Given two integers $m,n$ such that $n < m$, it is easy to construct a ring with global dimension $m$ or weak dimension $n$. But I wonder whether there exists a ring satisfying both the conditions?


1 Answer 1


If $R$ is Noetherian then they are equal.

For $n=0$ one can use the fact that any Boolean ring has weak dimension $0$ (any module is flat), but a free Boolean ring on $\aleph_n$ generators have global dimension $n+1$, see the last paragraph of this paper.

For any given pair of $(m,n)$ one can perhaps use polynomial rings over the examples for $n=0$ case (The global dimensions do go up properly, but the behavior of weak dimensions seem to be trickier, may be someone who is a real expert can confirm this?)

  • $\begingroup$ Thank you, Hailong Dao. The example you provides works, since w.dimR[X] = w.dimR+1 for any ring R, where w.dimR denotes the weak dimension of R. This result can be found in Page 23 of Sarah Glaz's book-commutative coherent ring. $\endgroup$
    – TmobiusX
    Aug 15, 2010 at 5:54

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