## 55.13 Picard groups of models

Assume $R, K, k, \pi , C, X, n, C_1, \ldots , C_ n, m_1, \ldots , m_ n$ are as in Situation 55.9.3. In Lemma 55.9.5 we found an exact sequence

$0 \to \mathbf{Z} \to \mathbf{Z}^{\oplus n} \to \mathop{\mathrm{Pic}}\nolimits (X) \to \mathop{\mathrm{Pic}}\nolimits (C) \to 0$

We want to use this sequence to study the $\ell$-torsion in the Picard groups for suitable primes $\ell$.

Lemma 55.13.1. In Situation 55.9.3 let $d = \gcd (m_1, \ldots , m_ n)$. If $\mathcal{L}$ is an invertible $\mathcal{O}_ X$-module which

1. restricts to the trivial invertible module on $C$, and

2. has degree $0$ on each $C_ i$,

then $\mathcal{L}^{\otimes d} \cong \mathcal{O}_ X$.

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. $\square$

Lemma 55.13.2. In Situation 55.9.3 let $T$ be the numerical type associated to $X$. There exists a canonical map

$\mathop{\mathrm{Pic}}\nolimits (C) \to \mathop{\mathrm{Pic}}\nolimits (T)$

whose kernel is exactly those invertible modules on $C$ which are the restriction of invertible modules $\mathcal{L}$ on $X$ with $\deg _{C_ i}(\mathcal{L}|_{C_ i}) = 0$ for $i = 1, \ldots , n$.

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

$\frac{\deg }{w} : \mathop{\mathrm{Pic}}\nolimits (X) \to \mathbf{Z}^{\oplus n}, \quad \mathcal{L} \mapsto (\frac{\deg (\mathcal{L}|_{C_1})}{w_1}, \ldots , \frac{\deg (\mathcal{L}|_{C_ n})}{w_ n})$

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

$((C_ j \cdot C_1)/w_1, \ldots , (C_ j \cdot C_ n)/w_ n) = (a_{1j}/w_1, \ldots , a_{nj}/w_ n)$

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

$\xymatrix{ \mathbf{Z}^{\oplus n} \ar[r] \ar[d]_{\text{id}} & \mathop{\mathrm{Pic}}\nolimits (X) \ar[r] \ar[d]^{\frac{\deg }{w}} & \mathop{\mathrm{Pic}}\nolimits (C) \ar[r] \ar[d] & 0 \\ \mathbf{Z}^{\oplus n} \ar[r]^{(a_{ij}/w_ i)} & \mathbf{Z}^{\oplus n} \ar[r] & \mathop{\mathrm{Pic}}\nolimits (T) \ar[r] & 0 }$

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

Lemma 55.13.3. In Situation 55.9.3 let $d = \gcd (m_1, \ldots , m_ n)$ and let $T$ be the numerical type associated to $X$. Let $h \geq 1$ be an integer prime to $d$. There exists an exact sequence

$0 \to \mathop{\mathrm{Pic}}\nolimits (X)[h] \to \mathop{\mathrm{Pic}}\nolimits (C)[h] \to \mathop{\mathrm{Pic}}\nolimits (T)[h]$

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 Lemma 55.13.1 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 $\xi$. $\square$

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. $\square$

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