The Stacks project

Lemma 58.5.2. Let $X$ be a scheme. The category $\textit{FÉt}_ X$ has finite limits and finite colimits and for any morphism $X' \to X$ the base change functor $\textit{FÉt}_ X \to \textit{FÉt}_{X'}$ is exact.

Proof. Finite limits and left exactness. By Categories, Lemma 4.18.4 it suffices to show that $\textit{FÉt}_ X$ has a final object and fibred products. This is clear because the category of all schemes over $X$ has a final object (namely $X$) and fibred products. Also, fibred products of schemes finite étale over $X$ are finite étale over $X$. Moreover, it is clear that base change commutes with these operations and hence base change is left exact (Categories, Lemma 4.23.2).

Finite colimits and right exactness. By Categories, Lemma 4.18.7 it suffices to show that $\textit{FÉt}_ X$ has finite coproducts and coequalizers. Finite coproducts are given by disjoint unions (the empty coproduct is the empty scheme). Let $a, b : Z \to Y$ be two morphisms of $\textit{FÉt}_ X$. Since $Z \to X$ and $Y \to X$ are finite étale we can write $Z = \underline{\mathop{\mathrm{Spec}}}(\mathcal{C})$ and $Y = \underline{\mathop{\mathrm{Spec}}}(\mathcal{B})$ for some finite locally free $\mathcal{O}_ X$-algebras $\mathcal{C}$ and $\mathcal{B}$. The morphisms $a, b$ induce two maps $a^\sharp , b^\sharp : \mathcal{B} \to \mathcal{C}$. Let $\mathcal{A} = \text{Eq}(a^\sharp , b^\sharp )$ be their equalizer. If

\[ \underline{\mathop{\mathrm{Spec}}}(\mathcal{A}) \longrightarrow X \]

is finite étale, then it is clear that this is the coequalizer (after all we can write any object of $\textit{FÉt}_ X$ as the relative spectrum of a sheaf of $\mathcal{O}_ X$-algebras). This we may do after replacing $X$ by the members of an étale covering (Descent, Lemmas 35.23.23 and 35.23.29). Thus by Étale Morphisms, Lemma 41.18.3 we may assume that $Y = \coprod _{i = 1, \ldots , n} X$ and $Z = \coprod _{j = 1, \ldots , m} X$. Then

\[ \mathcal{C} = \prod \nolimits _{1 \leq j \leq m} \mathcal{O}_ X \quad \text{and}\quad \mathcal{B} = \prod \nolimits _{1 \leq i \leq n} \mathcal{O}_ X \]

After a further replacement by the members of an open covering we may assume that $a, b$ correspond to maps $a_ s, b_ s : \{ 1, \ldots , m\} \to \{ 1, \ldots , n\} $, i.e., the summand $X$ of $Z$ corresponding to the index $j$ maps into the summand $X$ of $Y$ corresponding to the index $a_ s(j)$, resp. $b_ s(j)$ under the morphism $a$, resp. $b$. Let $\{ 1, \ldots , n\} \to T$ be the coequalizer of $a_ s, b_ s$. Then we see that

\[ \mathcal{A} = \prod \nolimits _{t \in T} \mathcal{O}_ X \]

whose spectrum is certainly finite étale over $X$. We omit the verification that this is compatible with base change. Thus base change is a right exact functor. $\square$

Comments (4)

Comment #6532 by Tim Holzschuh on

In the proof of the existence of finite limits, it is written:

"... and fibred products and fibred products of schemes finite ..."

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