Lemma 10.32.4 (0AMG)—The Stacks project

Table of contents
Part 1: Preliminaries
Chapter 10: Commutative Algebra
Section 10.32: Locally nilpotent ideals
Lemma 10.32.4 (cite)

Lemma 10.32.4. Let $R$ be a ring and let $I \subset R$ be a locally nilpotent ideal. An element $x$ of $R$ is a unit if and only if the image of $x$ in $R/I$ is a unit.

Proof. If $x$ is a unit in $R$, then its image is clearly a unit in $R/I$. It remains to prove the converse. Assume the image of $y \in R$ in $R/I$ is the inverse of the image of $x$. Then $xy = 1 - z$ for some $z \in I$. This means that $1\equiv z$ modulo $xR$. Since $z$ lies in the locally nilpotent ideal $I$, we have $z^ N = 0$ for some sufficiently large $N$. It follows that $1 = 1^ N \equiv z^ N = 0$ modulo $xR$. In other words, $x$ divides $1$ and is hence a unit. $\square$

Comments (2)

Comment #3629 by Rene Schoof on October 18, 2018 at 11:45

$\ldots$ Then $xy=1-z$ for some $z\in I$ . This means that $1\equiv z$ modulo $xR$ . Since $z$ lies in the locally nilpotent ideal $I$ , we have $z^N=0$ for some sufficiently large $N$ . It follows that $1=1^N\equiv z^N=0$ modulo $xR$ . In other words, $x$ divides $1$ and is hence a unit.

Comment #3729 by Johan on October 23, 2018 at 18:23

Hahaha! Yes, this is quite a bit better. Thank you. Changes can be found here.

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