Definition 10.150.1. 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.
\[ \xymatrix{ S \ar[r] \ar@{-->}[rd] & A/I \\ R \ar[r] \ar[u] & A \ar[u] } \]
Clearly a ring map is formally étale if and only if it is both formally smooth and formally unramified.
Lemma 10.150.2. Let $R \to S$ be a formally étale map. Let $R \to R'$ be any ring map. Then the base change $S' = R' \otimes _ R S$ is formally étale over $R'$.
Lemma 10.150.3. 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 10.138.13. By Lemma 10.148.3 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 10.138.13. By Lemma 10.148.3 it is formally unramified. Hence $R \to S$ is formally étale. $\square$
Lemma 10.150.4. 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 10.150.1. 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 10.150.5. 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 $A \to A/I$. $\square$
Lemma 10.150.6. Let $R \to S$ be a ring map. Let $J \subset S$ be an ideal such that $R \to S/J$ is surjective; let $I \subset R$ be the kernel. If $R \to S$ is formally étale, then $R/I^ n \to S/J^ n$ is an isomorphism for all $n$ and $\bigoplus I^ n/I^{n + 1} \to \bigoplus J^ n/J^{n + 1}$ is an isomorphism of graded rings.
Proof. Using the lifting property inductively we find dotted arrows
The corresponding maps $S/J^ n \to R/I^ n$ are isomorphisms since the compositions $S/J^ n \to R/I^ n \to S/J^ n$ are (inductively) the identity by the uniqueness in the lifting property of formally étale ring maps. $\square$
Lemma 10.150.7. Let $R \to S \to S'$ be ring maps. Let $J$, resp. $J'$ be the kernel of the multiplication map $S \otimes _ R S \to S$, resp. $S' \otimes _{R'} S' \to S'$. If $S \to S'$ is formally étale, then the map is an isomorphism for all $k \geq 0$. In particular, the map $S' \otimes _ S \Omega _{S/R} \to \Omega _{S'/R}$ is an isomorphism.
\[ S' \otimes _ S \left((S \otimes _ R S)/J^{k + 1}\right) \longrightarrow (S' \otimes _ R S')/(J')^{k + 1} \]
Proof. Observe that $S' \otimes _ S (S \otimes _ R S) = S' \otimes _ R S$ and the ideal $J$ generates in $S' \otimes _ R S$ the kernel $I$ of the surjection $S' \otimes _ R S \to S'$. Whence the left hand side is equal to $S' \otimes _ R S/I^{k + 1}$. The map $S' \otimes _ R S \to S' \otimes _ R S'$ is formally étale by Lemma 10.150.2. Thus we conclude that the displayed arrow in the statement of the lemma is an isomorphism by Lemma 10.150.6. The final assertion follows from this and the fact that $\Omega _{S/R} = J/J^2$ and $\Omega _{S'/R} = J'/(J')^2$ by Lemma 10.131.13. $\square$
Lemma 10.150.8. Let $R \to S \to S'$ be ring maps with $S \to S'$ formally étale (for example étale). Let $M$ be an $S$-module. Set $M' = S' \otimes _ R M$. Then we have It follows that for any $S$-module $N$ and any finite order differential operator $D : M \to N$ there exists a unique extension $D' : M' \to S' \otimes _ S N$ of $D$ to a differential operator (of the same or lesser order).
\[ S' \otimes _ S P^ k_{S/R}(M) = P^ k_{S'/R}(M') \]
Proof. Let $J$ and $J'$ be as in the statement of Lemma 10.150.7. Then we have
The first and the last equalities are from Lemma 10.133.9. The third equality is Lemma 10.150.7. The final assertion holds because if $D$ corresponds to the linear map $\gamma : P^ k_{S/R}(M) \to N$, then we can let $D' : M' \to S' \otimes _ R N$ correspond to the linear map $1 \otimes \gamma : S' \otimes _ R P^ k_{S/R}(M) \to S' \otimes _ R N$. $\square$
Remark 10.150.9. Let $R \to S \to S'$ be ring maps with $S \to S'$ formally étale (for example étale). Let $M_ i$, $i = 1, 2, 3$ be $S$-modules and let $D_ i : M_ i \to M_{i + 1}$, $i = 1, 2$ be differential operators of finite order. Then if $D'_ i : M'_ i \to M'_{i + 1}$, $i = 1, 2$ are the extensions of $D_ i$ to $M'_ i = S' \otimes _ S M_ i$ as in Lemma 10.150.8, then $D'_2 \circ D'_1$ is the extension of $D_2 \circ D_1$. In particular, if $M$ is an $S$-module, then $M' = S' \otimes _ S M$ is a module over the $S$-algebra $\text{Diff}_{S/R}(M, M)$.
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