R noetherian implies that any direct limit of injective modules is injective 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 $R\to E_k\to L$. 

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. 
 A: $\def\Hom{\operatorname{Hom}}$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 Mi a directed system of injective modules.
Since each Mi 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.
A: I am not convinced that your argument is correct. It's written out a bit confusingly in spots, which makes the proof difficult to evaluate. One would usually write about taking an upper bound of the $k_j$ and not the max, for example. I am still not clear on what the $A_i$ are. You reserve $i$ for morphisms to the direct limit and then use it as an index for your directed system. That sort of thing.
Now for the bit about the $B_i$ specifically, I don't really understand the argument about surjectivity, but I suspect the following is a counterexample. Take the category of $\mathbb Z$-modules and consider $Z := \oplus_{\mathbb N} \mathbb Z$. Let $\sigma: Z \to Z$ be the left shift homomorphism (note that this erases the leftmost coordinate). We can regard this as a directed set $Z \overset{\sigma}{\to} Z \overset{\sigma}{\to} Z \overset{\sigma}{\to} \dots$, which has direct limit zero (since enough applications of $\sigma$ will erase all of the finitely many nonzero coordinates of any element), and for which every one of the directed system maps is surjective. We even have that $\mathbb Z$ is Noetherian, so the situation is quite similar, and still the modules do not 'stabilize' at zero.
Here is an approach of the 'hands on' sort you wanted to use: take your $a_1, \dots, a_n$ and pull $f(a_1), \dots, f(a_n)$ back to elements $y_1, \dots, y_n$ in some module $E_\alpha$ of your directed system. We would like to send $a_i \mapsto y_i$ and have this extend linearly to a map $\phi: I \to E_\alpha$. Unfortunately this extension is probably not well defined. You can check that if $\sum \lambda_i a_i = 0$ implies $\sum \lambda_i y_i = 0$, then $\phi$ is well defined.
Let us say that we transport an element $x$ of some $E_i$ to $E_j$ if we apply some $h^i_j$ to $x$ and regard $h^i_j(x)$ as the 'new' $x$.
Notice that if we have the insertion $\rho_\alpha: E_\alpha \to L$ with $\rho_\alpha(x) = 0$, then we can transport $x$ to some module where it will become $0$. Now the $\vec \lambda \in R^n$ such that $\sum \lambda_i a_i = 0$ form a submodule, which since $R$ is Noetherian is finitely generated, say by $\Lambda^1, \dots, \Lambda^z$ (these are indices, not exponents). Since $$\rho_\alpha(\sum \Lambda^1_i y_i) = \sum \Lambda^1_i \rho_\alpha(y_i) = \sum \Lambda^1_i f(a_i) = f(\sum \Lambda^1_i a_i) = 0$$
it must be that we can transport the $y_i$ so that $\sum \Lambda^1_i y_i = 0$. Repeat this with $\Lambda^2, \dots, \Lambda^z$ to find a module where the desired $\phi$ is well defined. You have already demonstrated you know how to complete the rest from here.
