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?