Example 33.30.4. Lemma 33.30.3 is not true without some condition on the scheme $X$ over the field $k$. Here is an example. Let $k$ be a field. Let $t \in \mathbf{P}^1_ k$ be a closed point. Set $X = \mathbf{P}^1 \setminus \{ t\}$. Then we have a surjection

$\mathbf{Z} = \mathop{\mathrm{Pic}}\nolimits (\mathbf{P}^1_ k) \longrightarrow \mathop{\mathrm{Pic}}\nolimits (X)$

The first equality by Divisors, Lemma 31.28.5 and surjective by Divisors, Lemma 31.28.3 (as $\mathbf{P}^1_ k$ is smooth of dimension $1$ over $k$ and hence all its local rings are discrete valuation rings). If $\mathcal{L}$ is in the kernel of the displayed map, then $\mathcal{L} \cong \mathcal{O}_{\mathbf{P}^1_ k}(nt)$ for some $n \in \mathbf{Z}$. We leave it to the reader to show that $\mathcal{O}_{\mathbf{P}^1_ k}(t) \cong \mathcal{O}_{\mathbf{P}^1_ k}(d)$ where $d = [\kappa (t) : k]$. Hence

$\mathop{\mathrm{Pic}}\nolimits (X) = \mathbf{Z}/d\mathbf{Z}$

Thus if $t$ is not a $k$-rational point, then $d > 1$ and this Picard group is nonzero. On the other hand, if we extend the ground field $k$ to any field extension $k'$ such that there exists a $k$-embedding $\kappa (t) \to k'$, then $\mathbf{P}^1_{k'} \setminus X_{k'}$ has a $k'$-rational point $t'$. Hence $\mathcal{O}_{\mathbf{P}^1_{k'}}(1) = \mathcal{O}_{\mathbf{P}^1_{k'}}(t')$ will be in the kernel of the map $\mathbf{Z} \to \mathop{\mathrm{Pic}}\nolimits (X_{k'})$ and it will follow in the same manner as above that $\mathop{\mathrm{Pic}}\nolimits (X_{k'}) = 0$.

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