The Stacks project

Lemma 65.49.5. Let $S$ be a scheme. For every algebraic space $X$ over $S$ satisfying the equivalent conditions of Lemma 65.49.1 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 65.49.1 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 65.49.1 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 65.49.1, 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 65.49.4 and More on Morphisms, Lemma 37.17.3.

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 65.49.1. 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 63.9.1. 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 63.10.5). 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 65.49.1 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. $\square$


Comments (0)


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 07U4. Beware of the difference between the letter 'O' and the digit '0'.