Lemma 66.48.5. Let $S$ be a scheme. Let $f : Y \to X$ be a quasi-compact and quasi-separated morphism of algebraic spaces over $S$. The factorization $f = \nu \circ f'$, where $\nu : X' \to X$ is the normalization of $X$ in $Y$ is characterized by the following two properties:

the morphism $\nu $ is integral, and

for any factorization $f = \pi \circ g$, with $\pi : Z \to X$ integral, there exists a commutative diagram

\[ \xymatrix{ Y \ar[d]_{f'} \ar[r]_ g & Z \ar[d]^\pi \\ X' \ar[ru]^ h \ar[r]^\nu & X } \]

for a unique morphism $h : X' \to Z$.

Moreover, in (2) the morphism $h : X' \to Z$ is the normalization of $Z$ in $Y$.

**Proof.**
Let $\mathcal{O}' \subset f_*\mathcal{O}_ Y$ be the integral closure of $\mathcal{O}_ X$ as in Definition 66.48.3. The morphism $\nu $ is integral by construction, which proves (1). Assume given a factorization $f = \pi \circ g$ with $\pi : Z \to X$ integral as in (2). By Definition 66.45.2 $\pi $ is affine, and hence $Z$ is the relative spectrum of a quasi-coherent sheaf of $\mathcal{O}_ X$-algebras $\mathcal{B}$. The morphism $g : X \to Z$ corresponds to a map of $\mathcal{O}_ X$-algebras $\chi : \mathcal{B} \to f_*\mathcal{O}_ Y$. Since $\mathcal{B}(U)$ is integral over $\mathcal{O}_ X(U)$ for every affine $U$ étale over $X$ (by Definition 66.45.2) we see from Lemma 66.48.1 that $\chi (\mathcal{B}) \subset \mathcal{O}'$. By the functoriality of the relative spectrum Lemma 66.20.7 this provides us with a unique morphism $h : X' \to Z$. We omit the verification that the diagram commutes.

It is clear that (1) and (2) characterize the factorization $f = \nu \circ f'$ since it characterizes it as an initial object in a category. The morphism $h$ in (2) is integral by Lemma 66.45.12. Given a factorization $g = \pi ' \circ g'$ with $\pi ' : Z' \to Z$ integral, we get a factorization $f = (\pi \circ \pi ') \circ g'$ and we get a morphism $h' : X' \to Z'$. Uniqueness implies that $\pi ' \circ h' = h$. Hence the characterization (1), (2) applies to the morphism $h : X' \to Z$ which gives the last statement of the lemma.
$\square$

## Comments (0)