Lemma 67.48.9. Let $S$ be a scheme. Let $f : X \to Y$ be a quasi-compact, quasi-separated and universally closed morphisms of algebraic spaces over $S$. Then $f_*\mathcal{O}_ X$ is integral over $\mathcal{O}_ Y$. In other words, the normalization of $Y$ in $X$ is equal to the factorization
of Remark 67.20.9.
Proof. The question is étale local on $Y$, hence we may reduce to the case where $Y = \mathop{\mathrm{Spec}}(R)$ is affine. Let $h \in \Gamma (X, \mathcal{O}_ X)$. We have to show that $h$ satisfies a monic equation over $R$. Think of $h$ as a morphism as in the following commutative diagram
Let $Z \subset \mathbf{A}^1_ Y$ be the scheme theoretic image of $h$, see Definition 67.16.2. The morphism $h$ is quasi-compact as $f$ is quasi-compact and $\mathbf{A}^1_ Y \to Y$ is separated, see Lemma 67.8.9. By Lemma 67.16.3 the morphism $X \to Z$ has dense image on underlying topological spaces. By Lemma 67.40.6 the morphism $X \to Z$ is closed. Hence $h(X) = Z$ (set theoretically). Thus we can use Lemma 67.40.7 to conclude that $Z \to Y$ is universally closed (and even proper). Since $Z \subset \mathbf{A}^1_ Y$, we see that $Z \to Y$ is affine and proper, hence integral by Lemma 67.45.7. Writing $\mathbf{A}^1_ Y = \mathop{\mathrm{Spec}}(R[T])$ we conclude that the ideal $I \subset R[T]$ of $Z$ contains a monic polynomial $P(T) \in R[T]$. Hence $P(h) = 0$ and we win. $\square$
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