# Does the Grothendieck ring of varieties contain torsion?

Let $K_0(Var_k)$ be the abelian group generated by the isomorphism classes of varieties over the field $k$ with the relations $$[X]=[U]+[X\setminus U]$$ for every variety $X$ and open subvariety $U$.

Is this abelian group torsion free? (Any positive or negative result for any particular field $k$ is welcome.)

• Just a remark: over an algebraically closed field of char. $0$, Larsen and Lunts show that $K_0(\text{Var}_k)/\mathbb{L}$ is torsion-free (here $\mathbb{L}=[\mathbb{A}^1]$). So any example of something annihilated by an integer $n$ has to be in the ideal $(\mathbb{L})$. Thus any example would give a proof that $\mathbb{L}$ is a zero-divisor; it is one, but this was open until very recently (it is a result of Borisov). So I think this is very likely to be open. – Daniel Litt Dec 8 '15 at 19:10
• @DanielLitt If in a day or two there aren't other answers, can I request that you upgrade your comment to an answer (especially with links to the work by Larsen--Lunts and Borisov)? I think it pretty much answers the question. – Theo Johnson-Freyd Dec 9 '15 at 1:55
• @TheoJohnson-Freyd: Sure, though hopefully someone has more to say than what I've written. – Daniel Litt Dec 9 '15 at 3:04

Larsen-Lunts show that if $k$ is algebraically closed of characteristic zero, then there is a natural isomorphisms $$K_0(\text{Var}_k)/(\mathbb{L})\overset{\sim}{\longrightarrow} \mathbb{Z}[SB],$$ where $SB$ is the monoid of stable birational equivalence classes of smooth birational varieties and $\mathbb{L}=[\mathbb{A}^1]$. (There is a way of making sense of this if $k$ is not algebraically closed, but the target will no longer be a monoid ring, so the argument I am making will break.)
$\mathbb{Z}[SB]$ is manifestly torsion-free; thus if $nX=0$ for $n\in \mathbb{Z}$ and $X\in K_0(\text{Var}_k)$, we must have $X\in (\mathbb{L})$. Thus either we would have an example where $\mathbb{L}$ is a zero divisor, or a proof that $$\bigcap_n (\mathbb{L}^n)\neq 0.$$ $\mathbb{L}$ was recently shown to be a zero divisor by Borisov, but this was open for a long time; as far as I know, the question of whether or not $$\bigcap_n (\mathbb{L}^n)=0$$ is open as well (I've thought about it a bit, because I used the $\mathbb{L}$-adic topology on $K_0(\text{Var}_k)$ in my paper "Symmetric Powers do not Stabilize," and I think it's quite hard). So this question is almost certain to be open for $k$ algebraically closed of characteristic zero.
• Hm. This is not completely convincing, is it? You're saying "If the Grothendieck ring contained torsion then this would resolve problems known to be hard, so the problem is surely open". But it's also possible that the Grothendieck ring does not contain torsion; this could very well be easier to show than Hausdorffness of the $\mathbb L$-adic topology, right? – Dan Petersen Dec 10 '15 at 9:25
• I seem to recall that Zakharevic had a theorem along the lines of: The kernel of multiplication by $\mathbb L$ is generated by examples like the one by Borisov. I wonder if that has any implications here. – Gjergji Zaimi Dec 10 '15 at 17:29