The Grothendieck Group. [on hold]












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Given a set A, how would you go about constructing the Grothendieck group K of M, the free commutative monoid on A?



Many thanks for any help.










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put on hold as off-topic by jgon, Dietrich Burde, KReiser, Cesareo, amWhy Jan 5 at 1:57


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – jgon, Dietrich Burde, KReiser, Cesareo, amWhy

If this question can be reworded to fit the rules in the help center, please edit the question.









  • 2




    Isn't it just the same as free abelian group on $A$?
    – Wojowu
    Jan 4 at 16:30






  • 1




    A simple google search will show to construct both the Grothendieck group K of M, and the free commutative monoid on A.
    – user458276
    Jan 4 at 17:53










  • I would "go about" reading the wikipedia article, which has further references.
    – Dietrich Burde
    Jan 4 at 19:15


















0














Given a set A, how would you go about constructing the Grothendieck group K of M, the free commutative monoid on A?



Many thanks for any help.










share|cite|improve this question













put on hold as off-topic by jgon, Dietrich Burde, KReiser, Cesareo, amWhy Jan 5 at 1:57


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – jgon, Dietrich Burde, KReiser, Cesareo, amWhy

If this question can be reworded to fit the rules in the help center, please edit the question.









  • 2




    Isn't it just the same as free abelian group on $A$?
    – Wojowu
    Jan 4 at 16:30






  • 1




    A simple google search will show to construct both the Grothendieck group K of M, and the free commutative monoid on A.
    – user458276
    Jan 4 at 17:53










  • I would "go about" reading the wikipedia article, which has further references.
    – Dietrich Burde
    Jan 4 at 19:15
















0












0








0







Given a set A, how would you go about constructing the Grothendieck group K of M, the free commutative monoid on A?



Many thanks for any help.










share|cite|improve this question













Given a set A, how would you go about constructing the Grothendieck group K of M, the free commutative monoid on A?



Many thanks for any help.







abstract-algebra monoid






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 4 at 16:27









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put on hold as off-topic by jgon, Dietrich Burde, KReiser, Cesareo, amWhy Jan 5 at 1:57


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – jgon, Dietrich Burde, KReiser, Cesareo, amWhy

If this question can be reworded to fit the rules in the help center, please edit the question.




put on hold as off-topic by jgon, Dietrich Burde, KReiser, Cesareo, amWhy Jan 5 at 1:57


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – jgon, Dietrich Burde, KReiser, Cesareo, amWhy

If this question can be reworded to fit the rules in the help center, please edit the question.








  • 2




    Isn't it just the same as free abelian group on $A$?
    – Wojowu
    Jan 4 at 16:30






  • 1




    A simple google search will show to construct both the Grothendieck group K of M, and the free commutative monoid on A.
    – user458276
    Jan 4 at 17:53










  • I would "go about" reading the wikipedia article, which has further references.
    – Dietrich Burde
    Jan 4 at 19:15
















  • 2




    Isn't it just the same as free abelian group on $A$?
    – Wojowu
    Jan 4 at 16:30






  • 1




    A simple google search will show to construct both the Grothendieck group K of M, and the free commutative monoid on A.
    – user458276
    Jan 4 at 17:53










  • I would "go about" reading the wikipedia article, which has further references.
    – Dietrich Burde
    Jan 4 at 19:15










2




2




Isn't it just the same as free abelian group on $A$?
– Wojowu
Jan 4 at 16:30




Isn't it just the same as free abelian group on $A$?
– Wojowu
Jan 4 at 16:30




1




1




A simple google search will show to construct both the Grothendieck group K of M, and the free commutative monoid on A.
– user458276
Jan 4 at 17:53




A simple google search will show to construct both the Grothendieck group K of M, and the free commutative monoid on A.
– user458276
Jan 4 at 17:53












I would "go about" reading the wikipedia article, which has further references.
– Dietrich Burde
Jan 4 at 19:15






I would "go about" reading the wikipedia article, which has further references.
– Dietrich Burde
Jan 4 at 19:15












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