Timeline for Double points in the Grothendieck ring
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| S Jul 18 at 4:34 | history | suggested | Andrew Stout | CC BY-SA 4.0 |
Put brackets to denote equivalence class in Grothendieck ring of varieties.
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| Jul 17 at 22:10 | review | Suggested edits | |||
| S Jul 18 at 4:34 | |||||
| Jul 17 at 22:07 | answer | added | Andrew Stout | timeline score: 2 | |
| Mar 11, 2020 at 15:02 | answer | added | Balazs | timeline score: 5 | |
| Mar 11, 2020 at 11:44 | comment | added | YCor | Maybe you want to ask whether there is some "enriched" Grothendieck ring mapping onto $K_0(V_k)$, which takes into account non-reducedness (so that the class of "$X-X_{\mathrm{red}}$", in some sense, would be meaningful and possibly nonzero...) I don't know how to formalize this properly. | |
| Mar 10, 2020 at 22:33 | history | edited | YCor |
edited tags
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| Mar 10, 2020 at 19:11 | comment | added | Devlin Mallory | Yes, that's exactly it (and more generally the same is true for the reduced subscheme $X_{\mathrm{red}}$ of any scheme $X$: the complement is empty, so the classes of $X$ and $X_{\mathrm{red}}$ agree in the Grothendieck ring). | |
| Mar 10, 2020 at 19:01 | comment | added | THC | @DevlinMallory: And by "definitionally," you mean that the complement of $\mathrm{Spec}(k[x]/(x^2))$ in $\mathrm{Spec}(k[x]/(x))$ is empty ? | |
| Mar 10, 2020 at 18:43 | comment | added | Devlin Mallory | Sure: $\mathrm{Spec}\, k[x]/x$ is a closed subscheme of $\mathrm{Spec}\, k[x]/x^2$ (corresponding to the surjection $k[x]/x^2\to k[x]/x$, and thus definitionally $[\mathrm{Spec}\, k[x]/x]=[\mathrm{Spec}\, k[x]/x^2]$ in the Grothendieck ring. Then, since for any $k$-scheme $X$ we have $X\times_k k \cong X$ we have that $[\mathrm{Spec}\,k]$ is the identity in the Grothendieck ring. | |
| Mar 10, 2020 at 18:24 | comment | added | THC | @DevlinMallory : could you be more explicit in the context of my question ? Thanks ! | |
| Mar 10, 2020 at 18:09 | comment | added | Devlin Mallory | $[X]=[X_{\mathrm{red}}]$ for any $k$-scheme $X$, just because the Grothendieck ring is defined by saying that for any closed subscheme $Z$ of $X$ we have $[Z]+[X-Z]=[X]$, including for $Z=X_{\mathrm{red}}$. | |
| Mar 10, 2020 at 18:07 | history | asked | THC | CC BY-SA 4.0 |