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Does the category of finite dimensional free modules over a principal ideal domain have all finite colimits?
A surjective homomorphism between finite free modules of the same rankSources on a category of ordinalsMonads on Set and their strengthFree objects in the category of dg modulesWhat does the type theory of a free-category-over-M look like?What's wrong or where is the AOC required in this proof? (free modules over a PID)Refinement of the structure theorem of finitely generated modules over a principal ideal domainWhy is $mathbb Z_p$ a free $mathbb Z$-module? Equivalently, why is $mathbb Q_p$ finite dimensional over $mathbb Q$?If ideal $I$ of domain $R$ is free $R$-module, then $I$ is principal ideal.Does the comma category of a set over category of graphs have initial and terminal objects?
$begingroup$
A paper I'm reading states this fact without proof, and I can't seem to find a proof for it.
category-theory free-modules
asked 5 hours ago
violetavioleta
387210
$endgroup$
add a comment |
$begingroup$
A paper I'm reading states this fact without proof, and I can't seem to find a proof for it.
category-theory free-modules
asked 5 hours ago
violetavioleta
387210
$endgroup$
add a comment |
$begingroup$
A paper I'm reading states this fact without proof, and I can't seem to find a proof for it.
category-theory free-modules
asked 5 hours ago
violetavioleta
387210
$endgroup$
A paper I'm reading states this fact without proof, and I can't seem to find a proof for it.
category-theory free-modules
category-theory free-modules
asked 5 hours ago
violetavioleta
387210
asked 5 hours ago
violetavioleta
387210
asked 5 hours ago
violetavioleta
387210
asked 5 hours ago
violetavioleta
387210
asked 5 hours ago
violetavioleta
387210
387210
add a comment |
add a comment |
1 Answer
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Yes. First, the category of finitely generated modules has all finite colimits (it is closed under finite colimits in the category of all modules). But the category of finite rank free modules is a reflective subcategory: the reflector just mods out the torsion submodule (here we use the fact that any finitely generated torsion-free module is free). It follows that the category of finite rank free modules also has all finite colimits.
Explicitly, given a finite diagram, you compute its colimit as usual in the category of modules, and then mod out the torsion submodule to get the colimit in the category of finite rank free modules. This still has the required universal property because any map to a free module will vanish on the torsion submodule.
answered 5 hours ago
Eric WofseyEric Wofsey
196k14226357
$endgroup$
$begingroup$
Interesting, do you have any reference on the proof that the category of finitely generated modules has all finite colimits?
$endgroup$
– violeta
4 hours ago
2
$begingroup$
A finite colimit (in the category of all modules) of finitely generated modules is finitely generated; this follows immediately from the explicit construction of colimits as a quotient of the coproduct of all the objects in the diagram (the coproduct in this case being a finite direct sum).
$endgroup$
– Eric Wofsey
4 hours ago
$begingroup$
Oh, I see! Thanks a lot :)
$endgroup$
– violeta
4 hours ago
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Yes. First, the category of finitely generated modules has all finite colimits (it is closed under finite colimits in the category of all modules). But the category of finite rank free modules is a reflective subcategory: the reflector just mods out the torsion submodule (here we use the fact that any finitely generated torsion-free module is free). It follows that the category of finite rank free modules also has all finite colimits.
Explicitly, given a finite diagram, you compute its colimit as usual in the category of modules, and then mod out the torsion submodule to get the colimit in the category of finite rank free modules. This still has the required universal property because any map to a free module will vanish on the torsion submodule.
answered 5 hours ago
Eric WofseyEric Wofsey
196k14226357
$endgroup$
$begingroup$
Interesting, do you have any reference on the proof that the category of finitely generated modules has all finite colimits?
$endgroup$
– violeta
4 hours ago
2
$begingroup$
A finite colimit (in the category of all modules) of finitely generated modules is finitely generated; this follows immediately from the explicit construction of colimits as a quotient of the coproduct of all the objects in the diagram (the coproduct in this case being a finite direct sum).
$endgroup$
– Eric Wofsey
4 hours ago
$begingroup$
Oh, I see! Thanks a lot :)
$endgroup$
– violeta
4 hours ago
add a comment |
$begingroup$
Yes. First, the category of finitely generated modules has all finite colimits (it is closed under finite colimits in the category of all modules). But the category of finite rank free modules is a reflective subcategory: the reflector just mods out the torsion submodule (here we use the fact that any finitely generated torsion-free module is free). It follows that the category of finite rank free modules also has all finite colimits.
Explicitly, given a finite diagram, you compute its colimit as usual in the category of modules, and then mod out the torsion submodule to get the colimit in the category of finite rank free modules. This still has the required universal property because any map to a free module will vanish on the torsion submodule.
answered 5 hours ago
Eric WofseyEric Wofsey
196k14226357
$endgroup$
$begingroup$
Interesting, do you have any reference on the proof that the category of finitely generated modules has all finite colimits?
$endgroup$
– violeta
4 hours ago
2
$begingroup$
A finite colimit (in the category of all modules) of finitely generated modules is finitely generated; this follows immediately from the explicit construction of colimits as a quotient of the coproduct of all the objects in the diagram (the coproduct in this case being a finite direct sum).
$endgroup$
– Eric Wofsey
4 hours ago
$begingroup$
Oh, I see! Thanks a lot :)
$endgroup$
– violeta
4 hours ago
add a comment |
$begingroup$
Yes. First, the category of finitely generated modules has all finite colimits (it is closed under finite colimits in the category of all modules). But the category of finite rank free modules is a reflective subcategory: the reflector just mods out the torsion submodule (here we use the fact that any finitely generated torsion-free module is free). It follows that the category of finite rank free modules also has all finite colimits.
Explicitly, given a finite diagram, you compute its colimit as usual in the category of modules, and then mod out the torsion submodule to get the colimit in the category of finite rank free modules. This still has the required universal property because any map to a free module will vanish on the torsion submodule.
answered 5 hours ago
Eric WofseyEric Wofsey
196k14226357
$endgroup$
Yes. First, the category of finitely generated modules has all finite colimits (it is closed under finite colimits in the category of all modules). But the category of finite rank free modules is a reflective subcategory: the reflector just mods out the torsion submodule (here we use the fact that any finitely generated torsion-free module is free). It follows that the category of finite rank free modules also has all finite colimits.
Explicitly, given a finite diagram, you compute its colimit as usual in the category of modules, and then mod out the torsion submodule to get the colimit in the category of finite rank free modules. This still has the required universal property because any map to a free module will vanish on the torsion submodule.
answered 5 hours ago
Eric WofseyEric Wofsey
196k14226357
edited 4 hours ago
answered 5 hours ago
Eric WofseyEric Wofsey
196k14226357
answered 5 hours ago
Eric WofseyEric Wofsey
196k14226357
answered 5 hours ago
Eric WofseyEric Wofsey
196k14226357
196k14226357
$begingroup$
Interesting, do you have any reference on the proof that the category of finitely generated modules has all finite colimits?
$endgroup$
– violeta
4 hours ago
2
$begingroup$
A finite colimit (in the category of all modules) of finitely generated modules is finitely generated; this follows immediately from the explicit construction of colimits as a quotient of the coproduct of all the objects in the diagram (the coproduct in this case being a finite direct sum).
$endgroup$
– Eric Wofsey
4 hours ago
$begingroup$
Oh, I see! Thanks a lot :)
$endgroup$
– violeta
4 hours ago
add a comment |
$begingroup$
Interesting, do you have any reference on the proof that the category of finitely generated modules has all finite colimits?
$endgroup$
– violeta
4 hours ago
2
$begingroup$
A finite colimit (in the category of all modules) of finitely generated modules is finitely generated; this follows immediately from the explicit construction of colimits as a quotient of the coproduct of all the objects in the diagram (the coproduct in this case being a finite direct sum).
$endgroup$
– Eric Wofsey
4 hours ago
$begingroup$
Oh, I see! Thanks a lot :)
$endgroup$
– violeta
4 hours ago
$begingroup$
Interesting, do you have any reference on the proof that the category of finitely generated modules has all finite colimits?
$endgroup$
– violeta
4 hours ago
$begingroup$
Interesting, do you have any reference on the proof that the category of finitely generated modules has all finite colimits?
$endgroup$
– violeta
4 hours ago
2
2
$begingroup$
A finite colimit (in the category of all modules) of finitely generated modules is finitely generated; this follows immediately from the explicit construction of colimits as a quotient of the coproduct of all the objects in the diagram (the coproduct in this case being a finite direct sum).
$endgroup$
– Eric Wofsey
4 hours ago
$begingroup$
A finite colimit (in the category of all modules) of finitely generated modules is finitely generated; this follows immediately from the explicit construction of colimits as a quotient of the coproduct of all the objects in the diagram (the coproduct in this case being a finite direct sum).
$endgroup$
– Eric Wofsey
4 hours ago
$begingroup$
Oh, I see! Thanks a lot :)
$endgroup$
– violeta
4 hours ago
$begingroup$
Oh, I see! Thanks a lot :)
$endgroup$
– violeta
4 hours ago
add a comment |
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