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. $\square$

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