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Let $k$ be a field, and let $K_0(V_k)$ be the Grothendieck ring of $k$-varieties. Let $K_0(V_k^0)$ be the Grothendieck ring of $0$-dimensional $k$-varieties. I also assume that $k$ is perfect.

(1) I have read that $K_0(V_k^0)$ is generated by the classes $[\mathrm{Spec}(K)]$ with $K$ a finite field extension of $k$.

(2) I have also read that $K_0(V_k^0)$ is the free $\mathbb{Z}$-module generated by the isomorphism classes of transitive permutation representations of $\mathrm{Gal}(\overline{k}/k)$.

As I am come from far outside Galois theory or cohomology theories, I wondered why both statements are true, and if someone could provide some clear references. What happens if $k$ is not perfect ?

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    $\begingroup$ (1) is easy: a $0$-dimensional scheme is a finite union of points, and $[Z^{\operatorname{red}}] = [Z] \in K_0(\mathbf{Var}_k)$ for any $k$-scheme $Z$. Since isomorphism classes of finite transitive $\operatorname{Gal}(\bar k/k)$-sets is the same thing as finite separable field extensions, at least this shows it's generated by such. I'm not sure why there are no relations. $\endgroup$
    – R. van Dobben de Bruyn
    Commented Jun 3, 2020 at 16:46
  • $\begingroup$ @R.vanDobbendeBruyn : "Since isomorphism classes of finite transitive Gal(𝑘¯/𝑘)-sets is the same thing as finite separable field extensions" -- could you explain this in more detail ? (Thanks !!) -- THC $\endgroup$
    – THC
    Commented Jun 3, 2020 at 17:05
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    $\begingroup$ This is Galois theory. To a separable field extension $k \to \ell$, associate the set $\operatorname{Hom}_k(\ell, \bar k)$, which has a natural Galois action by postcomposition. Conversely, a finite transitive $\operatorname{Gal}(\bar k/k)$-set is of the form $G/H$ for some closed subgroup $H \subseteq G = \operatorname{Gal}(\bar k/k)$, so take $\ell = (\bar k)^H$. Galois theory says that these constructions are inverses. $\endgroup$
    – R. van Dobben de Bruyn
    Commented Jun 3, 2020 at 17:07
  • $\begingroup$ @R.vanDobbendeBruyn : does $(\overline{k})^H$ denote the fixed elements subfield in $\overline{k}$, of $H$ ? $\endgroup$
    – THC
    Commented Jun 4, 2020 at 15:31
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    $\begingroup$ In characteristic zero, classes of fields are linearly independent by the Larsen-Lunts theorem, as in the answer here: mathoverflow.net/questions/354718/… $\endgroup$
    – Evgeny Shinder
    Commented Jun 5, 2020 at 22:12

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