If $R$ is a commutative ring with $1$ having a maximal ideal $m$ such that the local ring $R_m$ is a field, how could one check that $R/m$ is an injective $R$-module?
If we want to use Baer Lemma, we should prove that any $R$-module homomorphism from any ideal $I$ of $R$ to $R/m$ may be extended to $R$. I just know that if $I$ is not a subset of $R$ then $I+m=R$, so $I∩m$ is maximal in $I$....
Let $I$ be an ideal of $R$, and $f\in\operatorname{Hom}(I,R/M)$. Then $f_M\in\operatorname{Hom}(I_M,R_M/MR_M)$. If we take into account that $R/M\simeq R_M/MR_M$ we obtain a homomorphism $g\in\operatorname{Hom}(I_M,R/M)$. Since $R_M$ is a field we have $I_M=(0)$ or $R_M$. Clearly, in both cases, $g$ can be extended to $R_M$, and thus $f$ can be extended to $R$.