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.)

  • 16
    $\begingroup$ 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 $[E_1]=[E_2]$. $\endgroup$
    – Qing Liu
    Sep 9, 2010 at 7:41

1 Answer 1


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 $(p-1)/24$.

  • 3
    $\begingroup$ 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. $\endgroup$
    – Qing Liu
    Sep 27, 2010 at 20:37
  • 2
    $\begingroup$ Just a clarification: $A$ exists over some number field and hence over all larger fields. $\endgroup$ Sep 28, 2010 at 4:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.