Definition 35.18.1. Let $\mathcal{P}$ be a property of germs of schemes. We say that $\mathcal{P}$ is *étale local* (resp. *smooth local*) if for any étale (resp. smooth) morphism of germs $(U', u') \to (U, u)$ we have $\mathcal{P}(U, u) \Leftrightarrow \mathcal{P}(U', u')$.

## 35.18 Local properties of germs

Let $(X, x)$ be a germ of a scheme. The dimension of $X$ at $x$ is the minimum of the dimensions of open neighbourhoods of $x$ in $X$, and any small enough open neighbourhood has this dimension. Hence this is an invariant of the isomorphism class of the germ. We denote this simply $\dim _ x(X)$. The following lemma tells us that the assertion $\dim _ x(X) = d$ is an étale local property of germs.

Lemma 35.18.2. Let $f : U \to V$ be an étale morphism of schemes. Let $u \in U$ and $v = f(u)$. Then $\dim _ u(U) = \dim _ v(V)$.

**Proof.**
In the statement $\dim _ u(U)$ is the dimension of $U$ at $u$ as defined in Topology, Definition 5.10.1 as the minimum of the Krull dimensions of open neighbourhoods of $u$ in $U$. Similarly for $\dim _ v(V)$.

Let us show that $\dim _ v(V) \geq \dim _ u(U)$. Let $V'$ be an open neighbourhood of $v$ in $V$. Then there exists an open neighbourhood $U'$ of $u$ in $U$ contained in $f^{-1}(V')$ such that $\dim _ u(U) = \dim (U')$. Suppose that $Z_0 \subset Z_1 \subset \ldots \subset Z_ n$ is a chain of irreducible closed subschemes of $U'$. If $\xi _ i \in Z_ i$ is the generic point then we have specializations $\xi _ n \leadsto \xi _{n - 1} \leadsto \ldots \leadsto \xi _0$. This gives specializations $f(\xi _ n) \leadsto f(\xi _{n - 1}) \leadsto \ldots \leadsto f(\xi _0)$ in $V'$. Note that $f(\xi _ j) \not= f(\xi _ i)$ if $i \not= j$ as the fibres of $f$ are discrete (see Morphisms, Lemma 29.36.7). Hence we see that $\dim (V') \geq n$. The inequality $\dim _ v(V) \geq \dim _ u(U)$ follows formally.

Let us show that $\dim _ u(U) \geq \dim _ v(V)$. Let $U'$ be an open neighbourhood of $u$ in $U$. Note that $V' = f(U')$ is an open neighbourhood of $v$ by Morphisms, Lemma 29.25.10. Hence $\dim (V') \geq \dim _ v(V)$. Pick a chain $Z_0 \subset Z_1 \subset \ldots \subset Z_ n$ of irreducible closed subschemes of $V'$. Let $\xi _ i \in Z_ i$ be the generic point, so we have specializations $\xi _ n \leadsto \xi _{n - 1} \leadsto \ldots \leadsto \xi _0$. Since $\xi _0 \in f(U')$ we can find a point $\eta _0 \in U'$ with $f(\eta _0) = \xi _0$. Consider the map of local rings

which is a flat local ring map by Morphisms, Lemma 29.36.12. Note that the points $\xi _ i$ correspond to primes of the ring on the left by Schemes, Lemma 26.13.2. Hence by going down (see Algebra, Section 10.41) for the displayed ring map we can find a sequence of specializations $\eta _ n \leadsto \eta _{n - 1} \leadsto \ldots \leadsto \eta _0$ in $U'$ mapping to the sequence $\xi _ n \leadsto \xi _{n - 1} \leadsto \ldots \leadsto \xi _0$ under $f$. This implies that $\dim _ u(U) \geq \dim _ v(V)$. $\square$

Let $(X, x)$ be a germ of a scheme. The isomorphism class of the local ring $\mathcal{O}_{X, x}$ is an invariant of the germ. The following lemma says that the property $\dim (\mathcal{O}_{X, x}) = d$ is an étale local property of germs.

Lemma 35.18.3. Let $f : U \to V$ be an étale morphism of schemes. Let $u \in U$ and $v = f(u)$. Then $\dim (\mathcal{O}_{U, u}) = \dim (\mathcal{O}_{V, v})$.

**Proof.**
The algebraic statement we are asked to prove is the following: If $A \to B$ is an étale ring map and $\mathfrak q$ is a prime of $B$ lying over $\mathfrak p \subset A$, then $\dim (A_{\mathfrak p}) = \dim (B_{\mathfrak q})$. This is More on Algebra, Lemma 15.44.2.
$\square$

Let $(X, x)$ be a germ of a scheme. The isomorphism class of the local ring $\mathcal{O}_{X, x}$ is an invariant of the germ. The following lemma says that the property “$\mathcal{O}_{X, x}$ is regular” is an étale local property of germs.

Lemma 35.18.4. Let $f : U \to V$ be an étale morphism of schemes. Let $u \in U$ and $v = f(u)$. Then $\mathcal{O}_{U, u}$ is a regular local ring if and only if $\mathcal{O}_{V, v}$ is a regular local ring.

**Proof.**
The algebraic statement we are asked to prove is the following: If $A \to B$ is an étale ring map and $\mathfrak q$ is a prime of $B$ lying over $\mathfrak p \subset A$, then $A_{\mathfrak p}$ is regular if and only if $B_{\mathfrak q}$ is regular. This is More on Algebra, Lemma 15.44.3.
$\square$

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