Lemma 10.148.2. Let $R \to S$ be a ring map. The following are equivalent:

1. $R \to S$ is formally unramified,

2. 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 $\tau _1 = \tau _2$. $\square$

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