# An inverse problem for Grothendieck rings of varieties

Suppose $$A$$ is a given commutative ring, and suppose that one knows that $$A$$ is isomorphic to the Grothendieck ring of $$k$$-varieties for some unknown field $$k$$.

Can $$k$$ be recovered from $$A$$ ? If not, what about the characteristic of $$k$$ ?

A related question is obviously the following: if $$k$$ and $$k'$$ are nonisomorphic fields, can the Grothendiek rings $$K_0(V_k)$$ and $$K_0(V_{k'})$$ be isomorphic ?

• I think the field is countable (or finite) if and only if the Grothendieck ring is countable. – Evgeny Shinder Mar 2 '19 at 18:16
• A related question is whether every automorphism of $K_0(V_{\mathbb C})$ comes from that of $\mathbb C$. – Lev Borisov Mar 6 '19 at 0:42
• And another one: if $L/K$ is a Galois extension with Galois group $G$, what does $K_0(Var/L)^G$ have to do with $K_0(Var/K)$? – Evgeny Shinder Mar 7 '19 at 14:47
• I suspect that these Grothedieck rings are isomorphic for an arithmetically profinite extension of $\mathbb{Q}$ and for its fields of norms. – Mikhail Bondarko Jun 26 '19 at 9:21
• @MikhailBondarko : could you describe your ideas in an answer ? – THC Jun 27 '19 at 15:00

I haven't written a complete proof, but I expect your last (and therefore) first question to have a negative answer. Here is what I believe is a counterexample. Let $$k$$ be the algebraic closure of $$\mathbb{Q}(x_1,x_2,\ldots)$$ (a countable number of variables). We can view this as a subfield of $$\mathbb{C}$$. We have a homomorphism $$e:K_0(V_k)\to K_0(V_\mathbb{C})$$ given by sending the symbol of a variety $$[X]$$ to $$[X\otimes \mathbb{C}]$$. This is seen to be surjective because any complex variety can be defined over a finitely generated field, and therefore over $$k$$. I expect this to be injective as well. It would be enough to show that for (reducible) $$k$$-varieties $$X$$ and $$Y$$, $$[X]-[Y]=0$$ when it lies in the kernel of $$e$$. If $$e([X]-[Y])=0$$, we have finite partitions into locally closed sets $$X\otimes \mathbb{C}= \cup X_i$$ and $$Y\otimes \mathbb{C}= \cup Y_i$$ such that $$X_i\cong Y_i$$. As before, the partitions and isomorphisms should be definable over $$k$$, so $$[X]-[Y]=0$$.
• I don't think this will work, because each complex variety is defined over a different finitely generated field, and they won't all be contained in the image of $k$. – Will Sawin Mar 1 '19 at 14:39