Let $k$ be an algebraically closed field of characteristic zero.

Let $A$ be the $\mathbf{Z}$-subalgebra of the Grothendieck ring of $k$-varieties $K_0(\text{Var}_k)$ generated by classes of semi-abelian varieties.

Question 1. What can we say about the ring homomorphism $A\to K_0(\text{Var}_k)$?

It is obviously very far from being surjective. Mainly, is it integral/finite/flat/essentially of finite type?

Question 2. The class of smooth projective varieties over $k$ generates $K_0(\text{Var}_k)$ in the obvious sense. Is there a more manageable generating class of $k$-varieties, possibly containing semi-abelian varieties at the very least?

Question 3. In light of Question 2, does the class $\mathcal{P}$ of varieties over $k$ have any hope to be a generating class for $K_0(\text{Var}_k)$?

We call $\mathcal{P}_0$ the class of $k$-varieties whose objects are:

• abelian varieties
• semi-abelian varieties (extension of abelian varieties by a torus)
• smooth hypersurfaces in large projective spaces over $k$
• smooth projective varieties of dimension $\le 3$
• smooth projective toric varieties
• Calabi-Yau varieties
• products of the above

Let $\mathcal{P}_1$ be the class of $k$-varieties whose objects are obtained by blowing up an element of $\mathcal{P}_0$ along a closed $k$-subvariety that is again an element of $\mathcal{P}_0$.

Finally, let $\mathcal{P}$ be the class of all those $k$-varieties such that their class in the Grothendieck ring of varieties $K_0(\text{Var}_k)$ belongs to the subring $R\subset K_0(\text{Var}_k)$ generated by the classes of the elements of $\mathcal{P}_1$.

(1) Is the class of $X$ in the Grothendieck ring of $k$-varieties, the same as that of an element of $\mathcal{P}$? In other words, is $R = K_0(\text{Var}_k)$?

(2) Does there exist a surjection $Y\to X$ over $k$, with $Y$ in $\mathcal{P}$?

(3) Is any of the two questions above known to have a partial answer?

Examples. $\mathcal{P}$ contains all the classes mentioned above, and in addition, for instance, all rational $k$-varieties. Obviously abelian varieties are far from being enough, for (1) and (2) to have a positive answer.