# Is the Grothendieck ring of varieties reduced?

A neat construction of Bjorn Poonen shows that the Grothendieck ring of varieties (over a field of char. 0) is not a domain: http://arxiv.org/abs/math/0204306

Is the Grothendieck ring of varieties reduced? (My guess: the answer is yes, the proof is easy enough that several people have observed this without writing it up anywhere. But I don't know how to show it.)

• In positive characteristic $p$, if you take two supersingular elliptic curves $E_1, E_2$, then $E_i\times E_j$ is isomorphic to $E_1^2$ for any pair $i,j$, so $([E_1]-[E_2])^2=0$. But it is unclear for me why should $[E_1]=[E_2]$. – Qing Liu Sep 9 '10 at 7:41

Qing Liu's example probably works, only we don't know if an abelian variety in positive characteric is determined by its class in $K_0(\mathrm{Var}_k)$. However, we do know that in characteristic zero (this is what Bjorn Poonen uses in his examples) and the non-cancellation is a purely arithmetic phenomenon and hence can be realised in characteristic zero.
Hence, we let $\mathcal A$ be a maximal order in a definite (i.e., $\mathcal A\otimes\mathbb R$ is non-split) quaternion algebra over $\mathbb Q$. There is an abelian variety $A$ over some field $k$ of characteristic zero with $\mathcal A=\mathrm{End}(A)$ (Bjorn works hard to get his example defined over $\mathbb Q$, here I make no such claim). For any (right) f.g. projective (i.e., torsion free) $\mathcal A$-module $M$ we may define an abelian variety $M\bigotimes_{\mathcal A}A$ characterised by $\mathrm{Hom}(M\bigotimes_{\mathcal A}A,B)=\mathrm{Hom}_{\mathcal A}(M,\mathrm{Hom}(A,B))$ for all abelian varieties (concretely it is constructed by realising $M$ is the kernel of an idempotent of some $\mathcal A^n$ and then taking the kernel of the same idempotent acting on $A^n$). In any case we see that $M$ and $N$ are isomorphic precisely when $M\bigotimes_{\mathcal A}A$ is isomorphic to $N\bigotimes_{\mathcal A}A$.
Now (all the arithmetic results used below can be found in for instance Irving Reiner: Maximal orders, Academic Press, London-New York), the class group of $\mathcal A$ is equal to the ray class group of $\mathbb Q$ with respect to the infinite prime, i.e., the group of fractional ideals of $\mathbb Q$ modulo ideals with a strictly positive generators. As that is all ideals we find that the class group is trivial. Furthermore, we have the Eichler stability theorem which says that projective modules of rank $\geq2$ are determined by their rank and image in the class group and hence are determined by their rank (the rank condition comes in in that $\mathrm{M}_k(\mathcal A)$ is a central simple algebra which is indefinite at the infinite prime). In particular if $M_1$ and $M_2$ are two rank $1$ modules over $\mathcal A$ and $A_1$ and $A_2$ are the corresponding abelian varieties we get that $A_1\bigoplus A_2\cong A\bigoplus A$ as the left (resp. right) hand side is associated to $M_1\bigoplus M_2$ (resp. $\mathcal A^2$). Therefore, to get an example it is enough to give an example of an $\mathcal A$ for which there exist $M_1\not\cong M_2$. The number (or more easily the mass) of isomorphism classes of ideals can be computed using mass formulas and tends to infinity with the discriminant of $\mathcal A$. It is interesting to note that when the discriminant is a prime $p$ we can go backwards using supersingular elliptic curves: The mass is equal to the mass of supersingular elliptic curves in characteristic $p$ and the latter mass can be computed geometrically to be equal to $(p-1)/24$.
• Very nice example ! So in characteristic 0, the Grothendieck ring of varieties is not reduced (at least over some field $k$). Probably neither in postive characteristic. – Qing Liu Sep 27 '10 at 20:37
• Just a clarification: $A$ exists over some number field and hence over all larger fields. – Torsten Ekedahl Sep 28 '10 at 4:23