Let $k$ be a field. The naive Grothendieck ring of varieties $K_0(\text{Var})$ is generated by isomorphism classes of varieties over $k$ with the scissors relation $[X]=[X-Y]+[Y]$ for $Y$ a closed subvariety of $X$. We might naturally want to compare it with $K_0(\text{Mot})$, where here I want to remain flexible with what I mean by $\text{Mot}$ - for the generic situation, maybe pure effective motives over $k$ with either $\mathbb{Z}$ or $\mathbb{Q}$-coefficients - just as long as we have a symmetric monoidal category to apply $K_0$ to. If the situation is more understood after localizing or with a different equivalence relation, I'd be happy to consider that instead.

In characteristic zero, I can see how to define a map $K_0(\text{Var})\to K_0(\text{Mot})$; we just need "excision for motives," i.e. an agreement of the motives corresponding to $[X]-[Y]$ and $[W]-[Z]$ if $X-Y\cong W-Z$ as affine varieties. This follows from the fairly recent and difficult result of weak factorization in characteristic zero; we get a rational map from $X$ to $W$ which can be factored through series of blow-ups and blow-downs, and then Manin's computation of the motive of a blow-up does the rest. Singular varieties can be handled just fine for the same reason by Hironaka's algorithm. This map then, conjecturally, should be the semisimplification of the "mixed motive" associated to a general variety over $k$.

In characteristic $p$, we don't have resolution of singularities, nor weak factorization - so, is the very existence of this map out of reach?

(1) (How) can we define this map in positive characteristic?

Poonen proved that $K_0$ is not a domain for $k$ of characteristic zero; the case over $\mathbb{C}$ is easiest to illustrate: take two elliptic curves $E_1$ and $E_2$ with CM by $\mathcal{O}_K$ and a nonprincipal ideal $\mathcal{I}$ respectively, in a quadratic imaginary field $K$ of class number two; then $\mathcal{O}_K^2\cong \mathcal{I}^2$, so $([E_1]-[E_2])([E_1]+[E_2])=[E_1]^2-[E_2]^2=0$ exhibits zero divisors, as $[E_1]\ne [E_2]$ since the Albanese variety is a birational invariant, and elliptic curves are their own Albanese varieties.

I think this same argument shows that $K_0(\text{Mot})$ is not a domain for $\mathbb{Z}$-coefficients, and it shows that our map is not injective for $\mathbb{Q}$-coefficients.

(2) If we restrict to the classes of smooth varieties, and take motives with integral coefficients, is this map injective?

(3) If we take motives with rational coefficients, is the resulting Grothendieck group a domain?

derivedcategory of motives, then sending a variety to its Borel-Moore motive induces $K_0(Var) \to K_0(Mot)$ over any field. The point is that there is a triangle $M(Z) \to M(X) \to M(U)$. This works rationally always (all candidates for $Mot$ are equivalent then). Integrally it works with $Mot$ the category of $H\mathbb Z$-modules (the triangle is then essentially Bloch's localization theorem for higher Chow groups). $\endgroup$ – Marc Hoyois Feb 14 '17 at 3:03