R noetherian implies that any direct limit of injective modules is injective

Question

This is a problem in Rotman Homological Algebra 5.23.

$R$ is noetherian if and only if every direct limit of injective modules is injective.

If every direct limit of injective modules is injective, any countable direct sums of injective is injective by obvious directed systems with inclusion ordering. This implies $R$ is noetherian.

The question is that I am not sure about the other direction's proof being correct and uncertain about some argument. I want to use Baer criterion to conclude the proof. Let $I\subset R$ be an ideal and $f:I\to L$ where $R$ is noetherian and $L$ is the direct limit of some family of modules $E_i$ with $i_j:E_j\to L$ as insertion morphisms and $h_j^i:M_i\to M_j$ for $i\leq j$.

I followed spirit of Lam's Lecture on Modules and Rings and I did not fully understand some of his arguments.

Since $R$ is noetherian, $I$ is f.g. Let $I=(a_1,\dots, a_n)$. Since $L$ is a direct limit and $f:I\to L$, I can identify each element $f(a_j)$ of $L$ through $i_{k_j}(e_{k_j})$ for some $e_k\in E_k$ and hence, a f.g. submodule $A_i$ of $E_i$ which will be surjective onto $f(I)$ by taking $i=Max(k_j)$.

Starting from this point, I construct another directed system by $N_1=\cdots =N_{i-1}=\{0\},N_i=A_i,N_2=h^i_{i+1}(A_i),...$ and the morphism is simply restricted $h^i_j$. Let $L'$ be the direct limit of $N_i$. This system induces an injective map $L'\to L$.

Define $B_k$'s by $0\to B_k\to N_k\to f(I)\to 0$ exact sequence with $B_1=\cdots=B_{i-1}=\{0\}$. $B_i$ form a directed system. Now $B_i\to f(I)$ is just 0. And $B_i\to E_i$ injective maps induces another injective map between limits and this limit is 0. So take direct limit on exact sequence preserving exactness, and find $0\to LimB=0\to L'\to f(I)\to 0$. So $L'\cong f(I)$ and $L'$ is a submodule of $L$.

I hope I am correct so far.

This is the part I am very uncertain. Why $LimB=0$ implies there is some $k$ large so that $B_k=0$? Then extension map follows trivially as $L'\cong T\subset E_k$. So $I\to L'\to T\to E_k$ allows the extension map to $R\to E_k$ and thus $R\to E_k\to L$.

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I think I figured out why $LimB=0$ implies $B_k=0$ for some $k$. I will abuse notation for $h^i_j$ as the restricted map to $B_i$'s. Since any $B_k\to LimB$ is sent to 0, we have $h^i_j(b_i)=0$ for all $i\geq k$ and all $b_i\in B_i$. $h^i_j:B^i\to B_j$ is a surjective map from the definition of $A_i\to A_j$'s map. Thus $B_j=0$ for $j>i$. Hence $B_j=0$ eventually for some $j$. This solves my problem for uncertainty. It remains to check the proof being correct.

Answer 1

$\def\Hom{\operatorname{Hom}}$Here is one way to prove that a direct limit of injective modules over a Noetherian ring is injective:

Let R be my ring, I a left ideal of R, and M_i a directed system of injective modules.

Since each M_i is injective, there are surjections $$\Hom(R,M_i)\twoheadrightarrow \Hom(I,M_i).$$

Since taking direct limits is exact, there is a surjection $$\varinjlim\Hom(R,M_i)\twoheadrightarrow \varinjlim\Hom(I,M_i).$$

Since R is Noetherian, I is finitely presented, so $\Hom(I,-)$ commutes with direct limits. Therefore we have a surjection $$\Hom(R,\varinjlim M_i)\twoheadrightarrow \Hom(I,\varinjlim M_i).$$

By Baer's criterion, this suffices to prove that $\varinjlim M_i$ is injective, as required.