# The Stacks Project

## Tag 03GY

Lemma 36.45.1. 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:

1. the morphism $f'$ is universally closed, quasi-compact, quasi-separated, and surjective,
2. the morphism $\pi : S' \to S$ is integral,
3. we have $f'_*\mathcal{O}_X = \mathcal{O}_{S'}$,
4. we have $S' = \underline{\mathop{\rm Spec}}_S(f_*\mathcal{O}_X)$, and
5. $S'$ is the normalization of $S$ in $X$, see Morphisms, Definition 28.50.3.

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

Proof. By Morphisms, Lemma 28.39.10 the morphism $f$ is quasi-compact. Hence the normalization $S'$ of $S$ in $X$ is defined (Morphisms, Definition 28.50.3) and we have the factorization $X \to S' \to S$. By Morphisms, Lemma 28.50.11 we have (2), (4), and (5). The morphism $f'$ is universally closed by Morphisms, Lemma 28.39.7. It is quasi-compact by Schemes, Lemma 25.21.15 and quasi-separated by Schemes, Lemma 25.21.14.

To show the remaining statements we may assume the base scheme $S$ is affine, say $S = \mathop{\rm Spec}(R)$. Then $S' = \mathop{\rm Spec}(A)$ 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 25.24.1, 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 $V(I) \subset S' = \mathop{\rm Spec}(A)$. 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 radical ideal of $A$ and we conclude that $V(I) = S'$ as desired. $\square$

The code snippet corresponding to this tag is a part of the file more-morphisms.tex and is located in lines 12721–12743 (see updates for more information).

\begin{lemma}
\label{lemma-stein-universally-closed}
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:
\begin{enumerate}
\item the morphism $f'$ is universally closed, quasi-compact, quasi-separated,
and surjective,
\item the morphism $\pi : S' \to S$ is integral,
\item we have $f'_*\mathcal{O}_X = \mathcal{O}_{S'}$,
\item we have $S' = \underline{\Spec}_S(f_*\mathcal{O}_X)$, and
\item $S'$ is the normalization of $S$ in $X$, see
Morphisms, Definition \ref{morphisms-definition-normalization-X-in-Y}.
\end{enumerate}
Formation of the factorization $f = \pi \circ f'$ commutes
with flat base change.
\end{lemma}

\begin{proof}
By Morphisms, Lemma \ref{morphisms-lemma-universally-closed-quasi-compact}
the morphism $f$ is quasi-compact. Hence the normalization $S'$ of $S$ in
$X$ is defined (Morphisms, Definition
\ref{morphisms-definition-normalization-X-in-Y})
and we have the factorization $X \to S' \to S$. By
Morphisms, Lemma \ref{morphisms-lemma-normalization-in-universally-closed}
we have (2), (4), and (5). The morphism $f'$ is universally closed by
Morphisms, Lemma \ref{morphisms-lemma-image-proper-scheme-closed}.
It is quasi-compact by
Schemes, Lemma \ref{schemes-lemma-quasi-compact-permanence}
and quasi-separated by
Schemes, Lemma \ref{schemes-lemma-compose-after-separated}.

\medskip\noindent
To show the remaining statements we may assume the base scheme $S$ is affine,
say $S = \Spec(R)$. Then $S' = \Spec(A)$ 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
\ref{schemes-lemma-push-forward-quasi-coherent},
and hence equal to $\widetilde{A}$). This proves (3).

\medskip\noindent
Let us show that $f'$ is surjective. As $f'$ is universally closed (see above)
the image of $f'$ is a closed subset
$V(I) \subset S' = \Spec(A)$. 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 radical ideal of $A$ and we conclude
that $V(I) = S'$ as desired.
\end{proof}

Comment #695 by Kestutis Cesnavicius on June 17, 2014 a 1:50 am UTC

The quasi-compactness assumption is superfluous in the view of http://stacks.math.columbia.edu/tag/04XU

Comment #704 by Johan (site) on June 17, 2014 a 7:52 pm UTC

OK, I changed this and I also fixed the typo you pointed out later. See this commit.

Comment #2469 by A student of algebraic geometry on March 31, 2017 a 2:29 pm UTC

Tag 035H defines the normalisation of $S$ in $X$ to be the relative spectrum of the integral closer of $\mathcal{O}_S$ in $f_\ast\mathcal{O}_X$ over $S$. So, item (5) does not seem to hold by construction.

Comment #2503 by Johan (site) on April 14, 2017 a 12:01 am UTC

Dear student of algebraic geometry, you are right! Fixed here.

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