Lemma 44.6.5. Let k be a separably closed field. Let X be a smooth projective curve of genus g over k. Let K/k be a field extension and let \mathcal{L} be an invertible sheaf on X_ K. Then there exists an invertible sheaf \mathcal{L}_0 on X such that \dim _ K H^0(X_ K, \mathcal{L} \otimes _{\mathcal{O}_{X_ K}} \mathcal{L}_0|_{X_ K}) = 1 and \dim _ K H^1(X_ K, \mathcal{L} \otimes _{\mathcal{O}_{X_ K}} \mathcal{L}_0|_{X_ K}) = 0.
Proof. This proof is a variant of the proof of Varieties, Lemma 33.44.16. We encourage the reader to read that proof first.
First we pick an ample invertible sheaf \mathcal{L}_0 and we replace \mathcal{L} by \mathcal{L} \otimes _{\mathcal{O}_{X_ K}} \mathcal{L}_0^{\otimes n}|_{X_ K} for some n \gg 0. The result will be that we may assume that H^0(X_ K, \mathcal{L}) \not= 0 and H^1(X_ K, \mathcal{L}) = 0. Namely, we will get the vanishing by Cohomology of Schemes, Lemma 30.17.1 and the nonvanishing because the degree of the tensor product is \gg 0. We will finish the proof by descending induction on t = \dim _ K H^0(X_ K, \mathcal{L}). The base case t = 1 is trivial. Assume t > 1.
Observe that for a k-rational point x of X, the inverse image x_ K is a K-rational point of X_ K. Moreover, there are infinitely many k-rational points by Varieties, Lemma 33.25.6. Therefore the points x_ K form a Zariski dense collection of points of X_ K.
Let s \in H^0(X_ K, \mathcal{L}) be nonzero. From the previous paragraph we deduce there exists a k-rational point x such that s does not vanish in x_ K. Let \mathcal{I} be the ideal sheaf of i : x_ K \to X_ K as in Varieties, Lemma 33.43.8. Look at the short exact sequence
Observe that H^0(X_ K, i_*i^*\mathcal{L}) = H^0(x_ K, i^*\mathcal{L}) has dimension 1 over K. Since s does not vanish at x we conclude that
is surjective. Hence \dim _ K H^0(X_ K, \mathcal{I} \otimes _{\mathcal{O}_{X_ K}} \mathcal{L}) = t - 1. Finally, the long exact sequence of cohomology also shows that H^1(X_ K, \mathcal{I} \otimes _{\mathcal{O}_{X_ K}} \mathcal{L}) = 0 thereby finishing the proof of the induction step. \square
Comments (0)
There are also: