41.15 Topological invariance of the étale topology

Next, we present an extremely crucial theorem which, roughly speaking, says that étaleness is a topological property.

Theorem 41.15.1. Let $X$ and $Y$ be two schemes over a base scheme $S$. Let $S_0$ be a closed subscheme of $S$ with the same underlying topological space (for example if the ideal sheaf of $S_0$ in $S$ has square zero). Denote $X_0$ (resp. $Y_0$) the base change $S_0 \times _ S X$ (resp. $S_0 \times _ S Y$). If $X$ is étale over $S$, then the map

$\mathop{\mathrm{Mor}}\nolimits _ S(Y, X) \longrightarrow \mathop{\mathrm{Mor}}\nolimits _{S_0}(Y_0, X_0)$

is bijective.

Proof. After base changing via $Y \to S$, we may assume that $Y = S$. In this case the theorem states that any $S$-morphism $\sigma _0 : S_0 \to X$ actually factors uniquely through a section $S \to X$ of the étale structure morphism $f : X \to S$.

Uniqueness. Suppose we have two sections $\sigma , \sigma '$ through which $\sigma _0$ factors. Because $X \to S$ is étale we see that $\Delta : X \to X \times _ S X$ is an open immersion (Morphisms, Lemma 29.35.13). The morphism $(\sigma , \sigma ') : S \to X \times _ S X$ factors through this open because for any $s \in S$ we have $(\sigma , \sigma ')(s) = (\sigma _0(s), \sigma _0(s))$. Thus $\sigma = \sigma '$.

To prove existence we first reduce to the affine case (we suggest the reader skip this step). Let $X = \bigcup X_ i$ be an affine open covering such that each $X_ i$ maps into an affine open $S_ i$ of $S$. For every $s \in S$ we can choose an $i$ such that $\sigma _0(s) \in X_ i$. Choose an affine open neighbourhood $U \subset S_ i$ of $s$ such that $\sigma _0(U_0) \subset X_{i, 0}$. Note that $X' = X_ i \times _ S U = X_ i \times _{S_ i} U$ is affine. If we can lift $\sigma _0|_{U_0} : U_0 \to X'_0$ to $U \to X'$, then by uniqueness these local lifts will glue to a global morphism $S \to X$. Thus we may assume $S$ and $X$ are affine.

Existence when $S$ and $X$ are affine. Write $S = \mathop{\mathrm{Spec}}(A)$ and $X = \mathop{\mathrm{Spec}}(B)$. Then $A \to B$ is étale and in particular smooth (of relative dimension $0$). As $|S_0| = |S|$ we see that $S_0 = \mathop{\mathrm{Spec}}(A/I)$ with $I \subset A$ locally nilpotent. Thus existence follows from Algebra, Lemma 10.138.17. $\square$

From the proof of preceding theorem, we also obtain one direction of the promised functorial characterization of étale morphisms. The following theorem will be strengthened in Étale Cohomology, Theorem 59.45.2.

Theorem 41.15.2 (Une equivalence remarquable de catégories). Let $S$ be a scheme. Let $S_0 \subset S$ be a closed subscheme with the same underlying topological space (for example if the ideal sheaf of $S_0$ in $S$ has square zero). The functor

$X \longmapsto X_0 = S_0 \times _ S X$

defines an equivalence of categories

$\{ \text{schemes }X\text{ étale over }S \} \leftrightarrow \{ \text{schemes }X_0\text{ étale over }S_0 \}$

Proof. By Theorem 41.15.1 we see that this functor is fully faithful. It remains to show that the functor is essentially surjective. Let $Y \to S_0$ be an étale morphism of schemes.

Suppose that the result holds if $S$ and $Y$ are affine. In that case, we choose an affine open covering $Y = \bigcup V_ j$ such that each $V_ j$ maps into an affine open of $S$. By assumption (affine case) we can find étale morphisms $W_ j \to S$ such that $W_{j, 0} \cong V_ j$ (as schemes over $S_0$). Let $W_{j, j'} \subset W_ j$ be the open subscheme whose underlying topological space corresponds to $V_ j \cap V_{j'}$. Because we have isomorphisms

$W_{j, j', 0} \cong V_ j \cap V_{j'} \cong W_{j', j, 0}$

as schemes over $S_0$ we see by fully faithfulness that we obtain isomorphisms $\theta _{j, j'} : W_{j, j'} \to W_{j', j}$ of schemes over $S$. We omit the verification that these isomorphisms satisfy the cocycle condition of Schemes, Section 26.14. Applying Schemes, Lemma 26.14.2 we obtain a scheme $X \to S$ by glueing the schemes $W_ j$ along the identifications $\theta _{j, j'}$. It is clear that $X \to S$ is étale and $X_0 \cong Y$ by construction.

Thus it suffices to show the lemma in case $S$ and $Y$ are affine. Say $S = \mathop{\mathrm{Spec}}(R)$ and $S_0 = \mathop{\mathrm{Spec}}(R/I)$ with $I$ locally nilpotent. By Algebra, Lemma 10.143.2 we know that $Y$ is the spectrum of a ring $\overline{A}$ with

$\overline{A} = (R/I)[x_1, \ldots , x_ n]/(\overline{f}_1, \ldots , \overline{f}_ n)$

such that

$\overline{g} = \det \left( \begin{matrix} \partial \overline{f}_1/\partial x_1 & \partial \overline{f}_2/\partial x_1 & \ldots & \partial \overline{f}_ n/\partial x_1 \\ \partial \overline{f}_1/\partial x_2 & \partial \overline{f}_2/\partial x_2 & \ldots & \partial \overline{f}_ n/\partial x_2 \\ \ldots & \ldots & \ldots & \ldots \\ \partial \overline{f}_1/\partial x_ n & \partial \overline{f}_2/\partial x_ n & \ldots & \partial \overline{f}_ n/\partial x_ n \end{matrix} \right)$

maps to an invertible element in $\overline{A}$. Choose any lifts $f_ i \in R[x_1, \ldots , x_ n]$. Set

$A = R[x_1, \ldots , x_ n]/(f_1, \ldots , f_ n)$

Since $I$ is locally nilpotent the ideal $IA$ is locally nilpotent (Algebra, Lemma 10.32.3). Observe that $\overline{A} = A/IA$. It follows that the determinant of the matrix of partials of the $f_ i$ is invertible in the algebra $A$ by Algebra, Lemma 10.32.4. Hence $R \to A$ is étale and the proof is complete. $\square$

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