Lemma 16.7.2. Let $R$ be a Noetherian ring. Let $\Lambda $ be an $R$-algebra. Let $\pi \in R$ and assume that $\text{Ann}_ R(\pi ) = \text{Ann}_ R(\pi ^2)$ and $\text{Ann}_\Lambda (\pi ) = \text{Ann}_\Lambda (\pi ^2)$. Let $A \to \Lambda $ and $D \to \Lambda $ be $R$-algebra maps with $A$ and $D$ of finite presentation. Assume

$\pi $ is strictly standard in $A$ over $R$, and

there exists an $R$-algebra map $A/\pi ^4 A \to D/\pi ^4 D$ compatible with the maps to $\Lambda /\pi ^4 \Lambda $.

Then we can find an $R$-algebra map $B \to \Lambda $ with $B$ of finite presentation and $R$-algebra maps $A \to B$ and $D \to B$ compatible with the maps to $\Lambda $ such that $H_{D/R}B \subset H_{B/D}$ and $H_{D/R}B \subset H_{B/R}$.

**Proof.**
We apply Lemma 16.7.1 to

\[ D \longrightarrow A \otimes _ R D \longrightarrow \Lambda \]

and the image of $\pi $ in $D$. By Lemma 16.2.7 we see that $\pi $ is strictly standard in $A \otimes _ R D$ over $D$. As our section $\rho : (A \otimes _ R D)/\pi ^4 (A \otimes _ R D) \to D/\pi ^4 D$ we take the map induced by the map in (2). Thus Lemma 16.7.1 applies and we obtain a factorization $A \otimes _ R D \to B \to \Lambda $ with $B$ of finite presentation and $\mathfrak a B \subset H_{B/D}$ where

\[ \mathfrak a = \text{Ann}_ D(\text{Ann}_ D(\pi ^2)/\text{Ann}_ D(\pi )). \]

For any prime $\mathfrak q$ of $D$ such that $D_\mathfrak q$ is flat over $R$ we have $\text{Ann}_{D_\mathfrak q}(\pi ^2)/\text{Ann}_{D_\mathfrak q}(\pi ) = 0$ because annihilators of elements commutes with flat base change and we assumed $\text{Ann}_ R(\pi ) = \text{Ann}_ R(\pi ^2)$. Because $D$ is Noetherian we see that $\text{Ann}_ D(\pi ^2)/\text{Ann}_ D(\pi )$ is a finite $D$-module, hence formation of its annihilator commutes with localization. Thus we see that $\mathfrak a \not\subset \mathfrak q$. Hence we see that $D \to B$ is smooth at any prime of $B$ lying over $\mathfrak q$. Since any prime of $D$ where $R \to D$ is smooth is one where $D_\mathfrak q$ is flat over $R$ we conclude that $H_{D/R}B \subset H_{B/D}$. The final inclusion $H_{D/R}B \subset H_{B/R}$ follows because compositions of smooth ring maps are smooth (Algebra, Lemma 10.137.14).
$\square$

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