## 36.31 Étale neighbourhoods

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 36.31.1. 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 $\kappa (s) = \kappa (u)$.

If $f : (V, v) \to (U, u)$ is a morphism of étale neighbourhoods, then $f$ is automatically étale, see Morphisms, Lemma 28.34.18. Hence it turns $(V, v)$ into an étale neighbourhood of $(U, u)$. Of course, since the composition of étale morphisms is étale (Morphisms, Lemma 28.34.3) 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 28.34.9). We will use these remarks without further mention throughout this section.

Note that $\kappa (s) \subset \kappa (u)$ is a finite separable extension if $(U, u) \to (S, s)$ is an étale neighbourhood, see Morphisms, Lemma 28.34.15.

Lemma 36.31.2. Let $S$ be a scheme. Let $s \in S$. Let $\kappa (s) \subset k$ be a finite separable field extension. Then there exists an étale neighbourhood $(U, u) \to (S, s)$ such that the field extension $\kappa (s) \subset \kappa (u)$ is isomorphic to $\kappa (s) \subset k$.

**Proof.**
We may assume $S$ is affine. In this case the lemma follows from Algebra, Lemma 10.141.15.
$\square$

Lemma 36.31.3. 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 28.34.4. 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 25.17.5. For part (2), define $U''$ as the fibre product

\[ \xymatrix{ U'' \ar[r] \ar[d] & U \ar[d]^{(h_1, h_2)} \\ U' \ar[r]^-\Delta & U' \times _ S U'. } \]

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 28.34.4 and 28.34.3). Hence the vertical arrow $(h_1, h_2)$ is étale by Morphisms, Lemma 28.34.18. 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 36.31.4. 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 36.31.3.
$\square$

Lemma 36.31.5. Let $S$ be a scheme. Let $s \in S$. Then we have

\[ \mathcal{O}_{S, s}^ h = \mathop{\mathrm{colim}}\nolimits _{(U, u)} \mathcal{O}(U) \]

where the colimit is over the filtered category which is opposite to the category of elementary étale neighbourhoods $(U, u)$ of $(S, s)$.

**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 10.150.7.
$\square$

We can lift étale neighbourhoods of points on fibres to the total space.

slogan
Lemma 36.31.6. 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

\[ C = (B \otimes _ A \kappa (\mathfrak p)) \otimes _{B \otimes _ A (A/\mathfrak p)_ f} D \]

See Algebra, Lemma 10.141.3 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 10.141.10. This proves the lemma.
$\square$

## Comments (2)

Comment #4011 by Janos Kollar on

Comment #4016 by Alexander Schmidt on