Let $R$ be a Noetherian semi-perfect ring. Let $M$ be a finitely generated left $R$-module. Let $n \geq 0$ be an integer. If $\text{Ext}^{n+1}_R(M,S)=0$ for every simple left $R$-module $S$, then is it true that $M$ has projective dimension at most $n$? Can the assumption semi-perfect be weakened to semi-local (https://en.m.wikipedia.org/wiki/Semi-local_ring) ?
If this is well-known, I would be happy with a reference that also discusses the proof.
This is true for semi-local rings.
Suppose $R$ is semi-local, so $R/\operatorname{rad}R$ is semisimple. Suppose that $M$ is finitely generated with $\operatorname{Ext}_R^{n+1}(M,S)=0$ for every simple $R$-module $S$.
Take a projective resolution $$0\to X\to P_n\to\cdots\to P_1\to P_0\to M\to0$$ of $M$ by finitely generated projective modules, truncated after the $P_n$ term, so $M$ is the kernel of the differential $P_n\to P_{n-1}$.
Since $X$ is finitely generated and $R$ is semilocal, $X/\operatorname{rad}X$ is a direct sum of finitely many simple modules, and so $\operatorname{Ext}_R^{n+1}(M,X/\operatorname{rad}X)=0$.
Therefore the quotient map $X\to X/\operatorname{rad}X$ factors as $X\to P_n\xrightarrow{\alpha}X/\operatorname{rad}X$ for some $\alpha$. Since $P_n$ is projective, $\alpha$ lifts to a map $\tilde{\alpha}:P_n\to X$, and the composition $X\to P_n\to X$ must be surjective, since $\operatorname{rad}X$ is a superfluous submodule of $X$. But a surjective endomorphism $f$ of a Noetherian module $X$ is invertible, as otherwise $\ker(f)\subset\ker(f^2)\subset\ker(f^3)\subset\cdots$ would be an ascending chain of submodules.
So $X\to P_n\to X$ is an isomorphism, and so the inclusion $X\to P_n$ is the inclusion of a direct summand, and so if $Q_n$ is a complement of $X$ in $P_n$ then $$0\to Q_n\to P_{n-1}\to\cdots\to P_1\to P_0\to M\to0$$ is a projective resolution of $X$, and so $X$ has projective dimension at most $n$.