Grothendieck group of tensor products
$begingroup$
For two unital rings $R,S$, one can form the ring $Rotimes_mathbb{Z} S$
(this is the coproduct in the category of rings). I was wondering whether the following formula holds for the Grothendieck group of projective modules over these rings:
$K_0(Rotimes_mathbb{Z} S ) cong K_0(R) otimes_mathbb{Z} K_0(S)$ as abelian groups (or as rings if $R,S$ are commutative rings).
Since there are ring maps $R rightarrow Rotimes_mathbb{Z} S$ and $S rightarrow Rotimes_mathbb{Z} S$, there are induced maps $K_0(R) rightarrow
K_0(Rotimes_mathbb{Z} S ), K_0(R) rightarrow
K_0(Rotimes_mathbb{Z} S )$; if $R,S$ are commutative rings, these are maps of commutative rings and hence there is an induced map
$K_0(R) otimes_mathbb{Z} K_0(S) rightarrow K_0(Rotimes_mathbb{Z} S )$. Is this an isomorphism? Note that $K_0(R times S) cong K_0(R) times K_0(S)$ but I am interested in the tensor product of rings.
More generally, is there a description of $Proj(R otimes_mathbb{Z} S)$ in terms of $Proj(R otimes_mathbb{Z} S)$. In another direction, what about if there is an extra ring $T$: is $K_0(R otimes_T S) cong K_0(R) otimes_{K_0(T)} K_0(S)$?
ring-theory modules k-theory
asked Jan 23 at 19:54
user39598user39598
174213
$endgroup$
add a comment |
$begingroup$
For two unital rings $R,S$, one can form the ring $Rotimes_mathbb{Z} S$
(this is the coproduct in the category of rings). I was wondering whether the following formula holds for the Grothendieck group of projective modules over these rings:
$K_0(Rotimes_mathbb{Z} S ) cong K_0(R) otimes_mathbb{Z} K_0(S)$ as abelian groups (or as rings if $R,S$ are commutative rings).
Since there are ring maps $R rightarrow Rotimes_mathbb{Z} S$ and $S rightarrow Rotimes_mathbb{Z} S$, there are induced maps $K_0(R) rightarrow
K_0(Rotimes_mathbb{Z} S ), K_0(R) rightarrow
K_0(Rotimes_mathbb{Z} S )$; if $R,S$ are commutative rings, these are maps of commutative rings and hence there is an induced map
$K_0(R) otimes_mathbb{Z} K_0(S) rightarrow K_0(Rotimes_mathbb{Z} S )$. Is this an isomorphism? Note that $K_0(R times S) cong K_0(R) times K_0(S)$ but I am interested in the tensor product of rings.
More generally, is there a description of $Proj(R otimes_mathbb{Z} S)$ in terms of $Proj(R otimes_mathbb{Z} S)$. In another direction, what about if there is an extra ring $T$: is $K_0(R otimes_T S) cong K_0(R) otimes_{K_0(T)} K_0(S)$?
ring-theory modules k-theory
asked Jan 23 at 19:54
user39598user39598
174213
$endgroup$
add a comment |
$begingroup$
For two unital rings $R,S$, one can form the ring $Rotimes_mathbb{Z} S$
(this is the coproduct in the category of rings). I was wondering whether the following formula holds for the Grothendieck group of projective modules over these rings:
$K_0(Rotimes_mathbb{Z} S ) cong K_0(R) otimes_mathbb{Z} K_0(S)$ as abelian groups (or as rings if $R,S$ are commutative rings).
Since there are ring maps $R rightarrow Rotimes_mathbb{Z} S$ and $S rightarrow Rotimes_mathbb{Z} S$, there are induced maps $K_0(R) rightarrow
K_0(Rotimes_mathbb{Z} S ), K_0(R) rightarrow
K_0(Rotimes_mathbb{Z} S )$; if $R,S$ are commutative rings, these are maps of commutative rings and hence there is an induced map
$K_0(R) otimes_mathbb{Z} K_0(S) rightarrow K_0(Rotimes_mathbb{Z} S )$. Is this an isomorphism? Note that $K_0(R times S) cong K_0(R) times K_0(S)$ but I am interested in the tensor product of rings.
More generally, is there a description of $Proj(R otimes_mathbb{Z} S)$ in terms of $Proj(R otimes_mathbb{Z} S)$. In another direction, what about if there is an extra ring $T$: is $K_0(R otimes_T S) cong K_0(R) otimes_{K_0(T)} K_0(S)$?
ring-theory modules k-theory
asked Jan 23 at 19:54
user39598user39598
174213
$endgroup$
For two unital rings $R,S$, one can form the ring $Rotimes_mathbb{Z} S$
(this is the coproduct in the category of rings). I was wondering whether the following formula holds for the Grothendieck group of projective modules over these rings:
$K_0(Rotimes_mathbb{Z} S ) cong K_0(R) otimes_mathbb{Z} K_0(S)$ as abelian groups (or as rings if $R,S$ are commutative rings).
Since there are ring maps $R rightarrow Rotimes_mathbb{Z} S$ and $S rightarrow Rotimes_mathbb{Z} S$, there are induced maps $K_0(R) rightarrow
K_0(Rotimes_mathbb{Z} S ), K_0(R) rightarrow
K_0(Rotimes_mathbb{Z} S )$; if $R,S$ are commutative rings, these are maps of commutative rings and hence there is an induced map
$K_0(R) otimes_mathbb{Z} K_0(S) rightarrow K_0(Rotimes_mathbb{Z} S )$. Is this an isomorphism? Note that $K_0(R times S) cong K_0(R) times K_0(S)$ but I am interested in the tensor product of rings.
More generally, is there a description of $Proj(R otimes_mathbb{Z} S)$ in terms of $Proj(R otimes_mathbb{Z} S)$. In another direction, what about if there is an extra ring $T$: is $K_0(R otimes_T S) cong K_0(R) otimes_{K_0(T)} K_0(S)$?
ring-theory modules k-theory
ring-theory modules k-theory
asked Jan 23 at 19:54
user39598user39598
174213
asked Jan 23 at 19:54
user39598user39598
174213
edited Jan 23 at 20:03
user39598
asked Jan 23 at 19:54
user39598user39598
174213
asked Jan 23 at 19:54
user39598user39598
174213
asked Jan 23 at 19:54
user39598user39598
174213
174213
add a comment |
add a comment |
1 Answer
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No, this is not true in general. For instance, suppose $R=S=mathbb{Q}(i)$. Then $K_0(R)=K_0(S)congmathbb{Z}$, but $Rotimes Scong mathbb{Q}(i)times mathbb{Q}(i)$ so $K_0(Rotimes S)congmathbb{Z}^2$.
In that example, the natural map $K_0(R)otimes K_0(S)to K_0(Rotimes S)$ fails to be surjective. It can also fail to be injective. For instance, if $R$ and $S$ are fields of different characteristic, then $Rotimes S=0$ so $K(Rotimes S)=0$.
answered Jan 23 at 21:54
Eric WofseyEric Wofsey
188k14216346
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1 Answer
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$begingroup$
No, this is not true in general. For instance, suppose $R=S=mathbb{Q}(i)$. Then $K_0(R)=K_0(S)congmathbb{Z}$, but $Rotimes Scong mathbb{Q}(i)times mathbb{Q}(i)$ so $K_0(Rotimes S)congmathbb{Z}^2$.
In that example, the natural map $K_0(R)otimes K_0(S)to K_0(Rotimes S)$ fails to be surjective. It can also fail to be injective. For instance, if $R$ and $S$ are fields of different characteristic, then $Rotimes S=0$ so $K(Rotimes S)=0$.
answered Jan 23 at 21:54
Eric WofseyEric Wofsey
188k14216346
$endgroup$
add a comment |
$begingroup$
No, this is not true in general. For instance, suppose $R=S=mathbb{Q}(i)$. Then $K_0(R)=K_0(S)congmathbb{Z}$, but $Rotimes Scong mathbb{Q}(i)times mathbb{Q}(i)$ so $K_0(Rotimes S)congmathbb{Z}^2$.
In that example, the natural map $K_0(R)otimes K_0(S)to K_0(Rotimes S)$ fails to be surjective. It can also fail to be injective. For instance, if $R$ and $S$ are fields of different characteristic, then $Rotimes S=0$ so $K(Rotimes S)=0$.
answered Jan 23 at 21:54
Eric WofseyEric Wofsey
188k14216346
$endgroup$
add a comment |
$begingroup$
No, this is not true in general. For instance, suppose $R=S=mathbb{Q}(i)$. Then $K_0(R)=K_0(S)congmathbb{Z}$, but $Rotimes Scong mathbb{Q}(i)times mathbb{Q}(i)$ so $K_0(Rotimes S)congmathbb{Z}^2$.
In that example, the natural map $K_0(R)otimes K_0(S)to K_0(Rotimes S)$ fails to be surjective. It can also fail to be injective. For instance, if $R$ and $S$ are fields of different characteristic, then $Rotimes S=0$ so $K(Rotimes S)=0$.
answered Jan 23 at 21:54
Eric WofseyEric Wofsey
188k14216346
$endgroup$
No, this is not true in general. For instance, suppose $R=S=mathbb{Q}(i)$. Then $K_0(R)=K_0(S)congmathbb{Z}$, but $Rotimes Scong mathbb{Q}(i)times mathbb{Q}(i)$ so $K_0(Rotimes S)congmathbb{Z}^2$.
In that example, the natural map $K_0(R)otimes K_0(S)to K_0(Rotimes S)$ fails to be surjective. It can also fail to be injective. For instance, if $R$ and $S$ are fields of different characteristic, then $Rotimes S=0$ so $K(Rotimes S)=0$.
answered Jan 23 at 21:54
Eric WofseyEric Wofsey
188k14216346
answered Jan 23 at 21:54
Eric WofseyEric Wofsey
188k14216346
answered Jan 23 at 21:54
Eric WofseyEric Wofsey
188k14216346
answered Jan 23 at 21:54
Eric WofseyEric Wofsey
188k14216346
188k14216346
add a comment |
add a comment |
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