
Lemma 10.141.3. Results on étale ring maps.

1. The ring map $R \to R_ f$ is étale for any ring $R$ and any $f \in R$.

2. Compositions of étale ring maps are étale.

3. A base change of an étale ring map is étale.

4. The property of being étale is local: Given a ring map $R \to S$ and elements $g_1, \ldots , g_ m \in S$ which generate the unit ideal such that $R \to S_{g_ j}$ is étale for $j = 1, \ldots , m$ then $R \to S$ is étale.

5. Given $R \to S$ of finite presentation, and a flat ring map $R \to R'$, set $S' = R' \otimes _ R S$. The set of primes where $R' \to S'$ is étale is the inverse image via $\mathop{\mathrm{Spec}}(S') \to \mathop{\mathrm{Spec}}(S)$ of the set of primes where $R \to S$ is étale.

6. An étale ring map is syntomic, in particular flat.

7. If $S$ is finite type over a field $k$, then $S$ is étale over $k$ if and only if $\Omega _{S/k} = 0$.

8. Any étale ring map $R \to S$ is the base change of an étale ring map $R_0 \to S_0$ with $R_0$ of finite type over $\mathbf{Z}$.

9. Let $A = \mathop{\mathrm{colim}}\nolimits A_ i$ be a filtered colimit of rings. Let $A \to B$ be an étale ring map. Then there exists an étale ring map $A_ i \to B_ i$ for some $i$ such that $B \cong A \otimes _{A_ i} B_ i$.

10. Let $A$ be a ring. Let $S$ be a multiplicative subset of $A$. Let $S^{-1}A \to B'$ be étale. Then there exists an étale ring map $A \to B$ such that $B' \cong S^{-1}B$.

Proof. In each case we use the corresponding result for smooth ring maps with a small argument added to show that $\Omega _{S/R}$ is zero.

Proof of (1). The ring map $R \to R_ f$ is smooth and $\Omega _{R_ f/R} = 0$.

Proof of (2). The composition $A \to C$ of smooth maps $A \to B$ and $B \to C$ is smooth, see Lemma 10.135.14. By Lemma 10.130.7 we see that $\Omega _{C/A}$ is zero as both $\Omega _{C/B}$ and $\Omega _{B/A}$ are zero.

Proof of (3). Let $R \to S$ be étale and $R \to R'$ be arbitrary. Then $R' \to S' = R' \otimes _ R S$ is smooth, see Lemma 10.135.4. Since $\Omega _{S'/R'} = S' \otimes _ S \Omega _{S/R}$ by Lemma 10.130.12 we conclude that $\Omega _{S'/R'} = 0$. Hence $R' \to S'$ is étale.

Proof of (4). Assume the hypotheses of (4). By Lemma 10.135.13 we see that $R \to S$ is smooth. We are also given that $\Omega _{S_{g_ i}/R} = (\Omega _{S/R})_{g_ i} = 0$ for all $i$. Then $\Omega _{S/R} = 0$, see Lemma 10.22.2.

Proof of (5). The result for smooth maps is Lemma 10.135.17. In the proof of that lemma we used that $\mathop{N\! L}\nolimits _{S/R} \otimes _ S S'$ is homotopy equivalent to $\mathop{N\! L}\nolimits _{S'/R'}$. This reduces us to showing that if $M$ is a finitely presented $S$-module the set of primes $\mathfrak q'$ of $S'$ such that $(M \otimes _ S S')_{\mathfrak q'} = 0$ is the inverse image of the set of primes $\mathfrak q$ of $S$ such that $M_{\mathfrak q} = 0$. This follows from Lemma 10.39.6.

Proof of (6). Follows directly from the corresponding result for smooth ring maps (Lemma 10.135.10).

Proof of (7). Follows from Lemma 10.138.3 and the definitions.

Proof of (8). Lemma 10.136.14 gives the result for smooth ring maps. The resulting smooth ring map $R_0 \to S_0$ satisfies the hypotheses of Lemma 10.129.8, and hence we may replace $S_0$ by the factor of relative dimension $0$ over $R_0$.

Proof of (9). Follows from (8) since $R_0 \to A$ will factor through $A_ i$ for some $i$ by Lemma 10.126.3.

Proof of (10). Follows from (9), (1), and (2) since $S^{-1}A$ is a filtered colimit of principal localizations of $A$. $\square$

Comment #260 by David Holmes on

Typo: In (1) in the statement of the Lemma, I think the 'If' should be removed.

Comment #2812 by Dario Weißmann on

In the proof of (5), we omit the last part. But this is actually (a version of) lemma 10.39.6.

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