The Stacks project

41.20 Descending étale morphisms

In order to understand the language used in this section we encourage the reader to take a look at Descent, Section 35.34. Let $f : X \to S$ be a morphism of schemes. Consider the pullback functor

41.20.0.1
\begin{equation} \label{etale-equation-descent-etale} \text{schemes }U\text{ étale over }S \longrightarrow \begin{matrix} \text{descent data }(V, \varphi )\text{ relative to }X/S \\ \text{ with }V\text{ étale over }X \end{matrix} \end{equation}

sending $U$ to the canonical descent datum $(X \times _ S U, can)$.

Proof. Let $a, b : U_1 \to U_2$ be two morphisms between schemes étale over $S$. Assume the base changes of $a$ and $b$ to $X$ agree. We have to show that $a = b$. By Proposition 41.6.3 it suffices to show that $a$ and $b$ agree on points and residue fields. This is clear because for every $u \in U_1$ we can find a point $v \in X \times _ S U_1$ mapping to $u$. $\square$

Lemma 41.20.2. Assume $f : X \to S$ is submersive and any étale base change of $f$ is submersive. Then the functor (41.20.0.1) is fully faithful.

Proof. By Lemma 41.20.1 the functor is faithful. Let $U_1 \to S$ and $U_2 \to S$ be étale morphisms and let $a : X \times _ S U_1 \to X \times _ S U_2$ be a morphism compatible with canonical descent data. We will prove that $a$ is the base change of a morphism $U_1 \to U_2$.

Let $U'_2 \subset U_2$ be an open subscheme. Consider $W = a^{-1}(X \times _ S U'_2)$. This is an open subscheme of $X \times _ S U_1$ which is compatible with the canonical descent datum on $V_1 = X \times _ S U_1$. This means that the two inverse images of $W$ by the projections $V_1 \times _{U_1} V_1 \to V_1$ agree. Since $V_1 \to U_1$ is surjective (as the base change of $X \to S$) we conclude that $W$ is the inverse image of some subset $U'_1 \subset U_1$. Since $W$ is open, our assumption on $f$ implies that $U'_1 \subset U_1$ is open.

Let $U_2 = \bigcup U_{2, i}$ be an affine open covering. By the result of the preceding paragraph we obtain an open covering $U_1 = \bigcup U_{1, i}$ such that $X \times _ S U_{1, i} = a^{-1}(X \times _ S U_{2, i})$. If we can prove there exists a morphism $U_{1, i} \to U_{2, i}$ whose base change is the morphism $a_ i : X \times _ S U_{1, i} \to X \times _ S U_{2, i}$ then we can glue these morphisms to a morphism $U_1 \to U_2$ (using faithfulness). In this way we reduce to the case that $U_2$ is affine. In particular $U_2 \to S$ is separated (Schemes, Lemma 26.21.13).

Assume $U_2 \to S$ is separated. Then the graph $\Gamma _ a$ of $a$ is a closed subscheme of

\[ V = (X \times _ S U_1) \times _ X (X \times _ S U_2) = X \times _ S U_1 \times _ S U_2 \]

by Schemes, Lemma 26.21.10. On the other hand the graph is open for example because it is a section of an étale morphism (Proposition 41.6.1). Since $a$ is a morphism of descent data, the two inverse images of $\Gamma _ a \subset V$ under the projections $V \times _{U_1 \times _ S U_2} V \to V$ are the same. Hence arguing as in the second paragraph of the proof we find an open and closed subscheme $\Gamma \subset U_1 \times _ S U_2$ whose base change to $X$ gives $\Gamma _ a$. Then $\Gamma \to U_1$ is an étale morphism whose base change to $X$ is an isomorphism. This means that $\Gamma \to U_1$ is universally bijective, hence an isomorphism by Theorem 41.14.1. Thus $\Gamma $ is the graph of a morphism $U_1 \to U_2$ and the base change of this morphism is $a$ as desired. $\square$

Lemma 41.20.3. Let $f : X \to S$ be a morphism of schemes. In the following cases the functor (41.20.0.1) is fully faithful:

  1. $f$ is surjective and universally closed (e.g., finite, integral, or proper),

  2. $f$ is surjective and universally open (e.g., locally of finite presentation and flat, smooth, or etale),

  3. $f$ is surjective, quasi-compact, and flat.

Proof. This follows from Lemma 41.20.2. For example a closed surjective map of topological spaces is submersive (Topology, Lemma 5.6.5). Finite, integral, and proper morphisms are universally closed, see Morphisms, Lemmas 29.44.7 and 29.44.11 and Definition 29.41.1. On the other hand an open surjective map of topological spaces is submersive (Topology, Lemma 5.6.4). Flat locally finitely presented, smooth, and étale morphisms are universally open, see Morphisms, Lemmas 29.25.10, 29.34.10, and 29.36.13. The case of surjective, quasi-compact, flat morphisms follows from Morphisms, Lemma 29.25.12. $\square$

Lemma 41.20.4. Let $f : X \to S$ be a morphism of schemes. Let $(V, \varphi )$ be a descent datum relative to $X/S$ with $V \to X$ étale. Let $S = \bigcup S_ i$ be an open covering. Assume that

  1. the pullback of the descent datum $(V, \varphi )$ to $X \times _ S S_ i/S_ i$ is effective,

  2. the functor (41.20.0.1) for $X \times _ S (S_ i \cap S_ j) \to (S_ i \cap S_ j)$ is fully faithful, and

  3. the functor (41.20.0.1) for $X \times _ S (S_ i \cap S_ j \cap S_ k) \to (S_ i \cap S_ j \cap S_ k)$ is faithful.

Then $(V, \varphi )$ is effective.

Proof. (Recall that pullbacks of descent data are defined in Descent, Definition 35.34.7.) Set $X_ i = X \times _ S S_ i$. Denote $(V_ i, \varphi _ i)$ the pullback of $(V, \varphi )$ to $X_ i/S_ i$. By assumption (1) we can find an étale morphism $U_ i \to S_ i$ which comes with an isomorphism $X_ i \times _{S_ i} U_ i \to V_ i$ compatible with $can$ and $\varphi _ i$. By assumption (2) we obtain isomorphisms $\psi _{ij} : U_ i \times _{S_ i} (S_ i \cap S_ j) \to U_ j \times _{S_ j} (S_ i \cap S_ j)$. By assumption (3) these isomorphisms satisfy the cocycle condition so that $(U_ i, \psi _{ij})$ is a descend datum for the Zariski covering $\{ S_ i \to S\} $. Then Descent, Lemma 35.35.10 (which is essentially just a reformulation of Schemes, Section 26.14) tells us that there exists a morphism of schemes $U \to S$ and isomorphisms $U \times _ S S_ i \to U_ i$ compatible with $\psi _{ij}$. The isomorphisms $U \times _ S S_ i \to U_ i$ determine corresponding isomorphisms $X_ i \times _ S U \to V_ i$ which glue to a morphism $X \times _ S U \to V$ compatible with the canonical descent datum and $\varphi $. $\square$

Lemma 41.20.5. Let $(A, I)$ be a henselian pair. Let $U \to \mathop{\mathrm{Spec}}(A)$ be a quasi-compact, separated, étale morphism such that $U \times _{\mathop{\mathrm{Spec}}(A)} \mathop{\mathrm{Spec}}(A/I) \to \mathop{\mathrm{Spec}}(A/I)$ is finite. Then

\[ U = U_{fin} \amalg U_{away} \]

where $U_{fin} \to \mathop{\mathrm{Spec}}(A)$ is finite and $U_{away}$ has no points lying over $Z$.

Proof. By Zariski's main theorem, the scheme $U$ is quasi-affine. In fact, we can find an open immersion $U \to T$ with $T$ affine and $T \to \mathop{\mathrm{Spec}}(A)$ finite, see More on Morphisms, Lemma 37.43.3. Write $Z = \mathop{\mathrm{Spec}}(A/I)$ and denote $U_ Z \to T_ Z$ the base change. Since $U_ Z \to Z$ is finite, we see that $U_ Z \to T_ Z$ is closed as well as open. Hence by More on Algebra, Lemma 15.11.6 we obtain a unique decomposition $T = T' \amalg T''$ with $T'_ Z = U_ Z$. Set $U_{fin} = U \cap T'$ and $U_{away} = U \cap T''$. Since $T'_ Z \subset U_ Z$ we see that all closed points of $T'$ are in $U$ hence $T' \subset U$, hence $U_{fin} = T'$, hence $U_{fin} \to \mathop{\mathrm{Spec}}(A)$ is finite. We omit the proof of uniqueness of the decomposition. $\square$

Proposition 41.20.6. Let $f : X \to S$ be a surjective integral morphism. The functor (41.20.0.1) induces an equivalence

\[ \begin{matrix} \text{schemes quasi-compact,} \\ \text{separated, étale over }S \end{matrix} \longrightarrow \begin{matrix} \text{descent data }(V, \varphi )\text{ relative to }X/S\text{ with} \\ V\text{ quasi-compact, separated, étale over }X \end{matrix} \]

Proof. By Lemma 41.20.3 the functor (41.20.0.1) is fully faithful and the same remains the case after any base change $S \to S'$. Let $(V, \varphi )$ be a descent data relative to $X/S$ with $V \to X$ quasi-compact, separated, and étale. We can use Lemma 41.20.4 to see that it suffices to prove the effectivity Zariski locally on $S$. In particular we may and do assume that $S$ is affine.

If $S$ is affine we can find a directed set $\Lambda $ and an inverse system $X_\lambda \to S_\lambda $ of finite morphisms of affine schemes of finite type over $\mathop{\mathrm{Spec}}(\mathbf{Z})$ such that $(X \to S) = \mathop{\mathrm{lim}}\nolimits (X_\lambda \to S_\lambda )$. See Algebra, Lemma 10.127.15. Since limits commute with limits we deduce that $X \times _ S X = \mathop{\mathrm{lim}}\nolimits X_\lambda \times _{S_\lambda } X_\lambda $ and $X \times _ S X \times _ S X = \mathop{\mathrm{lim}}\nolimits X_\lambda \times _{S_\lambda } X_\lambda \times _{S_\lambda } X_\lambda $. Observe that $V \to X$ is a morphism of finite presentation. Using Limits, Lemmas 32.10.1 we can find an $\lambda $ and a descent datum $(V_\lambda , \varphi _\lambda )$ relative to $X_\lambda /S_\lambda $ whose pullback to $X/S$ is $(V, \varphi )$. Of course it is enough to show that $(V_\lambda , \varphi _\lambda )$ is effective. Note that $V_\lambda $ is quasi-compact by construction. After possibly increasing $\lambda $ we may assume that $V_\lambda \to X_\lambda $ is separated and étale, see Limits, Lemma 32.8.6 and 32.8.10. Thus we may assume that $f$ is finite surjective and $S$ affine of finite type over $\mathbf{Z}$.

Consider an open $S' \subset S$ such that the pullback $(V', \varphi ')$ of $(V, \varphi )$ to $X' = X \times _ S S'$ is effective. Below we will prove, that $S' \not= S$ implies there is a strictly larger open over which the descent datum is effective. Since $S$ is Noetherian (and hence has a Noetherian underlying topological space) this will finish the proof. Let $\xi \in S$ be a generic point of an irreducible component of the closed subset $Z = S \setminus S'$. If $\xi \in S'' \subset S$ is an open over which the descent datum is effective, then the descent datum is effective over $S' \cup S''$ by the glueing argument of the first paragraph. Thus in the rest of the proof we may replace $S$ by an affine open neighbourhood of $\xi $.

After a first such replacement we may assume that $Z$ is irreducible with generic point $Z$. Let us endow $Z$ with the reduced induced closed subscheme structure. After another shrinking we may assume $X_ Z = X \times _ S Z = f^{-1}(Z) \to Z$ is flat, see Morphisms, Proposition 29.27.1. Let $(V_ Z, \varphi _ Z)$ be the pullback of the descent datum to $X_ Z/Z$. By More on Morphisms, Lemma 37.57.1 this descent datum is effective and we obtain an étale morphism $U_ Z \to Z$ whose base change is isomorphic to $V_ Z$ in a manner compatible with descent data. Of course $U_ Z \to Z$ is quasi-compact and separated (Descent, Lemmas 35.23.1 and 35.23.6). Thus after shrinking once more we may assume that $U_ Z \to Z$ is finite, see Morphisms, Lemma 29.51.1.

Let $S = \mathop{\mathrm{Spec}}(A)$ and let $I \subset A$ be the prime ideal corresponding to $Z \subset S$. Let $(A^ h, IA^ h)$ be the henselization of the pair $(A, I)$. Denote $S^ h = \mathop{\mathrm{Spec}}(A^ h)$ and $Z^ h = V(IA^ h) \cong Z$. We claim that it suffices to show effectivity after base change to $S^ h$. Namely, $\{ S^ h \to S, S' \to S\} $ is an fpqc covering ($A \to A^ h$ is flat by More on Algebra, Lemma 15.12.2) and by More on Morphisms, Lemma 37.57.1 we have fpqc descent for separated étale morphisms. Namely, if $U^ h \to S^ h$ and $U' \to S'$ are the objects corresponding to the pullbacks $(V^ h, \varphi ^ h)$ and $(V', \varphi ')$, then the required isomorphisms

\[ U^ h \times _ S S^ h \to S^ h \times _ S V^ h \quad \text{and}\quad U^ h \times _ S S' \to S^ h \times _ S U' \]

are obtained by the fully faithfulness pointed out in the first paragraph. In this way we reduce to the situation described in the next paragraph.

Here $S = \mathop{\mathrm{Spec}}(A)$, $Z = V(I)$, $S' = S \setminus Z$ where $(A, I)$ is a henselian pair, we have $U' \to S'$ corresponding to the descent datum $(V', \varphi ')$ and we have a finite étale morphism $U_ Z \to Z$ corresponding to the descent datum $(V_ Z, \varphi _ Z)$. We no longer have that $A$ is of finite type over $\mathbf{Z}$; but the rest of the argument will not even use that $A$ is Noetherian. By More on Algebra, Lemma 15.13.2 we can find a finite étale morphism $U_{fin} \to S$ whose restriction to $Z$ is isomorphic to $U_ Z \to Z$. Write $X = \mathop{\mathrm{Spec}}(B)$ and $Y = V(IB)$. Since $(B, IB)$ is a henselian pair (More on Algebra, Lemma 15.11.8) and since the restriction $V \to X$ to $Y$ is finite (as base change of $U_ Z \to Z$) we see that there is a canonical disjoint union decomposition

\[ V = V_{fin} \amalg V_{away} \]

were $V_{fin} \to X$ is finite and where $V_{away}$ has no points lying over $Y$. See Lemma 41.20.5. Using the uniqueness of this decomposition over $X \times _ S X$ we see that $\varphi $ preserves it and we obtain

\[ (V, \varphi ) = (V_{fin}, \varphi _{fin}) \amalg (V_{away}, \varphi _{away}) \]

in the category of descent data. By More on Algebra, Lemma 15.13.2 there is a unique isomorphism

\[ X \times _ S U_{fin} \longrightarrow V_{fin} \]

compatible with the given isomorphism $Y \times _ Z U_ Z \to V \times _ X Y$ over $Y$. By the uniqueness we see that this isomorphism is compatible with descent data, i.e., $(X \times _ S U_{fin}, can) \cong (V_{fin}, \varphi _{fin})$. Denote $U'_{fin} = U_{fin} \times _ S S'$. By fully faithfulness we obtain a morphism $U'_{fin} \to U'$ which is the inclusion of an open (and closed) subscheme. Then we set $U = U_{fin} \amalg _{U'_{fin}} U'$ (glueing of schemes as in Schemes, Section 26.14). The morphisms $X \times _ S U_{fin} \to V$ and $X \times _ S U' \to V$ glue to a morphism $X \times _ S U \to V$ which is the desired isomorphism. $\square$


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