Lemma 29.53.4 (035I)—The Stacks project

Lemma 29.53.4. Let $f : Y \to X$ be a quasi-compact and quasi-separated morphism of schemes. 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 some unique morphism $h : X' \to Z$.

Moreover, the morphism $f' : Y \to X'$ is dominant and 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 29.53.3. 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 29.44.1 $\pi $ is affine, and hence $Z$ is the relative spectrum of a quasi-coherent sheaf of $\mathcal{O}_ X$-algebras $\mathcal{B}$. The morphism $g : Y \to Z$ 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 open $U \subset X$ (by Definition 29.44.1) we see from Lemma 29.53.1 that $\chi (\mathcal{B}) \subset \mathcal{O}'$. By the functoriality of the relative spectrum Lemma 29.11.5 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.

From the universal property in (2) we see that $f'$ does not factor through a proper closed subscheme of $X'$. Hence the scheme theoretic image of $f'$ is $X'$. Since $f'$ is quasi-compact (by Schemes, Lemma 26.21.14 and the fact that $\nu $ is separated as an affine morphism) we see that $f'(Y)$ is dense in $X'$. Hence $f'$ is dominant.

Observe that $g$ is quasi-compact and quasi-separated by Schemes, Lemmas 26.21.13 and 26.21.14. Thus the last statement of the lemma makes sense. The morphism $h$ in (2) is integral by Lemma 29.44.14. 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 final assertion of the lemma. $\square$

Comments (7)

Comment #4280 by ykm on June 30, 2019 at 18:42

minor typo in line 4 of the proof: " The morphism $g:X \to Z$ corresponds ..." : the domain of g is Y, not X

Comment #4445 by Johan on September 01, 2019 at 10:06

THanks and fixed here.

Comment #5028 by Tongmu He (何通木) on April 11, 2020 at 12:58

In 29.53.4, it is useful to remark that $f':Y \to X'$ is dominant, since, by the universal property, the normalization $X'$ is the scheme theoretic image of the quasi-compact morphism $f'$ .

Comment #5262 by Johan on June 16, 2020 at 10:07

@#5028. Yes, this is true and I added it. However more can be said of course. For example the proof you give shows that $X'$ is the scheme theoretic image of $Y$ by $f'$ . So some of the properties of the situation will be implicitly given by what happens in the proof of the lemma and this holds more generally of course. Thanks and fixed here.

Comment #8449 by Elías Guisado on June 01, 2023 at 09:25

Since we've defined the normalization only for a quasi-compact and quasi-separated morphism, shouldn't one remark at some point during the proof that this is indeed the case for $g$ ? (It follows from 26.21.13 and 26.21.14.)

Also, instead of "with a unique morphism $h:X'\to Z$ . We omit the verification that the diagram commutes," one could write "with a unique morphism $h:X'\to Z$ over $X$ . On the other hand, by https://stacks.math.columbia.edu/tag/01LQ#comment-8448 , we have $h\circ f'=g$ ."

Comment #8493 by Elías Guisado on June 15, 2023 at 08:00

In case it is worth of being included, here's the proof of the uniqueness of $h$ : Suppose $\tilde{h}:X'\to Z$ makes also the diagram commute. In particular, $\tilde{h}$ is a morphism over $X$ , so it is uniquely determined by $\pi_*(\mathcal{O}_X\to\tilde{h}_*\mathcal{O}_{X'})=\pi_*(\tilde{h}^\sharp):\mathcal{B}\to\mathcal{O}'$ (see Constructions, Section 27.4). Since $h\circ f'=g$ , the composite $\mathcal{O}_Z\xrightarrow{\tilde{h}^\sharp}\tilde{h}_*\mathcal{O}_{X'}\to g_*\mathcal{O}_X$ equals $g^\sharp$ . Mapping this composite through $\pi_*$ gives $\mathcal{B}\xrightarrow{\pi_*(\tilde{h}^\sharp)}\mathcal{O}'\to f_*\mathcal{O}_Y$ , i.e., $\pi_*(g^\sharp)=\chi$ . Since $\mathcal{O}'\to f_*\mathcal{O}_Y$ is injective, it follows that $\pi_*(\tilde{h}^\sharp)=\pi_*(h^\sharp)$ , whence $\tilde{h}=h$ .

Comment #9069 by Stacks project on May 06, 2024 at 15:23

OK, I added the check that $g$ is qcqs here. Thanks.

There are also:

4 comment(s) on Section 29.53: Relative normalization

Comments (7)

Post a comment