Consider the Grothendieck ring $K_0(\mathcal{V}_k)$ of $k$-varieties. Then for any $k$-variety $X$, and closed subset $C$ of $X$, we have the relation $$[X] = [X \setminus C] + [C],$$ where $[X]$ means the isomorphism class of $X$. Now suppose, for example, that $\mathbf{P}$ is a projective $k$-space of some dimension $n \geq 2$, and let $C$ and $C'$ be finite sets of closed points in $\mathbf{P}$ of the same order. Then we have $$[\mathbf{P} \setminus C] + [C] = [\mathbf{P} \setminus C'] + [C'].$$

Are $C$ and $C'$ considered to be isomorphic $k$-varieties ? If so, we obtain

$$[\mathbf{P} \setminus C] = [\mathbf{P} \setminus C'],$$

which seems pretty strange, taken that, for instance, $C$ could be a set of rational points on some line, and $C'$ a set of rational points in general position. Or should the isomorphism class of (for instance) a finite point set be understood differently ?

And if the answer of the first question above is "yes," is there a way to distinguish between such sets in the context of Grothendieck rings ? What would be a/the cohomological interpretation of $$[\mathbf{P} \setminus C] = [\mathbf{P} \setminus C']?$$

  • 2
    $\begingroup$ If $C$ is isomorphic to $C'$, then for every field extension $L/k$, then the set of $L$-rational points $C(L)$ is in bijection with $C'(L)$. Even if $C$ and $C'$ have the same length as $k$-schemes, it can easily happen that for some $L/k$, the sets $C(L)$ and $C'(L)$ do not have the same number of points. Also, the classes of $C$ and $C'$ depend only on the underlying reduced subscheme, so that is another aspect of this. $\endgroup$ – Jason Starr Sep 13 '17 at 15:52

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