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

Definition: a (not necessarily commutative) left and right Noetherian ring $R$ is said to be Auslander-Gorenstein if

(i) $R$ has finite left and right injective dimension (in which case it turns out they are equal)

(ii) For any f.g. $R$-module $M$ and submodule $N$ of $Ext^j(M, R)$, $Ext^i(N, R) = 0$ whenever $i < j$

There are two results which relate this to Krull dimension:

  1. If $R$ is commutative Noetherian and has finite injective dimension $n$ then (a) it is Auslander-Gorenstein and (b) $Kdim(R) = n$. [1]

  2. If $R$ is Auslander-Gorenstein of injective dimension $n$ then $Kdim(R) \leq n$. [2]

So the question is: does there exist a (left and right) Noetherian ring of finite injective dimension $n$ with $Kdim(R) > n$? That is, does the second result actually require $R$ to be Auslander-Gorenstein, or did it only need the injective-dimension-finite part?

Obviously, those two results mean that such an $R$ cannot be commutative or Auslander-Gorenstein. Unfortunately, the commonly quoted example of a Noetherian ring which is not Auslander-Gorenstein - $\left( \begin{array}{cc} k & V \\ 0 & k \end{array} \right)$ for $V$ a $k$-vector space of dimension at least 2 - is Artinian, so doesn't help.

[1]: Hyman Bass. On the ubiquity of Gorenstein rings, 1963.

[2]: K. Ajitabh, S. P. Smith, and J. J Zhang. Auslander-Gorenstein rings, 1998.

share|improve this question
    
It seems that your first result says that the answer is no. –  Fernando Muro Apr 12 '13 at 17:09
    
Fernando's right, you must have mis-typed something. Are you actually trying to ask about a ring $R$ which is Auslander-Gorenstein and has Krull dimension not equal to injective dimension (hence $R$ cannot be Noetherian)? –  David White Apr 12 '13 at 21:32
    
Oh, I misread this. The condition you're trying to weaken is commutativity. But then do you have to distinguish between right injective dimension and left injective dimension? –  David White Apr 13 '13 at 1:11
    
@David, there's no nice Krull dimension for noncommutative rings. –  Fernando Muro Apr 13 '13 at 12:25
2  
@David - fortunately, no, there's a result of Zaks which says that if right and left injective dimension are finite then they're equal. @Fernando - the definition of Krull dimension for non-commutative rings may not be nice, but it certainly exists and is studied! @all - sorry for being unclear about definitions and non-commutativity. I've added some more background that hopefully makes things clearer. –  Christopher Fish Apr 15 '13 at 12:14
add comment

Your Answer

 
discard

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

Browse other questions tagged or ask your own question.