Lemma 66.48.9. 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 66.20.9.

**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 66.16.2. The morphism $h$ is quasi-compact as $f$ is quasi-compact and $\mathbf{A}^1_ Y \to Y$ is separated, see Lemma 66.8.9. By Lemma 66.16.3 the morphism $X \to Z$ has dense image on underlying topological spaces. By Lemma 66.40.6 the morphism $X \to Z$ is closed. Hence $h(X) = Z$ (set theoretically). Thus we can use Lemma 66.40.7 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 66.45.7. 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.
$\square$

## Comments (0)