Prove that the following module is injective Let $R$ be a commutative Noetherian ring with identity. Prove that if $I$ is an ideal of $R$ and $E$ an injective $R$-module, then $\bigcup_{n\geq 1}(0:_{E}I^{n})$ is an injective $R$-module. Please help me.
 A: For noetherian rings one knows that $E\simeq \bigoplus_{\mathfrak p\in\operatorname{Spec}R} E_R(R/\mathfrak p)^{\mu(\mathfrak p,E)}$. Since the local cohomology commutes with the direct sums and a direct sum of injective modules is injective (for noetherian rings), one can reduce the problem to the following: show that $H_I^0(E_R(R/\mathfrak p))$ is an injective module for all $\mathfrak p\in\operatorname{Spec}R$. But this is easy, since $H_I^0(E_R(R/\mathfrak p))=0$ if $I\nsubseteq \mathfrak p$, and $H_I^0(E_R(R/\mathfrak p))=E_R(R/\mathfrak p)$ when $I\subseteq\mathfrak p$.
Edit. Since the OP wants "introduction methods" let me try again by using Baer's criterion: let $J$ be an ideal of $R$ and $f:J\to E'$ an $R$-homomorphism, where $E'=\bigcup_{n\geq 1}(0:_{E}I^{n})$. We want to extend $f$ to $R$. By taking the advantage of living in a noetherian world from $f(a)\in E'$ for all $a\in J$ we get an $n\ge 1$ such that $f(I^nJ)=0$. By Artin-Rees lemma there exists $m\ge n$ such that $I^m\cap J\subseteq I^nJ$ hence, in particular $f(I^m\cap J)=0$. This shows that $f$ factors through $J/I^m\cap J\simeq (I^m+J)/I^m$ and (since $E$ is injective) this one can be extended to a homomorphism $g:R/I^m\to E$. But the image of $g$ is contained in $E'$, so the composition of the canonical surjection $R\to R/I^m$ with $g$ is the desired extension of $f$ to $R$.
