Self-injective ring but not semisimple? It is well-known that if $K$ is a field, then $K[x]/(f(x))$ is a self-injective ring for any polynomial $f(x)$ in $K[x]$. On the other hand, we know that a ring $R$ being semisimple is equivalent to the fact that every finitely generated $R$-module is quasi-injective. 
My question is that could one replace "finitely generated" by "cyclic"? 
The ring $R=\mathbb Q[x]/(x^2)$ is not semisimple since it is not von Neumann regular. If each cyclic $R$-module is quasi-injective, then the answer to my question is "No". But, the last sentence is not clear to me, and I thank to any person helping.
 A: So to be sure, the question in the OP appears to be

If all cyclic $R$ modules are quasi-injective, does that imply $R$ is semisimple?

And then you are working on a counterexample to disprove the statement.
The point is that the cyclic $\Bbb Q[x]/(x^2)$ modules are all quotients of $\Bbb Q[x]/(x^2)$, and there are very few (only three) of them. Let $R:=\Bbb Q[x]/(x^2)$.
$R/R$ is trivially quasi-injective.
If $I$ is the ideal generated by $(x+(x^2))$ in $R$, then $R/I$ is simple, so it's also  quasi-injective.
The only question is if $R$ is quasi-injective as a module over itself. Any module homomorphism from $\{0\}$ or $R$ trivially extends to $R$, so the only question is if any module homomorphism from $I$ to $R$ can be extended to be from $R$ to $R$.
Can you see the fairly obvious method of how to extend such maps? Ensure your choice makes a well-defined mapping.
Alternatively, if you already know $R$ is self-injective, you can immediately explain that $R$ is already quasi-injective.
This establishes that all cyclic modules of $R$ are quasi-injective. But $R$ is clearly not semisimple since commutative semisimple rings do not have nonzero nilpotent elements ($R$ has the nilpotent element $x+(x^2)$.) 

If you really wanted to build a toy, you could replace $\Bbb Q$ with the field of two elements. Then $R$ has only four elements, so it would be very easy to check all maps from $I$ to $R$!
