Submodule of an injective $R$-module annihilated by an ideal $I$ of $R$ is an injective $R/I$-module. I want to solve this, but I have no idea: > Let $Q$ be an injective left $R$-module and $I\subset R$ is an ideal. Show that the submodule of $Q$ annihilated by $I$ is injective as a $R/I$-module. Is there any hint? Thank you a lot.

Hint: use Baer's Criterion: a left $R$-module $Q$ is injective if and only if any homomorphism $g: J → Q$ defined on a left ideal $J$ of $R$ can be extended to all of R.

More detail:

Let $S$ denote the submodule of $Q$ annihilated by $I$. We'll try to use Baer's Criterion. So let $J$ be a left ideal of $R/I$, and let $g: J \to S$ be a homomorphism.

What do we know? We know $Q$ is injective, so we know something about homomorphisms from $K$ to $Q$ where $K$ is a left ideal of $R$. How can we use this? Well, $J$ is a left ideal of $R/I$, so there's a corresponding left ideal $K$ of $R$ with $I \subset K$. And to get a homomorphism $K \to Q$, we can just define $h(k) = g(k + I) \in S \subset Q$.

Now we use Baer's criterion: $h$ extends to a homomorphism $R \to Q$. We're almost there; we just need to show that $h$ induces a homomorphism $R/I \to S$. I'll let you fill in those details.