**Proof.**
Write $X = \bar X - \{ x_1, \ldots , x_ r\} $. Then $\mathop{\mathrm{Pic}}\nolimits (X) = \mathop{\mathrm{Pic}}\nolimits (\bar X)/ R$, where $R$ is the subgroup generated by $\mathcal{O}_{\bar X}(x_ i)$, $1 \leq i \leq r$. Since $r \geq 1$, we see that $\mathop{\mathrm{Pic}}\nolimits ^0(X) \to \mathop{\mathrm{Pic}}\nolimits (X)$ is surjective, hence $\mathop{\mathrm{Pic}}\nolimits (X)$ is divisible. Applying the Kummer sequence, we get (1) and (3). For (2), recall that

\begin{align*} H_{\acute{e}tale}^1(X, \mu _ n) & = \{ (\mathcal L, \alpha ) | \mathcal L \in \mathop{\mathrm{Pic}}\nolimits (X), \alpha : \mathcal{L}^{\otimes n} \to \mathcal{O}_ X\} /\cong \\ & = \{ (\bar{\mathcal L},\ D,\ \bar\alpha )\} /\tilde{R} \end{align*}

where $\bar{\mathcal L} \in \mathop{\mathrm{Pic}}\nolimits ^0(\bar X)$, $D$ is a divisor on $\bar X$ supported on $\left\{ x_1, \ldots , x_ r\right\} $ and $ \bar{\alpha }: \bar{\mathcal L}^{\otimes n} \cong \mathcal{O}_{\bar{X}}(D)$ is an isomorphism. Note that $D$ must have degree 0. Further $\tilde{R}$ is the subgroup of triples of the form $(\mathcal{O}_{\bar X}(D'), n D', 1^{\otimes n})$ where $D'$ is supported on $\left\{ x_1, \ldots , x_ r\right\} $ and has degree 0. Thus, we get an exact sequence

\[ 0 \longrightarrow H_{\acute{e}tale}^1(\bar X, \mu _ n) \longrightarrow H_{\acute{e}tale}^1(X, \mu _ n) \longrightarrow \bigoplus _{i = 1}^ r \mathbf{Z}/n\mathbf{Z} \xrightarrow {\ \sum \ } \mathbf{Z}/n\mathbf{Z} \longrightarrow 0 \]

where the middle map sends the class of a triple $(\bar{ \mathcal L}, D, \bar\alpha )$ with $D = \sum _{i = 1}^ r a_ i (x_ i)$ to the $r$-tuple $(a_ i)_{i = 1}^ r$. It now suffices to use Lemma 57.68.1 to count ranks.
$\square$

## Comments (1)

Comment #4982 by Noah Olander on