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 $k$, constructed below, have any hope to be a generating class for $K_0(\text{Var}_k)$, and if not, upon calling $B$ the $\mathbf{Z}$-subalgebra of $K_0(\text{Var}_k)$ it generates, same question as in Question 1, for the ring homomorphism $B\to K_0(\text{Var}_k)$.

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 $B\subset K_0(\text{Var}_k)$ generated by the classes of the elements of $\mathcal{P}_1$.

**Question 4.** Does there at least exist a morphism $f: Y\to X$ over $k$, with $Y$ in $\mathcal{P}$, and such that $f$ is surjective/flat/such that $f^*\omega_X\simeq\omega_Y$?

(This last question is expected to have negative answer for $f$ required to be surjective, and a good strategy to see this is suggested in the comments.)