An $R$-module I is called faithfully injective if it is injective and the functor $Hom\_R(-, I)$ has the image of a complex being exact if and only if the original complex is exact.


I wonder if it is known when, for a finite dimensional algebra, there exists a faithfully injective module which is also a projective module?



Faithfully injective projective modules
There are certainly finite dimensional algebras for which there are no projective modules that are also injective. An example is the $D\_4$ quiver algebra with two incoming and two outgoing arrows to the central vertex. $\begin{smallmatrix}\bullet\\\downarrow\\\bullet\to\bullet\to\bullet\\\downarrow\\\bullet\end{smallmatrix}$


On the other hand, if $I$ is faithfully injective module and $S$ is a simple module then $\operatorname{Hom}(S,I)$ has to be non-zero, so $I$ has the injective hull of $S$ as a summand. Thus the number of isomorphism classes of summands of $I$ is at least (and therefore exactly) the number of isomorphism classes of simple modules. If $I$ is also projective, then since the number of isomorphism classes of projective indecomposables is the same as the number of simples, every projective indecomposable is isomorphic to a direct summand of $I$, and then it follows that every projective module is injective. In particular, the regular representation of $R$ is injective, so $R$ is self-injective. On the other hand, if $R$ is self-injective then $R$ itself is a faithfully injective module that is also projective.


So the answer to your question is that the existence of a faithfully injective module that is also projective is equivalent to $R$ being self-injective.


