Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
    
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
1  
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
add comment

2 Answers 2

up vote 7 down vote accepted

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 $n$.

share|improve this answer
    
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
add comment

Apologies for responding to a year-old question- I'm certainly not attempting any sort of necromancy- but I didn't see the point in making a new question when this is close to what I would like to know.

Anywho, an example of a ring $R$ with Findim $\neq$ r.gl.dim is $\mathcal{O}G$, with $\mathcal{O}$ a complete discrete valuation ring and $G$ a finite group. Since $\mathcal{O}$ is hereditary, the $\mathcal{O}G$-modules induced from $\mathcal{O}$-modules have projective dimension $\leq 1$, and the non-projective $\mathcal{O}G$-lattices (free as $\mathcal{O}$-modules) have infinite projective dimension, and it is not hard to see that no other projective dimensions are possible ($\mathcal{O}G$ is nearly self-injective).

My question is (and I strongly believe the answer to be yes): Are the only $kG$-modules (where $k$ is the residue field of $\mathcal{O}$, algebraically closed of charactistic $p$ dividing $|G|$) with finite projective dimension (as $\mathcal{O}G$-modules) the projective $kG$-modules? I know for sure that the simple modules have infinite projective dimension, and it feels like this is certainly known (if not well-known).

I should also point out that an $\mathcal{O}G$ module can have projective dimension 1, but not be induced from the trivial subgroup.

Thanks for any help!

share|improve this answer
    
This is a new question (but more suitable for math.stackexchange.com). For an answer see math.stackexchange.com/questions/136110/… since kG is selfinjective. –  Julian Kuelshammer Sep 5 '12 at 11:42
    
This doesn't really help- I want to know about projective dimensions of kG modules viewed as OG modules. –  user26223 Oct 30 '12 at 12:32
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.