TITLE: Is there a finite dimensional algebra with left finitistic dimension different from its right finitistic dimension?
QUESTION [5 upvotes]: Let $\Lambda$ be finite dimensional algebra over a field $k$. The (left) finitistic dimension of a finite dimensional algebra is defined as
$$\operatorname{findim}(\Lambda)=\sup\{\operatorname{pd}M | M \in \operatorname{mod}\Lambda,\operatorname{pd}M < \infty\}$$
where $\operatorname{mod}\Lambda$ is the category of finitely generated left $\Lambda$-modules, and $\operatorname{pd}M$ is the projective dimension of $M$. The right finitistic dimension is simply $\operatorname{findim}(\Lambda^{op})$, which by applying the duality $\operatorname{Hom}_k(-, k)$ is equal to
$$\operatorname{findim}(\Lambda^{op})=\sup\{\operatorname{id}M | M \in \operatorname{mod}\Lambda,\operatorname{id}M < \infty\}.$$
I have seen people mention that $\operatorname{findim}(\Lambda)$ doesn't necessarily equal $\operatorname{findim}(\Lambda^{op})$, but I could not find an example by searching online.
I believe I have managed to find an example myself, so I'm answering my own question here in case anyone else is looking for an example in the future. If anyone has more examples or more information about the difference between the two dimensions, I will appreciate your answers.

REPLY [7 votes]: The left and right finitistic dimension can take any pair of values. This is shown in Example 2.3 of
Green, Edward L.; Kirkman, Ellen; Kuzmanovich, James, Finitistic dimensions of finite dimensional monomial algebras, J. Algebra 136, No. 1, 37-50 (1991). ZBL0727.16003.
The simplest example where they differ is slightly simpler than the example given in Jacob FG's answer. Let $Q$ be the quiver with two vertices, a loop at the first vertex, and one arrow from the first to the second vertex. Then $\Lambda=kQ/\operatorname{rad}^2(kQ)$ has $\operatorname{findim}(\Lambda)=1$ and $\operatorname{findim}(\Lambda^{\text{op}})=0$. The examples in the cited paper are generalizations of this.