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*](http://books.google.com/books?id=r9VoYbk-8c4C&q=iff+there+exists+a+right#v=onepage&q=%2522iff%2520there%2520exists%2520a%2520right%2522%2520%2522M%2520such%2520that%2522&f=false), 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](http://www.ams.org/journals/proc/1990-109-04/S0002-9939-1990-1027096-4/S0002-9939-1990-1027096-4.pdf). 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](http://www.ams.org/journals/tran/1960-095-03/S0002-9947-1960-0157984-8/S0002-9947-1960-0157984-8.pdf) it is defined exactly as little finitistic dimension above.



> 
> Are these two terms always interchangeable?
> 
> 
> 



An example where finitistic dimension does not equal right global dimension?
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 $n$.


