Endomorphisms of an ideal If $R$ is a ring (with identity, but not necessarily commutative), and $I$ is an ideal of $R$, then thinking of $I$ as an $R$-module, what is
$$ \mathrm{End}_R(I) $$
Is there some way to see what this endomorphism ring is isomorphic to in general?
I have been thinking about representation theory, and I am specifically wondering about the following case: $G$ is a finite group, $\Bbb{K}$ a field, and $\Bbb{K}[G]$ the group ring. If we think of $\Bbb{K}[G]$ as a $\Bbb{K}[G]$-module (under multiplication), then any submodule $M \leq \Bbb{K}[G]$ is just an ideal of the group ring, and I would like to know how to view
$$ \mathrm{End}_{\Bbb{K}[G]}(M) $$
 A: In general I don't think there is a comprehensive answer. It's good that you mentioned your context though, because you do have a decent answer in that case.
In your K[G] situation, the groupring is always quasi-Frobenius, hence right self-injective. Given any ideal $I$, you can view an endomorphism of $I$ as a map from $I\to R$. Injectivity of R extends this to a map $R\to R$ which is given by left multiplication by an element of R. Thus all endomorphisms of $I$ exist as restrictions of left multiplication endomorphisms of R. 
You would like to map R onto the endomorphism ring of $I$. The problem is that two such endomorphisms will restrict to the same endomorphism of $I$. If you think about it a bit, you will see why the kernel of your map should be $\ell.ann(I)$, the left annihilator of $I$. So, $End(I_R)\cong R/\ell.ann(I)$.
Without this nice structure, you'd have to know more about the module structure of I. There are many theorems on how to find the endomorphism rings of special modules. The case when I is a simple module or semisimple module would allow you to compute the endomorphism ring directly.
If $I$ is injective as a right R module, then it is a direct summand of R, hence of the form $eR$ for an idempotent e. It's well known that $End(eR)\cong eRe$ as rings.
A version of the also holds if R is projective as a right module: the endomorphism ring is $eSe$ for some idempotent e is a matrix ring S over R. Much more complicated to compute, but in principle it can be done.
