Lemma 10.138.12. Let $R \to S$ be a ring map. Let $I \subset R$ be an ideal. Assume

1. $I^2 = 0$,

2. $R \to S$ is flat, and

3. $R/I \to S/IS$ is formally smooth.

Then $R \to S$ is formally smooth.

Proof. Assume (1), (2) and (3). Let $P = R[\{ x_ t\} _{t \in T}] \to S$ be a surjection of $R$-algebras with kernel $J$. Thus $0 \to J \to P \to S \to 0$ is a short exact sequence of flat $R$-modules. This implies that $I \otimes _ R S = IS$, $I \otimes _ R P = IP$ and $I \otimes _ R J = IJ$ as well as $J \cap IP = IJ$. We will use throughout the proof that

$\Omega _{(S/IS)/(R/I)} = \Omega _{S/R} \otimes _ S (S/IS) = \Omega _{S/R} \otimes _ R R/I = \Omega _{S/R} / I\Omega _{S/R}$

and similarly for $P$ (see Lemma 10.131.12). By Lemma 10.138.7 the sequence

10.138.12.1
\begin{equation} \label{algebra-equation-split} 0 \to J/(IJ + J^2) \to \Omega _{P/R} \otimes _ P S/IS \to \Omega _{S/R} \otimes _ S S/IS \to 0 \end{equation}

is split exact. Of course the middle term is $\bigoplus _{t \in T} S/IS \text{d}x_ t$. Choose a splitting $\sigma : \Omega _{P/R} \otimes _ P S/IS \to J/(IJ + J^2)$. For each $t \in T$ choose an element $f_ t \in J$ which maps to $\sigma (\text{d}x_ t)$ in $J/(IJ + J^2)$. This determines a unique $S$-module map

$\tilde\sigma : \Omega _{P/R} \otimes _ R S = \bigoplus S\text{d}x_ t \longrightarrow J/J^2$

with the property that $\tilde\sigma (\text{d}x_ t) = f_ t$. As $\sigma$ is a section to $\text{d}$ the difference

$\Delta = \text{id}_{J/J^2} - \tilde\sigma \circ \text{d}$

is a self map $J/J^2 \to J/J^2$ whose image is contained in $(IJ + J^2)/J^2$. In particular $\Delta ((IJ + J^2)/J^2) = 0$ because $I^2 = 0$. This means that $\Delta$ factors as

$J/J^2 \to J/(IJ + J^2) \xrightarrow {\overline{\Delta }} (IJ + J^2)/J^2 \to J/J^2$

where $\overline{\Delta }$ is a $S/IS$-module map. Using again that the sequence (10.138.12.1) is split, we can find a $S/IS$-module map $\overline{\delta } : \Omega _{P/R} \otimes _ P S/IS \to (IJ + J^2)/J^2$ such that $\overline{\delta } \circ d$ is equal to $\overline{\Delta }$. In the same manner as above the map $\overline{\delta }$ determines an $S$-module map $\delta : \Omega _{P/R} \otimes _ P S \to J/J^2$. After replacing $\tilde\sigma$ by $\tilde\sigma + \delta$ a simple computation shows that $\Delta = 0$. In other words $\tilde\sigma$ is a section of $J/J^2 \to \Omega _{P/R} \otimes _ P S$. By Lemma 10.138.7 we conclude that $R \to S$ is formally smooth. $\square$

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