Given a smooth, projective (complex) varieties $X$, is it true that the grothendieck group $K\_0(X)$ of equivalence classes of coherent sheaves on $X$, is generated by clases of invertible sheaves i.e., any class of a coherent sheaf on $X$ can be written as a linear combination of classes of invertible sheaves on $X$? Any reference/idea will be most welcome.



Grothendieck group generated by classes of invertible sheaves
Here is a different, perhaps more elementary example.


Consider the Grassmannian $\mathrm{Gr}(2,4)$ of lines in $\mathbf{P}^3$, and let $\mathscr{Q}$ be the universal quotient bundle. The line bundle $\mathrm{det}(\mathscr{Q})$ generates the Picard group of $\mathrm{Gr}(2,4)$. If $[\mathscr{Q}]$ were in the subgroup of $\mathrm{K}^0$ generated by line bundles, then we would have $[\mathscr{Q}]=\sum\_i n\_i [\mathrm{det}(\mathscr{Q})^{\otimes r\_{i}}]$ for some integers $n\_i, r\_i$. Taking Chern classes implies
$$c\_2(\mathscr{Q})= m c\_1(\mathscr{Q})^2$$
for some integer $m$, which cannot be true. Indeed, fix a point $p$ and a line $L$ in $\mathbf{P}^3$; the Chern class $c\_1(\mathscr{Q})$ is geometrically represented by the Schubert cycle $\Sigma\_1(L)=\{L' \ | \ L\cap L'\neq\emptyset\}$, while $c\_2(\mathscr{Q})$ is represented by $\Sigma\_2(p)=\{L' \ | \ p\in L'\}$. Then
$$1=\int\_{\mathrm{Gr}(2,4)} c\_2(\mathscr{Q})^2\neq m^2\int\_{\mathrm{Gr}(2,4)} c\_1(\mathscr{Q})^4=2m^2.$$
The integral on the left is the number of lines through two distinct points in $\mathbf{P}^3$, while the integral on the right is the number of lines meeting four given (general) lines in $\mathbf{P}^3$.


I think only few varieties have the property which you want. It is certainly true for curves: if you have a locally free sheaf $\mathscr{E}$ on a curve, then you can find an exact sequence of locally free sheaves
$$0\rightarrow\mathscr{E}'\rightarrow\mathscr{E}\rightarrow\mathscr{E}''\rightarrow 0,$$
where the ranks of $\mathscr{E}'$ and $\mathscr{E}''$ are strictly smaller than $\mathscr{E}$ (unless, of course, $\mathscr{E}$ is already of rank $1$).
The same property holds if $\mathscr{E}$ is a locally free sheaf of rank $>2$ on a surface. Indeed, after twisting with an invertible sheaf, we may assume that $\mathscr{E}$ is globally generated; by a standard result (e.g. Exercise 8.2 in Chapter 2 of Hartshorne) we can then find a nonzero section $s:\mathscr{O}\rightarrow\mathscr{E}$ whose cokernel is locally free. The K-theory of surfaces can thus be generated by locally free sheaves of rank $1$ and $2$, and analogous statements hold in higher dimension.


