If $K(X) \cong \mathbb{Z}^n$ what is $n$? Let $X$ be a smooth projective algebraic variety over $\mathbb{C}$, suppose Grothendieck group $K(X)$ is a free abelian group $K(X) \cong \mathbb{Z}^n$. Is it true that rank $n$ is the topological Euler characteristic $\chi(X)$ of $X$?
I saw this claim in a paper for two dimensional case $\dim X = 2$, but it was without proof. 
 A: Here is a partial answer.
Define an affine stratification of a variety $X$ as a decomposition of $X$ into a finite disjoint union of affine spaces $X_i=\Bbb A^{n_i}$ so that the closure of $X_i$ is again a disjoint union of $X_j$s. Then it is known result that any $X$ which admits an affine stratification has a free Grothendieck group, freely generated by the classes of $\mathcal{O}_{\overline{X_i}}$. (I'm pretty sure this argument is in Eisenbud & Harris' book on intersection theory, but my copy is packed away in a box right now and I can't get at it).
I don't know that this proposition is an equivalence (going backwards seems difficult, I would think), but in the case where your variety does in fact have an affine stratification, it provides exactly the result you're looking for. Calculate the topological Euler characteristic by treating each $\Bbb A^{n_i}$ as a cell homeomorphic to $\Bbb R^{2n_i}$, then check that the free generating set is in bijective correspondence with the set of even-dimensional cells. Since there are no odd-dimensional cells, one sees that the Euler characteristic exactly matches the number of generators.
