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Daniel Litt
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The answer is "no" in general, even if one asks if the classes of invertible sheaves generated $K_0(X)$ as a ring; K3 surfaces provide a counterexample. For a K3 surface $X$ over $\mathbb{C}$, the subring of $$K_0(X)_\mathbb{Q}$$ has uncountable dimensional as a $\mathbb{Q}$-vector space; this follows, for example, from the fact that this vector space is isomorphic to the rational Chow ring of $X$ (which was proven to be very large by Mumford). But the subspace generated by line bundles has countable dimension (as the Picard group of $X$ is a finitely generated abelian group of rank at most 20).

Sorry forEDIT: Here's a slightly different argument, with references. Consider any K3 surface $X/k$, with $k$ algebraically closed. Let $K=k(X)$ be the very brief answerfunction field of -- will try to track down references later$X$. Then by Lemma 2.9 here, the map $$K_0(X)_{\mathbb{Q}}\to K_0(X_K)_{\mathbb{Q}}$$ is injective; by Proposition 2.10, it is not surjective. But the span of the invertible sheaves is contained in its image (again by the discreteness of $\text{Pic}(X)$), so we're done. If you'd like an example over an algebraically closed field, you can extend to the algebraic closure of $K$ (which again works by Lemma 2.9 of the Huybrechts reference above).

The answer is "no" in general, even if one asks if the classes of invertible sheaves generated $K_0(X)$ as a ring; K3 surfaces provide a counterexample. For a K3 surface $X$ over $\mathbb{C}$, the subring of $$K_0(X)_\mathbb{Q}$$ has uncountable dimensional as a $\mathbb{Q}$-vector space; this follows, for example, from the fact that this vector space is isomorphic to the rational Chow ring of $X$ (which was proven to be very large by Mumford). But the subspace generated by line bundles has countable dimension (as the Picard group of $X$ is a finitely generated abelian group of rank at most 20).

Sorry for the very brief answer -- will try to track down references later...

The answer is "no" in general, even if one asks if the classes of invertible sheaves generated $K_0(X)$ as a ring; K3 surfaces provide a counterexample. For a K3 surface $X$ over $\mathbb{C}$, the subring of $$K_0(X)_\mathbb{Q}$$ has uncountable dimensional as a $\mathbb{Q}$-vector space; this follows, for example, from the fact that this vector space is isomorphic to the rational Chow ring of $X$ (which was proven to be very large by Mumford). But the subspace generated by line bundles has countable dimension (as the Picard group of $X$ is a finitely generated abelian group of rank at most 20).

EDIT: Here's a slightly different argument, with references. Consider any K3 surface $X/k$, with $k$ algebraically closed. Let $K=k(X)$ be the function field of $X$. Then by Lemma 2.9 here, the map $$K_0(X)_{\mathbb{Q}}\to K_0(X_K)_{\mathbb{Q}}$$ is injective; by Proposition 2.10, it is not surjective. But the span of the invertible sheaves is contained in its image (again by the discreteness of $\text{Pic}(X)$), so we're done. If you'd like an example over an algebraically closed field, you can extend to the algebraic closure of $K$ (which again works by Lemma 2.9 of the Huybrechts reference above).

Source Link
Daniel Litt
  • 23k
  • 5
  • 84
  • 144

The answer is "no" in general, even if one asks if the classes of invertible sheaves generated $K_0(X)$ as a ring; K3 surfaces provide a counterexample. For a K3 surface $X$ over $\mathbb{C}$, the subring of $$K_0(X)_\mathbb{Q}$$ has uncountable dimensional as a $\mathbb{Q}$-vector space; this follows, for example, from the fact that this vector space is isomorphic to the rational Chow ring of $X$ (which was proven to be very large by Mumford). But the subspace generated by line bundles has countable dimension (as the Picard group of $X$ is a finitely generated abelian group of rank at most 20).

Sorry for the very brief answer -- will try to track down references later...