When limits preserve projectivity

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Are there examples of rings $R$ with nonzero global dimension over which inverse limits of projectives are projective?



If not, are there any $R$'s over which products of projectives are projective (equivalently, $R^Bbb N$ projective)?







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    Are there examples of rings $R$ with nonzero global dimension over which inverse limits of projectives are projective?



    If not, are there any $R$'s over which products of projectives are projective (equivalently, $R^Bbb N$ projective)?







    share|cite|improve this question






















      up vote
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      down vote

      favorite
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      up vote
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      down vote

      favorite
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      Are there examples of rings $R$ with nonzero global dimension over which inverse limits of projectives are projective?



      If not, are there any $R$'s over which products of projectives are projective (equivalently, $R^Bbb N$ projective)?







      share|cite|improve this question












      Are there examples of rings $R$ with nonzero global dimension over which inverse limits of projectives are projective?



      If not, are there any $R$'s over which products of projectives are projective (equivalently, $R^Bbb N$ projective)?









      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Aug 9 at 16:55









      xsnl

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          A theorem of Chase states that projective (right) $R$-modules are closed under arbitrary products if and only if $R$ is right perfect and left coherent.



          If (and only if) $R$ also has global dimension at most two, then projective modules are also closed under kernels, and hence under arbitrary limits.



          If by inverse limits you mean only directed limits, there may be more examples, but I don’t know.






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          • Thanks! After good sleep, I've cooked up several (obvious) artinian examples; but right perfect left coherent ring looks for me very close to artinian — am I right that in commutative case those conditions are equivalent? — so what's in the gap?
            – xsnl
            Aug 10 at 10:28











          • As for inverse limits, I'd be happy to understand conditions (maybe on both ring and ideal) under which projectives are closed under adic completion, but I have a feeling that it's bad question (in a sense that no good answer exists).
            – xsnl
            Aug 10 at 10:28










          • @xsnl Yes, for commutative rings, perfect and coherent implies artinian.
            – Jeremy Rickard
            Aug 10 at 11:33










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          1 Answer
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          active

          oldest

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          1 Answer
          1






          active

          oldest

          votes









          active

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          active

          oldest

          votes








          up vote
          1
          down vote



          accepted










          A theorem of Chase states that projective (right) $R$-modules are closed under arbitrary products if and only if $R$ is right perfect and left coherent.



          If (and only if) $R$ also has global dimension at most two, then projective modules are also closed under kernels, and hence under arbitrary limits.



          If by inverse limits you mean only directed limits, there may be more examples, but I don’t know.






          share|cite|improve this answer




















          • Thanks! After good sleep, I've cooked up several (obvious) artinian examples; but right perfect left coherent ring looks for me very close to artinian — am I right that in commutative case those conditions are equivalent? — so what's in the gap?
            – xsnl
            Aug 10 at 10:28











          • As for inverse limits, I'd be happy to understand conditions (maybe on both ring and ideal) under which projectives are closed under adic completion, but I have a feeling that it's bad question (in a sense that no good answer exists).
            – xsnl
            Aug 10 at 10:28










          • @xsnl Yes, for commutative rings, perfect and coherent implies artinian.
            – Jeremy Rickard
            Aug 10 at 11:33














          up vote
          1
          down vote



          accepted










          A theorem of Chase states that projective (right) $R$-modules are closed under arbitrary products if and only if $R$ is right perfect and left coherent.



          If (and only if) $R$ also has global dimension at most two, then projective modules are also closed under kernels, and hence under arbitrary limits.



          If by inverse limits you mean only directed limits, there may be more examples, but I don’t know.






          share|cite|improve this answer




















          • Thanks! After good sleep, I've cooked up several (obvious) artinian examples; but right perfect left coherent ring looks for me very close to artinian — am I right that in commutative case those conditions are equivalent? — so what's in the gap?
            – xsnl
            Aug 10 at 10:28











          • As for inverse limits, I'd be happy to understand conditions (maybe on both ring and ideal) under which projectives are closed under adic completion, but I have a feeling that it's bad question (in a sense that no good answer exists).
            – xsnl
            Aug 10 at 10:28










          • @xsnl Yes, for commutative rings, perfect and coherent implies artinian.
            – Jeremy Rickard
            Aug 10 at 11:33












          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          A theorem of Chase states that projective (right) $R$-modules are closed under arbitrary products if and only if $R$ is right perfect and left coherent.



          If (and only if) $R$ also has global dimension at most two, then projective modules are also closed under kernels, and hence under arbitrary limits.



          If by inverse limits you mean only directed limits, there may be more examples, but I don’t know.






          share|cite|improve this answer












          A theorem of Chase states that projective (right) $R$-modules are closed under arbitrary products if and only if $R$ is right perfect and left coherent.



          If (and only if) $R$ also has global dimension at most two, then projective modules are also closed under kernels, and hence under arbitrary limits.



          If by inverse limits you mean only directed limits, there may be more examples, but I don’t know.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Aug 10 at 7:52









          Jeremy Rickard

          15.6k11540




          15.6k11540











          • Thanks! After good sleep, I've cooked up several (obvious) artinian examples; but right perfect left coherent ring looks for me very close to artinian — am I right that in commutative case those conditions are equivalent? — so what's in the gap?
            – xsnl
            Aug 10 at 10:28











          • As for inverse limits, I'd be happy to understand conditions (maybe on both ring and ideal) under which projectives are closed under adic completion, but I have a feeling that it's bad question (in a sense that no good answer exists).
            – xsnl
            Aug 10 at 10:28










          • @xsnl Yes, for commutative rings, perfect and coherent implies artinian.
            – Jeremy Rickard
            Aug 10 at 11:33
















          • Thanks! After good sleep, I've cooked up several (obvious) artinian examples; but right perfect left coherent ring looks for me very close to artinian — am I right that in commutative case those conditions are equivalent? — so what's in the gap?
            – xsnl
            Aug 10 at 10:28











          • As for inverse limits, I'd be happy to understand conditions (maybe on both ring and ideal) under which projectives are closed under adic completion, but I have a feeling that it's bad question (in a sense that no good answer exists).
            – xsnl
            Aug 10 at 10:28










          • @xsnl Yes, for commutative rings, perfect and coherent implies artinian.
            – Jeremy Rickard
            Aug 10 at 11:33















          Thanks! After good sleep, I've cooked up several (obvious) artinian examples; but right perfect left coherent ring looks for me very close to artinian — am I right that in commutative case those conditions are equivalent? — so what's in the gap?
          – xsnl
          Aug 10 at 10:28





          Thanks! After good sleep, I've cooked up several (obvious) artinian examples; but right perfect left coherent ring looks for me very close to artinian — am I right that in commutative case those conditions are equivalent? — so what's in the gap?
          – xsnl
          Aug 10 at 10:28













          As for inverse limits, I'd be happy to understand conditions (maybe on both ring and ideal) under which projectives are closed under adic completion, but I have a feeling that it's bad question (in a sense that no good answer exists).
          – xsnl
          Aug 10 at 10:28




          As for inverse limits, I'd be happy to understand conditions (maybe on both ring and ideal) under which projectives are closed under adic completion, but I have a feeling that it's bad question (in a sense that no good answer exists).
          – xsnl
          Aug 10 at 10:28












          @xsnl Yes, for commutative rings, perfect and coherent implies artinian.
          – Jeremy Rickard
          Aug 10 at 11:33




          @xsnl Yes, for commutative rings, perfect and coherent implies artinian.
          – Jeremy Rickard
          Aug 10 at 11:33












           

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