Lemma 33.38.7. Let $f : X \to Y$ be a finite morphism of schemes. Assume there exists an open $V \subset Y$ such that $f^{-1}(V) \to V$ is an isomorphism and $Y \setminus V$ is a discrete space. Then every invertible $\mathcal{O}_ X$-module is the pullback of an invertible $\mathcal{O}_ Y$-module.

Proof. We will use that $\mathop{\mathrm{Pic}}\nolimits (X) = H^1(X, \mathcal{O}_ X^*)$, see Cohomology, Lemma 20.6.1. Consider the Leray spectral sequence for the abelian sheaf $\mathcal{O}_ X^*$ and $f$, see Cohomology, Lemma 20.13.4. Consider the induced map

$H^1(X, \mathcal{O}_ X^*) \longrightarrow H^0(Y, R^1f_*\mathcal{O}_ X^*)$

Divisors, Lemma 31.17.1 says exactly that this map is zero. Hence Leray gives $H^1(X, \mathcal{O}_ X^*) = H^1(Y, f_*\mathcal{O}_ X^*)$. Next we consider the map

$f^\sharp : \mathcal{O}_ Y^* \longrightarrow f_*\mathcal{O}_ X^*$

By assumption the kernel and cokernel of this map are supported on the closed subset $T = Y \setminus V$ of $Y$. Since $T$ is a discrete topological space by assumption the higher cohomology groups of any abelian sheaf on $Y$ supported on $T$ is zero (follows from Cohomology, Lemma 20.20.1, Modules, Lemma 17.6.1, and the fact that $H^ i(T, \mathcal{F}) = 0$ for any $i > 0$ and any abelian sheaf $\mathcal{F}$ on $T$). Breaking the displayed map into short exact sequences

$0 \to \mathop{\mathrm{Ker}}(f^\sharp ) \to \mathcal{O}_ Y^* \to \mathop{\mathrm{Im}}(f^\sharp ) \to 0,\quad 0 \to \mathop{\mathrm{Im}}(f^\sharp ) \to f_*\mathcal{O}_ X^* \to \mathop{\mathrm{Coker}}(f^\sharp ) \to 0$

we first conclude that $H^1(Y, \mathcal{O}_ Y^*) \to H^1(Y, \mathop{\mathrm{Im}}(f^\sharp ))$ is surjective and then that $H^1(Y, \mathop{\mathrm{Im}}(f^\sharp )) \to H^1(Y, f_*\mathcal{O}_ X^*)$ is surjective. Combining all the above we find that $H^1(Y, \mathcal{O}_ Y^*) \to H^1(X, \mathcal{O}_ X^*)$ is surjective as desired. $\square$

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