Timeline for Geometric vs combinatorial motives over Spec Z
Current License: CC BY-SA 3.0
3 events
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| Feb 17, 2015 at 2:20 | comment | added | Daniel Litt | That said, the kernel is certainly non-empty. Take two isogenous elliptic curves over $\mathcal{O}_K[1/N]$, where $K$ is some number field. Then their difference will map to zero. But $[E_1]\neq [E_2]$ because their generic fibers are not stably birational (here I am using a result of Larsen and Lunts that the Grothendieck ring distinguishes non-stably birational varieties in characteristic zero). | |
| Feb 16, 2015 at 23:00 | comment | added | David E Speyer | Question: Let $X$ be a scheme with non-trivial Brauer group, so there are maps $E \to X$ which are $\mathbb{P}^r$ bundles in the etale topology but not in the Zariski topology. Then I think $\# E(\mathbb{F}_q) = \# \left( X \times \mathbb{P}^r \right)(\mathbb{F}_q)$. Are the two sides also equal in the Grothendieck ring? | |
| Feb 16, 2015 at 21:21 | history | asked | Andreas Holmstrom | CC BY-SA 3.0 |