The Stacks project

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 37.55.1 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 37.55.1 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. $\square$


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