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Let $X$ and $Y$ be two smooth projective varieties over $k$. Consider the Chow motives $M(X\times\mathbb{P}^n)\simeq M(X)\otimes M(\mathbb{P}^n)$ and $M(Y\times\mathbb{P}^n)\simeq M(Y)\otimes M(\mathbb{P}^n)$.

We suppose that $M(X)\otimes M(\mathbb{P}^n)$ is a direct summand of $M(Y)\otimes M(\mathbb{P}^n)$. Suppose furthermore that $M(X)$ is not direct summand of $M(\mathbb{P}^n)$ and $M(\mathbb{P}^n)$ is not a direct summand of $M(Y)$.

Do we have in this case that $M(X)$ is a direct summand of $M(Y)$?

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    $\begingroup$ For Chow motives this seems pretty hard, because we do not in general have Künneth components. In other categories of motives, I think it should be straightforward to prove. Anyway, observe that $h(\mathbb{P}^{n}) \cong \bigoplus_{i=0}^{n} \mathbb{L}^{\otimes i}$, and thus for any motive $M$, we have $M \otimes h(\mathbb{P}^{n}) \cong \bigoplus_{i=0}^{n} M(-i)$. However, this is not enough, as far as I can see. $\endgroup$
    – jmc
    Commented Mar 11, 2014 at 8:10

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Let $A$ and $B$ be two central simple algebras of the same degree which generate the same subgroup in the Brauer group of a field.

The projections from the product $\mathrm{SB}(A)\times \mathrm{SB}(B)$ of the corresponding Severi-Brauer varieties to each of the factors are projective bundles, hence you have

$$M(\mathrm{SB}(A)\times \mathrm{SB}(B))\simeq M(\mathrm{SB}(A))\otimes M(\mathbb{P}^n)\simeq M(\mathrm{SB}(B))\otimes M(\mathbb{P}^n)$$

On the other hand if $B$ is not the opposite algebra of $A$, the motives of the corresponding Severi-Brauer varieties are not isomorphic.

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  • $\begingroup$ Why are they not isomorphic? $\endgroup$
    – Will Sawin
    Commented Jul 11, 2015 at 16:18
  • $\begingroup$ This is Criteria of motivic equivalence for quadratic forms and central simple algebras, prop 7.10, by Karpenko. $\endgroup$
    – Louis La Brocante
    Commented Jul 11, 2015 at 17:39

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