Artinian ring with zero finitistic dimension Let $R$ be a left artinian ring with identity.
Suppose $R$ contains copies of all its simple right $R$-modules.
Is it true that every left $R$-module of finite projective dimension is projective (so the big left finitistic projective dimension of $R$ is zero)?
Or is the above only true for finitely generated left $R$-modules?
I know that some version of this must be true according to the literature. This must be easy to understand but I am not seeing why it is true.
Can anyone help me with this? Thanks a lot.
 A: Just a partial answer
In what follows the word "module" means, by default, "left module". This is based on what I found in "Finitistic dimension and a homological generalization of semi-primary rings", by H. Bass. According to the article, the proof given would show that an Artin algebra containing copies of all its simple right $A$-modules has big finitistic left dimension zero, but I do not understand why. I only see why this works for the small finitistic dimension.
Let $A$ be an Artin $R$-algebra, i.e., suppose $A$ is an $R$-algebra where $R$ a commutative artinian ring with identity and $A$ finitely generated as an $R$-module. 
Claim: If $A$ contains copies of all its simple right $A$-modules then $A$ has small left finitistic dimension zero (that is, all finitely generated $A$-modules of finite projective dimension are projective).
Proof:
(Step 0) In this case there is a duality between $A-mod$ and $mod-A$ that generalises the standard duality,
$$(D:A-mod \longrightarrow (mod-A)^{op}, D': (mod-A)^{op} \longrightarrow A-mod).$$
Hence, if $S$ is any simple $A$-module then $D(S)$ is a simple right $A$-module (this is a "if and only if") and, by assumption, $D(S)$ can be included in $A_A$. By the duality, we can assure that there is an epimorphism in $A-mod$
$$f: D'(A_A)\longrightarrow S,$$
and because $A$ is a projective right $A$-module, then $D'(A_A)=:I$ is an injective $A$-module.
(Step 1) Let $M$ be a left $A$-module of finite projective dimension and suppose by contradiction that $M$ is not projective. We have the short exact sequence in $A-mod$
$$
0 \longrightarrow Ker (p) \longrightarrow P \longrightarrow M \longrightarrow 0,
$$
where $p:P \longrightarrow M $ is the projective cover of $M$ - note that (left) artinian rings with identity are (left) perfect rings. We may suppose without loss of generality that $Ker(p)$ is projective. In fact, by successive application of classic inequalities relating  homological dimensions and short exact sequences, we can always find some left $A$-module of projective dimension 1.
So $Ker(p)$ is a non-zero projective module. If we suppose that $M$ is finitely generated over $A$ then its projective cover is a finitely generated module, and hence $Ker(p)$ is finitely generated. It follows that $Top(Ker(p))$ is a non-zero semisimple left $A$-module, so we can find an epimorphism in $A-mod$
$$g:Ker(p) \longrightarrow S,$$ 
where $S$ is some simple $A$-module. By step 0 and because $Ker(p)$ is projective there is $h_1:Ker(p) \longrightarrow I$ such that $f \circ h_1 =g$.
By $I$'s injectiveness there is $h_2:P \longrightarrow I$ such that $h_2 \circ ker(p)=h_1$.
It follows that $f \circ h_2 \circ \ker(p)= g$. Because $g \neq 0$ then $f \circ h_2 \neq 0$. Since $S$ is simple, then $f \circ h_2$ is epic. But then $rad(P) \subseteq Ker(f \circ h_2)$, since $P/ Rad(P)$ is the largest semisimple factor of $P$. On the other hand, $p$ is an essential epic, hence $Ker(p) \subseteq Rad(P)$. So $Ker(p) \subseteq Ker(f \circ h_2)$ and, at the same time, $f \circ h_2 \circ ker(p)= g \neq 0$, which is a contradiction.
