Let $I$ be an injective module containing a module $M$, let $M_1$ be a submodule of $I$ maximal with respect to the property that $M_1∩M=0$, and let $M_2$ be a submodule of $I$ containing $M$ maximal with respect to the property that $M_1∩M_2=0$. I want to prove that $M_2$ is the injective hull of $M$.
I first tried $M∩(N⊕M_1)$, where $N$ is a submodule of $M_2$ with $N∩M=0$, to prove that $N=0$ and deduce that $M$ is essential in $M_2$. Is this direct sum decomposed into the direct sum of $M∩N$ and $M∩M_1$? If so, by the condition imposed on $M_1$, we would have $N⊕M_1=M_1$ leading to $N$ a submodule of $M_1$. Then....?
Could anybody help me? Thanks.
As I commented above, $M_1⊕M$ and $M_1⊕M_2$ are essential in $I$ (since $M_1$ is a complement of $M$, and $M_2$ is a complement of $M_1$). So, the essential hull of both direct sums is the essential hull of $I$ which is $I$ itself. But then we could write $E(M_1)⊕E(M)=E(M_1)⊕E(M_2)$. Now, since $M_2$ (and hence $E(M_2)$) contains $M$, we deduce that $E(M_2)=E(M)⊕S$ for some module $S$. Replacing this in the above we get $E(M_1)⊕E(M)=E(M_1)⊕E(M)⊕S$ which, by taking element from both sides and the concept of direct sum, we have $S=0$, so $E(M)=E(M_2)$. Therefore $M$ is essential in $E(M_2)$ and hence in $M_2$. Now, it is readily seen that $M_2$ is a maximal essential extension of $M$. For if $M_3$ is a submodule of $I$ strictly containing $M_2$ with $M⊂_eM_3$, then since, by hypothesis, $M∩(M_3∩M_1)=0$ we infer that $M_3∩M_1=0$, opposing to the property of $M_2$. Done.