Timeline for Can the Grothendieck ring of varities over a field $k$ be defined for non separated schemes?
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| Sep 21, 2014 at 23:38 | comment | added | schn93 | 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. | |
| Sep 16, 2014 at 13:51 | comment | added | Ariyan Javanpeykar | Sorry I misread the question. | |
| Sep 16, 2014 at 13:15 | vote | accept | Manuel Mérida Angulo | ||
| Sep 16, 2014 at 13:00 | answer | added | bananastack | timeline score: 3 | |
| Sep 16, 2014 at 12:50 | history | edited | Manuel Mérida Angulo | CC BY-SA 3.0 |
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| Sep 16, 2014 at 12:42 | comment | added | Manuel Mérida Angulo | 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. | |
| Sep 16, 2014 at 9:15 | comment | added | Ariyan Javanpeykar | 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...) | |
| Sep 16, 2014 at 8:46 | review | Close votes | |||
| Sep 16, 2014 at 19:06 | |||||
| Sep 16, 2014 at 8:26 | review | First posts | |||
| Sep 16, 2014 at 8:30 | |||||
| Sep 16, 2014 at 8:18 | history | asked | Manuel Mérida Angulo | CC BY-SA 3.0 |