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