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

$I^2 = 0$,

$R \to S$ is flat, and

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

Then $R \to S$ is formally smooth.

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

$I^2 = 0$,

$R \to S$ is flat, and

$R/I \to S/IS$ 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. $\square$

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)

There are also: