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

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?

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I tagged this representation theory because a paper I'm reading says "The Finitistic Dimension Conjecture is one of the main open problems in the representation theory of algebras" so I figured some representation theorists might be familiar with finitistic dimension. – David White Jun 27 '11 at 18:39
I learned after asking this question that $k[t]/(t^2)$ modules are exactly $k^n \oplus (k[t]/(t^2))^m$ where $0\leq n,m \leq \infty$. So if there are any $k$ factors the projective dimension is $\infty$, and otherwise the projective dimension is $0$. – David White Jul 6 '11 at 20:28

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 $n$.

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Ah, this fact about quasi-Frobenius rings was exactly what I needed (it's also in Lam, but in section 15 which I haven't read yet). Thanks for another excellent answer to one of my simple algebra questions. You've been really helpful for me as I try to learn the algebra necessary to do my algebraic topology research. – David White Jun 27 '11 at 19:14