I want to prove that if the category of left $R$-modules has a simple cogenerator $M$, then $R$ is a simple ring. (Here, $R$ is an arbitrary ring with $1$.)
My try is to take an ideal $I$ of $R$ and consider $IM$ which is a submodule of $M$. Since $M$ is simple, so $IM=0$ or $IM=M$. But, $M$ is a cogenerator, so it is faithful. Hence, the former case leads to $I=0$. For the latter case $IM=M$ I am stuck and appreciate anybody helping me. Also, I am aware of the well-known fact that $M$ is a cogenerator if and only if the injective hull of every simple left $R$-module is isomorphic to a direct summand of $M$. Applying this to the simple $M$, one infers that $M$ could be an injective $R$-module. Then...!