This is related to Ueno, Algebraic Geometry 2, Chpt 6, sec 1's proof of vanishing of cohomology of quasicoherent sheaves over noetherian affine schemes.
$R$ is a noetherian ring and $I$ is an injective $R$-module. Let $J\subset R$ be an ideal. Then $I_1=\{x\in I\mid\exists n,J^nx=0\}$ is injective.
$\textbf{Q:}$ Is there any particular obvious reason to expect $I_1$ to be injective? Since $I$ may very likely to be non-finitely generated, there is no way to choose $n$ s.t. $n$ uniformly s.t. $J^Nx=0$ to define $I$. The proof roughly goes through as the following. Consider ideal $A\subset R$. Given $\phi:A\to I_1$, I want to extend the map to $R$. In order to do so, f.g. of $A$ gives a power of $J$ s.t. $J^n\phi(A)=0$ identically. Hence application of Artin-Rees yields $J^{n-m}\phi(J^m\cap A)=0$. This indicates $\phi:A\to I_1$ descends to $\phi:A/(J^m\cap A)\to I_1$ for $m$ large. In particular, I have $\phi:(A+J^m)/J^m\to I_1\subset I$. Now consider $(A+J^m)/J^m\subset R/J^m$ and extend the map to $I$ first. Then realize $J^m$ is quotient out. Hence the extended map has image in $I_1$. Therefore, there is an extension which gives injectivity of $I_1$. The critical steps are obtaining quotient map of $(A+J^m)/J^m\to I$ and realization of this map identical to original $A\to I_1$ map.
A general case:
Proof: Let $B$ be an $\bar{R}$-module, $A$ be a submodule of $B$, and $f : A \rightarrow M_1$ be a homomorphism. Consider the following diagram: $$ \require{AMScd} \begin{CD} 0@>{}>> A @>{}>> B\\ \ @V{f}VV \\ 0@>{}>> M_1 @>{}>> M \end{CD}. $$
Since $M$ is an injective $R$-module, there exists $g : B \rightarrow M$ which completes the above diagram as a commutative diagram. It is easy to see that $Im(g) \subseteq M_1$. Therefore $M_1$ is an injective $\bar{R}$-module.