Let $R$ be a ring with $1$, which is Noetherian and Artinian as a left $R$-module. Let $M$ be a left $R$-module and let $N$ be a finitely generated $R$-module. I am asked to show that if $\mathrm{Ext}^n_R (M,R/\mathrm{rad}(R))=0$ for all $n\geq 1$, then $\mathrm{Ext}^n_R (M,N)=0$ for all $n\geq 1$.
The submodule $\mathrm{rad}(R)$ of $R$ is by definition the intersection all maximal submodules of the $R$-module $R$. I think I should find a relation between of $N$ and $R/\mathrm{rad}(R)$, and then use a long exact sequence of $\mathrm{Ext}$ or something, but I can't see how to start. Any hints?
If $R$ is left Artinian, then (it is automatically left Noetherian and) $R/\mathrm{rad}(R)$ is semisimple, as a ring and, consequently, as a left $R$-module (this is a consequence of Hopkins-Levitzki theorem: A ring $R$ is left Artinian iff it is left Noetherian, $\mathrm{rad}(R)$ is nilpotent and $R/\mathrm{rad}(R)$ is semisimple). Since $\mathrm{rad}(R)$ can be expressed as intersection of all maximal left ideals, it follows that it admits a surjection to every simple left $R$-module, $R/\mathrm{rad}(R)\twoheadrightarrow S \simeq R/M$. Since $R/\mathrm{rad}(R)$ is semisimple, this surjection has to split. Thus, every simple submodule is a direct summand of $R/\mathrm{rad}(R)$. Consequently, by $\mathrm{Ext}_R^n(M, A\oplus B)=\mathrm{Ext}_R^n(M, A)\oplus \mathrm{Ext}_R^n(M, B)$, from $\mathrm{Ext}_R^n(M, R/\mathrm{rad}(R))=0$ it follows that $\mathrm{Ext}_R^n(M, S)=0$ for every simple left $S$-module. The rest is done as in your comment.