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Determinant trick for Nakayama

Statement

Let $R$ be a commutative ring, let $I ⊴ R$ be an ideal, and let $M$ be a finitely generated left $R$ -module. If $I M = M$ , then there exists $a \in I$ such that

(1 - a) M = 0.

Facts & Assumptions

Given: A commutative ring $R$ , an ideal $I ⊴ R$ , and a finitely generated left $R$ -module $M$ with $I M = M$ .

[L1]

A finitely generated module has a finite generating set (Generated submodule, cyclic and finitely generated modules, module basis and free module).

[L2]

The submodule $I M$ consists of finite sums of products $i m$ with $i \in I$ and $m \in M$ (The submodule $I M$ generated by products of elements of an ideal $I$ with elements of a module $M$ ).

[L3]

For a positive-size square matrix $A$ over a commutative ring, $A adj (A) = \det (A) I$ (For every positive-sized square matrix over a commutative ring, $A adj (A) = adj (A) A = \det (A) I$ ).

Proof

technique · direct

1.1

If $M = 0$ , then $a = 0 \in I$ satisfies $(1 - a) M = 0$ . So assume $M \neq 0$ and choose generators $m_{1}, \dots, m_{n}$ with $n \geq 1$ by [L1].

L1givenchoose

2.1

Since $I M = M$ , each generator has the form $m_{i} = \sum_{j} a_{i j} m_{j}$ with $a_{i j} \in I$ . Writing $A = (a_{i j})$ and $m = (m_{1}, \dots, m_{n})^{T}$ , this is $(I_{n} - A) m = 0$ .

L2step 1.1algebra

3.1

Multiply the relation of step 2.1 by $adj (I_{n} - A)$ . By [L3], this gives $\det (I_{n} - A) m = 0$ , so $\det (I_{n} - A)$ annihilates every generator and hence all of $M$ .

L3step 2.1algebra

3.2

Expanding $\det (I_{n} - A)$ , the identity permutation contributes $1$ , and every other term contains at least one entry of $A$ , hence lies in $I$ . Therefore $\det (I_{n} - A) = 1 - a$ for some $a \in I$ .

step 2.1algebra

4.1

Step 3.1 and step 3.2 give $(1 - a) M = 0$ for some $a \in I$ .

step 3.1step 3.2∎

Depends on

Used by

Assuming the Axiom of Choice, Nakayama's lemma Theorem

Dependency tree · two levels

13 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

A. Altman and S. Kleiman, A Term of Commutative Algebra, 13th ed., Exercise 10.12 (standard reference, not scraped)
The Stacks Project, Section 10.19: Nakayama's Lemma (standard reference, not scraped)