Let $X$ be a smooth projective variety (e.g. Grassmannians). Since $X$ is smooth, the groups $G_0(X):=K_0(CohX)$ and $K_0(X):=K_0(VectX)$, the Grothendieck groups of coherent sheaves of modules on $X$ and the Grothendieck group of vector bundles on $X$, are isomorphic.
My question, when is $K_0(X)$ free? For example, for projective space of dimension $n$, this group is free and of rank $n$.
Even for elliptic curves this isn't true -- for example there are line bundles $\mathcal{L}$ whose square is the structure sheaf (in fact, these are classified by an étale cohomology group, so there's some interesting math going on here). I am no expert, but I think this sort of thing can happen widely in general.
One interesting case where this does not happen is when $X$ has an "affine stratification", that is, a decomposition $X = \bigsqcup X_i$, where each $X_i$ is an affine space $\mathbb{A}^{n_i}$, and the closure of each $X_i$ is a union of other $X_i$'s. This is the case for projective spaces, Grassmannians, flag varieties, and products and blowups of these -- so a fair number of common (but simple) spaces. In this setting, $K_0(X)$ is freely generated by the classes of the structure sheaves $\mathcal{O}_{\overline{X_i}}$.