Lemma 37.53.1. Let $S$ be a scheme. Let $f : X \to S$ be a universally closed and quasi-separated morphism. There exists a factorization

\[ \xymatrix{ X \ar[rr]_{f'} \ar[rd]_ f & & S' \ar[dl]^\pi \\ & S & } \]

with the following properties:

the morphism $f'$ is universally closed, quasi-compact, quasi-separated, and surjective,

the morphism $\pi : S' \to S$ is integral,

we have $f'_*\mathcal{O}_ X = \mathcal{O}_{S'}$,

we have $S' = \underline{\mathop{\mathrm{Spec}}}_ S(f_*\mathcal{O}_ X)$, and

$S'$ is the normalization of $S$ in $X$, see Morphisms, Definition 29.53.3.

Formation of the factorization $f = \pi \circ f'$ commutes with flat base change.

**Proof.**
By Morphisms, Lemma 29.41.8 the morphism $f$ is quasi-compact. Hence the normalization $S'$ of $S$ in $X$ is defined (Morphisms, Definition 29.53.3) and we have the factorization $X \to S' \to S$. By Morphisms, Lemma 29.53.11 we have (2), (4), and (5). The morphism $f'$ is universally closed by Morphisms, Lemma 29.41.7. It is quasi-compact by Schemes, Lemma 26.21.14 and quasi-separated by Schemes, Lemma 26.21.13.

To show the remaining statements we may assume the base scheme $S$ is affine, say $S = \mathop{\mathrm{Spec}}(R)$. Then $S' = \mathop{\mathrm{Spec}}(A)$ with $A = \Gamma (X, \mathcal{O}_ X)$ an integral $R$-algebra. Thus it is clear that $f'_*\mathcal{O}_ X$ is $\mathcal{O}_{S'}$ (because $f'_*\mathcal{O}_ X$ is quasi-coherent, by Schemes, Lemma 26.24.1, and hence equal to $\widetilde{A}$). This proves (3).

Let us show that $f'$ is surjective. As $f'$ is universally closed (see above) the image of $f'$ is a closed subset $V(I) \subset S' = \mathop{\mathrm{Spec}}(A)$. Pick $h \in I$. Then $h|_ X = f^\sharp (h)$ is a global section of the structure sheaf of $X$ which vanishes at every point. As $X$ is quasi-compact this means that $h|_ X$ is a nilpotent section, i.e., $h^ n|X = 0$ for some $n > 0$. But $A = \Gamma (X, \mathcal{O}_ X)$, hence $h^ n = 0$. In other words $I$ is contained in the Jacobson radical ideal of $A$ and we conclude that $V(I) = S'$ as desired.
$\square$

## Comments (4)

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