Is there a coherent sheaf which is not a quotient of locally free sheaf?

Suppose $X$ is an algebraic variety, is there a coherent sheaf $\mathcal{F}$ on $X$ which is not a quotient of locally free sheaf？

(Hartshorne II Cor 5.18 showed that on every projective variety, coherent sheaves are quotient of locally free sheaves of finite rank, by extension of coherent sheaves, the same holds for quasi-projective varieties.)

There are smooth schemes which don't have this property, usually called resolution property. A simple example is $X=\mathbb{A}^n \cup_{\mathbb{A}^n \setminus \{0\}} \mathbb{A}^n$, the affine $n$-space with a double origin, and $n \geq 2$. A locally free sheaf on $X$ pulls back to two locally free sheaves on $\mathbb{A}^n$ which become isomorphic when restricted to $\mathbb{A}^n \setminus \{0\}$. Since every locally free sheaf on $\mathbb{A}^n$ is free (Quillen-Suslin Theorem), the isomorphism corresponds to an invertible matrix of global sections of $\mathbb{A}^n \setminus \{0\}$, which by Hartog's Lemma coincide with the global sections of $\mathbb{A}^n$ (here we use $n \geq 2$). Hence, the isomorphism of the two free sheaves extends on $\mathbb{A}^n$, which means that the original locally free sheaf on $X$ is free. It follows that not every coherent sheaf on $X$ is a quotient of a locally free i.e. free sheaf, because otherwise $\mathcal{O}_X$ would be ample and therefore $X$ would be separated.
• Sorry I have a small question: If every coherent sheaf on $X$ is a quotient of a locally free, how does it imply $\mathcal{O}_X$ be ample? Thanks! – Qixiao Jun 29 '14 at 6:24