Lemma 70.4.6. Let $S$ be a scheme. Let $X$ be a normal integral algebraic space over $S$. Then $\Gamma (X, \mathcal{O}_ X)$ is a normal domain.

Proof. Set $R = \Gamma (X, \mathcal{O}_ X)$. Then $R$ is a domain by Lemma 70.4.4. Let $f = a/b$ be an element of the fraction field of $R$ which is integral over $R$. For any $U \to X$ étale with $U$ a scheme there is at most one $f_ U \in \Gamma (U, \mathcal{O}_ U)$ with $b|_ U f_ U = a|_ U$. Namely, $U$ is reduced and the generic points of $U$ map to the generic point of $X$ which implies that $b|_ U$ is a nonzerodivisor. For every $x \in |X|$ we choose $U \to X$ as in Lemma 70.4.5. Then there is a unique $f_ U \in \Gamma (U, \mathcal{O}_ U)$ with $b|_ U f_ U = a|_ U$ because $\Gamma (U, \mathcal{O}_ U)$ is a normal domain by Properties, Lemma 28.7.9. By the uniqueness mentioned above these $f_ U$ glue and define a global section $f$ of the structure sheaf, i.e., of $R$. $\square$

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).