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







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    $begingroup$


    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







    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      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







      share|cite|improve this question









      $endgroup$




      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






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      asked Jan 8 at 19:28









      user39598user39598

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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 $G_0(R)$.






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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 $G_0(R)$.






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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 $G_0(R)$.






              share|cite|improve this answer









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                3





                $begingroup$

                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 $G_0(R)$.






                share|cite|improve this answer









                $endgroup$



                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 $G_0(R)$.







                share|cite|improve this answer












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                share|cite|improve this answer










                answered Jan 8 at 21:54









                Eric WofseyEric Wofsey

                193k14221352




                193k14221352






























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