Example 58.5.1. Let $k$ be an algebraically closed field and $X = \mathop{\mathrm{Spec}}(k)$. In this case $\textit{FÉt}_ X$ is equivalent to the category of finite sets. This works more generally when $k$ is separably algebraically closed. The reason is that a scheme étale over $k$ is the disjoint union of spectra of fields finite separable over $k$, see Morphisms, Lemma 29.36.7.

## 58.5 Finite étale morphisms

In this section we prove enough basic results on finite étale morphisms to be able to construct the étale fundamental group.

Let $X$ be a scheme. We will use the notation $\textit{FÉt}_ X$ to denote the category of schemes finite and étale over $X$. Thus

an object of $\textit{FÉt}_ X$ is a finite étale morphism $Y \to X$ with target $X$, and

a morphism in $\textit{FÉt}_ X$ from $Y \to X$ to $Y' \to X$ is a morphism $Y \to Y'$ making the diagram

\[ \xymatrix{ Y \ar[rr] \ar[rd] & & Y' \ar[ld] \\ & X } \]commute.

We will often call an object of $\textit{FÉt}_ X$ a *finite étale cover* of $X$ (even if $Y$ is empty). It turns out that there is a stack $p : \textit{FÉt} \to \mathit{Sch}$ over the category of schemes whose fibre over $X$ is the category $\textit{FÉt}_ X$ just defined. See Examples of Stacks, Section 94.6.

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

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

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

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. $\square$

Remark 58.5.3. Let $X$ be a scheme. Consider the natural functors $F_1 : \textit{FÉt}_ X \to \mathit{Sch}$ and $F_2 : \textit{FÉt}_ X \to \mathit{Sch}/X$. Then

The functors $F_1$ and $F_2$ commute with finite colimits.

The functor $F_2$ commutes with finite limits,

The functor $F_1$ commutes with connected finite limits, i.e., with equalizers and fibre products.

The results on limits are immediate from the discussion in the proof of Lemma 58.5.2 and Categories, Lemma 4.16.2. It is clear that $F_1$ and $F_2$ commute with finite coproducts. By the dual of Categories, Lemma 4.23.2 we need to show that $F_1$ and $F_2$ commute with coequalizers. In the proof of Lemma 58.5.2 we saw that coequalizers in $\textit{FÉt}_ X$ look étale locally like this

which is certainly a coequalizer in the category of schemes. Hence the statement follows from the fact that being a coequalizer is fpqc local as formulated precisely in Descent, Lemma 35.13.8.

Lemma 58.5.4. Let $X$ be a scheme. Given $U, V$ finite étale over $X$ there exists a scheme $W$ finite étale over $X$ such that

and such that the same remains true after any base change.

**Proof.**
By More on Morphisms, Lemma 37.66.4 there exists a scheme $W$ representing $\mathit{Mor}_ X(U, V)$. (Use that an étale morphism is locally quasi-finite by Morphisms, Lemmas 29.36.6 and that a finite morphism is separated.) This scheme clearly satisfies the formula after any base change. To finish the proof we have to show that $W \to X$ is finite étale. This we may do after replacing $X$ by the members of an étale covering (Descent, Lemmas 35.23.23 and 35.23.6). Thus by Étale Morphisms, Lemma 41.18.3 we may assume that $U = \coprod _{i = 1, \ldots , n} X$ and $V = \coprod _{j = 1, \ldots , m} X$. In this case $W = \coprod _{\alpha : \{ 1, \ldots , n\} \to \{ 1, \ldots , m\} } X$ by inspection (details omitted) and the proof is complete.
$\square$

Let $X$ be a scheme. A *geometric point* of $X$ is a morphism $\mathop{\mathrm{Spec}}(k) \to X$ where $k$ is algebraically closed. Such a point is usually denoted $\overline{x}$, i.e., by an overlined small case letter. We often use $\overline{x}$ to denote the scheme $\mathop{\mathrm{Spec}}(k)$ as well as the morphism, and we use $\kappa (\overline{x})$ to denote $k$. We say $\overline{x}$ *lies over* $x$ to indicate that $x \in X$ is the image of $\overline{x}$. We will discuss this further in Étale Cohomology, Section 59.29. Given $\overline{x}$ and an étale morphism $U \to X$ we can consider

Since $U_{\overline{x}}$ as a scheme over $\overline{x}$ is a disjoint union of copies of $\overline{x}$ (Morphisms, Lemma 29.36.7) we can also describe this set as

The assignment $U \mapsto |U_{\overline{x}}|$ is a functor which is often denoted $F_{\overline{x}}$.

Lemma 58.5.5. Let $X$ be a connected scheme. Let $\overline{x}$ be a geometric point. The functor

defines a Galois category (Definition 58.3.6).

**Proof.**
After identifying $\textit{FÉt}_{\overline{x}}$ with the category of finite sets (Example 58.5.1) we see that our functor $F_{\overline{x}}$ is nothing but the base change functor for the morphism $\overline{x} \to X$. Thus we see that $\textit{FÉt}_ X$ has finite limits and finite colimits and that $F_{\overline{x}}$ is exact by Lemma 58.5.2. We will also use that finite limits in $\textit{FÉt}_ X$ agree with the corresponding finite limits in the category of schemes over $X$, see Remark 58.5.3.

If $Y' \to Y$ is a monomorphism in $\textit{FÉt}_ X$ then we see that $Y' \to Y' \times _ Y Y'$ is an isomorphism, and hence $Y' \to Y$ is a monomorphism of schemes. It follows that $Y' \to Y$ is an open immersion (Étale Morphisms, Theorem 41.14.1). Since $Y'$ is finite over $X$ and $Y$ separated over $X$, the morphism $Y' \to Y$ is finite (Morphisms, Lemma 29.44.14), hence closed (Morphisms, Lemma 29.44.11), hence it is the inclusion of an open and closed subscheme of $Y$. It follows that $Y$ is a connected objects of the category $\textit{FÉt}_ X$ (as in Definition 58.3.6) if and only if $Y$ is connected as a scheme. Then it follows from Topology, Lemma 5.7.7 that $Y$ is a finite coproduct of its connected components both as a scheme and in the sense of Definition 58.3.6.

Let $Y \to Z$ be a morphism in $\textit{FÉt}_ X$ which induces a bijection $F_{\overline{x}}(Y) \to F_{\overline{x}}(Z)$. We have to show that $Y \to Z$ is an isomorphism. By the above we may assume $Z$ is connected. Since $Y \to Z$ is finite étale and hence finite locally free it suffices to show that $Y \to Z$ is finite locally free of degree $1$. This is true in a neighbourhood of any point of $Z$ lying over $\overline{x}$ and since $Z$ is connected and the degree is locally constant we conclude. $\square$

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