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?

  • 2
    $\begingroup$ If you replace $Mot$ by a suitable derived category 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). $\endgroup$ Feb 14, 2017 at 3:03
  • $\begingroup$ Dear Marc, I have looked at your comment with suspicion for some time. Now I have realized that the arrows in your triangle should actually be reversed. Thus it makes sense to pass to (Verdier) duals to obtain the usual triangle of motives with compact support. $\endgroup$ Feb 15, 2017 at 18:40

2 Answers 2


(1) As pointed out by Marc Hoyois, there is a natural homomorphism $K_0(Var)\to K_0(DM^{gm})$, where $DM^{gm}$ is the category of geometric Voevodsky motives with coefficients in any (commutative unital) ring $R$; you should only assume (at our current level of knowledge on the resolution of singularities) that the base field characteristic $p$ is invertible in $R$ (to put Borel-Moore motives into $DM^{gm}$). Next, the Voevodsky embedding $Chow\to DM^{gm}$ induces an isomorphism on (the corresponding versions of) their $K_0$-groups; see Proposition 2.3.3 of my "$\mathbb{Z}[1/p]$-motivic resolution of singularities".

(3) I don't know.:)

On the one hand, one can easily prove that the "(more or less) standard" motivic conjectures predict that $K_0(Chow)$ is the free abelian group generated by isomorphism classes of indecomposable numerical motives. Next one can try to proceed using the Tannkian formalism.

On the other hand, one can possibly find zero divisors using an argument similar to Lemma 3 in Poonen's https://arxiv.org/pdf/math/0204306.pdf.

  • $\begingroup$ Thanks a lot! So your result defines the morphism to the motivic Grothendieck group in very high generality, then. I'm not very familiar with motivic homotopy, but I'll give it a read. I tried thinking about what the polarized Tannakian structure buys us, but struggled; it's hard to see how anything can come easily from the formalism, since certainly twisting by a general simple motive will not be full, and trying to detect whether there is an isomorphism seems to require real geometry. I'm not sure if an abelian varieties counterexample can work; as isogenies become isomorphisms. $\endgroup$
    – peterx
    Feb 16, 2017 at 2:22
  • $\begingroup$ You probably don't get an example with abelian varieties for rational coefficients. Yet you can try to think whether you can have $X^2\cong Y^2$ with $X\not\cong Y$ in a Tannakian category. $\endgroup$ Feb 17, 2017 at 9:55

Answer to question #2 is no. (Also I don't know what I meant by "restrict to classes of smooth varieties" since these generate the ring in the only case where we know how to define the map.) $[\mathbb{A}^1]$ is a zero-divisor in $K_0(\text{Var})$, due I think to Borisov (though I'd also like to take the excuse to mention Zakharevich's incredible paper where she computes the kernel of multiplication by $[\mathbb{A}^1]$ using her K-theory of assemblers). On the other hand, Lefschetz twisting is fully faithful, which is a classical result that can be verified explicitly; a reference is Manin's early paper on the motive of a blow-up.


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