Lemma 58.8.3. Let $X \subset X'$ be a thickening of schemes. The functor

$\textit{FÉt}_{X'} \longrightarrow \textit{FÉt}_ X,\quad U' \longmapsto U' \times _{X'} X$

is an equivalence of categories.

Proof. For a discussion of thickenings see More on Morphisms, Section 37.2. Let $U' \to X'$ be an étale morphism such that $U = U' \times _{X'} X \to X$ is finite étale. Then $U' \to X'$ is finite étale as well. This follows for example from More on Morphisms, Lemma 37.3.4. Now, if $X \subset X'$ is a finite order thickening then this remark combined with Étale Morphisms, Theorem 41.15.2 proves the lemma. Below we will prove the lemma for general thickenings, but we suggest the reader skip the proof.

Let $X' = \bigcup X_ i'$ be an affine open covering. Set $X_ i = X \times _{X'} X_ i'$, $X_{ij}' = X'_ i \cap X'_ j$, $X_{ij} = X \times _{X'} X_{ij}'$, $X_{ijk}' = X'_ i \cap X'_ j \cap X'_ k$, $X_{ijk} = X \times _{X'} X_{ijk}'$. Suppose that we can prove the theorem for each of the thickenings $X_ i \subset X'_ i$, $X_{ij} \subset X_{ij}'$, and $X_{ijk} \subset X_{ijk}'$. Then the result follows for $X \subset X'$ by relative glueing of schemes, see Constructions, Section 27.2. Observe that the schemes $X_ i'$, $X_{ij}'$, $X_{ijk}'$ are each separated as open subschemes of affine schemes. Repeating the argument one more time we reduce to the case where the schemes $X'_ i$, $X_{ij}'$, $X_{ijk}'$ are affine.

In the affine case we have $X' = \mathop{\mathrm{Spec}}(A')$ and $X = \mathop{\mathrm{Spec}}(A'/I')$ where $I'$ is a locally nilpotent ideal. Then $(A', I')$ is a henselian pair (More on Algebra, Lemma 15.11.2) and the result follows from Lemma 58.8.2 (which is much easier in this case). $\square$

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