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The Krull intersection is the $(1 - a)$ -torsion submodule, and it vanishes in the Jacobson-radical case

Statement

The first clause below is choice-free; the second uses the published Jacobson-radical unit criterion and therefore inherits its Axiom-of-Choice boundary.

Let $R$ be a Noetherian commutative ring, let $I \subset R$ be an ideal, and let $M$ be a finite $R$ -module. Put $K : = ⋂_{n \geq 0} I^{n} M .$ Then:

$K$ is exactly the set of elements $m \in M$ for which $(1 - a) m = 0$ for some $a \in I$ ;
if $I \subseteq J (R)$ , then $K = 0$ .

Facts & Assumptions

Given: A Noetherian commutative ring $R$ , an ideal $I \subset R$ , a finite $R$ -module $M$ , and $K = ⋂_{n \geq 0} I^{n} M$ .

[L1]

Artin-Rees applies to the submodule $K \subseteq M$ (Artin-Rees controls intersections of submodules with high ideal powers).

[L2]

If a finite module $N$ satisfies $I N = N$ , then $(1 - a) N = 0$ for some $a \in I$ (Determinant trick for Nakayama).

[L3]

Assuming the Axiom of Choice, $a \in J (R)$ exactly when $1 - r a$ is a unit for every $r \in R$ (Assuming the Axiom of Choice, an element lies in the Jacobson radical exactly when one minus any multiple is a unit).

[L4]

A submodule of a finite module over a Noetherian ring is finite (Over a Noetherian ring a module is Noetherian exactly when it is finitely generated, exactly when it is finitely presented).

Proof

technique · direct

1.1L1given

Since $K \subseteq I^{n} M$ for every $n$ , Artin-Rees gives some $c$ such that for all $n \geq c$ , $K = I^{n} M \cap K = I^{n - c} (I^{c} M \cap K) = I^{n - c} K .$ In particular $I K = K$ .

1.2givenalgebra

Conversely, if $(1 - a) m = 0$ for some $a \in I$ , then $m = a m \in I M$ . Iterating gives $m = a^{n} m \in I^{n} M$ for every $n \geq 0$ , hence $m \in K$ .

1.3algebra

Steps 2.1 and 1.2 identify $K$ with the set of $(1 - a)$ -torsion elements claimed in part 1.

2.1L2L4step 1.1algebra

The submodule $K \subseteq M$ is finite by [L4]. Applying [L2] to $K$ and the equality $I K = K$ from step 1.1 gives $a \in I$ with $(1 - a) K = 0.$ So every element of $K$ satisfies the displayed torsion condition.

3.1L3step 2.1algebra

Assume now $I \subseteq J (R)$ . For any $m \in K$ , step 2.1 gives some $a \in I \subseteq J (R)$ with $(1 - a) m = 0$ . By [L3], $1 - a$ is a unit, so multiplying by its inverse gives $m = 0$ . Thus $K = 0$ .

4.1algebra∎

Therefore both stated conclusions hold.

Depends on

Used by

Dependency tree · two levels

18 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

Allen B. Altman and Steven L. Kleiman, A Term of Commutative Algebra, Exercise (20.19) (standard reference, not scraped)
J. S. Milne, A Primer of Commutative Algebra, §24 (standard reference, not scraped)

The Krull intersection is the (1−a)(1−a)-torsion submodule, and it vanishes in the Jacobson-radical case