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Let $K$ be a field, $\alpha, \beta \in \mathrm{Br}(K)$, let $X,Y$ be their Brauer-Severi Varieties, is there a way to calculate $A^*(X\times Y)$?

For example, if $\alpha,\beta$ both has degree $5$, $2\alpha=\beta$, then $A^*(X\times Y)$ is a subring (Will two non-rational equivalents cycles become rational equivalent after base change?) of $A^*(X\times Y_{\overline{K}})=A^*(\mathbb{P^4}\times\mathbb{P^4})=Z[H_1,H_2]/(H_1^5,H_2^5)$. Then $5H_i\in A^1$, $H_i\notin A^1$, but $4H_1+3H_2\in A^1$, as it is the first Chern class of the vector bundle bundle $O(1)\boxtimes (\Omega\otimes O(2))$ on $X\times Y$.

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In general I think it is fairly subtle.

There are some partial results though. If $\beta$ is in the subgroup of $\text{Br}(K)$ generated by $\alpha$ then the ring $\text{CH}(X\times Y)$ is isomorphic to a direct sum of shifted copies of $\text{CH}(X)$. In this case, the claim is due to the fact the product of Severi-Brauer varieties $X\times Y$ is isomorphic to a projective bundle over $X$.

Let me assume $\alpha\neq 0$. Let $A$ be the algebra representing $\alpha$ and $B$ be the algebra representing $\beta$. A complete calculation of the ring in this case can be obtained as follows: for a prime degree algebra $A$ the Chow ring is generated by chern classes of the tautological vector bundle of rank the degree of this algebra. So, denoting $X=\text{SB}(A)$, a description $\text{CH}(X)=\mathbf{Z}\oplus \mathbf{Z}e\oplus \mathbf{Z}c_2\oplus \mathbf{Z}c_3\oplus \mathbf{Z}c_4$ can be obtained where under the pullback to an algebraic closure, say $\text{CH}(\mathbf{P}^4)$, the elements $e,c_2,c_3,c_4$ map to $5h,5h^2,5h^3,5h^4$ respectively. By the projective bundle formula $\text{CH}(X\times Y)$ is isomorphic with $\text{CH}(X)[\xi]/(\xi^5+ s_1\xi^4 + s_2\xi^3 + s_3\xi^2 + s_4\xi + s_5)$ where $s_i$ are the $i$th Chern classes of a bundle realizing this product as a projective bundle over $X$. It just remains to find the $s_i$.

Let $\mathcal{T}_A$ be the tautological bundle of rank the degree of $A$ on $Y=\text{SB}(A)$. It's square $\mathcal{T}_A^{\otimes 2}$ is rank 25 bundle that splits into the sum of 5 indecomposable (and isomorphic) rank 5 bundles, which I'll call $\eta_2$. The projective bundle $X\times Y\rightarrow X$ is isomorphic with $\mathbf{P}(\eta_2)\rightarrow X$ and the Chern classes are determined as $s_i=c_i(\eta_2)$. The pullback of $c_i(\eta_2)$ to $\text{CH}(\mathbf{P}^4)$ is mapped to $c_i(\mathcal{O}(2)^{\oplus 5})= \binom{5}{i}2^i h^i$ so that $s_1= 2e$, $s_2= 8c_2$, $s_3= 16c_3$, $s_4= 16c_4$, and $s_5=0$.

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  • $\begingroup$ 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? $\endgroup$ – Qixiao Dec 30 '17 at 13:22
  • $\begingroup$ @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 $c_1(\mathcal{O}(1)\boxtimes (\Omega\otimes \mathcal{O}(2)))=4c_1(\pi_1^*\mathcal{O}(1))+c_1(\pi_2^*(\Omega\otimes \mathcal{O}(2)))= 4\pi_1^*c_1(\mathcal{O}(1))+\pi_2^*c_1(\Omega\otimes \mathcal{O}(2))=20H + \pi_2^*c_1(\Omega)+4\pi_2^*c_1(\mathcal{O}(2))=...$. $\endgroup$ – Eoin Dec 30 '17 at 20:11
  • $\begingroup$ 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 $\alpha+2\beta=5\alpha=0$ $\endgroup$ – Qixiao Dec 30 '17 at 20:23
  • $\begingroup$ @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 $10H_1+3(H_2-2H_1)=4H_1+3H_2$. $\endgroup$ – Eoin Dec 30 '17 at 21:05

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