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:

1. the morphism $\nu$ is integral, and

2. 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. $\square$

Comment #4280 by ykm on

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

Comment #5028 by Tongmu He (何通木) on

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 on

@#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 on

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 on

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 $\tilde{h}=h$.

There are also:

• 4 comment(s) on Section 29.53: Relative normalization

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 035I. Beware of the difference between the letter 'O' and the digit '0'.