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The Grothendieck ring of varieties over a field $k$ is the abelian group generated by isomorphim classes $[X]$ of separated, reduced $k$-schemes $X$ of finite type with the relation

$[X]=[Y] + [X\setminus Y]$

for any $Y \subset X$ a closed immersion and with the product structure given by

$[X\times Y]= [X]\cdot[Y]$.

My question is related to the condition of separateness. Why do we need the schemes to be separated? I have been searching in different papers and notes and I couldn't find anything about it. Could the Grothendieck ring be extended to non separated schemes?

Thank you very much.

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  • $\begingroup$ Have a look at mathoverflow.net/questions/25122/… for instance. Of course, you can define the grothendieck group without the condition of separatedness. I think noetherian (or just locally noetherian) should be enough; see Definition 1.4 in math.leidenuniv.nl/scripties/MasterJavanpeykar.pdf (I dont recommend you read that text too thoroughly...) $\endgroup$
    – Ariyan Javanpeykar
    Commented Sep 16, 2014 at 9:15
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    $\begingroup$ Thank you for both comments. Yes, the question is about the Grothendieck ring, defined as isomorphims classes of varieties modulo some relations. I should have written the definition in the post, I will edit it so that it is clearer. $\endgroup$
    – Manuel Mérida Angulo
    Commented Sep 16, 2014 at 12:42
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    $\begingroup$ Sorry I misread the question. $\endgroup$
    – Ariyan Javanpeykar
    Commented Sep 16, 2014 at 13:51
  • $\begingroup$ This is not an answer to your question, but let me note that schemes need not be reduced neither. If you apply the relation $[X]=[Y]+[X-Y]$ to the inclusion $X_{red} \hookrightarrow X$, them you get $[X]=[X_{red}]$ since the underlying topological space of $X-X_{red}$ is empty. So you can erase "reduced" in your definition of the Grothendieck ring and you get the same thing. $\endgroup$
    – schn93
    Commented Sep 21, 2014 at 23:38

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If I understand the setup of the question correctly, the answer is yes. Using non separated schemes makes no difference.

This is explained here.

In brief, whether you do the Grothendieck ring of "integral, finite type, separated" or "finite type" or even "finite type algebraic spaces" it's the same. When you pass to stacks it's a whole other mess of course.

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