The Stacks project

Proof. By the projection formula (Cohomology, Lemma 20.54.2) we see that $f_*f^*\mathcal{N} \cong \mathcal{N}$ for $\mathcal{N} \in \mathop{\mathrm{Pic}}\nolimits (Y)$. We claim that for $\mathcal{L} \in \mathop{\mathrm{Pic}}\nolimits (X)$ with $\mathcal{L}|_{X_ y} \cong \mathcal{O}_{X_ y}$ for all $y \in Y$ we have $\mathcal{N} = f_*\mathcal{L}$ is invertible and $\mathcal{L} \cong f^*\mathcal{N}$. This will finish the proof.

The $\mathcal{O}_ Y$-module $\mathcal{N} = f_*\mathcal{L}$ is coherent by Cohomology of Schemes, Proposition 30.19.1. Thus to see that it is an invertible $\mathcal{O}_ Y$-module, it suffices to check on stalks (Algebra, Lemma 10.78.2). Since the map from a Noetherian local ring to its completion is faithfully flat, it suffices to check the completion $(f_*\mathcal{L})_ y^\wedge$ is free (see Algebra, Section 10.97 and Lemma 10.78.6). For this we will use the theorem of formal functions as formulated in Cohomology of Schemes, Lemma 30.20.7. Since $f_*\mathcal{O}_ X = \mathcal{O}_ Y$ and hence $(f_*\mathcal{O}_ X)_ y^\wedge \cong \mathcal{O}_{Y, y}^\wedge$, it suffices to show that $\mathcal{L}|_{X_ n} \cong \mathcal{O}_{X_ n}$ for each $n$ (compatibly for varying $n$. By Lemma 37.4.1 we have an exact sequence

with notation as in the theorem on formal functions. Observe that we have a surjection

for some integers $r_ n \geq 0$. Since $\dim (X_ y) \leq 1$ this surjection induces a surjection on first cohomology groups (by the vanishing of cohomology in degrees $\geq 2$ coming from Cohomology, Proposition 20.20.7). Hence the $H^1$ in the sequence is zero and the transition maps $\mathop{\mathrm{Pic}}\nolimits (X_{n + 1}) \to \mathop{\mathrm{Pic}}\nolimits (X_ n)$ are injective as desired.

We still have to show that $f^*\mathcal{N} \cong \mathcal{L}$. This is proved by the same method and we omit the details. $\square$