I've been trying to find an (easy) example to show that there exists an Artin algebra $A$ such that $\operatorname{right.fin.dim}(A)\neq\operatorname{left.fin.dim}(A)$, where $\operatorname{left.fin.dim}(A)$ (respectively right) is the supremum of the projective dimensions of all finitely generated left (right) $A$-modules that have finite projective dimension.
By "easy" I mean as elementary as possible, particularly something that doesn't use Auslander-Reiten quivers.
With that aim, I found in the notes "Finitistic dimensions of monomial algebras" by Simon Rubinstein-Salzedo the following statement (he quotes it as "Auxiliary fact 2" and makes a reference to Bass'article "Finitistic dimension and a homological generalization of semi-primary rings"):
$\operatorname{left.fin.dim}(A)=0$ if and only if $\operatorname{soc}(A_A)$ contains an isomorphic copy of every simple right $A$-module.
Try as I may, I can't find the proof of that statement in Bass's article. Could someone provide a proof of the fact? (a more specific reference would be enough).
This is Lemma 6.2 in Bass's article.
The condition he actually states is that every simple left module is a quotient of an injective module. But for Artin algebras, finitely generated injective left modules are dual to finitely generated projective right modules, so by duality it's equivalent to the condition that Rubinstein-Salzedo quotes.