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Let $k$ be a field, and consider the Grothendieck ring of $k$-varieties, $K_0(V_k)$. Let $K/k$ and $K'/k$ be field extensions of $k$. We view $\mathrm{Spec}(K)$ and $\mathrm{Spec}(K')$ as $k$-schemes and consider their classes $[\mathrm{Spec}(K)]$ and $[\mathrm{Spec}(K')]$ in $K_0(V_k)$.

I have two general questions:

(A) what are the most interesting properties/criteria which lead to the equality $[\mathrm{Spec}(K)] = [\mathrm{Spec}(K')]$ ?

(B) when can we decide that $[\mathrm{Spec}(K)] \ne [\mathrm{Spec}(K')]$ ?

I understand that some of the first properties will be pretty easy or expected (but still interesting to mention), but also that the "next generation" of properties could be highly interesting.

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    $\begingroup$ You're using the notation $[]$ without defining it, and you're defining $K_0$ without using it. $\endgroup$ – YCor Mar 11 at 16:58
  • $\begingroup$ @YCor : I have no idea what you mean ... $\endgroup$ – THC Mar 11 at 18:07
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    $\begingroup$ I removed the tag "spectral theory" and you added it again. Have you seen what questions are asked with this tag? Have you seen what are the arxiv papers with this tag? The usage guidance of this tag is Schrodinger operators, operators on manifolds, general differential operators, numerical studies, integral operators, discrete models, resonances, non-self-adjoint operators, random operators/matrices. Putting this tag just spams people not interested with this stuff... $\endgroup$ – YCor Mar 12 at 0:34
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    $\begingroup$ I can try to explain why the equality of classes implies that the (Artin) motives of these fields over k are isomorphic. Unfortunately, it does not follow that the fields itself are isomorphic; thus I doubt that the converse implication is vald. Are you interested? $\endgroup$ – Mikhail Bondarko Mar 12 at 19:00
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    $\begingroup$ Actually, if k is of characteristic 0 then my claim is an easy application of the results of Gillet and Soule. Would you like me to turn this into an answer? I can also prove the statement if k is of characteristic $p>0$; yet then I will have to cite my own results.:) $\endgroup$ – Mikhail Bondarko Mar 12 at 19:15
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In characteristic zero $[\mathrm{Spec}(K)] = [\mathrm{Spec}(K')]$ for finite field extensions of $k$ implies that $K$ and $K'$ are isomorphic.

Indeed, by the Larsen-Lunts theorem for smooth projective connected schemes of finite type $[X] = [Y]$ implies that $X$ and $Y$ are stably birational; this applies to $\mathrm{Spec}(K)$, $\mathrm{Spec}(K')$ as they are connected smooth and projective. Now if $\mathrm{Spec}(K)$, $\mathrm{Spec}(K')$ are stably birational, then they are isomorphic. This is because if $X$ is smooth projective and stably birational to $\mathrm{Spec}(K)$, then $K = \Gamma(X, \mathcal{O}_X)$. Same argument applies to products of fields that is to reduced zero-dimensional finite $k$-schemes: here the Larsen-Lunts theorem will match up the stable birational types of the connected components.

In characteristic $p > 0$ the result may be still true but hopeless to prove without resolution of singularities.

UPDATE: the result above about fields is given in Proposition 5 in a paper by Liu and Sebag where they study what $[X] = [Y]$ implies in general in characteristic zero, using Larsen-Lunts Theorem: https://link.springer.com/article/10.1007/s00209-009-0518-7

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  • $\begingroup$ One question I would like to know the answer myself is: what if $\mathbf{L}^n\cdot([\mathrm{K}]-[\mathrm{K'}]) = 0$? Are $K$, $K'$ isomorphic then? Here $\mathbf{L} = [\mathbf{A}^1]$. $\endgroup$ – Evgeny Shinder Mar 12 at 21:15
  • $\begingroup$ I would guess that this follows by looking at the etale Euler characteristic. $\endgroup$ – naf Mar 13 at 3:36
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    $\begingroup$ @ulrich: What exactly do you mean? I think all the `cohomological' data (Artin motives, Galois modules) which can be extracted from $[X]$ will not be able to distinguish fields. $\endgroup$ – Evgeny Shinder Mar 13 at 8:10
  • $\begingroup$ OK, I suppose I was wrong. What I was suggesting (looking at the Artin motive) works if one of the fields is Galois, but not otherwise. $\endgroup$ – naf Mar 15 at 4:47

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