Baer's criterion

Remarks

The condition says that for every such $f$ there is $e \in E$ with $f(a) = ae$ for all $a \in I$; take $e = g(1)$. One direction is the definition applied to the inclusion $I \subseteq R$. For the other, let $M \subseteq N$ and $f \colon M \to E$; by Zorn's lemma there is a maximal extension $(M', f')$ of $f$ to a submodule $M' \subseteq N$. If $x \in N$ is not in $M'$, then $I = \set{a \in R : ax \in M'}$ is a left ideal, the map $a \mapsto f'(ax)$ extends to $R$ by hypothesis, and this extends $f'$ to $M' + Rx$, a contradiction; so $M' = N$.

Example: over a principal ideal domain $R$, a divisible module $E$ is injective. A nonzero ideal is $I = (r)$ with $r \ne 0$, and a map $f \colon I \to E$ is fixed by $f(r)$; choose $e$ with $re = f(r)$, then $a \mapsto ae$ extends $f$. For instance $\Q$ and $\Q/\Z$ are injective $\Z$-modules.

Going deeper

For a student

The criterion is the main tool for proving that a given module is injective. Over a principal ideal domain it shows that injective means divisible, which gives the injective $\Z$-modules $\Q$ and $\Q/\Z$. From this one shows that every module embeds in an injective one, first for abelian groups and then for modules over any ring via $\Hom_\Z(R, -)$. This existence result is what makes injective resolutions, and so right derived functors such as $\operatorname{Ext}$, available.