There is a well-known result by Quillen stating that if $X_A$ is the Severi-Brauer variety of a central simple algebra $A$ of degree $d$ over a field $k$, then its (Quillen) K-theory decomposes as $$K_\bullet(X_A) \simeq \bigoplus_{i=0}^{d-1}K_\bullet(A^{\otimes i}).$$

In every text I've seen mentioning it, the decomposition is stated at the level of $K_n$, and is just taken as a decomposition of abelian groups. But since $X_A$ is a scheme, $K_\bullet(X_A)$ is a graded-commutative ring, and this induces a multiplicative structure on $\bigoplus_{i=0}^{d-1} K_\bullet(A^{\otimes i})$.

It seems almost compelling that this structure should come from $$K_p(A^{\otimes i})\otimes K_q(A^{\otimes j})\to K_{p+q}(A^{\otimes i+j})\to K_{p+q}(A^{\otimes r})$$ where $r\equiv i+j$ modulo $d$, and the last map is the isomorphism given by Morita equivalence.

Nonetheless, I couldn't see any reference to that fact in the literature I've read. Is it a "folklore" result? Is is actually stated/proved somewhere and I've missed it?

Actually, even in the split case, where we get $K_\bullet(\mathbb{P}^{d-1} _k)\simeq K_\bullet(k)^d$, I haven't seen stated that this gives an isomorphism with the group ring $K_\bullet(k)[\mathbb{Z}/d\mathbb{Z}]$.

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    $\begingroup$ I don't think it is true: for the projective space the decomposition is given by the fact that $K(\mathbb{P}^n_S)$ is a free $K(S)$-module with basis $\{\mathcal{O}(i)\}_{i=0,\dots,d-1}$ and it's not true that $[\mathcal{O}(d)]=1$ in $K_0(\mathbb{P}^n_S)$ (rather, it is given by a polynomial in $[\mathcal{O}(1)]$). Presumably the case of a general Severi-Brauer variety is similar. $\endgroup$ Feb 11, 2020 at 18:32
  • $\begingroup$ @DenisNardin You are perfectly right, thank you. Actually, looking at the Koszul sequence $0\to \mathcal{O}\to \dots\to \mathcal{O}(i)^{\binom{d}{i}}\to\dots\to \mathcal{O}(d)\to 0$ shows that $\mathcal{O}(1)$ satisfies $(X-1)^d=0$, so $K_0(\mathbb{P}_k^d)\simeq \mathbb{Z}[X]/(X-1)^d$ is not reduced, whereas $K_0(k)[\mathbb{Z}/d\mathbb{Z}]$ is reduced (so they are really not isomorphic at all, this is not just a matter of a wrong basis choice). $\endgroup$ Feb 13, 2020 at 16:30


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