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On rings whose $K_0$ has nice properties

2

$begingroup$

Let $R$ be a commutative, reduced ring. It can be seen that $K_0(R) cong mathbb Z$ as groups if and only if every finitely generated projective module is stably free. My question is, are there similar interpretations when one in the following list happens or at least some broad class of rings for which one of the following happens:

(1) $K_0(R)$ is a finitely generated, free abelian group.

(2) $K_0(R)$ is finitely generated.

(3) $K_0(R)$ is free of infinite/uncountable rank.

algebraic-geometry reference-request commutative-algebra projective-module algebraic-k-theory

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edited Apr 8 at 0:43

user

asked Apr 4 at 22:14

useruser

1,533422

$endgroup$

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  1
  

  
  
  
  
  
  

  
  $begingroup$
  3 and 4 can not happen, since there is always a surjective map $K_0(R)tomathbbZ$. 1 and 2 are very strong restrictions. For `general' rings it is not true. 5 also happens rarely, since most rings coming from geometry, $K_0(R)$ has a non-trivial torsion free divisible group in it.
  $endgroup$
  – Mohan
  Apr 5 at 2:05
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Mohan: I just realized that there is always an injective map $mathbb Z to K_0(R) $ whenever $R$ has invariant basis number property , in particular in my case where $R$ is commutative ... which also takes care of 3 and 4 as I listed previously ...
  $endgroup$
  – user
  Apr 5 at 2:34
  

  
  
  
  

add a comment | 

2

$begingroup$

Let $R$ be a commutative, reduced ring. It can be seen that $K_0(R) cong mathbb Z$ as groups if and only if every finitely generated projective module is stably free. My question is, are there similar interpretations when one in the following list happens or at least some broad class of rings for which one of the following happens:

(1) $K_0(R)$ is a finitely generated, free abelian group.

(2) $K_0(R)$ is finitely generated.

(3) $K_0(R)$ is free of infinite/uncountable rank.

algebraic-geometry reference-request commutative-algebra projective-module algebraic-k-theory

share|cite|improve this question

edited Apr 8 at 0:43

user

asked Apr 4 at 22:14

useruser

1,533422

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  3 and 4 can not happen, since there is always a surjective map $K_0(R)tomathbbZ$. 1 and 2 are very strong restrictions. For `general' rings it is not true. 5 also happens rarely, since most rings coming from geometry, $K_0(R)$ has a non-trivial torsion free divisible group in it.
  $endgroup$
  – Mohan
  Apr 5 at 2:05
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Mohan: I just realized that there is always an injective map $mathbb Z to K_0(R) $ whenever $R$ has invariant basis number property , in particular in my case where $R$ is commutative ... which also takes care of 3 and 4 as I listed previously ...
  $endgroup$
  – user
  Apr 5 at 2:34
  

  
  
  
  

add a comment | 

$begingroup$

Let $R$ be a commutative, reduced ring. It can be seen that $K_0(R) cong mathbb Z$ as groups if and only if every finitely generated projective module is stably free. My question is, are there similar interpretations when one in the following list happens or at least some broad class of rings for which one of the following happens:

(1) $K_0(R)$ is a finitely generated, free abelian group.

(2) $K_0(R)$ is finitely generated.

(3) $K_0(R)$ is free of infinite/uncountable rank.

algebraic-geometry reference-request commutative-algebra projective-module algebraic-k-theory

share|cite|improve this question

edited Apr 8 at 0:43

user

asked Apr 4 at 22:14

useruser

1,533422

$endgroup$

share|cite|improve this question

edited Apr 8 at 0:43

user

asked Apr 4 at 22:14

useruser

1,533422

share|cite|improve this question

edited Apr 8 at 0:43

user

asked Apr 4 at 22:14

useruser

1,533422

asked Apr 4 at 22:14

useruser

1,533422

asked Apr 4 at 22:14

useruser

1,533422