$M$ is projective iff $\operatorname{Ext}_R ^1 (M,P)= 0 $ for $P$ projective?

Question

We know that an $R$-module $M$ is projective iff $\operatorname{Ext}_R ^1 (M,N)= 0 $ for every $R$- module $N$.

Is it true that:

$M$ is projective iff $\operatorname{Ext}_R ^1 (M,P)= 0 $ for every projective $R$- module $P$ ?

Answer 1

No. Let $R$ be a self-injective noetherian ring. Then any projective module is also injective.

In particular, for any $M$ and any projective $P$, $\mathrm{Ext}^1_R(M,P) = 0$, as $P$ is in particular injective.

However, $R$ can have non projective modules: for instance take $R = k[C_p]$ where $C_p$ is the cyclic group of order $p$ (prime) and $k$ has characteristic $p$. Then $R$ is self-injective, but $k$ (as an $R$-module, via the augmentation $k[C_p]\to k$) is not projective.

($R$ is self-injective because for any finite group $G$ and any field $k$, $k[G]$ is self-injective)