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)?
abstract-algebra ring-theory projective-module
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up vote
1
down vote
favorite
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)?
abstract-algebra ring-theory projective-module
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up vote
1
down vote
favorite
up vote
1
down vote
favorite
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)?
abstract-algebra ring-theory projective-module
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)?
abstract-algebra ring-theory projective-module
asked Aug 9 at 16:55
xsnl
1,360418
1,360418
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1 Answer
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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 donâÂÂt know.
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 â am I right that in commutative case those conditions are equivalent? â so what's in the gap?
â 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).
â xsnl
Aug 10 at 10:28
@xsnl Yes, for commutative rings, perfect and coherent implies artinian.
â Jeremy Rickard
Aug 10 at 11:33
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
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 donâÂÂt know.
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 â am I right that in commutative case those conditions are equivalent? â so what's in the gap?
â 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).
â xsnl
Aug 10 at 10:28
@xsnl Yes, for commutative rings, perfect and coherent implies artinian.
â Jeremy Rickard
Aug 10 at 11:33
add a comment |Â
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 donâÂÂt know.
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 â am I right that in commutative case those conditions are equivalent? â so what's in the gap?
â 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).
â xsnl
Aug 10 at 10:28
@xsnl Yes, for commutative rings, perfect and coherent implies artinian.
â Jeremy Rickard
Aug 10 at 11:33
add a comment |Â
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 donâÂÂt know.
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 donâÂÂt know.
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 â am I right that in commutative case those conditions are equivalent? â so what's in the gap?
â 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).
â xsnl
Aug 10 at 10:28
@xsnl Yes, for commutative rings, perfect and coherent implies artinian.
â Jeremy Rickard
Aug 10 at 11:33
add a comment |Â
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 â am I right that in commutative case those conditions are equivalent? â so what's in the gap?
â 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).
â xsnl
Aug 10 at 10:28
@xsnl Yes, for commutative rings, perfect and coherent implies artinian.
â 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 â am I right that in commutative case those conditions are equivalent? â so what's in the gap?
â 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 â am I right that in commutative case those conditions are equivalent? â so what's in the gap?
â 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).
â 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).
â xsnl
Aug 10 at 10:28
@xsnl Yes, for commutative rings, perfect and coherent implies artinian.
â Jeremy Rickard
Aug 10 at 11:33
@xsnl Yes, for commutative rings, perfect and coherent implies artinian.
â Jeremy Rickard
Aug 10 at 11:33
add a comment |Â
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