The Stacks project

Proof. Observe that $X \times _{\mathop{\mathrm{Spec}}(R)} \mathop{\mathrm{Spec}}(R/\pi ^ n)$ is a finite order thickening of $(X_ k)_{red}$ (this follows for example from Cohomology of Schemes, Lemma 30.10.2). Thus the canonical map $\mathop{\mathrm{Pic}}\nolimits (X \times _{\mathop{\mathrm{Spec}}(R)} \mathop{\mathrm{Spec}}(R/\pi ^ n)) \to \mathop{\mathrm{Pic}}\nolimits ((X_ k)_{red})$ identifies $h$ torsion by More on Morphisms, Lemma 37.4.2 and our assumption on $h$. Thus if $\mathcal{L}$ is an $h$-torsion invertible sheaf on $X$ which restricts to the trivial sheaf on $(X_ k)_{red}$ then $\mathcal{L}$ restricts to the trivial sheaf on $X \times _{\mathop{\mathrm{Spec}}(R)} \mathop{\mathrm{Spec}}(R/\pi ^ n)$ for all $n$. We find

using the theorem on formal functions (Cohomology of Schemes, Theorem 30.20.5) for the first and last equality and for example More on Algebra, Lemma 15.100.5 for the middle isomorphism. Since $H^0(X, \mathcal{L})$ is a finite $R$-module and $R$ is a discrete valuation ring, this means that $H^0(X, \mathcal{L})$ is free of rank $1$ as an $R$-module. Let $s \in H^0(X, \mathcal{L})$ be a basis element. Then tracing back through the isomorphisms above we see that $s|_{X \times _{\mathop{\mathrm{Spec}}(R)} \mathop{\mathrm{Spec}}(R/\pi ^ n)}$ is a trivialization for all $n$. Since the vanishing locus of $s$ is closed in $X$ and $X \to \mathop{\mathrm{Spec}}(R)$ is proper we conclude that the vanishing locus of $s$ is empty as desired. $\square$