It turns out to be logically more efficient to define the notion of a formally unramified map before introducing the notion of a formally étale one.
Definition 10.148.1. Let R \to S be a ring map. We say S is formally unramified over R if for every commutative solid diagram
where I \subset A is an ideal of square zero, there exists at most one dotted arrow making the diagram commute.
Lemma 10.148.2. Let R \to S be a ring map. The following are equivalent:
R \to S is formally unramified,
the module of differentials \Omega _{S/R} is zero.
Proof. Let J = \mathop{\mathrm{Ker}}(S \otimes _ R S \to S) be the kernel of the multiplication map. Let A_{univ} = S \otimes _ R S/J^2. Recall that I_{univ} = J/J^2 is isomorphic to \Omega _{S/R}, see Lemma 10.131.13. Moreover, the two R-algebra maps \sigma _1, \sigma _2 : S \to A_{univ}, \sigma _1(s) = s \otimes 1 \bmod J^2, and \sigma _2(s) = 1 \otimes s \bmod J^2 differ by the universal derivation \text{d} : S \to \Omega _{S/R} = I_{univ}.
Assume R \to S formally unramified. Then we see that \sigma _1 = \sigma _2. Hence \text{d}(s) = 0 for all s \in S. Hence \Omega _{S/R} = 0.
Assume that \Omega _{S/R} = 0. Let A, I, R \to A, S \to A/I be a solid diagram as in Definition 10.148.1. Let \tau _1, \tau _2 : S \to A be two dotted arrows making the diagram commute. Consider the R-algebra map A_{univ} \to A defined by the rule s_1 \otimes s_2 \mapsto \tau _1(s_1)\tau _2(s_2). We omit the verification that this is well defined. Since A_{univ} \cong S as I_{univ} = \Omega _{S/R} = 0 we conclude that \tau _1 = \tau _2. \square
Lemma 10.148.3. Let R \to S be a ring map. The following are equivalent:
R \to S is formally unramified,
R \to S_{\mathfrak q} is formally unramified for all primes \mathfrak q of S, and
R_{\mathfrak p} \to S_{\mathfrak q} is formally unramified for all primes \mathfrak q of S with \mathfrak p = R \cap \mathfrak q.
Proof. We have seen in Lemma 10.148.2 that (1) is equivalent to \Omega _{S/R} = 0. Similarly, by Lemma 10.131.8 we see that (2) and (3) are equivalent to (\Omega _{S/R})_{\mathfrak q} = 0 for all \mathfrak q. Hence the equivalence follows from Lemma 10.23.1. \square
Lemma 10.148.4. Let A \to B be a formally unramified ring map.
For S \subset A a multiplicative subset, S^{-1}A \to S^{-1}B is formally unramified.
For S \subset B a multiplicative subset, A \to S^{-1}B is formally unramified.
Proof. Follows from Lemma 10.148.3. (You can also deduce it from Lemma 10.148.2 combined with Lemma 10.131.8.) \square
Lemma 10.148.5. 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 unramified, then S = \mathop{\mathrm{colim}}\nolimits _{i \in I} S_ i is formally unramified over R
Proof. Consider a diagram as in Definition 10.148.1. By assumption there exists at most one R-algebra map S_ i \to A lifting the compositions S_ i \to S \to A/I. Since every element of S is in the image of one of the maps S_ i \to S we see that there is at most one map S \to A fitting into the diagram. \square
Comments (0)