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Does the category of finite dimensional free modules over a principal ideal domain have all finite colimits?

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2

$begingroup$

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

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asked May 10 at 19:03

violetavioleta

387210

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• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  link the article?
  $endgroup$
  – Alexander Gruber♦
  2 days ago
  

  
  
  
  

add a comment | 

2

$begingroup$

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

share|cite|improve this question

asked May 10 at 19:03

violetavioleta

387210

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  link the article?
  $endgroup$
  – Alexander Gruber♦
  2 days ago
  

  
  
  
  

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

share|cite|improve this question

asked May 10 at 19:03

violetavioleta

387210

$endgroup$

share|cite|improve this question

asked May 10 at 19:03

violetavioleta

387210

share|cite|improve this question

asked May 10 at 19:03

violetavioleta

387210

asked May 10 at 19:03

violetavioleta

387210

asked May 10 at 19:03

violetavioleta

387210

1 Answer
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5

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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.

share|cite|improve this answer

edited May 10 at 19:22

answered May 10 at 19:08

Eric WofseyEric Wofsey

197k14229360

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Interesting, do you have any reference on the proof that the category of finitely generated modules has all finite colimits?
  $endgroup$
  – violeta
  May 10 at 19:21
  

  
  
  
  


• 
  
  
  

  
  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
  May 10 at 19:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh, I see! Thanks a lot :)
  $endgroup$
  – violeta
  May 10 at 19:46
  

  
  
  
  

add a comment | 

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5

$begingroup$

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.

share|cite|improve this answer

edited May 10 at 19:22

answered May 10 at 19:08

Eric WofseyEric Wofsey

197k14229360

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Interesting, do you have any reference on the proof that the category of finitely generated modules has all finite colimits?
  $endgroup$
  – violeta
  May 10 at 19:21
  

  
  
  
  


• 
  
  
  

  
  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
  May 10 at 19:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh, I see! Thanks a lot :)
  $endgroup$
  – violeta
  May 10 at 19:46
  

  
  
  
  

add a comment | 

5

$begingroup$

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.

share|cite|improve this answer

edited May 10 at 19:22

answered May 10 at 19:08

Eric WofseyEric Wofsey

197k14229360

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Interesting, do you have any reference on the proof that the category of finitely generated modules has all finite colimits?
  $endgroup$
  – violeta
  May 10 at 19:21
  

  
  
  
  


• 
  
  
  

  
  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
  May 10 at 19:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh, I see! Thanks a lot :)
  $endgroup$
  – violeta
  May 10 at 19:46
  

  
  
  
  

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.

share|cite|improve this answer

edited May 10 at 19:22

answered May 10 at 19:08

Eric WofseyEric Wofsey

197k14229360

$endgroup$

share|cite|improve this answer

edited May 10 at 19:22

answered May 10 at 19:08

Eric WofseyEric Wofsey

197k14229360

answered May 10 at 19:08

Eric WofseyEric Wofsey

197k14229360

answered May 10 at 19:08

Eric WofseyEric Wofsey

197k14229360