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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.

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  • $\begingroup$ It seems that your first result says that the answer is no. $\endgroup$
    – Fernando Muro
    Commented Apr 12, 2013 at 17:09
  • $\begingroup$ 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)? $\endgroup$
    – David White
    Commented Apr 12, 2013 at 21:32
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    $\begingroup$ 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? $\endgroup$
    – David White
    Commented Apr 13, 2013 at 1:11
  • $\begingroup$ @David, there's no nice Krull dimension for noncommutative rings. $\endgroup$
    – Fernando Muro
    Commented Apr 13, 2013 at 12:25
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    $\begingroup$ @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. $\endgroup$
    – Christopher
    Commented Apr 15, 2013 at 12:14

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