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.)



Does the Grothendieck ring of varieties contain torsion?
As per Theo Johnson-Freyd's request, I'm converting my comment to an answer.


Larsen-Lunts [show](http://arxiv.org/pdf/math/0110255.pdf) 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](http://arxiv.org/abs/1412.6194), 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](http://arxiv.org/abs/1209.4708)," and I think it's quite hard). So this question is almost certain to be open for $k$ algebraically closed of characteristic zero.


