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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 ?

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  • $\begingroup$ I think the field is countable (or finite) if and only if the Grothendieck ring is countable. $\endgroup$ – Evgeny Shinder Mar 2 '19 at 18:16
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    $\begingroup$ A related question is whether every automorphism of $K_0(V_{\mathbb C})$ comes from that of $\mathbb C $. $\endgroup$ – Lev Borisov Mar 6 '19 at 0:42
  • $\begingroup$ 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 $K_0(Var/K)$? $\endgroup$ – Evgeny Shinder Mar 7 '19 at 14:47
  • $\begingroup$ I suspect that these Grothedieck rings are isomorphic for an arithmetically profinite extension of $\mathbb{Q}$ and for its fields of norms. $\endgroup$ – Mikhail Bondarko Jun 26 '19 at 9:21
  • $\begingroup$ @MikhailBondarko : could you describe your ideas in an answer ? $\endgroup$ – THC Jun 27 '19 at 15:00
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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 $[X]-[Y]=0$.

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    $\begingroup$ 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 $k$. $\endgroup$ – Will Sawin Mar 1 '19 at 14:39
  • $\begingroup$ You're probably right that there may be inconsistencies in the definig fields. But I've leave my answer up in case there is a way to fix it. $\endgroup$ – Donu Arapura Mar 1 '19 at 14:46

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