## 65.48 Relative normalization of algebraic spaces

This section is the analogue of Morphisms, Section 29.53.

Lemma 65.48.1. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Let $\mathcal{A}$ be a quasi-coherent sheaf of $\mathcal{O}_ X$-algebras. There exists a quasi-coherent sheaf of $\mathcal{O}_ X$-algebras $\mathcal{A}' \subset \mathcal{A}$ such that for any affine object $U$ of $X_{\acute{e}tale}$ the ring $\mathcal{A}'(U) \subset \mathcal{A}(U)$ is the integral closure of $\mathcal{O}_ X(U)$ in $\mathcal{A}(U)$.

Proof. By Properties of Spaces, Lemma 64.18.5 it suffices to prove that the rule given above defines a quasi-coherent module on $X_{affine, {\acute{e}tale}}$. To see this it suffices to show the following: Let $U_1 \to U_2$ be a morphism of affine objects of $X_{\acute{e}tale}$. Say $U_ i = \mathop{\mathrm{Spec}}(R_ i)$. Say $\mathcal{A}|_{(U_1)_{\acute{e}tale}}$ is the quasi-coherent sheaf associated to the $R_2$-algebra $A$. Let $A' \subset A$ be the integral closure of $R_2$ in $A$. Then $A' \otimes _{R_2} R_1$ is the integral closure of $R_1$ in $A \otimes _{R_2} R_1$. This is Algebra, Lemma 10.147.2. $\square$

Definition 65.48.2. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Let $\mathcal{A}$ be a quasi-coherent sheaf of $\mathcal{O}_ X$-algebras. The integral closure of $\mathcal{O}_ X$ in $\mathcal{A}$ is the quasi-coherent $\mathcal{O}_ X$-subalgebra $\mathcal{A}' \subset \mathcal{A}$ constructed in Lemma 65.48.1 above.

We will apply this in particular when $\mathcal{A} = f_*\mathcal{O}_ Y$ for a quasi-compact and quasi-separated morphism of algebraic spaces $f : Y \to X$ (see Lemma 65.11.2). We can then take the relative spectrum of the quasi-coherent $\mathcal{O}_ X$-algebra (Lemma 65.20.7) to obtain the normalization of $X$ in $Y$.

Definition 65.48.3. Let $S$ be a scheme. Let $f : Y \to X$ be a quasi-compact and quasi-separated morphism of algebraic spaces over $S$. Let $\mathcal{O}'$ be the integral closure of $\mathcal{O}_ X$ in $f_*\mathcal{O}_ Y$. The normalization of $X$ in $Y$ is the morphism of algebraic spaces

$\nu : X' = \underline{\mathop{\mathrm{Spec}}}_ X(\mathcal{O}') \to X$

over $S$. It comes equipped with a natural factorization

$Y \xrightarrow {f'} X' \xrightarrow {\nu } X$

of the initial morphism $f$.

To get the factorization, use Remark 65.20.9 and functoriality of the $\underline{\mathop{\mathrm{Spec}}}$ construction.

Lemma 65.48.4. Let $S$ be a scheme. Let $f : Y \to X$ be a quasi-compact and quasi-separated morphism of algebraic spaces over $S$. Let $Y \to X' \to X$ be the normalization of $X$ in $Y$.

1. If $W \to X$ is an étale morphism of algebraic spaces over $S$, then $W \times _ X X'$ is the normalization of $W$ in $W \times _ X Y$.

2. If $Y$ and $X$ are representable, then $Y'$ is representable and is canonically isomorphic to the normalization of the scheme $X$ in the scheme $Y$ as constructed in Morphisms, Section 29.54.

Proof. It is immediate from the construction that the formation of the normalization of $X$ in $Y$ commutes with étale base change, i.e., part (1) holds. On the other hand, if $X$ and $Y$ are schemes, then for $U \subset X$ affine open, $f_*\mathcal{O}_ Y(U) = \mathcal{O}_ Y(f^{-1}(U))$ and hence $\nu ^{-1}(U)$ is the spectrum of exactly the same ring as we get in the corresponding construction for schemes. $\square$

Here is a characterization of this construction.

Lemma 65.48.5. 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:

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 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 65.48.3. 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 65.45.2 $\pi$ 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 65.45.2) we see from Lemma 65.48.1 that $\chi (\mathcal{B}) \subset \mathcal{O}'$. By the functoriality of the relative spectrum Lemma 65.20.7 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 65.45.12. 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. $\square$

Lemma 65.48.6. Let $S$ be a scheme. Let $f : Y \to X$ be a quasi-compact and quasi-separated morphism of algebraic spaces over $S$. Let $X' \to X$ be the normalization of $X$ in $Y$. If $Y$ is reduced, so is $X'$.

Proof. This follows from the fact that a subring of a reduced ring is reduced. Some details omitted. $\square$

Lemma 65.48.7. Let $S$ be a scheme. Let $f : Y \to X$ be a quasi-compact and quasi-separated morphism of schemes. Let $X' \to X$ be the normalization of $X$ in $Y$. If $x' \in |X'|$ is a point of codimension $0$ (Properties of Spaces, Definition 64.10.2), then $x'$ is the image of some $y \in |Y|$ of codimension $0$.

Proof. By Lemma 65.48.4 and the definitions, we may assume that $X = \mathop{\mathrm{Spec}}(A)$ is affine. Then $X' = \mathop{\mathrm{Spec}}(A')$ where $A'$ is the integral closure of $A$ in $\Gamma (Y, \mathcal{O}_ Y)$ and $x'$ corresponds to a minimal prime of $A'$. Choose a surjective étale morphism $V \to Y$ where $V = \mathop{\mathrm{Spec}}(B)$ is affine. Then $A' \to B$ is injective, hence every minimal prime of $A'$ is the image of a minimal prime of $B$, see Algebra, Lemma 10.30.5. The lemma follows. $\square$

Lemma 65.48.8. Let $S$ be a scheme. Let $f : Y \to X$ be a quasi-compact and quasi-separated morphism of algebraic spaces over $S$. Suppose that $Y = Y_1 \amalg Y_2$ is a disjoint union of two algebraic spaces. Write $f_ i = f|_{Y_ i}$. Let $X_ i'$ be the normalization of $X$ in $Y_ i$. Then $X_1' \amalg X_2'$ is the normalization of $X$ in $Y$.

Proof. Omitted. $\square$

Lemma 65.48.9. Let $S$ be a scheme. Let $f : X \to Y$ be a quasi-compact, quasi-separated and universally closed morphisms of algebraic spaces over $S$. Then $f_*\mathcal{O}_ X$ is integral over $\mathcal{O}_ Y$. In other words, the normalization of $Y$ in $X$ is equal to the factorization

$X \longrightarrow \underline{\mathop{\mathrm{Spec}}}_ Y(f_*\mathcal{O}_ X) \longrightarrow Y$

of Remark 65.20.9.

Proof. The question is étale local on $Y$, hence we may reduce to the case where $Y = \mathop{\mathrm{Spec}}(R)$ is affine. Let $h \in \Gamma (X, \mathcal{O}_ X)$. We have to show that $h$ satisfies a monic equation over $R$. Think of $h$ as a morphism as in the following commutative diagram

$\xymatrix{ X \ar[rr]_ h \ar[rd]_ f & & \mathbf{A}^1_ Y \ar[ld] \\ & Y & }$

Let $Z \subset \mathbf{A}^1_ Y$ be the scheme theoretic image of $h$, see Definition 65.16.2. The morphism $h$ is quasi-compact as $f$ is quasi-compact and $\mathbf{A}^1_ Y \to Y$ is separated, see Lemma 65.8.9. By Lemma 65.16.3 the morphism $X \to Z$ has dense image on underlying topological spaces. By Lemma 65.40.6 the morphism $X \to Z$ is closed. Hence $h(X) = Z$ (set theoretically). Thus we can use Lemma 65.40.7 to conclude that $Z \to Y$ is universally closed (and even proper). Since $Z \subset \mathbf{A}^1_ Y$, we see that $Z \to Y$ is affine and proper, hence integral by Lemma 65.45.7. Writing $\mathbf{A}^1_ Y = \mathop{\mathrm{Spec}}(R[T])$ we conclude that the ideal $I \subset R[T]$ of $Z$ contains a monic polynomial $P(T) \in R[T]$. Hence $P(h) = 0$ and we win. $\square$

Lemma 65.48.10. Let $S$ be a scheme. Let $f : Y \to X$ be an integral morphism of algebraic spaces over $S$. Then the integral closure of $X$ in $Y$ is equal to $Y$.

Proof. By Lemma 65.45.7 this is a special case of Lemma 65.48.9. $\square$

Lemma 65.48.11. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Assume that

1. $Y$ is Nagata,

2. $f$ is quasi-separated of finite type,

3. $X$ is reduced.

Then the normalization $\nu : Y' \to Y$ of $Y$ in $X$ is finite.

Proof. The question is étale local on $Y$, see Lemma 65.48.4. Thus we may assume $Y = \mathop{\mathrm{Spec}}(R)$ is affine. Then $R$ is a Noetherian Nagata ring and we have to show that the integral closure of $R$ in $\Gamma (X, \mathcal{O}_ X)$ is finite over $R$. Since $f$ is quasi-compact we see that $X$ is quasi-compact. Choose an affine scheme $U$ and a surjective étale morphism $U \to X$ (Properties of Spaces, Lemma 64.6.3). Then $\Gamma (X, \mathcal{O}_ X) \subset \Gamma (U, \mathcal{O}_ X)$. Since $R$ is Noetherian it suffices to show that the integral closure of $R$ in $\Gamma (U, \mathcal{O}_ U)$ is finite over $R$. As $U \to Y$ is of finite type this follows from Morphisms, Lemma 29.53.15. $\square$

## Post a comment

Your email address will not be published. Required fields are marked.

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.

All contributions are licensed under the GNU Free Documentation License.

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 0BAZ. Beware of the difference between the letter 'O' and the digit '0'.