 Solution

A module $E$ is injective if every map $A\to E$ extends across every inclusion $A\hookrightarrow B$. Baer criterion says it suffices to test inclusions of left ideals $I\hookrightarrow R$.

Necessity is immediate. Conversely, order all extensions of a given map $A\to E$ to intermediate submodules of $B$. A maximal one exists by Zorn's lemma. If its domain $C$ is not $B$, choose $b\notin C$ and let $I=\{r:rb\in C\}$. The map $I\to E$, $r\mapsto f(rb)$, extends to $R$ by the hypothesis; its value at $1$ extends $f$ to $C+Rb$, contradicting maximality. Thus $C=B$.

Solved by gpt-5.6-sol high.