Detecting affine varieties in the Grothendieck ring

Is there anything special about the classes of affine varieties in the Grothendieck ring of varieties (over $$\mathbb{C})$$?. Is there some specialisation that allows us to distinguish classes of affine varieties from general classes?

After R. van Dobben de Bruyn's answer, it might be more interesting to consider the modified version: is there anything special about classes that can be represented by an irreducible smooth affine variety?

• Jouanolou's trick implies (together with the reasoning given by @deBruyn) that the class of any quasi-projective irreducible scheme times the class of affine space is the class of an irreducible affine scheme. Hence, if you invert the class of an affine line, then "everything is affine". May 18, 2020 at 12:05

Being affine is not invariant under scissors relations. In other words, it is possible that $$[X] = [Y]$$ in the Grothendieck ring, where $$X$$ is affine and $$Y$$ is not.

For example, the diagonal $$\Delta \subseteq \mathbf P^1 \times \mathbf P^1$$ is ample, so $$X = \mathbf P^1 \times \mathbf P^1 \setminus \Delta$$ is affine. But $$[X] = [\mathbf P^1 \times \mathbf P^1] - [\mathbf P^1] = (\mathbf L + 1)^2 - (\mathbf L + 1) = \mathbf L^2 + \mathbf L = [\mathbf P^2 - p],$$ and $$\mathbf P^2 - p$$ is not affine for any point $$p \in \mathbf P^2$$.

If you allow $$k$$-schemes with multiple components, every effective class has an affine representative. For example, a class of the form $$[X]$$ can be made affine by cutting $$X$$ into locally closed pieces that are affine (e.g. using Noetherian induction).

It could still be an interesting question which classes have a representative that is irreducible and affine (or irreducible smooth affine, or ...).

• Yes I see, this was a trivial question. Thanks. May 17, 2020 at 16:37
• Is it actually L^2+L on the left hand side and \Bbb{P}^2 - pt on the right?
– Eoin
May 17, 2020 at 16:53
• @Eoin: you're absolutely right! Fixed. May 17, 2020 at 16:56