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Is there any ring $R$ of infinite global dimension such that any $R$-module is a retract (i.e. direct summand) of some $\oplus_{i\in I}M_i$ where each $M_i$ has finite projective dimension?

I ask this because in the easy examples of rings of infinite global dimension I have in mind, there is always a simple $R$-module with infinite projective dimension. I wanted to know if this happens in general, or if pathological examples exist. In the situation I ask for above, the global dimension would be infinite, morally for asymptotic reasons.

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  • $\begingroup$ I seem to vaguely recall that by a result of Auslander the global dimension is the sup of the projective dimensions of all cyclic R-modules. This would imply what you want can't happen if I am recalling correctly $\endgroup$
    – Benjamin Steinberg
    Commented May 21, 2015 at 17:59
  • $\begingroup$ @BenjaminSteinberg yes, the projective dimension can be computed on cyclic modules, but I don't see why this answers negatively my question. What am I missing? $\endgroup$
    – Fernando Muro
    Commented May 21, 2015 at 18:04
  • $\begingroup$ If a cyclic R-modules is a retract of a direct sum then it is a retract of a finite direct sum because the splitting takes the generator into finitely many of the summands. $\endgroup$
    – Benjamin Steinberg
    Commented May 21, 2015 at 18:09
  • $\begingroup$ Since Ext commutes with direct sums you get that a summand in a finite direct sun of modules of finite projective dimension has finite projective dimension. $\endgroup$
    – Benjamin Steinberg
    Commented May 21, 2015 at 18:10
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    $\begingroup$ @BenjaminSteinberg thanks, I just didn't come up with the idea of using that f.g. modules are 'compact' w.r.t. direct sums. $\endgroup$
    – Fernando Muro
    Commented May 21, 2015 at 18:36

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This doesn't quite answer the question but shows that the real question is for an example where all cyclic modules have finite but unbounded projective dimension. So I will leave this here.

In https://projecteuclid.org/download/pdf_1/euclid.nmj/1118799684 Auslander shows that the global dimension of $R$ is the supremum of the projective dimensions of cyclic modules.

If a cyclic module is a retract of a direct sum of modules of finite projective dimension, then it would be a retract of a finite direct sum of such modules and hence have finite projective dimension. So what you want can't happen: there is no such $R$.

Added for clarity: the projective dimension of a module $M$ is the largest $n$ such that $Ext^n(M,-)$ is non-zero. Since Ext commutes with direct sums a direct summand of a finite direct sum of modules of finite projective dimension has finite projective dimension.

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    $\begingroup$ Sorry, your argument convinced me at a first glance, but I don't see the contradiction. There might be a sequence of cyclic modules with finite but divergent projective dimension. $\endgroup$
    – Fernando Muro
    Commented May 22, 2015 at 10:47
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    $\begingroup$ Let me think. But in any event this shows that if your hypothesis is true then all cyclic modules have finite projective dimension and your direct sum condition doesn't help. So what you really want is an example where each cyclic module has finite projective dimension which is unbounded. $\endgroup$
    – Benjamin Steinberg
    Commented May 22, 2015 at 11:32
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    $\begingroup$ I think you can find such examples by googling but from what I read it is unknown if there is a ring which is both left and right noetherian whose simple modules all have finite projective dimension but who has infinite global dimension $\endgroup$
    – Benjamin Steinberg
    Commented May 22, 2015 at 12:22
  • $\begingroup$ Thanks for your comments. I also think this is maybe a difficult question. Strange examples are... strange. $\endgroup$
    – Fernando Muro
    Commented May 22, 2015 at 12:58

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