Is there a finite dimensional algebra with left finitistic dimension different from its right finitistic dimension?

Question

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.

Answer 1

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.

Answer 2

Let $\Lambda$ be the path algebra of the quiver

with relations $(a^2, ac, ba, cbc)$. Then I claim $\operatorname{findim}(\Lambda) \geq 1$, while $\operatorname{findim}(\Lambda^{op})=0$.

The projectives are

$$P_1 = \matrix{&1&\\ 1&&2\\ &&1\\&&2},\quad P_2=\matrix{2\\1\\2}$$

we have an inejctive map $P_2 \to P_1$, so its cokernel has projective dimension 1, in particular $\operatorname{findim}(\Lambda) \geq 1$.

The injectives are

$$I_1 = \matrix{ &&1\\ 1&&2\\ &1& }, \quad I_2 = \matrix{ 1\\ 2\\ 1\\ 2 }$$

Assume for the sake of contradiction that $\operatorname{findim}(\Lambda^{op}) > 0$. Then there is a module $M$ with injective dimension 1. Let $$M \to I_M^0 \to I_M^1 \to 0$$ be a minimal injective resolution of $M$. Then because of minimality no direct summand of $I_M^0$ maps isomorphically into $I_M^1$. This means that if we compose with a projection onto one of the indecomposable summands of $I_M^1$ we would get an epimorphism from $I_M^0$ that is not split. If we can show that any epimorphism from an injective onto $I_1$ or $I_2$ is split we would have our contradiction.

Let $f:I \to I_1$ be an epimorphism, and let $x \in I_1$ be an element not killed by $a$. Since $f$ is an epimorphism there is an element $y\in I$ such that $f(y)=x$. Since $f(ay) = ax$ we have that $ay$ is non-zero. Since $ay$ is killed by the radical of $\Lambda$ it is in the socle. Mapping the generator, $ax$, of the socle of $I_1$ to $ay$ gives a splitting for $f$.

For $I_2$ we use the same argument except we use $bcb$ instead of $a$.

So any epimorphism from an injective to $I_1$ or $I_2$ is split, and thus $\operatorname{findim}(\Lambda^{op})=0$.

Answer 3

Consider the algebras $A_n$ given by the following quiver: with the relation that the composition of any two arrows is zero, i.e., $r^2 = 0$.

We can then conclude that $\text{findim} A_n^{\text{op}} = \text{Findim} A_n^{\text{op}} = 0 \leqslant n-1 = \text{findim} A_n = \text{Findim} A_n$.