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
X \longrightarrow \underline{\mathop{\mathrm{Spec}}}_ Y(f_*\mathcal{O}_ X) \longrightarrow Y
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
\xymatrix{ X \ar[rr]_ h \ar[rd]_ f & & \mathbf{A}^1_ Y \ar[ld] \\ & Y & }
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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