## 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 33.43.10). 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 15.98.5 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. $\square$

## Post a comment

Your email address will not be published. Required fields are marked.

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0CAA. Beware of the difference between the letter 'O' and the digit '0'.