The Stacks project

Proof. By Lemma 55.9.5 we have $\mathcal{L} \cong \mathcal{O}_ X(\sum a_ i C_ i)$ for some $a_ i \in \mathbf{Z}$. The degree of $\mathcal{L}|_{C_ j}$ is $\sum _ j a_ j(C_ i \cdot C_ j)$. In particular $(\sum a_ i C_ i \cdot \sum a_ i C_ i) = 0$. Hence we see from Lemma 55.9.7 that $(a_1, \ldots , a_ n) = q(m_1, \ldots , m_ n)$ for some $q \in \mathbf{Q}$. Thus $\mathcal{L} = \mathcal{O}_ X(lD)$ for some $l \in \mathbf{Z}$ where $D = \sum (m_ i/d) C_ i$ is as in Lemma 55.9.8 and we conclude. $\square$

Proof. Recall that $w_ i = [\kappa _ i : k]$ where $\kappa _ i = H^0(C_ i, \mathcal{O}_{C_ i)})$ and recall that the degree of any invertible module on $C_ i$ is divisible by $w_ i$ (Varieties, Lemma 33.44.10). Thus we can consider the map

The image of $\mathcal{O}_ X(C_ j)$ under this map is

which is exactly the image of the $j$th basis vector under the map $(a_{ij}/w_ i) : \mathbf{Z}^{\oplus n} \to \mathbf{Z}^{\oplus n}$ defining the Picard group of $T$, see Definition 55.4.1. Thus the canonical map of the lemma comes from the commutative diagram

with exact rows (top row by Lemma 55.9.5). The description of the kernel is clear. $\square$

Proof. Taking $h$-torsion in the exact sequence of Lemma 55.9.5 we obtain the exactness of $0 \to \mathop{\mathrm{Pic}}\nolimits (X)[h] \to \mathop{\mathrm{Pic}}\nolimits (C)[h]$ because $h$ is prime to $d$. Using the map of Lemma 55.13.2 we get a map $\mathop{\mathrm{Pic}}\nolimits (C)[h] \to \mathop{\mathrm{Pic}}\nolimits (T)[h]$ which annihilates elements of $\mathop{\mathrm{Pic}}\nolimits (X)[h]$. Conversely, if $\xi \in \mathop{\mathrm{Pic}}\nolimits (C)[h]$ maps to zero in $\mathop{\mathrm{Pic}}\nolimits (T)[h]$, then we can find an invertible $\mathcal{O}_ X$-module $\mathcal{L}$ with $\deg (\mathcal{L}|_{C_ i}) = 0$ for all $i$ whose restriction to $C$ is $\xi$. Then $\mathcal{L}^{\otimes h}$ is $d$-torsion by Lemma 55.13.1. Let $d'$ be an integer such that $dd' \equiv 1 \bmod h$. Such an integer exists because $h$ and $d$ are coprime. Then $\mathcal{L}^{\otimes dd'}$ is an $h$-torsion invertible sheaf on $X$ whose restriction to $C$ is $\xi$. $\square$

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. $\square$