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Primary submodules of finite modules are characterized by a singleton associated-prime set

Statement

Assume the Axiom of Choice.

Let $R$ be a Noetherian commutative ring, let $M$ be a finitely generated left $R$ -module, let $Q ⊊ M$ be a proper submodule, and let $p$ be a prime ideal. Put $N = M / Q$ . Then the following are equivalent:

$Q$ is $p$ -primary.
${Ass}_{R} (N) = {p}$ .
Every $a \notin p$ acts injectively on $N$ , and there exists $n \geq 1$ with $p^{n} N = 0$ .

Facts & Assumptions

Given: The Axiom of Choice, a Noetherian commutative ring $R$ , a finitely generated left $R$ -module $M$ , a proper submodule $Q ⊊ M$ , a prime ideal $p$ , and the quotient $N = M / Q$ .

[L1]

Zero divisors on a module over a Noetherian ring are exactly the elements lying in its associated primes (Zero divisors on a module over a Noetherian ring are the union of its associated primes).

[L2]

For a finitely generated module, support is $V$ of the annihilator (For a finite module, support is the set of primes containing the annihilator).

[L3]

Minimal primes in the support of a finite module are associated (Minimal support primes of a finite module are associated).

[L4]

A proper submodule is $p$ -primary exactly when it is primary and $\sqrt{{Ann}_{R} (N)} = p$ (Primary submodules and primary ideals).

[L5]

A localization or quotient of a Noetherian ring is Noetherian (Every quotient and every localisation of a Noetherian ring is Noetherian).

[L6]

The nilradical of a Noetherian ring is nilpotent (The nilradical of a Noetherian ring is nilpotent).

[L7]

The support of a finite module is the union of the $V (q)$ over its associated primes $q$ (The support is the union of the closures of the associated primes).

[L8]

Assuming the Axiom of Choice, the radical of an ideal is the intersection of the primes containing it (The radical of an ideal is the intersection of the prime ideals containing it).

Proof

technique · direct

1.1L2L3L4L8givenalgebra

Assume $Q$ is $p$ -primary. Then [L4] gives $\sqrt{{Ann}_{R} (N)} = p$ , so every prime in ${Supp}_{R} (N) = V ({Ann}_{R} (N))$ contains $p$ by [L2] and [L8]. Let $q \in {Ass}_{R} (N)$ . Every $a \in q$ is a zero divisor on $N$ , so primaryity makes $a$ act nilpotently on $N$ and hence some power of $a$ lies in ${Ann}_{R} (N)$ . Thus $a \in \sqrt{{Ann}_{R} (N)} = p$ , which proves $q \subseteq p$ . Since $p$ is minimal in $V ({Ann}_{R} (N))$ , fact [L3] gives $p \in {Ass}_{R} (N)$ . Now every associated prime contains $p$ and is contained in $p$ , so ${Ass}_{R} (N) = {p}$ .

1.2L1L2L5L6L7L8givenalgebra

Assume ${Ass}_{R} (N) = {p}$ . By [L1], the zero divisors on $N$ are exactly the elements of $p$ , so every $a \notin p$ acts injectively on $N$ . Also [L7] gives ${Supp}_{R} (N) = V (p)$ . Combining this with [L2] shows $V ({Ann}_{R} (N)) = V (p)$ , and then [L8] yields $\sqrt{{Ann}_{R} (N)} = p$ . Since $R / {Ann}_{R} (N)$ is Noetherian by [L5], fact [L6] applied to that quotient ring shows that its nilradical $p / {Ann}_{R} (N)$ is nilpotent. Hence $p^{n} \subseteq {Ann}_{R} (N)$ for some $n \geq 1$ , that is, $p^{n} N = 0$ .

1.3L4givenalgebra

Assume condition 3. If $a$ is a zero divisor on $N$ and $a \notin p$ , then multiplication by $a$ would be both noninjective and injective, impossible. So every zero divisor lies in $p$ . Since $p^{n} N = 0$ , each $a \in p$ acts nilpotently on $N$ , namely $a^{n} N = 0$ . Therefore $Q$ is primary. Also $p^{n} N = 0$ gives $p \subseteq \sqrt{{Ann}_{R} (N)}$ . Conversely, if $a \in \sqrt{{Ann}_{R} (N)}$ , then some power of $a$ kills $N$ . If $a \notin p$ , multiplication by $a$ is injective, hence every power of $a$ is injective, so a nonzero power cannot annihilate the nonzero module $N$ . Thus $a \in p$ . Therefore $\sqrt{{Ann}_{R} (N)} = p$ , and [L4] shows that $Q$ is $p$ -primary.

2.1step 1.1step 1.2step 1.3∎

Steps 1.1, 1.2, and 1.3 prove the equivalence of the three conditions.

Depends on

Used by

Dependency tree · two levels

35 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., Theorem (18.4) (standard reference, not scraped)
J. S. Milne, A Primer of Commutative Algebra, v4.03, Propositions 19.5 and 19.16 (standard reference, not scraped)