Lemma 72.4.4. Let S be a scheme. Let X be an integral algebraic space over S. Then \Gamma (X, \mathcal{O}_ X) is a domain.
Proof. Set R = \Gamma (X, \mathcal{O}_ X). If f, g \in R are nonzero and fg = 0 then X = V(f) \cup V(g) where V(f) denotes the closed subspace of X cut out by f. Since X is irreducible, we see that either V(f) = X or V(g) = X. Then either f = 0 or g = 0 by Properties of Spaces, Lemma 66.21.4. \square
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