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
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
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.
Comments (0)