Lemma 70.14.4. 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 66.20.1 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 10.24.1). However, we have seen that $K = H^0(X, \mathcal{O}_ X)$ is a field in Lemma 70.14.3. Thus $st$ is everywhere nonzero and we see that $s : \mathcal{O}_ X \to \mathcal{L}$ is an isomorphism.
$\square$

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