Decomposition of the injective hulls

I think this question is easy but I just cannot see how to solve it.

Let $R$ be a ring and $M$ an $R$-module. Suppose $0=\bigcap_{i=1}^n N_i$ is a decomposition of $0$ with irreducible submodules of $M$. Then $E(M)\cong \bigoplus_{i=1}^n E(M/N_i)$. ($E(\_)$ denotes the injective hull).

Of course each $E(M/N_i)$ is indecomposable by one of my previous question: When the injective hull is indecomposable .

But I cannot see how to prove this generalization, any hints?

In my previous question they called irreducible meet-irreducible and indecomposable directly-indecomposable, I'm not familiar with this terminology but this is what I mean:

$N$ is irreducible in $M$ if $N=N_1\cap N_2$ with $N_1,N_2\subset M$ then one of them is $N$.

$N$ is indecomposable if $N=N_1\oplus N_2$ implies one of them is 0.

POSSIBLE BEGINNING OF A PROOF:

Let's consider the following maps:

$\varphi_i:M\rightarrow M/N_i$

$j_i:M/N_i\rightarrow E(M/N_i)$

$\psi_i=j_i\varphi_i$

$\psi:M\rightarrow\bigoplus E(M/N_i)\;\;\;\;\;$ $\psi=(\psi_i)_i$

Let's prove that $\psi$ is ingective: $\mathrm{ker}\;\psi=\bigcap\mathrm{ker}\;\psi_i=\bigcap\mathrm{ker}\;\varphi_i=\bigcap N_i=0$.

So the only thing I have to prove is that $\psi$ is an essential extension, i.e. if $N\subset\bigoplus E(M/N_i)$ with $N\cap M=0$ then $N=0$. I made a couple of observations that may be useful, first since finite direct sum of injectives is injective we have a $N^\prime$ such that $\bigoplus E(M/N_i)=N\oplus N^\prime$, and also an $N^{\prime\prime}$ such that $\bigoplus E(M/N_i)=E(N)\oplus N^{\prime\prime}$. And by my previous question (the one linked above) $E(M/N_i)$ is indecomposable for every $i$. But now I don't know how to continue, any hints?

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Those two terms are more specific names for exactly what you are talking about. "irreducible" and "indecomposable" have dozens of meanings in mathematics, and it's really better to add some specific adjectives that are used in some algebra texts. "Meet" here refers to "intersection" and "directly" here refers to direct sums. – rschwieb Oct 3 '12 at 12:30

Hint: Consider the $M\to M/N_i \to E(M/N_i)$ homomorphisms, these induce an $M\to\bigoplus_i E(M/N_i)$, and try to prove its universal property.