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. $\square$
Post a comment
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$
). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)