The Stacks project

37.76 Completely decomposed morphisms

Nishnevich studied the notion of a completely decomposed family of ├ętale morphisms, in order to define what is now called the Nishnevich topology, see for example [Nishnevich].

Definition 37.76.1. A morphism $f : X \to Y$ of schemes is said to be completely decomposed1 if for all points $y \in Y$ there is a point $x \in X$ with $f(x) = y$ such that the field extension $\kappa (x)/\kappa (y)$ is trivial. A family of morphisms $\{ f_ i : X_ i \to Y\} _{i \in I}$ of schemes with fixed target is said to be completely decomposed if $\coprod f_ i : \coprod Y_ i \to X$ is completely decomposed.

We start with some basic lemmas.

Lemma 37.76.2. The composition of two completely decomposed morphisms of schemes is completely decomposed. If $\{ f_ i : X_ i \to Y\} _{i \in I}$ is completely decomposed and for each $i$ we have a family $\{ X_{ij} \to X_ i\} _{j \in J_ i}$ which is completely decomposed, then the family $\{ X_{ij} \to Y\} _{i \in I, j \in J_ i}$ is completely decomposed.

Proof. Omitted. $\square$

Lemma 37.76.3. The base change of a completely decomposed morphism of schemes is completely decomposed. If $\{ f_ i : X_ i \to Y\} _{i \in I}$ is completely decomposed and $Y' \to Y$ is a morphism of schemes, then $\{ X_ i \times _ Y Y' \to Y'\} _{i \in I}$ is completely decomposed.

Proof. Let $f : X \to Y$ and $g : Y' \to Y$ be morphisms of schemes. Let $y' \in Y'$ be a point with image $y = g(y')$ in $Y$. If $x \in X$ is a point such that $f(x) = y$ and $\kappa (x) = \kappa (y)$, then there exists a unique point $x' \in X' = X \times _ Y Y'$ which maps to $y'$ in $Y'$ and to $x$ in $X$ and moreover $\kappa (x') = \kappa (y')$, see Schemes, Lemma 26.17.5. From this fact the lemma follows easily; we omit the details. $\square$

reference

Lemma 37.76.4. Let $f : X \to Y$ be a morphism of schemes. Assume $f$ is completely decomposed, $f$ is locally of finite presentation, and $Y$ is quasi-compact and quasi-separated. Then there exist $n \geq 0$ and morphisms $Z_ i \to Y$, $i = 1, \ldots , n$ with the following properties

  1. $\coprod Z_ i \to Y$ is surjective,

  2. $Z_ i \to Y$ is an immersion for all $i$,

  3. $Z_ i \to Y$ is of finite presentation for all $i$, and

  4. the base change $X \times _ Y Z_ i \to Z_ i$ has a section for all $i$.

Proof. Let $y \in Y$. By assumption there is a morphism $\sigma : \mathop{\mathrm{Spec}}(\kappa (y)) \to X$ over $Y$. We can write $\mathop{\mathrm{Spec}}(\kappa (y))$ as a directed limit of affine schemes $Z$ over $Y$ such that $Z \to Y$ is an immersion of finite presentation. Namely, choose an affine open $y \in \mathop{\mathrm{Spec}}(A) \subset Y$ and say $y$ corresponds to the prime ideal $\mathfrak p$ of $A$. Then $\kappa (\mathfrak p)$ is the filtered colimit of the rings $(A/I)_ f$ where $I \subset \mathfrak p$ is a finitely generated ideal and $f \in A$, $f \not\in \mathfrak p$. The morphisms $Z = \mathop{\mathrm{Spec}}((A/I)_ f) \to Y$ are immersions of finite presentation; quasi-compactness of $Z \to Y$ follows as $Y$ is quasi-separated, see Schemes, Lemma 26.21.14. By Limits, Proposition 32.6.1 for some such $Z$ there is a morphism $\sigma ' : Z \to X$ over $Y$ agreeing with $\sigma $ on the spectrum of $\kappa (\mathfrak p)$. Since $\sigma '$ is a morphism over $Y$, we obtain a section of the projection $X \times _ Y Z \to Z$

We conclude that $Y$ is the union of the images of immersions $Z \to Y$ of finite presentation such that $X \times _ Y Z \to Z$ has a section. Since the image of $Z \to Y$ is constructible (Morphisms, Lemma 29.22.2) and since $Y$ is compact in the constructible topology (Properties, Lemma 28.2.4 and Topology, Lemma 5.23.2), we see that a finite number of these suffice. $\square$

Lemma 37.76.5. Let $S = \mathop{\mathrm{lim}}\nolimits _{\lambda \in \Lambda } S_\lambda $ be a limit of a directed system of schemes with affine transition morphisms. Let $0 \in \Lambda $ and let $f_0 : X_0 \to Y_0$ be a morphism of schemes over $S_0$. For $\lambda \geq 0$ let $f_\lambda : X_\lambda \to Y_\lambda $ be the base change of $f_0$ to $S_\lambda $ and let $f : X \to Y$ be the base change of $f_0$ to $S$. If

  1. $f$ is completely decomposed,

  2. $Y_0$ is quasi-compact and quasi-separated, and

  3. $f_0$ is locally of finite presentation,

then there exists an $\lambda \geq 0$ such that $f_\lambda $ is completely decomposed.

Proof. Since $Y_0$ is quasi-compact and quasi-separated, the scheme $Y$, which is affine over $Y_0$, is quasi-compact and quasi-separated. Choose $n \geq 0$ and $Z_ i \to Y$, $i = 1, \ldots , n$ as in Lemma 37.76.4. Denote $\sigma _ i : Z_ i \to X$ morphisms over $Y$ which exist by our choice of $Z_ i$. After increasing $0 \in \Lambda $ we may assume there exist morphisms $Z_{i, 0} \to Y_0$ of finite presentation whose base changes to $S$ are the morphisms $Z_ i \to Y$, see Limits, Lemma 32.10.1. By Limits, Lemma 32.8.13 we may assume, after possibly increasing $0$, that $Z_{i, 0} \to Y_0$ is an immersion. Since $\coprod Z_ i \to Y$ is surjective, we may assume, after possibly increasing $0$, that $\coprod Z_{i, 0} \to Y_0$ is surjective, see Limits, Lemma 32.8.15. Observe that $Z_ i = \mathop{\mathrm{lim}}\nolimits _{\lambda \geq 0} Z_{i, \lambda }$ where $Z_{i, \lambda } = Y_\lambda \times _{Y_0} Z_{i, 0}$. Let us view the compositions

\[ Z_ i \xrightarrow {\sigma _ i} X \to X_0 \]

as morphisms over $Y_0$. Since $f_0$ is locally of finite presentation by Limits, Proposition 32.6.1 we can find a $\lambda \geq 0$ such that there exist morphisms $\sigma '_{i, \lambda } : Z_{i, \lambda } \to X_0$ over $Y_0$ whose precomposition with $Z_ i \to Z_{i, \lambda }$ are the displayed arrows. Of course, then $\sigma '_{i, \lambda }$ determines a morphism $\sigma _{i, \lambda } : Z_{i, \lambda } \to X_\lambda = X_0 \times _{Y_0} Y_\lambda $ over $Y_\lambda $. Since $\coprod Z_{i, \lambda } \to Y_\lambda $ is surjective we conclude that $X_\lambda \to Y_\lambda $ is completely decomposed. $\square$

[1] This may be nonstandard terminology.

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