Let $B$ be the ring of $2$-adic integers, and $Q$ be its field of fractions, and consider the trivial extension $R=B\ltimes (Q/B)$, that is, the set $B\times (Q/B)$ with coordinatewise addition and multiplication given by $(a,x)(b,y)=(ab, ay+bx)$.
I found this example given without justification in one of those Handbook of Algebra volumes as a self-injective ring (with other properties) that I'm interested in.
It's clearly commutative and local, since $B$ is, and it is also non-Noetherian. What's a good way to see it is self-injective?
I would apply the Baer Criterion to $R_R$ if I felt confident I had foreseen all the ideals of $R$. Even then I'm not sure I see the trick to extend homomorphisms.
I've mainly been chasing this though: assume the injective hull $E=E(R_R)$ is strictly larger, and attempt to draw a contradiction. Suppose $x\in E\setminus R$. Then since $R$ is essential in $E$, there's an $r\in R$ such that $xr\in R\setminus\{0\}$. Obviously $r$ can't be among the units of $R$.
If $x(b,q)$ is my nonzero element of $R$, and $b$ is nonzero, then it can be factored into $x(b,0)(1, \frac qb)$, and after cancelling off the unit you have $x(b,0)\in R$.
If $b=0$ you have $x(0,q)\in R$ straightaway. So that narrows things down to two cases, but I don't see what else to do.
Can I proceed from there? Until now I suppose I haven't been using the fact that $B$ is uniserial, and that it gives $Q/B$ a special structure too. Maybe the key is there... I just haven't spotted it.
P. S. I found some references to papers by Sakano that seem like they might carry this as a special case, but I was unable to unearth an answer.
The key facts needed about $Q/B$ are:
(1) Every $B$-module endomorphism of $Q/B$ (in fact, every abelian group endomorphism) is of the form $z\mapsto az$ for some $a\in B$, so $B\cong\operatorname{Hom}_B(Q/B,Q/B)$ as $B$-modules by the map $a\mapsto[z\mapsto az]$. Also, more obviously, $Q/B\cong\operatorname{Hom}_B(B,Q/B)$ as $B$-modules by the map $x\mapsto[c\mapsto cx]$.
(2) $Q/B$ is injective as a $B$-module, so $\operatorname{Hom}_B(R,Q/B)$ is injective as an $R$-module.
Now define a map $$R=B\ltimes Q/B\to \operatorname{Hom}_B(B\oplus (Q/B),Q/B)=\operatorname{Hom}_B(R,Q/B)$$ by $$(a,x)\mapsto[(c,z)\mapsto cx+az)].$$
By (1), this is an isomorphism of $B$-modules, and a straightforward calculation shows that it is an $R$-module homomorphism. So, by (2), $R$ is injective as an $R$-module.