Definition 10.146.1. Let $R \to S$ be a ring map. We say $S$ is *formally étale over $R$* if for every commutative solid diagram

where $I \subset A$ is an ideal of square zero, there exists a unique dotted arrow making the diagram commute.

Definition 10.146.1. Let $R \to S$ be a ring map. We say $S$ is *formally étale over $R$* if for every commutative solid diagram

\[ \xymatrix{ S \ar[r] \ar@{-->}[rd] & A/I \\ R \ar[r] \ar[u] & A \ar[u] } \]

where $I \subset A$ is an ideal of square zero, there exists a unique dotted arrow making the diagram commute.

Clearly a ring map is formally étale if and only if it is both formally smooth and formally unramified.

Lemma 10.146.2. Let $R \to S$ be a ring map of finite presentation. The following are equivalent:

$R \to S$ is formally étale,

$R \to S$ is étale.

**Proof.**
Assume that $R \to S$ is formally étale. Then $R \to S$ is smooth by Proposition 10.136.13. By Lemma 10.144.2 we have $\Omega _{S/R} = 0$. Hence $R \to S$ is étale by definition.

Assume that $R \to S$ is étale. Then $R \to S$ is formally smooth by Proposition 10.136.13. By Lemma 10.144.2 it is formally unramified. Hence $R \to S$ is formally étale. $\square$

Lemma 10.146.3. Let $R$ be a ring. Let $I$ be a directed set. Let $(S_ i, \varphi _{ii'})$ be a system of $R$-algebras over $I$. If each $R \to S_ i$ is formally étale, then $S = \mathop{\mathrm{colim}}\nolimits _{i \in I} S_ i$ is formally étale over $R$

**Proof.**
Consider a diagram as in Definition 10.146.1. By assumption we get unique $R$-algebra maps $S_ i \to A$ lifting the compositions $S_ i \to S \to A/I$. Hence these are compatible with the transition maps $\varphi _{ii'}$ and define a lift $S \to A$. This proves existence. The uniqueness is clear by restricting to each $S_ i$.
$\square$

Lemma 10.146.4. Let $R$ be a ring. Let $S \subset R$ be any multiplicative subset. Then the ring map $R \to S^{-1}R$ is formally étale.

**Proof.**
Let $I \subset A$ be an ideal of square zero. What we are saying here is that given a ring map $\varphi : R \to A$ such that $\varphi (f) \mod I$ is invertible for all $f \in S$ we have also that $\varphi (f)$ is invertible in $A$ for all $f \in S$. This is true because $A^*$ is the inverse image of $(A/I)^*$ under the canonical map $A \to A/I$.
$\square$

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## Comments (2)

Comment #864 by Keenan Kidwell on

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