Prove that $\text{Hom}_R (I(M)/M, I(M)) = 0$?

Question

Problem: Suppose $I(M)$ is an injective hull of $R-$module $M$. Prove that each automorphism $\varphi \colon I(M) \rightarrow I(M)$ has this property if for all $x \in M, \varphi(x)=x$ then $\varphi$ is the identity function $\Longleftrightarrow$ $\text{Hom}_R (I(M)/M, I(M)) = 0$?

Could you give me some hint to solve this problem. Thank all! In this case, we work with the injective hull $I(M)$, should we use the homomorphism theorem to solve this?

Answer 1

In one direction, suppose that all automorphisms $\varphi : I(M)\to I(M)$ that restrict to the identity on $M$ are the identity on $I(M)$.

Now suppose $f:I(M)/M\to I(M)$. Compose with the projection to get a morphism $g:I(M)\to I(M)$ that vanishes on $M$. Thus $h=g+id$ is the identity on $M$.

Now use that $I(M)$ is the injective hull of $M$ to get that $h$ is an automorphism.

Conclude.

The other direction has nothing to do with injective hulls and is always true : if $\hom(I(M)/M,I(M))=0$ and $\varphi$ is an automorphism that is the identity on $M$, what can you say about $\varphi-id$ ? Etc.