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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