Timeline for Chow ring of product of Brauer-Severi Varieties

9 events

when toggle format	what		by	license	comment
Jan 1, 2018 at 4:22	vote	accept	CommunityBot
Dec 30, 2017 at 21:05	comment	added	Eoin		@Qixiao Ahh, okay. My bad. I think the map $\text{CH}(X\times Y)\rightarrow \text{CH}((X\times Y)_{\overline{K}})=\mathbf{Z}[H_1,H_2]/(H_1^5,H_2^5)$ is defined by $5H\mapsto 5H_1$ and $5(\xi+2H) \mapsto 5H_2$. So $\xi\mapsto H_2-2H_1$. And I guess the class your looking for is $10H_1+3(H_2-2H_1)=4H_1+3H_2$.
Dec 30, 2017 at 20:23	comment	added	user39380		Sorry I should clarify, geometrically we have $O(1)$ on $X_{\overline{k}}$, also geometrically we have $O(2)$, neither of them descend, but their box product descend as $\alpha+2\beta=5\alpha=0$
Dec 30, 2017 at 20:11	comment	added	Eoin		@Qixiao I don't know what you mean by $\mathcal{O}(1)$ on $X$ (or $Y$). If you mean a generator for $\text{Pic}(X)$ (or $Y$), then $c_1(\mathcal{O}(1)\boxtimes (\Omega\otimes \mathcal{O}(2)))=4c_1(\pi_1^\mathcal{O}(1))+c_1(\pi_2^(\Omega\otimes \mathcal{O}(2)))= 4\pi_1^c_1(\mathcal{O}(1))+\pi_2^c_1(\Omega\otimes \mathcal{O}(2))=20H + \pi_2^c_1(\Omega)+4\pi_2^c_1(\mathcal{O}(2))=...$.
Dec 30, 2017 at 13:22	comment	added	user39380		Thank you! I am a bit confused, the ring will be $\mathbb{Z}[5H, \xi]/\xi^5+10H\xi^4+\dots+80H^4\xi$, can we see $4H+3\xi$ is an element in this ring?
Dec 30, 2017 at 0:40	history	undeleted	Eoin
Dec 30, 2017 at 0:40	history	edited	Eoin	CC BY-SA 3.0	deleted 1 character in body
Dec 30, 2017 at 0:35	history	deleted	Eoin		via Vote
Dec 30, 2017 at 0:33	history	answered	Eoin	CC BY-SA 3.0