Lemma 15.43.1. 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 Noetherian if and only if $B_{\mathfrak q}$ is Noetherian.

## 15.43 Permanence of properties under étale maps

In this section we consider an étale ring map $\varphi : A \to B$ and we study which properties of $A$ are inherited by $B$ and which properties of the local ring of $B$ at $\mathfrak q$ are inherited by the local ring of $A$ at $\mathfrak p = \varphi ^{-1}(\mathfrak q)$. Basically, this section reviews and collects earlier results and does not add any new material.

We will use without further mention that an étale ring map is flat (Algebra, Lemma 10.142.3) and that a flat local homomorphism of local rings is faithfully flat (Algebra, Lemma 10.38.17).

**Proof.**
Since $A_\mathfrak p \to B_\mathfrak q$ is faithfully flat we see that $B_\mathfrak q$ Noetherian implies that $A_\mathfrak p$ is Noetherian, see Algebra, Lemma 10.162.1. Conversely, if $A_\mathfrak p$ is Noetherian, then $B_\mathfrak q$ is Noetherian as it is a localization of a finite type $A_\mathfrak p$-algebra.
$\square$

Lemma 15.43.2. If $A \to B$ is an étale ring map and $\mathfrak q$ is a prime of $B$ lying over $\mathfrak p \subset A$, then $\dim (A_{\mathfrak p}) = \dim (B_{\mathfrak q})$.

**Proof.**
Namely, because $A_{\mathfrak p} \to B_{\mathfrak q}$ is flat we have going down, and hence the inequality $\dim (A_{\mathfrak p}) \leq \dim (B_{\mathfrak q})$, see Algebra, Lemma 10.111.1. On the other hand, suppose that $\mathfrak q_0 \subset \mathfrak q_1 \subset \ldots \subset \mathfrak q_ n$ is a chain of primes in $B_{\mathfrak q}$. Then the corresponding sequence of primes $\mathfrak p_0 \subset \mathfrak p_1 \subset \ldots \subset \mathfrak p_ n$ (with $\mathfrak p_ i = \mathfrak q_ i \cap A_{\mathfrak p}$) is chain also (i.e., no equalities in the sequence) as an étale ring map is quasi-finite (see Algebra, Lemma 10.142.6) and a quasi-finite ring map induces a map of spectra with discrete fibres (by definition). This means that $\dim (A_{\mathfrak p}) \geq \dim (B_{\mathfrak q})$ as desired.
$\square$

Lemma 15.43.3. 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.

**Proof.**
By Lemma 15.43.1 we may assume both $A_\mathfrak p$ and $B_\mathfrak q$ are Noetherian in order to prove the equivalence. Let $x_1, \ldots , x_ t \in \mathfrak pA_\mathfrak p$ be a minimal set of generators. As $A_\mathfrak p \to B_\mathfrak q$ is faithfully flat we see that the images $y_1, \ldots , y_ t$ in $B_\mathfrak q$ form a minimal system of generators for $\mathfrak pB_\mathfrak q = \mathfrak q B_\mathfrak q$ (Algebra, Lemma 10.142.5). Regularity of $A_\mathfrak p$ by definition means $t = \dim (A_\mathfrak p)$ and similarly for $B_\mathfrak q$. Hence the lemma follows from the equality $\dim (A_\mathfrak p) = \dim (B_\mathfrak q)$ of Lemma 15.43.2.
$\square$

Lemma 15.43.4. If $A \to B$ is an étale ring map and $A$ is a Dedekind domain, then $B$ is a finite product of Dedekind domains. In particular, the localizations $B_\mathfrak q$ for $\mathfrak q \subset B$ maximal are discrete valuation rings.

**Proof.**
The statement on the local rings follows from Lemmas 15.43.2 and 15.43.3 and Algebra, Lemma 10.118.7. It follows that $B$ is a Noetherian normal ring of dimension $1$. By Algebra, Lemma 10.36.16 we conclude that $B$ is a finite product of normal domains of dimension $1$. These are Dedekind domains by Algebra, Lemma 10.119.17.
$\square$

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