Torsion-free coherent modules on a smooth curve are locally free

Local freeness implies torsion-freeness. Suppose $\mathcal{M}$ is finite locally free and let $U$, $m\mathrm{\ne }0$, $a\mathrm{\ne }0$ satisfy $am=0$. Since $m\mathrm{\ne }0$, choose $x\in U$ with $m_x\mathrm{\ne }0$. By [F1], the nonzero section $a$ maps to a nonzero element of $k(C)$, and its germ $a_x$ maps to that same element, so $a_x\mathrm{\ne }0$. Shrink to an open neighbourhood of $x$ on which $\mathcal{M}$ is free. The relation gives $a_x{m}_x=0$ in a free module over the domain ${\mathcal{O}}_{C,x}$; multiplication by the nonzero scalar $a_x$ is injective coordinatewise, forcing $m_x=0$, a contradiction. So a finite locally free module has no such $a,m$.

Torsion-freeness implies free stalks. Suppose $\mathcal{M}$ has no relation $am=0$ as in the statement, and let $x\in C$. The stalk ${\mathcal{M}}_x$ is a finitely generated module over the principal ideal domain ${\mathcal{O}}_{C,x}$ (a discrete valuation ring when $x$ is closed, the field $k(C)$ when $x$ is the generic point) [F1], and it is torsion-free: if $a_x{m}_x=0$ in ${\mathcal{M}}_x$ with $a_x\mathrm{\ne }0$ and $m_x\mathrm{\ne }0$, then by [F3] the relation is represented over an open $U$ by nonzero sections $a,m$ with $am=0$, contradicting the hypothesis. By [F2] the stalk ${\mathcal{M}}_x$ is free, say of rank $r_x\ge 0$.

Part (2). Suppose $am=0$ with $U$ open, $m\in \mathcal{M}(U)$ nonzero and $a\in {\mathcal{O}}_C(U)$ nonzero. Choose $x\in U$ with $m_x\mathrm{\ne }0$, and then an affine open neighbourhood $W=\mathrm{Spec}A$ with $x\in W\subseteq U$. The restriction $a{\mathrm{\mid }}_W$ is nonzero by [F1]. Let $Z=\{y\in W:a_y\in {\mathfrak{m}}_y\}$, the vanishing locus of the residue of $a{\mathrm{\mid }}_W$; it is the proper closed subset $V(a{\mathrm{\mid }}_W)$ of $W$, proper because $a{\mathrm{\mid }}_W$ has nonzero image in the function field [F1]. Since $W$ is an integral curve, [F7] says that $Z$ is a finite set of closed points of $W$. None is the generic point of $C$, and [F7] also says every such point is closed in $C$, so $Z$ is a finite closed subset of $C$. Put $V=C\setminus Z$. Then $V$ is open and $W\cup V=C$. On $W\cap V=W\setminus Z$, the residue of $a$ is nonzero at every point, hence $a$ is a unit in every stalk and $m{\mathrm{\mid }}_{W\cap V}=0$. The sections $m{\mathrm{\mid }}_W$ and $0$ over $V$ agree on the intersection, so by [F5] they glue to a global section of $\mathcal{M}$, which is nonzero because $m_x\mathrm{\ne }0$.

Conclusion of (1). If $\mathcal{M}$ is torsion-free then by step 1.2 every stalk is free; by [F4] each point has an open neighbourhood on which $\mathcal{M}$ is free of the rank of its stalk, so $\mathcal{M}$ is finite locally free. With step 1.1 this proves the equivalence (1).

Part (3). A subsheaf $\mathcal{N}\subseteq \mathcal{E}$ of a locally free module $\mathcal{E}$ is torsion-free: a relation $am=0$ with $m\mathrm{\ne }0$ and $a\mathrm{\ne }0$ in $\mathcal{N}$ is also a relation in $\mathcal{E}$, which is impossible by step 1.1; and a coherent torsion-free module is finite locally free by step 2.1.

Conclusion. Step 2.1 proves (1), step 1.3 proves (2) and step 3.1 proves (3). The points chosen in steps 1.1 and 1.3 exist because the corresponding sections are nonzero; the Axiom of Choice enters only through the suppliers recorded in [F6].