---
tag: A8HRT9H3XA
kind: statement
role: lemma
name: Determinant trick
msc: [13C05, 13A15]
uses: [0310REJJY8, 05QY1VEBYD, 096Z9MP2ME, XA0DCNH7WY]
about: [05QY1VEBYD]
sources:
  - url: https://en.wikipedia.org/wiki/Nakayama's_lemma
---

## Statement

Let $R$ be a [[commutative ring|0310REJJY8]], let $I$ be an [[ideal|096Z9MP2ME]] of $R$, and let $M$ be a [[finitely generated|05QY1VEBYD]] $R$-module with $IM = M$.
Then there is an element $r \in R$ with $r - 1 \in I$ and $rM = 0$.

## Remarks

Apply the [[Cayley–Hamilton theorem for modules|SH3R5VTMZH]] to $\varphi = \id_M$, which satisfies $\varphi(M) = M = IM$.
It gives $a_1, \ldots, a_n \in I$ with $(1 + a_1 + \cdots + a_n) \id_M = 0$, so $r = 1 + a_1 + \cdots + a_n$ works.
The name comes from the determinant of the matrix $I_n - (a_{ij})$ used in that proof.
If $I$ lies in the [[Jacobson radical|XA0DCNH7WY]] of $R$, then $r$ is a unit, so $M = 0$; this is [[Nakayama's lemma|15JHBYR6KH]].
The hypothesis that $M$ is finitely generated is needed: for $R = \Z$, $I = 2\Z$ and $M = \Q$ one has $IM = M$, but the only $r$ with $r\Q = 0$ is $r = 0$, and $0 - 1 \notin 2\Z$.

## Explanations

### L1 human

If multiplying a finitely generated module by an ideal does not make it smaller, then the module is killed by some element that is congruent to $1$ modulo the ideal.
It turns a statement about modules into a statement about a single ring element.

### L2 human

This is the most flexible form of [[Nakayama's lemma|15JHBYR6KH]].
It gives Nakayama's lemma for local rings and for the Jacobson radical at once.
It also shows that a finitely generated ideal $I$ with $I^2 = I$ is generated by an idempotent.
Indeed, there is $r \in 1 + I$ with $rI = 0$.
Then $e = 1 - r \in I$ satisfies $e x = x$ for $x \in I$, so $e^2 = e$ and $I = eR$.