Lemma 31.11.11. Let $X$ be an integral regular scheme of dimension $\leq 1$. Let $\mathcal{F}$ be a coherent $\mathcal{O}_ X$-module. The following are equivalent

1. $\mathcal{F}$ is torsion free,

2. $\mathcal{F}$ is finite locally free.

Proof. It is clear that a finite locally free module is torsion free. For the converse, we will show that if $\mathcal{F}$ is torsion free, then $\mathcal{F}_ x$ is a free $\mathcal{O}_{X, x}$-module for all $x \in X$. This is enough by Algebra, Lemma 10.78.2 and the fact that $\mathcal{F}$ is coherent. If $\dim (\mathcal{O}_{X, x}) = 0$, then $\mathcal{O}_{X, x}$ is a field and the statement is clear. If $\dim (\mathcal{O}_{X, x}) = 1$, then $\mathcal{O}_{X, x}$ is a discrete valuation ring (Algebra, Lemma 10.119.7) and $\mathcal{F}_ x$ is torsion free. Hence $\mathcal{F}_ x$ is free by More on Algebra, Lemma 15.22.11. $\square$

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