66.49 Normalization

This section is the analogue of Morphisms, Section 29.54.

Lemma 66.49.1. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. The following are equivalent

1. there is a surjective étale morphism $U \to X$ where $U$ is as scheme such that every quasi-compact open of $U$ has finitely many irreducible components,

2. for every scheme $U$ and every étale morphism $U \to X$ every quasi-compact open of $U$ has finitely many irreducible components,

3. for every quasi-compact algebraic space $Y$ étale over $X$ the set of codimension $0$ points of $Y$ (Properties of Spaces, Definition 65.10.2) is finite, and

4. for every quasi-compact algebraic space $Y$ étale over $X$ the space $|Y|$ has finitely many irreducible components.

If $X$ is representable this means that every quasi-compact open of $X$ has finitely many irreducible components.

Proof. The equivalence of (1) and (2) and the final statement follow from Descent, Lemma 35.16.3 and Properties of Spaces, Lemma 65.7.1. It is clear that (4) implies (1) and (2) by considering only those $Y$ which are schemes. Similarly, (3) implies (1) and (2) since for a scheme the codimension $0$ points are the generic points of its irreducible components, see for example Properties of Spaces, Lemma 65.11.1.

Conversely, assume (2) and let $Y \to X$ be an étale morphism of algebraic spaces with $Y$ quasi-compact. Then we can choose an affine scheme $V$ and a surjective étale morphism $V \to Y$ (Properties of Spaces, Lemma 65.6.3). Since $V$ has finitely many irreducible components by (2) and since $|V| \to |Y|$ is surjective and continuous, we conclude that $|Y|$ has finitely many irreducible components by Topology, Lemma 5.8.5. Thus (4) holds. Similarly, by Properties of Spaces, Lemma 65.11.1 the images of the generic points of the irreducible components of $V$ are the codimension $0$ points of $Y$ and we conclude that there are finitely many, i.e., (3) holds. $\square$

Lemma 66.49.2. Let $S$ be a scheme. Let $X$ be a locally Noetherian algebraic space over $S$. Then $X$ satisfies the equivalent conditions of Lemma 66.49.1.

Proof. If $U \to X$ is étale and $U$ is a scheme, then $U$ is a locally Noetherian scheme, see Properties of Spaces, Section 65.7. A locally Noetherian scheme has a locally finite set of irreducible components (Divisors, Lemma 31.26.1). Thus we conclude that $X$ passes condition (2) of the lemma. $\square$

Lemma 66.49.3. Let $S$ be a scheme. Let $f : X \to Y$ be a flat morphism of algebraic spaces over $S$. Then for $x \in |X|$ we have: $x$ has codimension $0$ in $X \Rightarrow f(x)$ has codimension $0$ in $Y$.

Proof. Via Properties of Spaces, Lemma 65.11.1 and étale localization this translates into the case of a morphism of schemes and generic points of irreducible components. Here the result follows as generalizations lift along flat morphisms of schemes, see Morphisms, Lemma 29.25.9. $\square$

Lemma 66.49.4. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Assume $f$ is flat and locally of finite type and assume $Y$ satisfies the equivalent conditions of Lemma 66.49.1. Then $X$ satisfies the equivalent conditions of Lemma 66.49.1 and for $x \in |X|$ we have: $x$ has codimension $0$ in $X \Rightarrow f(x)$ has codimension $0$ in $Y$.

Proof. The last statement follows from Lemma 66.49.3. Choose a surjective étale morphism $V \to Y$ where $V$ is a scheme. Choose a surjective étale morphism $U \to X \times _ Y V$ where $U$ is a scheme. It suffices to show that every quasi-compact open of $U$ has finitely many irreducible components. We will use the results of Properties of Spaces, Lemma 65.11.1 without further mention. By what we've already shown, the codimension $0$ points of $U$ lie above codimension $0$ points in $U$ and these are locally finite by assumption. Hence it suffices to show that for $v \in V$ of codimension $0$ the codimension $0$ points of the scheme theoretic fibre $U_ v = U \times _ V v$ are locally finite. This is true because $U_ v$ is a scheme locally of finite type over $\kappa (v)$, hence locally Noetherian and we can apply Lemma 66.49.2 for example. $\square$

Lemma 66.49.5. Let $S$ be a scheme. For every algebraic space $X$ over $S$ satisfying the equivalent conditions of Lemma 66.49.1 there exists a morphism of algebraic spaces

$\nu _ X : X^\nu \longrightarrow X$

with the following properties

1. if $X$ satisfies the equivalent conditions of Lemma 66.49.1 then $X^\nu$ is normal and $\nu _ X$ is integral,

2. if $X$ is a scheme such that every quasi-compact open has finitely many irreducible components, then $\nu _ X : X^\nu \to X$ is the normalization of $X$ constructed in Morphisms, Section 29.54,

3. if $f : X \to Y$ is a morphism of algebraic spaces over $S$ which both satisfy the equivalent conditions of Lemma 66.49.1 and every codimension $0$ point of $X$ is mapped by $f$ to a codimension $0$ point of $Y$, then there is a unique morphism $f^\nu : X^\nu \to Y^\nu$ of algebraic spaces over $S$ such that $\nu _ Y \circ f^\nu = f \circ \nu _ X$, and

4. if $f : X \to Y$ is an étale or smooth morphism of algebraic spaces and $Y$ satisfies the equivalent conditions of Lemma 66.49.1, then the hypotheses of (3) hold and the morphism $f^\nu$ induces an isomorphism $X^\nu \to X \times _ Y Y^\nu$.

Proof. Consider the category $\mathcal{C}$ whose objects are the schemes $U$ over $S$ such that every quasi-compact open of $U$ has finitely many irreducible components and whose morphisms are those morphisms $g : U \to V$ of schemes over $S$ such that every generic point of an irreducible component of $U$ is mapped to the generic point of an irreducible component of $V$. We have already shown that

1. for $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ we have a normalization morphism $\nu _ U : U^\nu \to U$ as in Morphisms, Definition 29.54.1,

2. for $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ the morphism $\nu _ U$ is integral and $U^\nu$ is a normal scheme, see Morphisms, Lemma 29.54.5,

3. for every $g : U \to V \in \text{Arrows}(\mathcal{C})$ there is a unique morphism $g^\nu : U^\nu \to V^\nu$ such that $\nu _ V \circ g^\nu = g \circ \nu _ U$, see Morphisms, Lemma 29.54.5 part (4) applied to the composition $X^\nu \to X \to Y$,

4. if $V \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ and $g : U \to V$ is étale or smooth, then $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ and $g \in \text{Arrows}(\mathcal{C})$ and the morphism $g^\nu$ induces an isomorphism $U^\nu \to U \times _ V V^\nu$, see Lemma 66.49.4 and More on Morphisms, Lemma 37.19.3.

Our task is to extend this construction to the corresponding category of algebraic spaces $X$ over $S$.

Let $X$ be an algebraic space over $S$ satisfying the equivalent conditions of Lemma 66.49.1. Let $U \to X$ be a surjective étale morphism where $U$ is a scheme. Set $R = U \times _ X U$ with projections $s, t : R \to U$ and $j = (t, s) : R \to U \times _ S U$ so that $X = U/R$, see Spaces, Lemma 64.9.1. Observe that $U$ and $R$ are objects of $\mathcal{C}$ by our assumptions on $X$ and that the morphisms $s$ and $t$ are étale morphisms of schemes over $S$. By (a) we have the normalization morphisms $\nu _ U : U^\nu \to U$ and $\nu _ R : R^\nu \to R$, by (d) we have morphisms $s^\nu : R^\nu \to U^\nu$, $t^\nu : R^\nu \to U^\nu$ which define isomorphisms $R^\nu \to R \times _{s, U} U^\nu$ and $R^\nu \to U^\nu \times _{U, t} R$. It follows that $s^\nu$ and $t^\nu$ are étale (as they are isomorphic to base changes of étale morphisms). The induced morphism $j^\nu = (t^\nu , s^\nu ) : R^\nu \to U^\nu \times _ S U^\nu$ is a monomorphism as it is equal to the composition

\begin{align*} R^\nu & \to (U^\nu \times _{U, t} R) \times _ R (R \times _{s, U} U^\nu ) \\ & = U^\nu \times _{U, t} R \times _{s, U} U^\nu \\ & \xrightarrow {j} U^\nu \times _ U (U \times _ S U) \times _ U U^\nu \\ & = U^\nu \times _ S U^\nu \end{align*}

The first arrow is the diagonal morphism of $\nu _ R$. (This tells us that $R^\nu$ is a subscheme of the restriction of $R$ to $U^\nu$.) A formal computation with fibre products using property (d) shows that $R^\nu \times _{s^\nu , U^\nu , t^\nu } R^\nu$ is the normalization of $R \times _{s, U, t} R$. Hence the étale morphism $c : R \times _{s, U, t} R \to R$ extends uniquely to $c^\nu$ by (d). The morphism $c^\nu$ is compatible with the projection $\text{pr}_{13} : U^\nu \times _ S U^\nu \times _ S U^\nu \to U^\nu \times _ S U^\nu$. Similarly, there are morphisms $i^\nu : R^\nu \to R^\nu$ compatible with the morphism $U^\nu \times _ S U^\nu \to U^\nu \times _ S U^\nu$ which switches factors and there is a morphism $e^\nu : U^\nu \to R^\nu$ compatible with the diagonal morphism $U^\nu \to U^\nu \times _ S U^\nu$. All in all it follows that $j^\nu : R^\nu \to U^\nu \times _ S U^\nu$ is an étale equivalence relation. At this point we may and do set $X^\nu = U^\nu /R^\nu$ (Spaces, Theorem 64.10.5). Then we see that we have $U^\nu = X^\nu \times _ X U$ by Groupoids, Lemma 39.20.7.

What have we shown in the previous paragraph is this: for every algebraic space $X$ over $S$ satisfying the equivalent conditions of Lemma 66.49.1 if we choose a surjective étale morphism $g : U \to X$ where $U$ is a scheme, then we obtain a cartesian diagram

$\xymatrix{ X^\nu \ar[d]_{\nu _ X} & U^\nu \ar[l]^{g^\nu } \ar[d]^{\nu _ U} \\ X & U \ar[l]_ g }$

of algebraic spaces. This immediately implies that $X^\nu$ is a normal algebraic space and that $\nu _ X$ is a integral morphism. This gives part (1) of the lemma.

We will show below that the morphism $\nu _ X : X^\nu \to X$ up to unique isomorphism is independent of the choice of $g$, but for now, if $X$ is a scheme, we choose $\text{id} : X \to X$ so that it is clear that we have part (2) of the lemma.

We still have to prove parts (3) and (4). Let $g : U \to X$ and $\nu _ X : X^\nu \to X$ and $g^\nu : U^\nu \to X^\nu$ be as above. Let $Z$ be a normal scheme and let $h : Z \to U$ and $a : Z \to X^\nu$ be morphisms over $S$ such that $g \circ h = \nu _ X \circ a$ and such that every irreducible compoent of $Z$ dominates an irreducible component of $U$ (via $h$). By Morphisms, Lemma 29.54.5 part (4) we obtain a unique morphism $h^\nu : Z \to U^\nu$ such that $h = \nu _ U \circ h^\nu$. Picture:

$\xymatrix{ X^\nu \ar[d]_{\nu _ X} & U^\nu \ar[l]^{g^\nu } \ar[d]^{\nu _ U} & Z \ar[l]^{h^\nu } \ar@/_1em/[ll]_ a \ar[dl]^ h \\ X & U \ar[l]_ g }$

Observe that $a = g^\nu \circ h^\nu$. Namely, since the square with corners $X^\nu$, $X$, $U^\nu$, $U$ is cartesian, this follows immediately from the fact that $h^\nu$ is unique (given $h$). In other words, given $h : Z \to U$ as above (and not $a$) there is a unique morphism $a : Z \to X^\nu$ with $\nu _ X \circ a = g \circ h$.

Let $f : X \to Y$ be as in part (3) of the statement of the lemma. Suppose we have chosen surjective étale morphisms $U \to X$ and $V \to Y$ where $U$ and $V$ are schemes such that $f$ lifts to a morphism $g : U \to V$. Then $g \in \text{Arrows}(\mathcal{C})$ and we obtain a unique morphism $g^\nu : U^\nu \to V^\nu$ compatible with $\nu _ U$ and $\nu _ V$. However, then the two morphisms

$R^\nu = U^\nu \times _{X^\nu } U^\nu \to U^\nu \to V^\nu \to Y^\nu$

must be the same by our comments in the previous paragraph (applied with $Y$ in stead of $X$). Since $X^\nu$ is constructed by taking the quotient of $U^\nu$ by $R^\nu$ it follows that we obtain a (unique) morphism $f^\nu : X^\nu \to Y^\nu$ as stated in (3).

To see that the construction of $X^\nu$ is independent of the choice of $g : U \to X$ surjective étale, apply the construction in the previous paragraph to $\text{id} : X \to X$ and a morphism $U' \to U$ between étale coverings of $X$. This is enough because given any two étale coverings of $X$ there is a third one which dominates both. The reader shows that the morphism between the two normalizations constructed using either $U' \to X$ or $U \to X$ becomes an isomorphism after base change to $U'$ and hence was an isomorphism. We omit the details.

We omit the proof of (4) which is similar; hint use part (d) above. $\square$

This leads us to the following definition.

Definition 66.49.6. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$ satisfying the equivalent conditions of Lemma 66.49.1. We define the normalization of $X$ as the morphism

$\nu _ X : X^\nu \longrightarrow X$

constructed in Lemma 66.49.5.

The definition applies to locally Noetherian algebraic spaces, see Lemma 66.49.2. Usually the normalization is defined only for reduced algebraic spaces. With the definition above the normalization of $X$ is the same as the normalization of the reduction $X_{red}$ of $X$.

Lemma 66.49.7. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$ satisfying the equivalent conditions of Lemma 66.49.1. The normalization morphism $\nu$ factors through the reduction $X_{red}$ and $X^\nu \to X_{red}$ is the normalization of $X_{red}$.

Proof. We may check this étale locally on $X$ and hence reduce to the case of schemes which is Morphisms, Lemma 29.54.2. Some details omitted. $\square$

Lemma 66.49.8. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$ satisfying the equivalent conditions of Lemma 66.49.1.

1. The normalization $X^\nu$ is normal.

2. The morphism $\nu : X^\nu \to X$ is integral and surjective.

3. The map $|\nu | : |X^\nu | \to |X|$ induces a bijection between the sets of points of codimension $0$ (Properties of Spaces, Definition 65.10.2).

4. Let $Z \to X$ be a morphism. Assume $Z$ is a normal algebraic space and that for $z \in |Z|$ we have: $z$ has codimension $0$ in $Z \Rightarrow f(z)$ has codimension $0$ in $X$. Then there exists a unique factorization $Z \to X^\nu \to X$.

Proof. Properties (1), (2), and (3) follow from the corresponding results for schemes (Morphisms, Lemma 29.54.5) combined with the fact that a point of a scheme is a generic point of an irreducible component if and only if the dimension of the local ring is zero (Properties, Lemma 28.10.4).

Let $Z \to X$ be a morphism as in (4). Let $U$ be a scheme and let $U \to X$ be a surjective étale morphism. Choose a scheme $V$ and a surjective étale morphism $V \to U \times _ X Z$. The condition on codimension $0$ points assures us that $V \to U$ maps generic points of irreducible components of $V$ to generic points of irreducible components of $U$. Thus we obtain a unique factorization $V \to U^\nu \to U$ by Morphisms, Lemma 29.54.5. The uniqueness guarantees us that the two maps

$V \times _{U \times _ X Z} V \to V \to U^\nu$

agree because these maps are the unique factorization of the map $V \times _{U \times _ X Z} V \to V \to U$. Since the algebraic space $U \times _ X Z$ is equal to the quotient $V/V \times _{U \times _ X Z} V$ (see Spaces, Section 64.9) we find a canonical morphism $U \times _ X Z \to U^\nu$. Picture

$\xymatrix{ U \times _ X Z \ar[r] \ar[d] & U^\nu \ar[r] \ar[d] & U \ar[d] \\ Z \ar@/_/[rr] \ar@{..>}[r] & X^\nu \ar[r] & X }$

To obtain the dotted arrow we note that the construction of the arrow $U \times _ X Z \to U^\nu$ is functorial in the étale morphism $U \to X$ (precise formulation and proof omitted). Hence if we set $R = U \times _ X U$ with projections $s, t : R \to U$, then we obtain a morphism $R \times _ X Z \to R^\nu$ commuting with $s, t : R \to U$ and $s^\nu , t^\nu : R^\nu \to U^\nu$. Recall that $X^\nu = U^\nu /R^\nu$, see proof of Lemma 66.49.5. Since $X = U/R$ a simple sheaf theoretic argument shows that $Z = (U \times _ X Z)/(R \times _ X Z)$. Thus the morphisms $U \times _ X Z \to U^\nu$ and $R \times _ X Z \to R^\nu$ define a morphism $Z \to X^\nu$ as desired. $\square$

Lemma 66.49.9. Let $S$ be a scheme. Let $X$ be a Nagata algebraic space over $S$. The normalization $\nu : X^\nu \to X$ is a finite morphism.

Proof. Since $X$ being Nagata is locally Noetherian, Definition 66.49.6 applies. By construction of $X^\nu$ in Lemma 66.49.5 we immediately reduce to the case of schemes which is Morphisms, Lemma 29.54.10. $\square$

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