Lemma 72.14.4. Let k be a field. Let X be a proper integral algebraic space over k. Let \mathcal{L} be an invertible \mathcal{O}_ X-module. If H^0(X, \mathcal{L}) and H^0(X, \mathcal{L}^{\otimes - 1}) are both nonzero, then \mathcal{L} \cong \mathcal{O}_ X.
Proof. Let s \in H^0(X, \mathcal{L}) and t \in H^0(X, \mathcal{L}^{\otimes - 1}) be nonzero sections. Let x \in |X| be a point in the support of s. Choose an affine étale neighbourhood (U, u) \to (X, x) such that \mathcal{L}|_ U \cong \mathcal{O}_ U. Then s|_ U corresponds to a nonzero regular function on the reduced (because X is reduced) scheme U and hence is nonvanishing in a generic point of an irreducible component of U. By Decent Spaces, Lemma 68.20.1 we conclude that the generic point \eta of |X| is in the support of s. The same is true for t. Then of course st must be nonzero because the local ring of X at \eta is a field (by aforementioned lemma the local ring has dimension zero, as X is reduced the local ring is reduced, and Algebra, Lemma 10.25.1). However, we have seen that K = H^0(X, \mathcal{O}_ X) is a field in Lemma 72.14.3. Thus st is everywhere nonzero and we see that s : \mathcal{O}_ X \to \mathcal{L} is an isomorphism. \square
Comments (0)