have nonzero annihilators making the action by $F[x]$ equivalent to the action of some quotient $F[x]/I$". – Karl Kronenfeld Dec 9 '13 at 4:30
• fair enough- thanks again! – xale Dec 9 '13 at 4:33

## 1 Answer

Consider a domain $R$ with a nontrivial maximal ideal $I$. All nonzero modules of $R/I$ are injective.

It looks like you are expecting that $M$ being $R/I$-injective implies $M$ is $R$-divisible. But that isn't true, considering you can take $M$ to be any nondivisible $R$ module and then look at $R/I$ for any maximal ideal to get an injective $R/I$ module.