Lemma 10.101.5 (051J)—The Stacks project

Processing math: 100%

Table of contents
Part 1: Preliminaries
Chapter 10: Commutative Algebra
Section 10.101: Flatness criteria over Artinian rings
Lemma 10.101.5 (cite)

numberstags

Lemma 10.101.5. Let $\varphi : R \to R'$ be a ring map. Let $I \subset R$ be an ideal. Let $M$ be an $R$ -module. Assume

$I$ is nilpotent,
$R \to R'$ is injective,
$M/IM$ is flat over $R/I$ , and
$R' \otimes _ R M$ is flat over $R'$ .

Then $M$ is flat over $R$ .

Proof. Define inductively $I_1 = I$ and $I_{n + 1} = \varphi ^{-1}(\varphi (I_ n)^2R')$ for $n \geq 1$ . Note that by Lemma 10.101.4 we find that $M/I_ nM$ is flat over $R/I_ n$ for each $n \geq 1$ . It is clear that $\varphi (I_ n) \subset \varphi (I)^{2^ n}R'$ . Since $I$ is nilpotent we see that $\varphi (I_ n) = 0$ for some $n$ . As $\varphi$ is injective we conclude that $I_ n = 0$ for some $n$ and we win. $\square$

Comments (1)

Comment #9807 by Branislav Sobot on October 19, 2024 at 05:26

It should be $\varphi(I_n)\subseteq\varphi(I)^{2^{n-1}}R'$ , but who cares...

There are also:

2 comment(s) on Section 10.101: Flatness criteria over Artinian rings

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).

Comments (1)

Post a comment