Lemma 66.48.1. 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 $\mathcal{A}(U)$.

## 66.48 Relative normalization of algebraic spaces

This section is the analogue of Morphisms, Section 29.53.

**Proof.**
Let $U$ be an object of $X_{\acute{e}tale}$. Then $U$ is a scheme. Denote $\mathcal{A}|_ U$ the restriction to the Zariski site. Then $\mathcal{A}|_ U$ is a quasi-coherent sheaf of $\mathcal{O}_ U$-algebras hence we can apply Morphisms, Lemma 29.53.1 to find a quasi-coherent subalgebra $\mathcal{A}'_ U \subset \mathcal{A}|_ U$ such that the value of $\mathcal{A}'_ U$ on any affine open $W \subset U$ is as given in the statement of the lemma. If $f : U' \to U$ is a morphism in $X_{\acute{e}tale}$, then $\mathcal{A}|_{U'} = f^*(\mathcal{A}|_ U)$ where $f^*$ means pullback by the morphism $f$ in the Zariski topology; this holds because $\mathcal{A}$ is quasi-coherent (see introduction to Properties of Spaces, Section 65.29 and the references to the discussion in the chapter on descent on schemes). Since $f$ is étale we find that More on Morphisms, Lemma 37.19.1 says that we get a canonical isomorphism $f^*(\mathcal{A}'_ U) = \mathcal{A}'_{U'}$. This immediately tells us that we obtain a sub presheaf $\mathcal{A}' \subset \mathcal{A}$ of $\mathcal{O}_ X$-algebras over $X_{\acute{e}tale}$ which is a sheaf for the Zariski topology and has the right values on affine objects. But the fact that each $\mathcal{A}'_ U$ is quasi-coherent on the scheme $U$ and that for $f : U' \to U$ étale we have $\mathcal{A}'_{U'} = f^*(\mathcal{A}'_ U)$ implies that $\mathcal{A}'$ is quasi-coherent on $X_{\acute{e}tale}$ as well (as this is a local property and we have the references above describing quasi-coherent modules on $U_{\acute{e}tale}$ in exactly this manner).
$\square$

Definition 66.48.2. 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 66.48.1 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 66.11.2). We can then take the relative spectrum of the quasi-coherent $\mathcal{O}_ X$-algebra (Lemma 66.20.7) to obtain the normalization of $X$ in $Y$.

Definition 66.48.3. 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

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

of the initial morphism $f$.

To get the factorization, use Remark 66.20.9 and functoriality of the $\underline{\mathop{\mathrm{Spec}}}$ construction.

Lemma 66.48.4. 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 $Y$.

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

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 66.48.5. 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:

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 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 66.48.3. 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 66.45.2 $\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 66.45.2) we see from Lemma 66.48.1 that $\chi (\mathcal{B}) \subset \mathcal{O}'$. By the functoriality of the relative spectrum Lemma 66.20.7 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 66.45.12. 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 66.48.6. 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 66.48.7. 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 65.10.2), then $x'$ is the image of some $y \in |Y|$ of codimension $0$.

**Proof.**
By Lemma 66.48.4 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 66.48.8. 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 66.48.9. 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

of Remark 66.20.9.

**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

Let $Z \subset \mathbf{A}^1_ Y$ be the scheme theoretic image of $h$, see Definition 66.16.2. The morphism $h$ is quasi-compact as $f$ is quasi-compact and $\mathbf{A}^1_ Y \to Y$ is separated, see Lemma 66.8.9. By Lemma 66.16.3 the morphism $X \to Z$ has dense image on underlying topological spaces. By Lemma 66.40.6 the morphism $X \to Z$ is closed. Hence $h(X) = Z$ (set theoretically). Thus we can use Lemma 66.40.7 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 66.45.7. 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 66.48.10. 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 66.45.7 this is a special case of Lemma 66.48.9.
$\square$

Lemma 66.48.11. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Assume that

$Y$ is Nagata,

$f$ is quasi-separated of finite type,

$X$ 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 66.48.4. 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 65.6.3). 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.
$\square$

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