Here's a problem that may ultimately require just simple algebraic-geometry skills to be solved, or perhaps it's very deep and will never be solved at all. From the comments, some literature and my memory it appears this was posed by Grothendieck as part of the big program of motives.
Consider classes of complex algebraic varieties X modulo relations
[X] - [Y] = [X\Y],
[X x Y] = [X] x [Y],
Also, if you're familiar with taking inverse of an affine line, let's do that too:
$$ \exists \mathbb A^{-1}\quad \text{such that}\quad [\mathbb A] \cdot [\mathbb A^{-1}] = [\mathbb A^0].$$
(+ if you want, you can also take idempotent completion and formal completion by A^-1).
It's not hard to see that you can add (formally) and multiply (geometric product as above) those things, so they form a ring. Let's denote this ring Mot (It's actually very close to what Grothendieck called baby motives.)
And for things that form a ring you can study their Spec. For example, you can talk about points of the ring — each point is by definition a homomorphism to complex numbers.
Question: what are the properties of
Spec Mot? How to describe its points?
For example, one point is Euler characteristics $\chi \in \text{Spec}\,\mathbf{Mot}$, since it's additive and multiplicative (it's even integral!) Any other homomorphism to complex numbers is thus sometimes called generalized Euler characteristics.
There's also a plane there given by mixed Hodge polynomials (that is, polynomials whose coefficients are weighted Hodge numbers $h^{p,q}_k$), since Hodge polynomial at a given point satisfies those relations too (see the references below).
As Ben says below, things would become even more interesting if we considered this ring for schemes over $\mathbb Z$, because then each $q$ would give a generalized Euler characteristic $\chi_q$ that counts points of $X(\mathbb F_q).$
Are there any other points? Any more information?
Best Answer
One interesting fact about Spec M is that it isn't integral; i.e., the ring M has zero divisors. This was proved by Poonen in 2002:
"The Grothendieck ring of varieties is not a domain"
Re points of Spec M: I suppose if you considered varieties over R instead of C, you would in addition have the map sending X to the Euler characteristic of X(R), though I've never seen this used.
Update: Oh, I've never seen this used because it's totally wrong. For instance, A^0(R) and A^1(R) have Euler characteristic 1 but P^1(R) doesn't have Euler characteristic 2. I think the mod-2 Euler characteristic would probably be OK here.