Lemma 55.13.4. In Situation 55.9.3 let $h$ be an integer prime to the characteristic of $k$. Then the map

$\mathop{\mathrm{Pic}}\nolimits (X)[h] \longrightarrow \mathop{\mathrm{Pic}}\nolimits ((X_ k)_{red})[h]$

is injective.

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

\begin{align*} H^0(X, \mathcal{L})^\wedge & = \mathop{\mathrm{lim}}\nolimits H^0(X \times _{\mathop{\mathrm{Spec}}(R)} \mathop{\mathrm{Spec}}(R/\pi ^ n), \mathcal{L}|_{X \times _{\mathop{\mathrm{Spec}}(R)} \mathop{\mathrm{Spec}}(R/\pi ^ n)}) \\ & \cong \mathop{\mathrm{lim}}\nolimits H^0(X \times _{\mathop{\mathrm{Spec}}(R)} \mathop{\mathrm{Spec}}(R/\pi ^ n), \mathcal{O}_{X \times _{\mathop{\mathrm{Spec}}(R)} \mathop{\mathrm{Spec}}(R/\pi ^ n)}) \\ & = R^\wedge \end{align*}

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. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).