direct sum of injective hull of two modules is equal to the injective hull of direct sum of those modules

Question

why is the direct sum of injective hull of two modules equal to the injective hull of direct sum of those modules? In other words, $E(M\oplus N)=E(M)\oplus E(N)$

Answer 1

To fix the definitions: if $A \subseteq B$ are modules, then $B$ is an essential extension of $A$ if whenever $C$ is a submodule of $B$ with $C \cap A=0$, we have $C=0$. An injective hull of a module $M$ is an injective module $E(M)$ together with an inclusion $M \hookrightarrow E(M)$ making $E(M)$ an essential extension of $M$.

Since $I$ is injective iff $\mathrm{Hom}(\cdot,I)$ is exact, finite sums of injective modules are injective. Thus $E(M) \oplus E(N)$ is injective. The inclusions $M \hookrightarrow E(M)$ and $N \hookrightarrow E(N)$ induce an inclusion $M \oplus N \hookrightarrow E(M) \oplus E(N)$. It remains to show that this is an essential extension of $M \oplus N$.

Here are the key points: since $E(M)$ is an essential extension of $M$, we obtain that $E(M) \oplus N$ is an essential extension of $M \oplus N$. Since $E(N)$ is an essential extension of $N$, we obtain that $E(M) \oplus E(N)$ is an essential extension of $E(M) \oplus N$. Now use the fact that if $A \subseteq B \subseteq C$ with $B$ an essential extension of $A$ and $C$ an essential extension of $B$ then $C$ is an essential extension of $A$.