An example where finitistic dimension does not equal right global dimension?

Question

The (right) big finitistic dimension of a ring is Findim$(R) =$ sup{proj.dim(M) | $M$ a right $R$-module of finite projective dimension}. The (right) little finitistic dimension findim$(R)$ is the sup over f.g. right modules of finite projective dimension.

The right global dimension of a ring is r.gl.dim$(R) =$ sup{proj.dim(M) | $M\in R$-mod}

It is clear that Findim$(R)\leq$ r.gl.dim$(R)$. According to T.Y. Lam's book Lectures on Modules and Rings, the right global dimension of $R$ is infinite iff there is some right $R$-module $M$ of infinite projective dimension. So if r.gl.dim$(R)<\infty$ then all $R$-modules have finite projective dimension and Findim$(R)=$r.gl.dim$(R)$. If r.gl.dim$(R)=\infty$ then there must be a module of infinite projective dimension but there could also be a chain of modules of finite but increasing projective dimension, i.e. big finitistic dimension can be finite or infinite. Clearly the only way to have Findim$(R)\neq$ r.gl.dim$(R)$ is to have infinite right global dim but finite big finitistic dimension.

What is an example of a ring with Findim$(R)\neq$ r.gl.dim$(R)$?

The classical example of a ring of infinite global dimension is $k[t]/(t^2)$ or really any ring with a nilpotent element. I suspect this ring has finite Findim, but I don't have a good enough sense of what $k[t]/(t^2)$-modules look like to easily pick out the ones of finite projective dimension.

If Findim$(R)=n$, can we find such an example for any $n$?

I have also seen the term "finitistic global dimension" running around. Here it is defined exactly as little finitistic dimension above.

Are these two terms always interchangeable?

Answer 1

A non-semisimple self-injective algebra like $k[t]/(t^2)$ has the property that its modules are either of infinite projective dimension or projective. So its Findim is actually zero, while its gldim is infinite.

For your second question, pick a ring $R$ with global dimension $n$ and consider the direct product ring $R\times k[t]/(t^2)$. Its global dimension is infinite, and its Findim is $n$.

Answer 2

Given $n \in \mathbb{N}$, there exists an algebra $A_n$ such that $\text{Findim}(A_n) = n - 1$ but $r.\text{gl.dim}(A_n) = \infty$.

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

Clearly, $A_n$ is a monomial algebra. According to the paper Finitistic dimensions of finite dimensional monomial algebras, it follows that $\text{Findim}(A_n) = n-1$. Moreover, it is straightforward to compute that $r.\text{gl.dim}(A_n) = \infty$.

Therefore, we have $\text{Findim}(A_n) = n-1 < \infty = r.\text{gl.dim}(A_n)$ for $n \geq 1$.

Specifically, when $n=1$, $A_1 \cong k[t]/(t^2)$, so $\text{Findim}(k[t]/(t^2)) = \text{Findim}(A_1) = 1-1 = 0$.