# When the injective hull is indecomposable

Let $R$ be a ring and $M$ an $R$-module. Denote by $E(M)$ the injective hull of $M$. I was trying to prove that the following conditions are equivalent:

1) $(0)$ is meet-irreducible in $M$;

2) $E(M)$ is directly indecomposable.

I was able to prove that 1 implies 2 but I'm having some difficulties in proving 2 implies 1, any suggestions?

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What does "0 inrreducible in M" mean? – rschwieb Oct 3 '12 at 0:46
@rschwieb: “$M$ has uniform dimension 1”, $M$ contains no direct sum of two submodules. – Jack Schmidt Oct 3 '12 at 0:53
@rschwieb: that was a typo, it's irreducible and it means if $(0)=N_1\cap N_2$ with $N_1,N_2\subset M$ then $N_1=0$ or $N_2=0$. – Chris Oct 3 '12 at 1:03
See page 84 of Lam's Lectures on Modules and Rings for a proof – Jack Schmidt Oct 3 '12 at 1:04
– Jack Schmidt Oct 3 '12 at 1:11
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## 1 Answer

It's elementary if you know the basic properties: "$E(M)$ is a maximal essential extension of $M$", and "injective submodules are direct summands".

Suppose $E(M)$ is indecomposable. Let $A$ and $B$ are nonzero submodules of $M$ such that $A\cap B=0$. Then $E(A)$ is an injective submodule of $E(M)$, and hence it is a direct factor. The only possibilities are $E(M)$ and $0$. Since $A$ is nonzero, it is not the latter, so $E(A)=E(M)$. But this means that $A$ is essential in $E(M)$, and so $A\cap B\neq 0$, a contradiction.

Now suppose $0$ is irreducible. Let $C\oplus D=E(M)$ with $C\neq 0$. Now $(C\cap M)\cap (D\cap M)=0$, and since $M$ is essential in $E(M)$, $M\cap C\neq 0$. By irreducibility of $0$, $M\cap D=0$, but again because $M$ is essential, this amounts to $D=0$. Thus, $E(M)$ is indecomposable.

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As you can see, commutativity does not come into play. – rschwieb Oct 3 '12 at 1:19
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