It turns out that some properties of morphisms are easier to study after doing an étale base change. It is convenient to introduce the following terminology.
Definition 37.35.1. Let $S$ be a scheme. Let $s \in S$ be a point.
An étale neighbourhood of $(S, s)$ is a pair $(U, u)$ together with an étale morphism of schemes $\varphi : U \to S$ such that $\varphi (u) = s$.
A morphism of étale neighbourhoods $f : (V, v) \to (U, u)$ of $(S, s)$ is simply a morphism of $S$-schemes $f : V \to U$ such that $f(v) = u$.
An elementary étale neighbourhood is an étale neighbourhood $\varphi : (U, u) \to (S, s)$ such that $\kappa (s) = \kappa (u)$.
The notion of an elementary étale neighbourhood has many different names in the literature, for example these are sometimes called “étale neighbourhoods” ([Page 36, Milne] or “strongly étale” ([Page 108, KPR]). Here we follow the convention of the paper [GruRay] by calling them elementary étale neighbourhoods.
If $f : (V, v) \to (U, u)$ is a morphism of étale neighbourhoods, then $f$ is automatically étale, see Morphisms, Lemma 29.36.18. Hence it turns $(V, v)$ into an étale neighbourhood of $(U, u)$. Of course, since the composition of étale morphisms is étale (Morphisms, Lemma 29.36.3) we see that conversely any étale neighbourhood $(V, v)$ of $(U, u)$ is an étale neighbourhood of $(S, s)$ as well. We also remark that if $U \subset S$ is an open neighbourhood of $s$, then $(U, s) \to (S, s)$ is an étale neighbourhood. This follows from the fact that an open immersion is étale (Morphisms, Lemma 29.36.9). We will use these remarks without further mention throughout this section.
Note that $\kappa (u)/\kappa (s)$ is a finite separable extension if $(U, u) \to (S, s)$ is an étale neighbourhood, see Morphisms, Lemma 29.36.15.
Lemma 37.35.2. Let $S$ be a scheme. Let $s \in S$. Let $k/\kappa (s)$ be a finite separable field extension. Then there exists an étale neighbourhood $(U, u) \to (S, s)$ such that the field extension $\kappa (u)/\kappa (s)$ is isomorphic to $k/\kappa (s)$.
Proof. We may assume $S$ is affine. In this case the lemma follows from Algebra, Lemma 10.144.3. $\square$
Lemma 37.35.3. Let $S$ be a scheme, and let $s$ be a point of $S$. The category of étale neighborhoods has the following properties:
Let $(U_ i, u_ i)_{i=1, 2}$ be two étale neighborhoods of $s$ in $S$. Then there exists a third étale neighborhood $(U, u)$ and morphisms $(U, u) \to (U_ i, u_ i)$, $i = 1, 2$.
Let $h_1, h_2: (U, u) \to (U', u')$ be two morphisms between étale neighborhoods of $s$. Assume $h_1$, $h_2$ induce the same map $\kappa (u') \to \kappa (u)$ of residue fields. Then there exist an étale neighborhood $(U'', u'')$ and a morphism $h : (U'', u'') \to (U, u)$ which equalizes $h_1$ and $h_2$, i.e., such that $h_1 \circ h = h_2 \circ h$.
Proof. For part (1), consider the fibre product $U = U_1 \times _ S U_2$. It is étale over both $U_1$ and $U_2$ because étale morphisms are preserved under base change, see Morphisms, Lemma 29.36.4. There is a point of $U$ mapping to both $u_1$ and $u_2$ for example by the description of points of a fibre product in Schemes, Lemma 26.17.5. For part (2), define $U''$ as the fibre product
Since $h_1$ and $h_2$ induce the same map of residue fields $\kappa (u') \to \kappa (u)$ there exists a point $u'' \in U''$ lying over $u'$ with $\kappa (u'') = \kappa (u')$. In particular $U'' \not= \emptyset $. Moreover, since $U'$ is étale over $S$, so is the fibre product $U'\times _ S U'$ (see Morphisms, Lemmas 29.36.4 and 29.36.3). Hence the vertical arrow $(h_1, h_2)$ is étale by Morphisms, Lemma 29.36.18. Therefore $U''$ is étale over $U'$ by base change, and hence also étale over $S$ (because compositions of étale morphisms are étale). Thus $(U'', u'')$ is a solution to the problem. $\square$
Lemma 37.35.4. Let $S$ be a scheme, and let $s$ be a point of $S$. The category of elementary étale neighborhoods of $(S, s)$ is cofiltered (see Categories, Definition 4.20.1).
Proof. This is immediate from the definitions and Lemma 37.35.3. $\square$
Lemma 37.35.5. Let $S$ be a scheme. Let $s \in S$. Then we have where the colimit is over the filtered category which is opposite to the category of elementary étale neighbourhoods $(U, u)$ of $(S, s)$.
\[ \mathcal{O}_{S, s}^ h = \mathop{\mathrm{colim}}\nolimits _{(U, u)} \mathcal{O}(U) \]
Proof. Let $\mathop{\mathrm{Spec}}(A) \subset S$ be an affine neighbourhood of $s$. Let $\mathfrak p \subset A$ be the prime ideal corresponding to $s$. With these choices we have canonical isomorphisms $\mathcal{O}_{S, s} = A_{\mathfrak p}$ and $\kappa (s) = \kappa (\mathfrak p)$. A cofinal system of elementary étale neighbourhoods is given by those elementary étale neighbourhoods $(U, u)$ such that $U$ is affine and $U \to S$ factors through $\mathop{\mathrm{Spec}}(A)$. In other words, we see that the right hand side is equal to $\mathop{\mathrm{colim}}\nolimits _{(B, \mathfrak q)} B$ where the colimit is over étale $A$-algebras $B$ endowed with a prime $\mathfrak q$ lying over $\mathfrak p$ with $\kappa (\mathfrak p) = \kappa (\mathfrak q)$. Thus the lemma follows from Algebra, Lemma 10.155.7. $\square$
We can lift étale neighbourhoods of points on fibres to the total space.
Lemma 37.35.6. Let $X \to S$ be a morphism of schemes. Let $x \in X$ with image $s \in S$. Let $(V, v) \to (X_ s, x)$ be an étale neighbourhood. Then there exists an étale neighbourhood $(U, u) \to (X, x)$ such that there exists a morphism $(U_ s, u) \to (V, v)$ of étale neighbourhoods of $(X_ s, x)$ which is an open immersion.
Proof. We may assume $X$, $V$, and $S$ affine. Say the morphism $X \to S$ is given by $A \to B$ the point $x$ by a prime $\mathfrak q \subset B$, the point $s$ by $\mathfrak p = A \cap \mathfrak q$, and the morphism $V \to X_ s$ by $B \otimes _ A \kappa (\mathfrak p) \to C$. Since $\kappa (\mathfrak p)$ is a localization of $A/\mathfrak p$ there exists an $f \in A$, $f \not\in \mathfrak p$ and an étale ring map $B \otimes _ A (A/\mathfrak p)_ f \to D$ such that
See Algebra, Lemma 10.143.3 part (9). After replacing $A$ by $A_ f$ and $B$ by $B_ f$ we may assume $D$ is étale over $B \otimes _ A A/\mathfrak p = B/\mathfrak p B$. Then we can apply Algebra, Lemma 10.143.10. This proves the lemma. $\square$
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