Two types of Grothendieck groups for rings
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For a Noetherian ring $R$, there seem to be two versions of zeroth K-theory one can associate to it: $K_0(R)$ the Grothendieck group of the exact category of projective modules and $G_0(R)$ the Grothendieck group of the abelian category of finitely-generated modules. There is a map $K_0(R) rightarrow G_0(R)$ and if $R$ is a regular ring, this map is an isomorphism. What is an example of a (non-regular) ring $R$ such that $K_0(R)$ and $G_0(R)$ are not isomorphic?
(I know that if $R = k[x]/(x^n)$, then $K_0(R), G_0(R)$ are both abstractly isomorphic to $mathbb{Z}$ but the map above is multiplication by $n$ and hence not an isomorphism. I would an example of $R$ where the two Grothendieck groups are not even abstractly isomorphic).
ring-theory homological-algebra k-theory
asked Jan 8 at 19:28
user39598user39598
169213
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add a comment |
$begingroup$
For a Noetherian ring $R$, there seem to be two versions of zeroth K-theory one can associate to it: $K_0(R)$ the Grothendieck group of the exact category of projective modules and $G_0(R)$ the Grothendieck group of the abelian category of finitely-generated modules. There is a map $K_0(R) rightarrow G_0(R)$ and if $R$ is a regular ring, this map is an isomorphism. What is an example of a (non-regular) ring $R$ such that $K_0(R)$ and $G_0(R)$ are not isomorphic?
(I know that if $R = k[x]/(x^n)$, then $K_0(R), G_0(R)$ are both abstractly isomorphic to $mathbb{Z}$ but the map above is multiplication by $n$ and hence not an isomorphism. I would an example of $R$ where the two Grothendieck groups are not even abstractly isomorphic).
ring-theory homological-algebra k-theory
asked Jan 8 at 19:28
user39598user39598
169213
$endgroup$
add a comment |
$begingroup$
For a Noetherian ring $R$, there seem to be two versions of zeroth K-theory one can associate to it: $K_0(R)$ the Grothendieck group of the exact category of projective modules and $G_0(R)$ the Grothendieck group of the abelian category of finitely-generated modules. There is a map $K_0(R) rightarrow G_0(R)$ and if $R$ is a regular ring, this map is an isomorphism. What is an example of a (non-regular) ring $R$ such that $K_0(R)$ and $G_0(R)$ are not isomorphic?
(I know that if $R = k[x]/(x^n)$, then $K_0(R), G_0(R)$ are both abstractly isomorphic to $mathbb{Z}$ but the map above is multiplication by $n$ and hence not an isomorphism. I would an example of $R$ where the two Grothendieck groups are not even abstractly isomorphic).
ring-theory homological-algebra k-theory
asked Jan 8 at 19:28
user39598user39598
169213
$endgroup$
For a Noetherian ring $R$, there seem to be two versions of zeroth K-theory one can associate to it: $K_0(R)$ the Grothendieck group of the exact category of projective modules and $G_0(R)$ the Grothendieck group of the abelian category of finitely-generated modules. There is a map $K_0(R) rightarrow G_0(R)$ and if $R$ is a regular ring, this map is an isomorphism. What is an example of a (non-regular) ring $R$ such that $K_0(R)$ and $G_0(R)$ are not isomorphic?
(I know that if $R = k[x]/(x^n)$, then $K_0(R), G_0(R)$ are both abstractly isomorphic to $mathbb{Z}$ but the map above is multiplication by $n$ and hence not an isomorphism. I would an example of $R$ where the two Grothendieck groups are not even abstractly isomorphic).
ring-theory homological-algebra k-theory
ring-theory homological-algebra k-theory
asked Jan 8 at 19:28
user39598user39598
169213
asked Jan 8 at 19:28
user39598user39598
169213
asked Jan 8 at 19:28
user39598user39598
169213
asked Jan 8 at 19:28
user39598user39598
169213
asked Jan 8 at 19:28
user39598user39598
169213
169213
add a comment |
add a comment |
1 Answer
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Let $k$ be a field and $R=k[[x,y]]/(xy)$. Then $R$ is local, so projective modules are free and $K_0(R)congmathbb{Z}$. I claim $G_0(R)$ is not cyclic, and in particular that the modules $M=R/(x)$ and $N=R/(y)$ are $mathbb{Z}$-linearly independent in $G_0(R)$. To prove this, note that localization gives a homomorphism $G_0(R)to G_0(R_x)times G_0(R_y)$. We have $M_x=0$ and $M_ycong R_y$, and $N_xcong R_x$ and $N_y=0$. It follows that the images of $M$ and $N$ in $G_0(R_x)times G_0(R_y)$ are linearly independent, and hence $M$ and $N$ are linearly independent in $G_0(R)$.
answered Jan 8 at 21:54
Eric WofseyEric Wofsey
187k14215344
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$begingroup$
Let $k$ be a field and $R=k[[x,y]]/(xy)$. Then $R$ is local, so projective modules are free and $K_0(R)congmathbb{Z}$. I claim $G_0(R)$ is not cyclic, and in particular that the modules $M=R/(x)$ and $N=R/(y)$ are $mathbb{Z}$-linearly independent in $G_0(R)$. To prove this, note that localization gives a homomorphism $G_0(R)to G_0(R_x)times G_0(R_y)$. We have $M_x=0$ and $M_ycong R_y$, and $N_xcong R_x$ and $N_y=0$. It follows that the images of $M$ and $N$ in $G_0(R_x)times G_0(R_y)$ are linearly independent, and hence $M$ and $N$ are linearly independent in $G_0(R)$.
answered Jan 8 at 21:54
Eric WofseyEric Wofsey
187k14215344
$endgroup$
add a comment |
$begingroup$
Let $k$ be a field and $R=k[[x,y]]/(xy)$. Then $R$ is local, so projective modules are free and $K_0(R)congmathbb{Z}$. I claim $G_0(R)$ is not cyclic, and in particular that the modules $M=R/(x)$ and $N=R/(y)$ are $mathbb{Z}$-linearly independent in $G_0(R)$. To prove this, note that localization gives a homomorphism $G_0(R)to G_0(R_x)times G_0(R_y)$. We have $M_x=0$ and $M_ycong R_y$, and $N_xcong R_x$ and $N_y=0$. It follows that the images of $M$ and $N$ in $G_0(R_x)times G_0(R_y)$ are linearly independent, and hence $M$ and $N$ are linearly independent in $G_0(R)$.
answered Jan 8 at 21:54
Eric WofseyEric Wofsey
187k14215344
$endgroup$
add a comment |
$begingroup$
Let $k$ be a field and $R=k[[x,y]]/(xy)$. Then $R$ is local, so projective modules are free and $K_0(R)congmathbb{Z}$. I claim $G_0(R)$ is not cyclic, and in particular that the modules $M=R/(x)$ and $N=R/(y)$ are $mathbb{Z}$-linearly independent in $G_0(R)$. To prove this, note that localization gives a homomorphism $G_0(R)to G_0(R_x)times G_0(R_y)$. We have $M_x=0$ and $M_ycong R_y$, and $N_xcong R_x$ and $N_y=0$. It follows that the images of $M$ and $N$ in $G_0(R_x)times G_0(R_y)$ are linearly independent, and hence $M$ and $N$ are linearly independent in $G_0(R)$.
answered Jan 8 at 21:54
Eric WofseyEric Wofsey
187k14215344
$endgroup$
Let $k$ be a field and $R=k[[x,y]]/(xy)$. Then $R$ is local, so projective modules are free and $K_0(R)congmathbb{Z}$. I claim $G_0(R)$ is not cyclic, and in particular that the modules $M=R/(x)$ and $N=R/(y)$ are $mathbb{Z}$-linearly independent in $G_0(R)$. To prove this, note that localization gives a homomorphism $G_0(R)to G_0(R_x)times G_0(R_y)$. We have $M_x=0$ and $M_ycong R_y$, and $N_xcong R_x$ and $N_y=0$. It follows that the images of $M$ and $N$ in $G_0(R_x)times G_0(R_y)$ are linearly independent, and hence $M$ and $N$ are linearly independent in $G_0(R)$.
answered Jan 8 at 21:54
Eric WofseyEric Wofsey
187k14215344
answered Jan 8 at 21:54
Eric WofseyEric Wofsey
187k14215344
answered Jan 8 at 21:54
Eric WofseyEric Wofsey
187k14215344
answered Jan 8 at 21:54
Eric WofseyEric Wofsey
187k14215344
187k14215344
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