Motivic class of mixed Tate motive

Question

Let $k$ be a field (of characteristic zero), $R$ be a ring and let $X\in DM(k;R)$ be a Tate motive. By definition, this means that $X$ is a summand of an object of the smallest strictly full triangulated subcategory of $DM(k;R)$ that contains $R(j)$ for all integers $j$.

It seems to be obvious that the motivic class of $X$ in $K_0(DM)$ is a rational function of $\mathbb L =[\mathbb A^1]$. However, I don't see why. By additivity, shouldn't we just have $[X]+[Y]=$ rational function of $\mathbb L$ (which is very little)?

Answer 1

Yes, the class of $X$ is a polynomial of $\mathbb L,\ \mathbb L^{-1}$. Here I assume that the coefficient ring $R$ is indecomposable (and so, contains no non-trivial idempotents), and you consider the "triangulated version" of $K_0(DM)$.

This statement certainly follows from the fact that the smallest strictly full triangulated subcategory $DMT(k;R)$ of $DM(k;R)$ that contains $R(j)$ is idempotent complete, i.e., contains all direct summands of its objects. Now, the latter statement is a consequence of the existence of a (Chow) weight structure on $DMT(k;R)$ whose heart consists on Chow-Tate motives; these are direct sums of $R(i)[2i]=\mathbb L^{\otimes i}$, and the point is that this heart is equivalent to the idempotent category of finitely generated graded free $R$-modules. This argument requires a very easy motivic calculation and a little of the theory of weight structures; you and read Corollary 2.1.2 and Remark 2.1.3 of our paper 'On constructing weight structures and extending them to idempotent completions"; see https://intlpress.com/site/pub/pages/journals/items/hha/content/vols/0020/0001/a003/index.html.