$K_0(R)$ is generated by $[R]$?
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Let $R$ be a unital associative ring. Let $F$ be the free abelian group on the set of all isomorphism classes $[P]$ of f.g. projective $R$-modules $P$. Let $K_0(R)$ be the quotient of $F$ modulo the subgroup $S$ spanned by $[P]+[Q]-[Poplus Q]$ for all projective $P$ and $Q$.
a) Suppose every f.g. projective of $R$ is free.
b) Suppose $R$ has the invariant basis property (i.e. the rank of a free module is unique).
If $R$ has property $a),b)$, then it is obvious that $K_0(R)$ is generated by $[R]$ as $mathbb{Z}$-algebra. (Here I assumed $K_0(R)$ is already endowed with ring structure via tensor product.)
We say that an $R$-module $M$ is stably free if $Moplus F=G$ for some f.g. free modules $F,G$ (thus, $[M]$ is the difference of 2 free modules).
Question: The book claims that if $a)$ is replaced by every f.g. projective mod being stably free and $b)$ holds, then $K_0(R)$ is generated by $[R]$. How do I deduce this when $M$ is stably free? In other words, I need to deduce $M$ free.
Ref. Milnor's Algebraic K-theory pg 5.(Question 1 and 2)
abstract-algebra commutative-algebra k-theory
edited Nov 25 at 1:33
darij grinberg
10.1k32961
asked Nov 24 at 20:35
user45765
2,4422720
add a comment |
up vote
0
down vote
favorite
Let $R$ be a unital associative ring. Let $F$ be the free abelian group on the set of all isomorphism classes $[P]$ of f.g. projective $R$-modules $P$. Let $K_0(R)$ be the quotient of $F$ modulo the subgroup $S$ spanned by $[P]+[Q]-[Poplus Q]$ for all projective $P$ and $Q$.
a) Suppose every f.g. projective of $R$ is free.
b) Suppose $R$ has the invariant basis property (i.e. the rank of a free module is unique).
If $R$ has property $a),b)$, then it is obvious that $K_0(R)$ is generated by $[R]$ as $mathbb{Z}$-algebra. (Here I assumed $K_0(R)$ is already endowed with ring structure via tensor product.)
We say that an $R$-module $M$ is stably free if $Moplus F=G$ for some f.g. free modules $F,G$ (thus, $[M]$ is the difference of 2 free modules).
Question: The book claims that if $a)$ is replaced by every f.g. projective mod being stably free and $b)$ holds, then $K_0(R)$ is generated by $[R]$. How do I deduce this when $M$ is stably free? In other words, I need to deduce $M$ free.
Ref. Milnor's Algebraic K-theory pg 5.(Question 1 and 2)
abstract-algebra commutative-algebra k-theory
edited Nov 25 at 1:33
darij grinberg
10.1k32961
asked Nov 24 at 20:35
user45765
2,4422720
What have you tried?
– Pedro Tamaroff♦
Nov 24 at 20:36
@PedroTamaroff I think I am stuck at the first step as I do not see how to deduce a non-trivial free module morphism to $M$. I would expect $G/F$ to be free module. However, I am very uncertain about this step. Then from here, I would expect localization kicks in to deduce the free module level isomorphism on every point and here I need use invariant basis property. Hence it induces isomorphism which concludes $M$ being free.
– user45765
Nov 24 at 20:38
@PedroTamaroff I think I might be heading towards wrong way of thinking as I am wondering whether every projective being free here. This is a fairly stringent condition on the ring property. Say $R$ being PID fits.
– user45765
Nov 24 at 20:40
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $R$ be a unital associative ring. Let $F$ be the free abelian group on the set of all isomorphism classes $[P]$ of f.g. projective $R$-modules $P$. Let $K_0(R)$ be the quotient of $F$ modulo the subgroup $S$ spanned by $[P]+[Q]-[Poplus Q]$ for all projective $P$ and $Q$.
a) Suppose every f.g. projective of $R$ is free.
b) Suppose $R$ has the invariant basis property (i.e. the rank of a free module is unique).
If $R$ has property $a),b)$, then it is obvious that $K_0(R)$ is generated by $[R]$ as $mathbb{Z}$-algebra. (Here I assumed $K_0(R)$ is already endowed with ring structure via tensor product.)
We say that an $R$-module $M$ is stably free if $Moplus F=G$ for some f.g. free modules $F,G$ (thus, $[M]$ is the difference of 2 free modules).
Question: The book claims that if $a)$ is replaced by every f.g. projective mod being stably free and $b)$ holds, then $K_0(R)$ is generated by $[R]$. How do I deduce this when $M$ is stably free? In other words, I need to deduce $M$ free.
Ref. Milnor's Algebraic K-theory pg 5.(Question 1 and 2)
abstract-algebra commutative-algebra k-theory
edited Nov 25 at 1:33
darij grinberg
10.1k32961
asked Nov 24 at 20:35
user45765
2,4422720
Let $R$ be a unital associative ring. Let $F$ be the free abelian group on the set of all isomorphism classes $[P]$ of f.g. projective $R$-modules $P$. Let $K_0(R)$ be the quotient of $F$ modulo the subgroup $S$ spanned by $[P]+[Q]-[Poplus Q]$ for all projective $P$ and $Q$.
a) Suppose every f.g. projective of $R$ is free.
b) Suppose $R$ has the invariant basis property (i.e. the rank of a free module is unique).
If $R$ has property $a),b)$, then it is obvious that $K_0(R)$ is generated by $[R]$ as $mathbb{Z}$-algebra. (Here I assumed $K_0(R)$ is already endowed with ring structure via tensor product.)
We say that an $R$-module $M$ is stably free if $Moplus F=G$ for some f.g. free modules $F,G$ (thus, $[M]$ is the difference of 2 free modules).
Question: The book claims that if $a)$ is replaced by every f.g. projective mod being stably free and $b)$ holds, then $K_0(R)$ is generated by $[R]$. How do I deduce this when $M$ is stably free? In other words, I need to deduce $M$ free.
Ref. Milnor's Algebraic K-theory pg 5.(Question 1 and 2)
abstract-algebra commutative-algebra k-theory
abstract-algebra commutative-algebra k-theory
edited Nov 25 at 1:33
darij grinberg
10.1k32961
asked Nov 24 at 20:35
user45765
2,4422720
edited Nov 25 at 1:33
darij grinberg
10.1k32961
asked Nov 24 at 20:35
user45765
2,4422720
edited Nov 25 at 1:33
darij grinberg
10.1k32961
edited Nov 25 at 1:33
darij grinberg
10.1k32961
edited Nov 25 at 1:33
darij grinberg
10.1k32961
10.1k32961
asked Nov 24 at 20:35
user45765
2,4422720
asked Nov 24 at 20:35
user45765
2,4422720
asked Nov 24 at 20:35
user45765
2,4422720
2,4422720
What have you tried?
– Pedro Tamaroff♦
Nov 24 at 20:36
@PedroTamaroff I think I am stuck at the first step as I do not see how to deduce a non-trivial free module morphism to $M$. I would expect $G/F$ to be free module. However, I am very uncertain about this step. Then from here, I would expect localization kicks in to deduce the free module level isomorphism on every point and here I need use invariant basis property. Hence it induces isomorphism which concludes $M$ being free.
– user45765
Nov 24 at 20:38
@PedroTamaroff I think I might be heading towards wrong way of thinking as I am wondering whether every projective being free here. This is a fairly stringent condition on the ring property. Say $R$ being PID fits.
– user45765
Nov 24 at 20:40
add a comment |
What have you tried?
– Pedro Tamaroff♦
Nov 24 at 20:36
@PedroTamaroff I think I am stuck at the first step as I do not see how to deduce a non-trivial free module morphism to $M$. I would expect $G/F$ to be free module. However, I am very uncertain about this step. Then from here, I would expect localization kicks in to deduce the free module level isomorphism on every point and here I need use invariant basis property. Hence it induces isomorphism which concludes $M$ being free.
– user45765
Nov 24 at 20:38
@PedroTamaroff I think I might be heading towards wrong way of thinking as I am wondering whether every projective being free here. This is a fairly stringent condition on the ring property. Say $R$ being PID fits.
– user45765
Nov 24 at 20:40
What have you tried?
– Pedro Tamaroff♦
Nov 24 at 20:36
What have you tried?
– Pedro Tamaroff♦
Nov 24 at 20:36
@PedroTamaroff I think I am stuck at the first step as I do not see how to deduce a non-trivial free module morphism to $M$. I would expect $G/F$ to be free module. However, I am very uncertain about this step. Then from here, I would expect localization kicks in to deduce the free module level isomorphism on every point and here I need use invariant basis property. Hence it induces isomorphism which concludes $M$ being free.
– user45765
Nov 24 at 20:38
@PedroTamaroff I think I am stuck at the first step as I do not see how to deduce a non-trivial free module morphism to $M$. I would expect $G/F$ to be free module. However, I am very uncertain about this step. Then from here, I would expect localization kicks in to deduce the free module level isomorphism on every point and here I need use invariant basis property. Hence it induces isomorphism which concludes $M$ being free.
– user45765
Nov 24 at 20:38
@PedroTamaroff I think I might be heading towards wrong way of thinking as I am wondering whether every projective being free here. This is a fairly stringent condition on the ring property. Say $R$ being PID fits.
– user45765
Nov 24 at 20:40
@PedroTamaroff I think I might be heading towards wrong way of thinking as I am wondering whether every projective being free here. This is a fairly stringent condition on the ring property. Say $R$ being PID fits.
– user45765
Nov 24 at 20:40
add a comment |
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
Assumption (b) is irrelevant, and you can conclude that $K_0(R)$ is generated by $[R]$ as a $mathbb{Z}$-module, not just as a $mathbb{Z}$-algebra. For any $[M]in K_0(R)$, there are finitely generated free modules $F$ and $G$ such that $Moplus F= G$ and so $[M]=[G]-[F]$. There are $m,ninmathbb{N}$ such that $Gcong R^m$ and $Fcong R^n$, and so $$[M]=[G]-[F]=[R^m]-[R^n]=m[R]-n[R]=(m-n)[R]$$ is an integer multiple of $[R]$.
Note in particular that proving $[M]$ is equal to a multiple of $[R]$ in $K_0(R)$ does not mean that $M$ itself is a free module.
answered Nov 24 at 21:01
Eric Wofsey
176k12202327
Maybe this is a dumb question. If $R$ does not have IBN, then say $R^mcong R^{m'}$ and $R^ncong R^{n'}$. So $(m-n)[R]=(m'-n')[R]$? How do I know $m-n=m'-n'$? Thanks.
– user45765
Nov 24 at 21:07
3
You don't. This doesn't change the fact that $[R]$ generates $K_0(R)$ as an abelian group.
– Eric Wofsey
Nov 24 at 21:08
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Assumption (b) is irrelevant, and you can conclude that $K_0(R)$ is generated by $[R]$ as a $mathbb{Z}$-module, not just as a $mathbb{Z}$-algebra. For any $[M]in K_0(R)$, there are finitely generated free modules $F$ and $G$ such that $Moplus F= G$ and so $[M]=[G]-[F]$. There are $m,ninmathbb{N}$ such that $Gcong R^m$ and $Fcong R^n$, and so $$[M]=[G]-[F]=[R^m]-[R^n]=m[R]-n[R]=(m-n)[R]$$ is an integer multiple of $[R]$.
Note in particular that proving $[M]$ is equal to a multiple of $[R]$ in $K_0(R)$ does not mean that $M$ itself is a free module.
answered Nov 24 at 21:01
Eric Wofsey
176k12202327
Maybe this is a dumb question. If $R$ does not have IBN, then say $R^mcong R^{m'}$ and $R^ncong R^{n'}$. So $(m-n)[R]=(m'-n')[R]$? How do I know $m-n=m'-n'$? Thanks.
– user45765
Nov 24 at 21:07
3
You don't. This doesn't change the fact that $[R]$ generates $K_0(R)$ as an abelian group.
– Eric Wofsey
Nov 24 at 21:08
add a comment |
up vote
3
down vote
accepted
Assumption (b) is irrelevant, and you can conclude that $K_0(R)$ is generated by $[R]$ as a $mathbb{Z}$-module, not just as a $mathbb{Z}$-algebra. For any $[M]in K_0(R)$, there are finitely generated free modules $F$ and $G$ such that $Moplus F= G$ and so $[M]=[G]-[F]$. There are $m,ninmathbb{N}$ such that $Gcong R^m$ and $Fcong R^n$, and so $$[M]=[G]-[F]=[R^m]-[R^n]=m[R]-n[R]=(m-n)[R]$$ is an integer multiple of $[R]$.
Note in particular that proving $[M]$ is equal to a multiple of $[R]$ in $K_0(R)$ does not mean that $M$ itself is a free module.
answered Nov 24 at 21:01
Eric Wofsey
176k12202327
Maybe this is a dumb question. If $R$ does not have IBN, then say $R^mcong R^{m'}$ and $R^ncong R^{n'}$. So $(m-n)[R]=(m'-n')[R]$? How do I know $m-n=m'-n'$? Thanks.
– user45765
Nov 24 at 21:07
3
You don't. This doesn't change the fact that $[R]$ generates $K_0(R)$ as an abelian group.
– Eric Wofsey
Nov 24 at 21:08
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Assumption (b) is irrelevant, and you can conclude that $K_0(R)$ is generated by $[R]$ as a $mathbb{Z}$-module, not just as a $mathbb{Z}$-algebra. For any $[M]in K_0(R)$, there are finitely generated free modules $F$ and $G$ such that $Moplus F= G$ and so $[M]=[G]-[F]$. There are $m,ninmathbb{N}$ such that $Gcong R^m$ and $Fcong R^n$, and so $$[M]=[G]-[F]=[R^m]-[R^n]=m[R]-n[R]=(m-n)[R]$$ is an integer multiple of $[R]$.
Note in particular that proving $[M]$ is equal to a multiple of $[R]$ in $K_0(R)$ does not mean that $M$ itself is a free module.
answered Nov 24 at 21:01
Eric Wofsey
176k12202327
Assumption (b) is irrelevant, and you can conclude that $K_0(R)$ is generated by $[R]$ as a $mathbb{Z}$-module, not just as a $mathbb{Z}$-algebra. For any $[M]in K_0(R)$, there are finitely generated free modules $F$ and $G$ such that $Moplus F= G$ and so $[M]=[G]-[F]$. There are $m,ninmathbb{N}$ such that $Gcong R^m$ and $Fcong R^n$, and so $$[M]=[G]-[F]=[R^m]-[R^n]=m[R]-n[R]=(m-n)[R]$$ is an integer multiple of $[R]$.
Note in particular that proving $[M]$ is equal to a multiple of $[R]$ in $K_0(R)$ does not mean that $M$ itself is a free module.
answered Nov 24 at 21:01
Eric Wofsey
176k12202327
answered Nov 24 at 21:01
Eric Wofsey
176k12202327
answered Nov 24 at 21:01
Eric Wofsey
176k12202327
answered Nov 24 at 21:01
Eric Wofsey
176k12202327
176k12202327
Maybe this is a dumb question. If $R$ does not have IBN, then say $R^mcong R^{m'}$ and $R^ncong R^{n'}$. So $(m-n)[R]=(m'-n')[R]$? How do I know $m-n=m'-n'$? Thanks.
– user45765
Nov 24 at 21:07
3
You don't. This doesn't change the fact that $[R]$ generates $K_0(R)$ as an abelian group.
– Eric Wofsey
Nov 24 at 21:08
add a comment |
Maybe this is a dumb question. If $R$ does not have IBN, then say $R^mcong R^{m'}$ and $R^ncong R^{n'}$. So $(m-n)[R]=(m'-n')[R]$? How do I know $m-n=m'-n'$? Thanks.
– user45765
Nov 24 at 21:07
3
You don't. This doesn't change the fact that $[R]$ generates $K_0(R)$ as an abelian group.
– Eric Wofsey
Nov 24 at 21:08
Maybe this is a dumb question. If $R$ does not have IBN, then say $R^mcong R^{m'}$ and $R^ncong R^{n'}$. So $(m-n)[R]=(m'-n')[R]$? How do I know $m-n=m'-n'$? Thanks.
– user45765
Nov 24 at 21:07
Maybe this is a dumb question. If $R$ does not have IBN, then say $R^mcong R^{m'}$ and $R^ncong R^{n'}$. So $(m-n)[R]=(m'-n')[R]$? How do I know $m-n=m'-n'$? Thanks.
– user45765
Nov 24 at 21:07
3
3
You don't. This doesn't change the fact that $[R]$ generates $K_0(R)$ as an abelian group.
– Eric Wofsey
Nov 24 at 21:08
You don't. This doesn't change the fact that $[R]$ generates $K_0(R)$ as an abelian group.
– Eric Wofsey
Nov 24 at 21:08
add a comment |
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What have you tried?
– Pedro Tamaroff♦
Nov 24 at 20:36
@PedroTamaroff I think I am stuck at the first step as I do not see how to deduce a non-trivial free module morphism to $M$. I would expect $G/F$ to be free module. However, I am very uncertain about this step. Then from here, I would expect localization kicks in to deduce the free module level isomorphism on every point and here I need use invariant basis property. Hence it induces isomorphism which concludes $M$ being free.
– user45765
Nov 24 at 20:38
@PedroTamaroff I think I might be heading towards wrong way of thinking as I am wondering whether every projective being free here. This is a fairly stringent condition on the ring property. Say $R$ being PID fits.
– user45765
Nov 24 at 20:40