$R$ left-Noetherian, so all left ideals are finitely generated. I am trying to prove that the category of left $R$-modules has enough injectives.
That is, given any left $R$ module $M$, there is an injective map $M \to I$ with $I$ an injective $R$- module.
I imagine I want to apply Baers criterion but I don’t even know what object $I$ to consider in the first place!
Edit: I have shown that over a PID an $R$-module is injective iff it is divisible, using Baer’s criterion.
Why not prove the general case directly? I think it is quite easy: firstly regarding $M$ as a $\mathbb Z$-module, for each nonzero $x\in M$, we may find a nonzero $\mathbb Zx\to\mathbb Q/\mathbb Z$ which extends to $f_x\colon M\to\mathbb Q/\mathbb Z$ as $\mathbb Q/\mathbb Z$ is $\mathbb Z$-injective. Then $M$ is embedded in $I:=\prod_{x\in M\setminus\{0\}}\mathbb Q/\mathbb Z$ as a $\mathbb Z$-module. It follows that $$ M\simeq\operatorname{Hom}_R(R,M)\hookrightarrow\operatorname{Hom}_{\mathbb Z}(R,M)\hookrightarrow\operatorname{Hom}_{\mathbb Z}(R, I),$$ where $\operatorname{Hom}_{\mathbb Z}(R, I)$ is an injective left $R$-module, as $$\operatorname{Hom}_R(-,\operatorname{Hom}_{\mathbb Z}(R, I))\simeq\operatorname{Hom}_{\mathbb Z}(R\otimes_R-,I)$$ and $R\otimes_R-$ and $\operatorname{Hom}_{\mathbb Z}(-,I)$ are exact.