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✓ 36 results · all verified · 28 also independently AI-judged

Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs. The 8 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

Associated Primes and Primary Decomposition

1 · Prerequisites

2 · Summary

This page develops the associated-prime language from annihilators of elements and cyclic submodules, then proves its behavior in short exact sequences, localization, prime filtrations, and zero-divisor questions for finite modules over Noetherian rings. Those results give the exact bridge from support and annihilator data to the finite set of primes that controls the module.

The second half adopts the quotient-zero-divisor definition of primary submodules, proves the equivalent finite-Noetherian characterizations, and then builds Lasker-Noether primary decomposition. The uniqueness results are stated at the right level: the radical set is intrinsic, isolated components are recovered by localization and contraction, and downward-closed intersections of components are canonical even though embedded components may vary.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-08-28Open item page →

Associated primes of a module

Definition

Let $R$ be a commutative ring and let $M$ be a left $R$ -module. A prime ideal $p ⊊ R$ is associated to $M$ when

p = {Ann}_{R} (m)

for some element $m \in M$ . The set of associated primes of $M$ is denoted

{Ass}_{R} (M) .

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-28Open item page →

A cyclic submodule is a residue module by its annihilator

Statement

Let $R$ be a commutative ring, let $M$ be a left $R$ -module, and let $m \in M$ . Then the cyclic submodule $R m$ is naturally isomorphic to

R / {Ann}_{R} (m) .

Facts & Assumptions

Given: A commutative ring $R$ , a left $R$ -module $M$ , and an element $m \in M$ .

[L1]

The annihilator of $m$ is ${Ann}_{R} (m) = {r \in R : r m = 0}$ (Annihilators, torsion elements and the torsion subset of a module).

Proof

technique · direct

1.1

Define $φ : R / {Ann}_{R} (m) \to R m$ by $φ (r + {Ann}_{R} (m)) = r m .$ If $r - r^{'} \in {Ann}_{R} (m)$ , then $(r - r^{'}) m = 0$ , so $r m = r^{'} m$ ; hence $φ$ is well-defined. Every element of $R m$ has the form $r m$ , so $φ$ is surjective.

L1givenconstruct

2.1

If $φ (r + {Ann}_{R} (m)) = 0$ , then $r m = 0$ , so $r \in {Ann}_{R} (m)$ by [L1]. Thus $r + {Ann}_{R} (m) = 0$ and $φ$ is injective.

L1step 1.1

3.1

Steps 1.1 and 2.1 show that $φ$ is the claimed natural isomorphism.

step 1.1step 2.1∎

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-28Open item page →

Associated primes are exactly primes of embedded cyclic residue modules

Statement

Let $R$ be a commutative ring, let $M$ be a left $R$ -module, and let $p$ be a prime ideal of $R$ . Then

p \in {Ass}_{R} (M)

if and only if there exists an injective $R$ -module homomorphism

R / p ↪ M .

Facts & Assumptions

Given: A commutative ring $R$ , a left $R$ -module $M$ , and a prime ideal $p \subseteq R$ .

[L1]

A prime ideal is associated to $M$ exactly when it is the annihilator of some element of $M$ (Associated primes of a module).

[L2]

For any $m \in M$ , the cyclic submodule $R m$ is naturally isomorphic to $R / {Ann}_{R} (m)$ (A cyclic submodule is a residue module by its annihilator).

Proof

technique · direct

1.1

Assume $p \in {Ass}_{R} (M)$ . By [L1], choose $m \in M$ with ${Ann}_{R} (m) = p$ . Then [L2] gives $R m ≅ R / p$ , and the inclusion $R m ↪ M$ composes with this isomorphism to give an embedding $R / p ↪ M$ .

L1L2choose

1.2

Conversely, let $j : R / p ↪ M$ be injective, and put $m = j (1 + p)$ . Every $r \in p$ kills $1 + p$ , so $r m = 0$ . If $r m = 0$ , then $j (r + p) = 0$ , and injectivity gives $r + p = 0$ , hence $r \in p$ . Therefore ${Ann}_{R} (m) = p$ , so $p \in {Ass}_{R} (M)$ by [L1].

L1givenalgebra

2.1

Steps 1.1 and 1.2 prove the equivalence.

step 1.1step 1.2∎

CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-28Open item page →

Associated primes of a cyclic quotient are colon primes

Statement

Let $R$ be a commutative ring and let $I ⊴ R$ be an ideal. Then

{Ass}_{R} (R / I) = {p \in Spec (R) : p = (I : r) for some r \notin I} .

Facts & Assumptions

Given: A commutative ring $R$ and an ideal $I ⊴ R$ .

[L1]

A prime ideal belongs to ${Ass}_{R} (M)$ exactly when it is the annihilator of some element of $M$ (Associated primes of a module).

[L2]

The quotient module $R / I$ consists of the cosets $r + I$ (Quotient module $M / N$ with scalar multiplication on additive cosets).

[L3]

The annihilator of an element $x$ is the set of scalars that kill $x$ (Annihilators, torsion elements and the torsion subset of a module).

Proof

technique · direct

1.1

For any $r \in R$ , the annihilator of $r + I$ in $R / I$ is ${Ann}_{R} (r + I) = {a \in R : a (r + I) = 0 + I} = {a \in R : a r \in I} = (I : r) .$

L2L3given

2.1

If $p \in {Ass}_{R} (R / I)$ , then [L1] gives a nonzero class $r + I$ with $p = {Ann}_{R} (r + I)$ . Since $r + I \neq 0$ , one has $r \notin I$ , and step 1.1 yields $p = (I : r)$ .

L1step 1.1

2.2

Conversely, if $p = (I : r)$ for some $r \notin I$ and if $p$ is prime, then step 1.1 gives $p = {Ann}_{R} (r + I)$ with $r + I \neq 0$ , so $p \in {Ass}_{R} (R / I)$ by [L1].

L1step 1.1

3.1

Steps 2.1 and 2.2 prove the stated description of ${Ass}_{R} (R / I)$ .

step 2.1step 2.2∎

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-28Open item page →

A nonzero module over a Noetherian ring has a maximal element annihilator

Statement

Assume Dependent Choice.

Let $R$ be a Noetherian commutative ring and let $M$ be a nonzero left $R$ -module. Then there exists a nonzero element $m \in M$ such that ${Ann}_{R} (m)$ is maximal, under inclusion, among the annihilators of nonzero elements of $M$ .

Facts & Assumptions

Given: Dependent Choice, a Noetherian commutative ring $R$ , and a nonzero left $R$ -module $M$ .

[L1]

Assuming Dependent Choice, in a Noetherian commutative ring every nonempty set of ideals has a maximal member (A commutative ring is Noetherian exactly when every ideal is finitely generated, exactly when its ideals satisfy the ascending chain condition, and exactly when every nonempty set of ideals has a maximal member).

[L2]

For $m \in M$ , the annihilator is the set ${Ann}_{R} (m) = {r \in R : r m = 0}$ (Annihilators, torsion elements and the torsion subset of a module).

Proof

technique · direct

1.1

For every $m \in M$ , the set ${Ann}_{R} (m)$ is an ideal: it contains $0$ , is closed under subtraction, and $a \in {Ann}_{R} (m)$ implies $r a \in {Ann}_{R} (m)$ for every $r \in R$ . Because $M \neq 0$ , choose $m_{0} \in M$ with $m_{0} \neq 0$ . Then $Σ = {{Ann}_{R} (m) : m \in M, m \neq 0}$ is a nonempty set of ideals of $R$ .

L2givenchoosealgebra

2.1

By [L1], the nonempty set $Σ$ has a maximal member. Thus some nonzero $m \in M$ satisfies that ${Ann}_{R} (m)$ is maximal among the annihilators of nonzero elements of $M$ .

L1step 1.1

3.1

This is exactly the required conclusion.

step 2.1∎

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-28Open item page →

A maximal element annihilator is prime

Statement

Let $R$ be a commutative ring, let $M$ be a left $R$ -module, and let $m \in M$ be nonzero. If ${Ann}_{R} (m)$ is maximal among the annihilators of nonzero elements of $M$ , then ${Ann}_{R} (m)$ is a prime ideal.

Facts & Assumptions

Given: A commutative ring $R$ , a left $R$ -module $M$ , and a nonzero element $m \in M$ such that ${Ann}_{R} (m)$ is maximal among the annihilators of nonzero elements of $M$ .

[L1]

The annihilator of an element $x \in M$ is ${Ann}_{R} (x) = {r \in R : r x = 0}$ (Annihilators, torsion elements and the torsion subset of a module).

Proof

technique · direct

1.1

Put $I = {Ann}_{R} (m)$ . Since $m \neq 0$ , one has $1 \notin I$ , so $I$ is proper. Let $a b \in I$ and assume $b \notin I$ . Then $b m \neq 0$ . Also $I \subseteq {Ann}_{R} (b m)$ because every $r \in I$ satisfies $r (b m) = b (r m) = 0$ . If $r \in {Ann}_{R} (b m)$ , then $(r b) m = 0$ , so $r b \in I$ by [L1].

L1givenalgebra

2.1

The element $b m$ is nonzero, so maximality of $I$ forces ${Ann}_{R} (b m) = I$ . Since $a b m = 0$ , the element $a$ lies in ${Ann}_{R} (b m) = I$ . Therefore $a b \in I$ and $b \notin I$ imply $a \in I$ , so $I$ is prime.

step 1.1

3.1

Thus ${Ann}_{R} (m)$ is a prime ideal.

step 2.1∎

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-28Open item page →

A nonzero module over a Noetherian ring has an associated prime

Statement

Let $R$ be a Noetherian commutative ring and let $M$ be a nonzero left $R$ -module. Then ${Ass}_{R} (M)$ is nonempty.

Facts & Assumptions

Given: A Noetherian commutative ring $R$ and a nonzero left $R$ -module $M$ .

[L1]

A prime ideal belongs to ${Ass}_{R} (M)$ exactly when it is the annihilator of some element of $M$ (Associated primes of a module).

[L2]

Some nonzero element of $M$ has annihilator maximal among annihilators of nonzero elements (A nonzero module over a Noetherian ring has a maximal element annihilator).

[L3]

An annihilator maximal among annihilators of nonzero elements is prime (A maximal element annihilator is prime).

Proof

technique · direct

1.1

By [L2], choose $m \in M$ with $m \neq 0$ such that ${Ann}_{R} (m)$ is maximal among the annihilators of nonzero elements of $M$ . Then [L3] shows that ${Ann}_{R} (m)$ is prime.

L2L3choose

2.1

By [L1], the prime ideal ${Ann}_{R} (m)$ belongs to ${Ass}_{R} (M)$ .

L1step 1.1

3.1

Therefore ${Ass}_{R} (M)$ is nonempty.

step 2.1∎

CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-28Open item page →

A module over a Noetherian ring has no associated primes exactly when it is zero

Statement

Let $R$ be a Noetherian commutative ring and let $M$ be a left $R$ -module. Then ${Ass}_{R} (M) = \emptyset ⟺ M = 0.$

Facts & Assumptions

Given: A Noetherian commutative ring $R$ and a left $R$ -module $M$ .

[L1]

Every nonzero $R$ -module has an associated prime (A nonzero module over a Noetherian ring has an associated prime).

Proof

technique · direct

1.1

If $M = 0$ , then the only element of $M$ is $0$ , whose annihilator is the whole ring $R$ . Since $R$ is not a prime ideal of itself, no associated prime can occur, so ${Ass}_{R} (M) = \emptyset$ .

givenalgebra

1.2

If $M \neq 0$ , then [L1] gives ${Ass}_{R} (M) \neq \emptyset$ .

L1given

2.1

Steps 1.1 and 1.2 prove the equivalence.

step 1.1step 1.2∎

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-28Open item page →

Associated primes of a submodule lie in those of the ambient module

Statement

0 ⟶ M^{'} ⟶ M ⟶ M^{''} ⟶ 0

is a short exact sequence of left $R$ -modules, then

{Ass}_{R} (M^{'}) \subseteq {Ass}_{R} (M) .

Facts & Assumptions

Given: A commutative ring $R$ and a short exact sequence $0 \to M^{'} \to M \to M^{''} \to 0$ of left $R$ -modules.

[L1]

A prime $p$ is associated to a module exactly when $R / p$ embeds in that module (Associated primes are exactly primes of embedded cyclic residue modules).

[L2]

In a short exact sequence, the left map is injective (Exact sequences and short exact sequences of modules).

Proof

technique · direct

1.1

Let $p \in {Ass}_{R} (M^{'})$ . By [L1], there is an embedding $R / p ↪ M^{'}$ . Composing with the injective map $M^{'} ↪ M$ from [L2] gives an embedding $R / p ↪ M$ .

L1L2

2.1

Applying [L1] again, the embedding from step 1.1 shows that $p \in {Ass}_{R} (M)$ . Therefore ${Ass}_{R} (M^{'}) \subseteq {Ass}_{R} (M)$ .

L1step 1.1∎

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-28Open item page →

Associated primes of the middle term lie in those of the ends

Statement

0 ⟶ M^{'} ⟶ M ⟶ M^{''} ⟶ 0

is a short exact sequence of left $R$ -modules, then

{Ass}_{R} (M) \subseteq {Ass}_{R} (M^{'}) \cup {Ass}_{R} (M^{''}) .

Facts & Assumptions

Given: A commutative ring $R$ and a short exact sequence $0 \to M^{'} \to M \to M^{''} \to 0$ of left $R$ -modules.

[L1]

In a short exact sequence, the image of the left map equals the kernel of the right map (Exact sequences and short exact sequences of modules).

Proof

technique · direct

1.1

Let $p \in {Ass}_{R} (M)$ , and choose $m \in M$ with ${Ann}_{R} (m) = p$ . If there exists $g \notin p$ with $g m \in M^{'}$ , then $g m \neq 0$ . Also $p \subseteq {Ann}_{R} (g m)$ , and if $a g m = 0$ then $a g \in p$ ; since $p$ is prime and $g \notin p$ , this gives $a \in p$ . Hence ${Ann}_{R} (g m) = p$ , so $p \in {Ass}_{R} (M^{'})$ .

givenalgebra

2.1

If no such $g$ exists, let $\overset{ˉ}{m}$ be the image of $m$ in $M^{''}$ . Then $\overset{ˉ}{m} \neq 0$ , for otherwise $m \in M^{'}$ and $g = 1$ would contradict the assumption. Every $a \in p$ kills $m$ , hence kills $\overset{ˉ}{m}$ . Conversely, if $a \overset{ˉ}{m} = 0$ , then $a m \in M^{'}$ . By the standing assumption, this forces $a \in p$ . Therefore ${Ann}_{R} (\overset{ˉ}{m}) = p$ , so $p \in {Ass}_{R} (M^{''})$ .

L1step 1.1algebra

3.1

Steps 1.1 and 2.1 show that every associated prime of $M$ lies in ${Ass}_{R} (M^{'}) \cup {Ass}_{R} (M^{''})$ .

step 1.1step 2.1∎

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-28Open item page →

Associated primes in a short exact sequence

Statement

0 ⟶ M^{'} ⟶ M ⟶ M^{''} ⟶ 0

is a short exact sequence of left $R$ -modules, then

{Ass}_{R} (M^{'}) \subseteq {Ass}_{R} (M) \subseteq {Ass}_{R} (M^{'}) \cup {Ass}_{R} (M^{''}) .

In particular,

{Ass}_{R} (M^{'} \oplus M^{''}) = {Ass}_{R} (M^{'}) \cup {Ass}_{R} (M^{''}) .

Facts & Assumptions

Given: A commutative ring $R$ and a short exact sequence $0 \to M^{'} \to M \to M^{''} \to 0$ of left $R$ -modules.

[L1]

In a short exact sequence, associated primes of the left term lie in those of the middle term (Associated primes of a submodule lie in those of the ambient module).

[L2]

In a short exact sequence, associated primes of the middle term lie in those of the outer terms (Associated primes of the middle term lie in those of the ends).

Proof

technique · direct

1.1

Facts [L1] and [L2] give ${Ass}_{R} (M^{'}) \subseteq {Ass}_{R} (M) \subseteq {Ass}_{R} (M^{'}) \cup {Ass}_{R} (M^{''}) .$

L1L2

2.1

Apply step 1.1 to the split exact sequence $0 ⟶ M^{'} ⟶ M^{'} \oplus M^{''} ⟶ M^{''} ⟶ 0.$ This gives ${Ass}_{R} (M^{'} \oplus M^{''}) \subseteq {Ass}_{R} (M^{'}) \cup {Ass}_{R} (M^{''})$ . The reverse inclusion follows by applying the left inclusion of step 1.1 to the two canonical injections $M^{'} ↪ M^{'} \oplus M^{''}$ and $M^{''} ↪ M^{'} \oplus M^{''}$ .

step 1.1

3.1

Steps 1.1 and 2.1 prove the theorem and its direct-sum corollary.

step 1.1step 2.1∎

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-28Open item page →

Associated primes localize forward

Statement

Let $R$ be a commutative ring, let $M$ be a left $R$ -module, let $S \subseteq R$ be multiplicative, and let $p \in {Ass}_{R} (M)$ with $p \cap S = \emptyset$ . Then

S^{- 1} p \in {Ass}_{S^{- 1} R} (S^{- 1} M) .

Facts & Assumptions

Given: A commutative ring $R$ , a left $R$ -module $M$ , a multiplicative subset $S \subseteq R$ , and a prime ideal $p \in {Ass}_{R} (M)$ with $p \cap S = \emptyset$ .

[L1]

A prime ideal is associated to a module exactly when its residue module embeds in that module (Associated primes are exactly primes of embedded cyclic residue modules).

[L2]

Injective module maps remain injective after localisation (Injective module maps remain injective after localisation).

[L3]

Localisation commutes with quotient modules, so $S^{- 1} (R / p) ≅ (S^{- 1} R) / (S^{- 1} p)$ (Localisation commutes with quotient modules and arbitrary direct sums).

[L4]

If $p \cap S = \emptyset$ , then $S^{- 1} p$ is a prime ideal of $S^{- 1} R$ (Prime ideals of a localization are exactly the primes disjoint from the denominator set).

Proof

technique · direct

1.1

By [L1], there is an injective $R$ -module map $R / p ↪ M$ . Applying [L2] yields an injective $S^{- 1} R$ -module map $S^{- 1} (R / p) ↪ S^{- 1} M$ .

L1L2

2.1

By [L3], the source identifies with $(S^{- 1} R) / (S^{- 1} p)$ , and by [L4] the ideal $S^{- 1} p$ is prime. Therefore [L1], applied over the ring $S^{- 1} R$ , shows that $S^{- 1} p$ is associated to $S^{- 1} M$ .

L1L3L4step 1.1

3.1

Hence associated primes localize forward.

step 2.1∎

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-28Open item page →

Associated primes of a localized finite module come from upstairs

Statement

Let $R$ be a Noetherian commutative ring, let $M$ be a finitely generated left $R$ -module, and let $S \subseteq R$ be multiplicative. If

q \in {Ass}_{S^{- 1} R} (S^{- 1} M),

then there exists $p \in {Ass}_{R} (M)$ with $p \cap S = \emptyset$ and $q = S^{- 1} p$ .

Facts & Assumptions

Given: A Noetherian commutative ring $R$ , a finitely generated left $R$ -module $M$ , a multiplicative subset $S \subseteq R$ , and a prime ideal $q \in {Ass}_{S^{- 1} R} (S^{- 1} M)$ .

[L1]

A prime ideal is associated to a module exactly when it is the annihilator of some element (Associated primes of a module).

[L2]

The annihilator of an element is the set of scalars that kill it (Annihilators, torsion elements and the torsion subset of a module).

[L3]

In a Noetherian commutative ring, every ideal is finitely generated (A commutative ring is Noetherian exactly when every ideal is finitely generated, exactly when its ideals satisfy the ascending chain condition, and exactly when every nonempty set of ideals has a maximal member).

[L4]

Prime ideals of $S^{- 1} R$ correspond exactly to primes of $R$ disjoint from $S$ , via $p \mapsto S^{- 1} p$ (Prime ideals of a localization are exactly the primes disjoint from the denominator set).

Proof

technique · direct

1.1

By [L1], choose $u \in S^{- 1} M$ with $u \neq 0$ and ${Ann}_{S^{- 1} R} (u) = q$ . Write $u = m / s$ with $m \in M$ and $s \in S$ . Since $1 / s$ is a unit of $S^{- 1} R$ , the element $m / 1 = (s / 1) u$ is also nonzero and has the same annihilator $q$ . By [L4], there is a prime ideal $p \subseteq R$ with $p \cap S = \emptyset$ and $q = S^{- 1} p$ . If $a m = 0$ , then $a / 1$ annihilates $m / 1$ , so $a / 1 \in q$ and therefore $a \in p$ . Thus ${Ann}_{R} (m) \subseteq p$ .

L1L2L4choosealgebra

2.1

By [L3], write $p = (f_{1}, \dots, f_{n})$ . Since each $f_{i} / 1 \in q = {Ann}_{S^{- 1} R} (m / 1)$ , there exists $s_{i} \in S$ with $s_{i} f_{i} m = 0$ in $M$ . Put $h = s_{1} \dots s_{n}$ . Then $h m \neq 0$ , because $h / 1$ is a unit and $(h / 1) (m / 1) = h m / 1$ . Also each generator $f_{i}$ kills $h m$ , so $p \subseteq {Ann}_{R} (h m)$ .

L2L3step 1.1choosealgebra

3.1

If $a \in {Ann}_{R} (h m)$ , then $a / 1$ annihilates $h m / 1$ . Since $h / 1$ is a unit, $a / 1$ also annihilates $m / 1$ , so $a / 1 \in q$ and therefore $a \in p$ by step 1.1. Hence ${Ann}_{R} (h m) \subseteq p$ , and step 2.1 gives ${Ann}_{R} (h m) = p$ .

L4step 1.1step 2.1algebra

4.1

The prime ideal $p$ is therefore the annihilator of the nonzero element $h m \in M$ , so $p \in {Ass}_{R} (M)$ by [L1]. Together with step 1.1, this proves the claim.

L1step 1.1step 3.1∎

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-28Open item page →

Associated primes commute with localization for finite modules

Statement

Let $R$ be a Noetherian commutative ring, let $M$ be a finitely generated left $R$ -module, and let $S \subseteq R$ be multiplicative. Then

{Ass}_{S^{- 1} R} (S^{- 1} M) = {S^{- 1} p : p \in {Ass}_{R} (M), p \cap S = \emptyset} .

Facts & Assumptions

Given: A Noetherian commutative ring $R$ , a finitely generated left $R$ -module $M$ , and a multiplicative subset $S \subseteq R$ .

[L1]

Associated primes of $M$ disjoint from $S$ localize to associated primes of $S^{- 1} M$ (Associated primes localize forward).

[L2]

Every associated prime of $S^{- 1} M$ is the localization of an associated prime of $M$ disjoint from $S$ (Associated primes of a localized finite module come from upstairs).

Proof

technique · direct

1.1

If $p \in {Ass}_{R} (M)$ and $p \cap S = \emptyset$ , then [L1] gives $S^{- 1} p \in {Ass}_{S^{- 1} R} (S^{- 1} M)$ .

1.2

Conversely, if $q \in {Ass}_{S^{- 1} R} (S^{- 1} M)$ , then [L2] produces $p \in {Ass}_{R} (M)$ with $p \cap S = \emptyset$ and $q = S^{- 1} p$ .

2.1

Steps 1.1 and 1.2 prove the stated equality of associated-prime sets.

step 1.1step 1.2∎

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-28Open item page →

Associated primes lie in the support

Statement

Let $R$ be a commutative ring and let $M$ be a left $R$ -module. Then

{Ass}_{R} (M) \subseteq {Supp}_{R} (M) .

Facts & Assumptions

Given: A commutative ring $R$ and a left $R$ -module $M$ .

[L1]

A prime ideal $p$ is associated to $M$ exactly when $p = {Ann}_{R} (m)$ for some $m \in M$ (Associated primes of a module).

[L2]

A prime ideal lies in ${Supp}_{R} (M)$ exactly when the localization $M_{p}$ is nonzero (Support of a module).

[L3]

The module localization $M_{p}$ consists of fractions $m / s$ with $s \in R ∖ p$ , and $m / 1 = 0$ exactly when some $s \notin p$ kills $m$ (Localisation of a module at a multiplicative subset, A localised module fraction is zero exactly when one denominator kills its numerator, Localisation at a prime ideal: $R_{p} = (R ∖ p)^{- 1} R$ ).

Proof

technique · direct

1.1

Let $p \in {Ass}_{R} (M)$ . By [L1], choose $m \in M$ with ${Ann}_{R} (m) = p$ . If $m / 1 = 0$ in $M_{p}$ , then by the definition of localization there exists $s \notin p$ with $s m = 0$ . Hence $s \in {Ann}_{R} (m) = p$ , a contradiction. So $m / 1 \neq 0$ in $M_{p}$ .

L1L3choosealgebra

2.1

By [L2], step 1.1 shows $p \in {Supp}_{R} (M)$ . Thus ${Ass}_{R} (M) \subseteq {Supp}_{R} (M)$ .

L2step 1.1∎

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Finite modules over Noetherian rings admit prime filtrations

Statement

Let $R$ be a Noetherian commutative ring and let $M$ be a finitely generated left $R$ -module. Then there exist submodules

0 = M_{0} \subset M_{1} \subset \dots \subset M_{n} = M

such that each quotient $M_{i} / M_{i - 1}$ is isomorphic to $R / p_{i}$ for some prime ideal $p_{i}$ of $R$ . When $M = 0$ , this is the empty filtration with $n = 0$ .

Facts & Assumptions

Given: A Noetherian commutative ring $R$ and a finitely generated left $R$ -module $M$ .

[L1]

A finitely generated module over a Noetherian ring is Noetherian (Finitely generated modules over a left Noetherian ring are Noetherian).

[L2]

In a Noetherian module, every nonempty family of submodules has a maximal member (Finite generation, ACC, and maximal-condition characterizations of Noetherian modules).

[L3]

Every nonzero module over a Noetherian ring has an associated prime (A nonzero module over a Noetherian ring has an associated prime).

[L4]

If $p$ is associated to a module, then $R / p$ embeds in that module (Associated primes are exactly primes of embedded cyclic residue modules).

[L5]

Quotient modules are formed from cosets $M / N$ (Quotient module $M / N$ with scalar multiplication on additive cosets).

Proof

technique · direct

1.1

If $M = 0$ , then the empty chain $0 = M$ is already a prime filtration.

given

2.1

Assume $M \neq 0$ . Let $Σ = {N \leq M : N admits a prime filtration} .$ By step 1.1, the zero submodule belongs to $Σ$ . Since $M$ is Noetherian by [L1], fact [L2] gives a maximal member $N$ of $Σ$ .

L1L2step 1.1construct

3.1

If $N \neq M$ , then $M / N$ is a nonzero quotient module by [L5]. Fact [L3] gives an associated prime $p$ of $M / N$ , and [L4] yields an embedded copy of $R / p$ in $M / N$ . Let $N^{'}$ be its preimage in $M$ . Then $N ⊊ N^{'} \leq M$ and $N^{'} / N ≅ R / p$ . Appending $N^{'}$ to a prime filtration of $N$ gives a prime filtration of $N^{'}$ , contradicting the maximality of $N$ in step 2.1. Therefore $N = M$ .

L3L4L5step 2.1choosealgebra

4.1

When $M \neq 0$ , step 3.1 shows that $M$ itself has a prime filtration; the zero case was handled in step 1.1.

step 1.1step 3.1∎

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Minimal support primes of a finite module are associated

Statement

Let $R$ be a Noetherian commutative ring and let $M$ be a finitely generated left $R$ -module. If $p$ is minimal in ${Supp}_{R} (M)$ , then

p \in {Ass}_{R} (M) .

Facts & Assumptions

Given: A Noetherian commutative ring $R$ , a finitely generated left $R$ -module $M$ , and a prime ideal $p$ minimal in ${Supp}_{R} (M)$ .

[L1]

The module $M$ admits a prime filtration $0 = M_{0} \subset M_{1} \subset \dots \subset M_{n} = M$ with $M_{i} / M_{i - 1} ≅ R / p_{i}$ for prime ideals $p_{i}$ (Finite modules over Noetherian rings admit prime filtrations).

[L2]

Support in a short exact sequence is the union of the supports of the outer terms (Support in a short exact sequence is the union of the outer supports).

[L3]

The support of $R / I$ is exactly $V (I)$ (The support of a cyclic quotient is its vanishing set).

Proof

technique · direct

1.1

Choose a prime filtration as in [L1]. Repeatedly applying [L2] to the short exact sequences $0 ⟶ M_{i - 1} ⟶ M_{i} ⟶ M_{i} / M_{i - 1} ⟶ 0$ and then using [L3] for $M_{i} / M_{i - 1} ≅ R / p_{i}$ gives ${Supp}_{R} (M) = V (p_{1}) \cup \dots \cup V (p_{n}) .$

L1L2L3algebra

2.1

By step 1.1, the prime $p$ contains some $p_{i}$ . Since $p_{i} \in V (p_{i}) \subseteq {Supp}_{R} (M)$ , the minimality of $p$ in the support forces $p_{i} = p$ . Choose the smallest such index $i$ , and choose $m \in M_{i}$ whose image in $M_{i} / M_{i - 1} ≅ R / p$ is nonzero. Then ${Ann}_{R} (m) \subseteq p$ , because any scalar killing $m$ kills its nonzero class in $R / p$ . Also $p_{j} M_{j} \subseteq M_{j - 1}$ for every $j$ , so $(p_{1} \dots p_{i}) m = 0$ .

L1step 1.1choosealgebra

3.1

For each $j < i$ , the minimal choice of $i$ gives $p_{j} \neq p$ and hence $p_{j} ⊈ p$ . Choose $a_{j} \in p_{j} ∖ p$ , and put $f = a_{1} \dots a_{i - 1}$ . Because $p$ is prime and no factor $a_{j}$ lies in $p$ , one has $f \notin p$ . Since ${Ann}_{R} (m) \subseteq p$ , this implies $f m \neq 0$ . Moreover $p f m = 0$ because $p = p_{i}$ and $(p_{1} \dots p_{i}) m = 0$ . If $b f m = 0$ , then $b f \in {Ann}_{R} (m) \subseteq p$ ; as $f \notin p$ and $p$ is prime, this forces $b \in p$ . Therefore ${Ann}_{R} (f m) = p$ , so $p \in {Ass}_{R} (M)$ .

step 2.1choosealgebra

4.1

Thus every support-prime minimal by inclusion is associated.

step 3.1∎

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The support is the union of the closures of the associated primes

Statement

Let $R$ be a Noetherian commutative ring and let $M$ be a finitely generated left $R$ -module. Then

{Supp}_{R} (M) = ⋃_{p \in {Ass}_{R} (M)} V (p) .

Facts & Assumptions

Given: A Noetherian commutative ring $R$ and a finitely generated left $R$ -module $M$ .

[L1]

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

[L2]

For a finitely generated module, ${Supp}_{R} (M) = V ({Ann}_{R} (M))$ (For a finite module, support is the set of primes containing the annihilator).

[L3]

The module $M$ admits a prime filtration, and support is the union of the supports of its prime-filtration quotients (Finite modules over Noetherian rings admit prime filtrations, Support in a short exact sequence is the union of the outer supports, The support of a cyclic quotient is its vanishing set).

[L4]

Every associated prime lies in the support (Associated primes lie in the support).

Proof

technique · direct

1.1

By [L3], write ${Supp}_{R} (M) = V (p_{1}) \cup \dots \cup V (p_{n})$ for the prime ideals $p_{i}$ occurring in a prime filtration of $M$ . Let $q \in {Supp}_{R} (M)$ . Then $p_{i} \subseteq q$ for some $i$ . Choose such an index $i$ with $p_{i}$ minimal under inclusion among the filtration primes contained in $q$ . If $r \in {Supp}_{R} (M)$ and $r \subseteq p_{i}$ , then [L3] gives some $p_{j} \subseteq r \subseteq p_{i}$ , so the minimal choice of $p_{i}$ forces $p_{j} = p_{i}$ and hence $r = p_{i}$ . Therefore $p_{i}$ is minimal in the support, so [L1] gives $p_{i} \in {Ass}_{R} (M)$ and $q \in V (p_{i})$ . This proves ${Supp}_{R} (M) \subseteq ⋃_{p \in {Ass}_{R} (M)} V (p)$ .

L1L3choosealgebra

1.2

Conversely, let $p \in {Ass}_{R} (M)$ and let $q \supseteq p$ . By [L4], the prime $p$ lies in the support; then [L2] gives ${Ann}_{R} (M) \subseteq p \subseteq q$ , so $q \in {Supp}_{R} (M)$ . Thus $V (p) \subseteq {Supp}_{R} (M)$ for every $p \in {Ass}_{R} (M)$ .

L2L4algebra

2.1

Steps 1.1 and 1.2 prove the support decomposition.

step 1.1step 1.2∎

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Finite modules over Noetherian rings have finitely many associated primes

Statement

Let $R$ be a Noetherian commutative ring and let $M$ be a finitely generated left $R$ -module. Then ${Ass}_{R} (M)$ is a finite set.

Facts & Assumptions

Given: A Noetherian commutative ring $R$ and a finitely generated left $R$ -module $M$ .

[L1]

The module $M$ admits a prime filtration $0 = M_{0} \subset M_{1} \subset \dots \subset M_{n} = M$ with $M_{i} / M_{i - 1} ≅ R / p_{i}$ (Finite modules over Noetherian rings admit prime filtrations).

[L2]

In a short exact sequence, associated primes of the middle term are contained in the union of those of the outer terms (Associated primes in a short exact sequence).

Proof

technique · direct

1.1

Choose a prime filtration as in [L1]. If $n = 0$ , then $M = 0$ and ${Ass}_{R} (M) = \emptyset$ , which is finite.

L1given

1.2

Assume $n \geq 1$ . Applying [L2] to $0 ⟶ M_{n - 1} ⟶ M ⟶ M / M_{n - 1} ⟶ 0$ gives ${Ass}_{R} (M) \subseteq {Ass}_{R} (M_{n - 1}) \cup {Ass}_{R} (M / M_{n - 1}) .$ Since $M / M_{n - 1} ≅ R / p_{n}$ , the class of $1 + p_{n}$ has annihilator $p_{n}$ , so ${Ass}_{R} (M / M_{n - 1}) = {p_{n}}$ . Repeating this argument down the filtration yields ${Ass}_{R} (M) \subseteq {p_{1}, \dots, p_{n}} .$

L1L2algebra

2.1

The right-hand side of step 1.2 is finite, so ${Ass}_{R} (M)$ is finite.

step 1.1step 1.2∎

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A zero divisor is contained in an associated prime

Statement

Let $R$ be a Noetherian commutative ring, let $M$ be a left $R$ -module, and let $x \in R$ be a zero divisor on $M$ . Then $x$ belongs to some prime ideal of ${Ass}_{R} (M)$ .

Facts & Assumptions

Given: A Noetherian commutative ring $R$ , a left $R$ -module $M$ , and an element $x \in R$ that is a zero divisor on $M$ .

[L1]

A prime ideal is associated to a module exactly when it is the annihilator of one of the module's elements (Associated primes of a module).

[L2]

Every nonzero module over a Noetherian ring has an associated prime (A nonzero module over a Noetherian ring has an associated prime).

Proof

technique · direct

1.1

Because $x$ is a zero divisor on $M$ , the set $N = {m \in M : x m = 0}$ is nonzero. It is a submodule: it contains $0$ , is closed under subtraction, and $x m = 0$ implies $x (r m) = r (x m) = 0$ for every $r \in R$ .

givenalgebra

2.1

By [L2], the nonzero module $N$ has an associated prime. By [L1], choose $m \in N$ such that $p = {Ann}_{R} (m)$ is prime. Since $m \in N$ , one has $x m = 0$ , so $x \in {Ann}_{R} (m) = p$ . The same equality and [L1] show $p \in {Ass}_{R} (M)$ as well, because $m$ is also an element of $M$ .

L1L2step 1.1choosealgebra

3.1

Therefore $x$ lies in an associated prime of $M$ .

step 2.1∎

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Zero divisors on a module over a Noetherian ring are the union of its associated primes

Statement

Let $R$ be a Noetherian commutative ring and let $M$ be a left $R$ -module. Then the set of zero divisors on $M$ is

⋃_{p \in {Ass}_{R} (M)} p .

If $M$ is finitely generated, this is a finite union.

Facts & Assumptions

Given: A Noetherian commutative ring $R$ and a left $R$ -module $M$ .

[L1]

Every zero divisor on $M$ lies in an associated prime of $M$ (A zero divisor is contained in an associated prime).

[L2]

If $M$ is finitely generated, then ${Ass}_{R} (M)$ is finite (Finite modules over Noetherian rings have finitely many associated primes).

Proof

technique · direct

1.1

If $x \in p$ for some $p \in {Ass}_{R} (M)$ , choose $m \neq 0$ in $M$ with ${Ann}_{R} (m) = p$ . Then $x m = 0$ , so $x$ is a zero divisor on $M$ .

givenalgebra

1.2

Conversely, every zero divisor on $M$ lies in an associated prime by [L1].

2.1

Steps 1.1 and 1.2 prove the union formula. When $M$ is finitely generated, fact [L2] makes that union finite.

L2step 1.1step 1.2∎

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Primary submodules and primary ideals

Definition

Let $R$ be a commutative ring, let $M$ be a left $R$ -module, and let $Q ⊊ M$ be a proper submodule. Then $Q$ is primary when every zero divisor on the quotient module $M / Q$ acts nilpotently on $M / Q$ ; equivalently, whenever multiplication by $a \in R$ on $M / Q$ has nontrivial kernel, there exists $n \geq 1$ with

a^{n} (M / Q) = 0.

\sqrt{{Ann}_{R} (M / Q)} = p,

then $Q$ is called $p$ -primary.

The radical here is well-defined: ${Ann}_{R} (M / Q)$ is an ideal of $R$ . Indeed, it contains $0$ , is closed under subtraction, and if $a$ annihilates $M / Q$ then so does $r a$ for every $r \in R$ , by the module axioms.

When $M = R$ , viewed as its regular module, a primary submodule is a primary ideal.

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The radical of a primary ideal is prime

Statement

Let $R$ be a commutative ring and let $Q ⊊ R$ be a primary ideal. Then $\sqrt{Q}$ is a prime ideal. In particular, $Q$ is $\sqrt{Q}$ -primary.

Facts & Assumptions

Given: A commutative ring $R$ and a primary ideal $Q ⊊ R$ .

[L1]

A primary ideal is a proper submodule whose quotient $R / Q$ has the property that every zero divisor acts nilpotently (Primary submodules and primary ideals).

[L2]

The radical $\sqrt{Q}$ consists of those $a \in R$ for which $a^{n} \in Q$ for some $n \geq 1$ (The radical of an ideal).

[L3]

A prime ideal is a proper ideal $P$ such that $a b \in P$ implies $a \in P$ or $b \in P$ (Prime ideals and maximal ideals in a commutative ring).

Proof

technique · direct

1.1

Let $a b \in \sqrt{Q}$ and assume $b \notin \sqrt{Q}$ . By [L2], some $n \geq 1$ has $(a b)^{n} \in Q$ , so in the quotient ring $A = R / Q$ one has $(\overset{ˉ}{a} \overset{ˉ}{b})^{n} = {\overset{ˉ}{a}}^{n} {\overset{ˉ}{b}}^{n} = 0.$ Because $b \notin \sqrt{Q}$ , the class $\overset{ˉ}{b}$ is not nilpotent in $A$ , so ${\overset{ˉ}{b}}^{n} \neq 0$ . Thus ${\overset{ˉ}{a}}^{n}$ kills the nonzero element ${\overset{ˉ}{b}}^{n}$ , which means that ${\overset{ˉ}{a}}^{n}$ is a zero divisor on $A$ .

L1L2givenalgebra

2.1

Since $Q$ is primary, [L1] makes every zero divisor on $A$ nilpotent. Therefore ${\overset{ˉ}{a}}^{n}$ is nilpotent, so $\overset{ˉ}{a}$ is nilpotent and hence $a \in \sqrt{Q}$ by [L2]. Thus $a b \in \sqrt{Q}$ and $b \notin \sqrt{Q}$ imply $a \in \sqrt{Q}$ , which is the primality condition from [L3].

L1L2L3step 1.1

3.1

Therefore $\sqrt{Q}$ is prime, and $Q$ is $\sqrt{Q}$ -primary by definition.

L1L2step 2.1∎

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Primary submodules are exactly quotients with nilpotent zero divisors

Statement

Let $R$ be a commutative ring, let $M$ be a left $R$ -module, and let $Q ⊊ M$ be a proper submodule. Then $Q$ is primary if and only if the following classical condition holds:

for every $a \in R$ and every $m \in M$ , if $a m \in Q$ and $m \notin Q$ , then there exists $n \geq 1$ such that

a^{n} M \subseteq Q .

Facts & Assumptions

Given: A commutative ring $R$ , a left $R$ -module $M$ , and a proper submodule $Q ⊊ M$ .

[L1]

A proper submodule $Q$ is primary exactly when every zero divisor on $M / Q$ acts nilpotently on $M / Q$ (Primary submodules and primary ideals).

[L2]

The quotient module $M / Q$ consists of cosets $m + Q$ (Quotient module $M / N$ with scalar multiplication on additive cosets).

Proof

technique · direct

1.1

Assume $Q$ is primary, and let $a m \in Q$ with $m \notin Q$ . Then $m + Q \neq 0$ in $M / Q$ by [L2], while $a (m + Q) = a m + Q = 0 + Q$ . Thus $a$ is a zero divisor on $M / Q$ . By [L1], some $n \geq 1$ satisfies $a^{n} (M / Q) = 0$ , which is equivalent to $a^{n} M \subseteq Q$ .

L1L2givenalgebra

1.2

Conversely, assume the displayed classical condition. Let $a$ be a zero divisor on $M / Q$ . Then there exists $m + Q \neq 0$ with $a (m + Q) = 0 + Q$ . By [L2], this means $m \notin Q$ and $a m \in Q$ . The hypothesis gives $n \geq 1$ with $a^{n} M \subseteq Q$ , equivalently $a^{n} (M / Q) = 0$ . Hence every zero divisor on $M / Q$ acts nilpotently, so $Q$ is primary by [L1].

L1L2given

2.1

Steps 1.1 and 1.2 prove the equivalence.

step 1.1step 1.2∎

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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

1.1

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 ${Ass}_{R} (N) = {p}$ .

L2L3L4L8givenalgebra

1.2

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, $p^{n} N = 0$ .

L1L2L5L6L7L8givenalgebra

1.3

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.

L4givenalgebra

2.1

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

step 1.1step 1.2step 1.3∎

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A finite intersection of primary submodules with one radical is primary

Statement

Let $R$ be a Noetherian commutative ring, let $M$ be a finitely generated left $R$ -module, and let $r \geq 1$ . Let $Q_{1}, \dots, Q_{r} ⊊ M$ be $p$ -primary submodules. Then

Q = Q_{1} \cap \dots \cap Q_{r}

is also $p$ -primary.

Facts & Assumptions

Given: A Noetherian commutative ring $R$ , a finitely generated left $R$ -module $M$ , an integer $r \geq 1$ , a prime ideal $p$ , and $p$ -primary submodules $Q_{1}, \dots, Q_{r} ⊊ M$ .

[L1]

A proper submodule is primary exactly when $a m \in Q^{'}$ and $m \notin Q^{'}$ imply $a^{n} M^{'} \subseteq Q^{'}$ for some $n \geq 1$ (Primary submodules are exactly quotients with nilpotent zero divisors).

[L2]

A primary submodule $Q^{'}$ is $p$ -primary when $\sqrt{{Ann}_{R} (M^{'} / Q^{'})} = p$ (Primary submodules and primary ideals).

Proof

technique · direct

1.1

Because $r \geq 1$ , one has $Q \subseteq Q_{1} ⊊ M$ , so $Q$ is proper. Suppose $a m \in Q$ and $m \notin Q$ . Choose $i$ with $m \notin Q_{i}$ . Since $Q_{i}$ is primary, [L1] gives $e \geq 1$ with $a^{e} M \subseteq Q_{i}$ , so $a \in \sqrt{{Ann}_{R} (M / Q_{i})} = p$ by [L2].

L1L2givenchoosealgebra

2.1

For each $j$ , step 1.1 gives $a \in p = \sqrt{{Ann}_{R} (M / Q_{j})}$ , so choose $e_{j} \geq 1$ with $a^{e_{j}} M \subseteq Q_{j}$ . The nonempty finite list has a maximum $E$ , and then $a^{E} M \subseteq Q_{j}$ for every $j$ , hence $a^{E} M \subseteq Q$ . By [L1], the proper submodule $Q$ is primary.

L1L2step 1.1choosealgebra

3.1

If $a \in p$ , the same finite-maximum argument as in step 2.1 gives a power of $a$ in ${Ann}_{R} (M / Q)$ , so $p \subseteq \sqrt{{Ann}_{R} (M / Q)}$ . Conversely, if a power of $a$ annihilates $M / Q$ , it also annihilates every $M / Q_{j}$ because $Q \subseteq Q_{j}$ , so $a \in \sqrt{{Ann}_{R} (M / Q_{j})} = p$ . Thus $\sqrt{{Ann}_{R} (M / Q)} = p$ , and [L2] makes $Q$ $p$ -primary.

L2step 2.1algebra∎

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Primary decompositions, minimality, and isolated components

Definition

Let $R$ be a commutative ring, let $M$ be a left $R$ -module, and let $N \subseteq M$ be a submodule.

A primary decomposition of $N$ in $M$ is an expression

N = Q_{1} \cap \dots \cap Q_{r}

with each $Q_{i} ⊊ M$ a primary submodule.

Such a decomposition is minimal when:

no component is redundant, so omitting any $Q_{i}$ changes the intersection;
the radicals $\sqrt{{Ann}_{R} (M / Q_{i})}$ are pairwise distinct.

These radicals are well-defined because Primary submodules and primary ideals establishes that each module annihilator ${Ann}_{R} (M / Q_{i})$ is an ideal of $R$ .

In a minimal decomposition, a component $Q_{i}$ is isolated when its radical is minimal, under inclusion, among the radicals occurring in the decomposition.

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An irreducible submodule of a Noetherian module is primary

Statement

Let $R$ be a commutative ring, let $M$ be a Noetherian left $R$ -module, and let $Q ⊊ M$ be irreducible, meaning that whenever

Q = A \cap B

with submodules $A, B \subseteq M$ , then $Q = A$ or $Q = B$ . Then $Q$ is primary.

Facts & Assumptions

Given: A commutative ring $R$ , a Noetherian left $R$ -module $M$ , and an irreducible proper submodule $Q ⊊ M$ .

[L1]

In a short exact sequence, a quotient of a Noetherian module is again Noetherian (Noetherian and Artinian conditions are each exact in short exact sequences).

[L2]

The quotient module $M / Q$ is formed from the cosets of $Q$ (Quotient module $M / N$ with scalar multiplication on additive cosets).

[L3]

A proper submodule is primary exactly when the classical power condition of the previous lemma holds (Primary submodules are exactly quotients with nilpotent zero divisors).

Proof

technique · direct

1.1

Let $N = M / Q$ . By [L1], the quotient module $N$ is Noetherian. The submodule $0 \subset N$ is irreducible: if $0 = A^{'} \cap B^{'}$ in $N$ , then taking inverse images in $M$ gives $Q = A \cap B$ with $Q \subseteq A, B$ , so the irreducibility of $Q$ forces $A = Q$ or $B = Q$ , hence $A^{'} = 0$ or $B^{'} = 0$ .

L1L2givenalgebra

2.1

Let $x \in R$ be a zero divisor on $N$ . Then $(0 :_{N} x) \neq 0$ . Because $N$ is Noetherian, the ascending chain $(0 :_{N} x) \subseteq (0 :_{N} x^{2}) \subseteq \dots$ stabilizes; choose $n \geq 1$ with $(0 :_{N} x^{n}) = (0 :_{N} x^{n + 1}) = (0 :_{N} x^{2 n})$ . If $z \in (0 :_{N} x^{n}) \cap x^{n} N$ , write $z = x^{n} y$ . Then $x^{n} z = 0$ , so $x^{2 n} y = 0$ and hence $y \in (0 :_{N} x^{2 n}) = (0 :_{N} x^{n})$ . Therefore $z = x^{n} y = 0$ , and $0 = (0 :_{N} x^{n}) \cap x^{n} N .$

step 1.1choosealgebra

3.1

Since $(0 :_{N} x) \subseteq (0 :_{N} x^{n})$ is nonzero, the irreducibility of $0 \subset N$ and the decomposition in step 2.1 force $x^{n} N = 0$ . Thus every zero divisor on $N$ acts nilpotently on $N$ . By [L3], this means $Q$ is primary.

L3step 2.1algebra

4.1

Hence every irreducible submodule of a Noetherian module is primary.

step 1.1step 3.1∎

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-28Open item page →

A finite primary decomposition can be stripped of redundant components

Statement

N = Q_{1} \cap \dots \cap Q_{r}

is a finite primary decomposition of a submodule $N \subseteq M$ , then some subfamily of the $Q_{i}$ has the same intersection and is irredundant.

Facts & Assumptions

Given: A commutative ring $R$ , a left $R$ -module $M$ , a submodule $N \subseteq M$ , and a finite primary decomposition $N = Q_{1} \cap \dots \cap Q_{r}$ .

[L1]

Minimality requires both that no component be redundant and that the component radicals be pairwise distinct (Primary decompositions, minimality, and isolated components).

Proof

technique · direct

1.1

If the displayed decomposition has no redundant component, then it is already irredundant, which is the first condition recorded in [L1]. Otherwise choose an index $j$ such that $Q_{j} \supseteq ⋂_{i \neq j} Q_{i}$ . Then $N = ⋂_{i = 1}^{r} Q_{i} = ⋂_{i \neq j} Q_{i},$ so removing $Q_{j}$ preserves the intersection.

L1givenchoosealgebra

2.1

Each removal in step 1.1 shortens the finite list of components by one. Repeating step 1.1 therefore terminates after finitely many deletions and produces a decomposition with the same intersection and no redundant component.

step 1.1algebra

3.1

This is the required irredundant subfamily.

step 2.1∎

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Equal-radical primary components can be combined

Statement

Let $R$ be a Noetherian commutative ring, let $M$ be a finitely generated left $R$ -module, and let

N = Q_{1} \cap \dots \cap Q_{r}

be a finite primary decomposition of a submodule $N \subseteq M$ , with each $Q_{i}$ $p_{i}$ -primary for a prime ideal $p_{i}$ . If several of the $p_{i}$ equal one prime $p$ , then replacing that whole group by the intersection of its components preserves the total intersection and produces one $p$ -primary component.

Facts & Assumptions

Given: A Noetherian commutative ring $R$ , a finitely generated left $R$ -module $M$ , and a finite primary decomposition $N = Q_{1} \cap \dots \cap Q_{r}$ with each $Q_{i}$ $p_{i}$ -primary for a prime ideal $p_{i}$ .

[L1]

For a prime ideal $p$ , a nonempty finite intersection of $p$ -primary submodules is again $p$ -primary (A finite intersection of primary submodules with one radical is primary).

[L2]

Minimality requires both irredundancy and pairwise distinct component radicals (Primary decompositions, minimality, and isolated components).

Proof

technique · direct

1.1

Fix a prime $p$ and let $I (p)$ be the set of indices $i$ for which $Q_{i}$ is $p$ -primary. If $I (p)$ is empty or has one element, there is nothing to combine. Otherwise set $Q (p) = ⋂_{i \in I (p)} Q_{i} .$ Fact [L1] shows that $Q (p)$ is again $p$ -primary.

L1givenconstruct

2.1

Replacing, for each prime $p$ occurring among the radicals, the whole block ${Q_{i} : i \in I (p)}$ by the single component $Q (p)$ does not change the total intersection, because intersections may be regrouped without changing their value. The resulting components have pairwise distinct radicals, which is the second minimality requirement recorded in [L2].

L2step 1.1algebra

3.1

Thus equal-radical components may be combined into one primary component with the same radical.

step 2.1∎

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-28Open item page →

Every submodule of a finite module over a Noetherian ring has a minimal primary decomposition

Statement

Assume Dependent Choice.

Let $R$ be a Noetherian commutative ring and let $M$ be a finitely generated left $R$ -module. Every submodule $N \subseteq M$ has a finite primary decomposition. After deleting redundant components and combining equal radicals, one obtains a minimal primary decomposition. When $N = M$ , the decomposition is the empty intersection, interpreted as $M$ . In particular, every ideal of a Noetherian ring has a minimal primary decomposition.

Facts & Assumptions

Given: Dependent Choice, a Noetherian commutative ring $R$ , a finitely generated left $R$ -module $M$ , and a submodule $N \subseteq M$ .

[L1]

A finitely generated module over a Noetherian ring is Noetherian (Finitely generated modules over a left Noetherian ring are Noetherian).

[L2]

Assuming Dependent Choice, a Noetherian module has the maximal condition on submodules (Finite generation, ACC, and maximal-condition characterizations of Noetherian modules).

[L3]

An irreducible submodule of a Noetherian module is primary (An irreducible submodule of a Noetherian module is primary).

[L4]

Finite primary decompositions can be made irredundant by deleting redundant components (A finite primary decomposition can be stripped of redundant components).

[L5]

In a finite primary decomposition of a submodule of a finitely generated module over a Noetherian commutative ring, equal-radical primary components can be combined into one primary component (Equal-radical primary components can be combined).

Proof

technique · direct

1.1

The submodule $M$ has the empty primary decomposition, whose intersection is interpreted as $M$ .

given

2.1

By [L1], the module $M$ is Noetherian. Suppose, toward contradiction, that some submodule of $M$ has no finite primary decomposition. Let $Σ$ be the set of such submodules. By step 1.1, $M \notin Σ$ . By [L2], choose a maximal element $Q$ of $Σ$ ; then $Q ⊊ M$ .

L1L2step 1.1givenconstruct

3.1

The submodule $Q$ is irreducible. Indeed, if $Q = A \cap B$ with $Q ⊊ A$ and $Q ⊊ B$ , then the maximality of $Q$ in step 2.1 forces finite primary decompositions of $A$ and $B$ . Intersecting those two finite decompositions gives a finite primary decomposition of $Q$ , contrary to $Q \in Σ$ .

step 2.1algebra

4.1

Because $Q$ is irreducible and proper and $M$ is Noetherian, [L3] shows that $Q$ is primary. But a primary submodule is already a one-term finite primary decomposition, again contradicting $Q \in Σ$ . Therefore $Σ$ is empty: every submodule of $M$ has a finite primary decomposition.

L3step 2.1step 3.1

5.1

If $N = M$ , the empty decomposition from step 1.1 is irredundant and has pairwise distinct radicals vacuously, so it is minimal. If $N ⊊ M$ , start from a finite primary decomposition supplied by step 4.1, apply [L4] to remove redundant components, and then apply [L5] to merge equal-radical blocks. The resulting decomposition is finite, has the same intersection, has no redundant component, and has pairwise distinct radicals; hence it is minimal.

L4L5step 1.1step 4.1

6.1

Taking $M = R$ recovers the ideal case, because ideals are precisely the submodules of the regular module over a commutative ring.

step 5.1algebra∎

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-28Open item page →

The radicals in a minimal primary decomposition are exactly the associated primes of the quotient

Statement

Let $R$ be a Noetherian commutative ring, let $M$ be a finitely generated left $R$ -module, and let

N = Q_{1} \cap \dots \cap Q_{r}

be a minimal primary decomposition in which each $Q_{i}$ is $p_{i}$ -primary. Assume each $p_{i}$ is a prime ideal. Then

{Ass}_{R} (M / N) = {p_{1}, \dots, p_{r}} .

Facts & Assumptions

Given: A Noetherian commutative ring $R$ , a finitely generated left $R$ -module $M$ , and a minimal primary decomposition $N = Q_{1} \cap \dots \cap Q_{r}$ with each $Q_{i}$ $p_{i}$ -primary for a prime ideal $p_{i}$ .

[L1]

In a minimal primary decomposition, the component radicals are pairwise distinct and no component is redundant (Primary decompositions, minimality, and isolated components).

[L2]

Associated primes of a submodule lie in those of the ambient module, and associated primes of a direct sum are the union of those of the summands (Associated primes in a short exact sequence).

[L3]

If $Q_{i}$ is $p_{i}$ -primary, then ${Ass}_{R} (M / Q_{i}) = {p_{i}}$ (Primary submodules of finite modules are characterized by a singleton associated-prime set).

[L4]

Every nonzero module over a Noetherian ring has an associated prime (A nonzero module over a Noetherian ring has an associated prime).

Proof

technique · direct

1.1

Let $δ : M / N ⟶ ⨁_{i = 1}^{r} M / Q_{i}$ be the diagonal map. Its kernel is zero, because $m + N$ maps to zero exactly when $m \in Q_{i}$ for every $i$ , that is, when $m \in N$ . Thus $δ$ is injective. By [L3] and the direct-sum part of [L2], ${Ass}_{R} ⁣ (⨁_{i = 1}^{r} M / Q_{i}) = {p_{1}, \dots, p_{r}} .$ Since $M / N$ is a submodule of that direct sum, the left-inclusion part of [L2] gives ${Ass}_{R} (M / N) \subseteq {p_{1}, \dots, p_{r}} .$

L2L3constructalgebra

1.2

Fix $i$ . Put $I_{i} = ⋂_{j \neq i} Q_{j}$ . By [L1], the decomposition is irredundant, so $I_{i} ⊈ Q_{i}$ and therefore $I_{i} / N \neq 0$ . The map $I_{i} / N ⟶ M / Q_{i}, m + N ⟼ m + Q_{i}$ is injective because its kernel is $(I_{i} \cap Q_{i}) / N = N / N = 0$ . Hence $I_{i} / N$ is a nonzero submodule of $M / Q_{i}$ . By [L2] and [L3], ${Ass}_{R} (I_{i} / N) \subseteq {Ass}_{R} (M / Q_{i}) = {p_{i}} .$ Fact [L4] makes ${Ass}_{R} (I_{i} / N)$ nonempty, so ${Ass}_{R} (I_{i} / N) = {p_{i}}$ . Applying [L2] again to the inclusion $I_{i} / N ↪ M / N$ yields $p_{i} \in {Ass}_{R} (M / N)$ .

L1L2L3L4choosealgebra

2.1

Step 1.2 shows every $p_{i}$ belongs to ${Ass}_{R} (M / N)$ , and step 1.1 gives the reverse inclusion. Therefore ${Ass}_{R} (M / N) = {p_{1}, \dots, p_{r}}$ .

step 1.1step 1.2∎

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The radicals in a minimal primary decomposition are intrinsic

Statement

Let $R$ be a Noetherian commutative ring, let $M$ be a finitely generated left $R$ -module, and let $N \subseteq M$ . Let $N = Q_{1} \cap \dots \cap Q_{r}$ be a minimal primary decomposition in which each $Q_{i}$ is $p_{i}$ -primary for a prime ideal $p_{i}$ . Then the set of component radicals is uniquely determined by $M / N$ and equals

{Ass}_{R} (M / N) .

Facts & Assumptions

Given: A Noetherian commutative ring $R$ , a finitely generated left $R$ -module $M$ , a submodule $N \subseteq M$ , and a minimal primary decomposition $N = Q_{1} \cap \dots \cap Q_{r}$ with each $Q_{i}$ $p_{i}$ -primary for a prime ideal $p_{i}$ .

[L1]

In the Noetherian finite-module setting, if a minimal primary decomposition has each component $p_{i}$ -primary for a prime ideal $p_{i}$ , then its radicals are exactly the associated primes of the quotient (The radicals in a minimal primary decomposition are exactly the associated primes of the quotient).

Proof

technique · direct

1.1

For the given minimal primary decomposition, [L1] gives ${p_{1}, \dots, p_{r}} = {Ass}_{R} (M / N) .$

L1given

2.1

The right-hand side depends only on the quotient $M / N$ , not on the chosen decomposition. Hence the set of component radicals is intrinsic.

step 1.1∎

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Localisation of a primary submodule either stays primary or becomes the whole module

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 $p$ -primary submodule for a prime ideal $p$ , and let $S \subseteq R$ be multiplicative.

If $S \cap p = \emptyset$ , then $S^{- 1} Q$ is an $S^{- 1} p$ -primary submodule of $S^{- 1} M$ .
If $S \cap p \neq \emptyset$ , then $S^{- 1} Q = S^{- 1} M$ .

Facts & Assumptions

Given: The Axiom of Choice, a Noetherian commutative ring $R$ , a finitely generated left $R$ -module $M$ , a prime ideal $p$ , a $p$ -primary submodule $Q ⊊ M$ , and a multiplicative subset $S \subseteq R$ .

[L1]

Assuming the Axiom of Choice, for a Noetherian commutative ring $A$ , a finitely generated left $A$ -module $P$ , a proper submodule $Q^{'} ⊊ P$ , and a prime ideal $r \subseteq A$ , the following are equivalent: $Q^{'}$ is $r$ -primary; ${Ass}_{A} (P / Q^{'}) = {r}$ ; every $a \notin r$ acts injectively on $P / Q^{'}$ , and some power of $r$ annihilates $P / Q^{'}$ (Primary submodules of finite modules are characterized by a singleton associated-prime set).

[L2]

Over a Noetherian commutative ring, associated primes of a finitely generated module localize exactly by extension of primes disjoint from the denominator set (Associated primes commute with localization for finite modules).

[L3]

Localisation commutes with quotient modules, so $S^{- 1} (M / Q) ≅ (S^{- 1} M) / (S^{- 1} Q)$ (Localisation commutes with quotient modules and arbitrary direct sums).

[L4]

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

Proof

technique · direct

1.1

Put $N = M / Q$ . Since $Q$ is $p$ -primary, [L1] gives $n \geq 1$ with $p^{n} N = 0$ . If $S \cap p \neq \emptyset$ , choose $s \in S \cap p$ . Then $s^{n} N = 0$ , so in the localization $S^{- 1} N$ the unit $s / 1$ annihilates every element. Hence $S^{- 1} N = 0$ , and [L3] shows $(S^{- 1} M) / (S^{- 1} Q) = 0$ , that is, $S^{- 1} Q = S^{- 1} M$ .

L1L3choosealgebra

1.2

Assume now that $S \cap p = \emptyset$ . Since $Q$ is $p$ -primary, the quotient $N$ has ${Ass}_{R} (N) = {p}$ by [L1]. Fact [L2] therefore gives ${Ass}_{S^{- 1} R} (S^{- 1} N) = {S^{- 1} p} .$ Also $p^{n} N = 0$ from [L1], so $(S^{- 1} p)^{n} S^{- 1} N = 0$ .

L1L2algebra

2.1

Let $u = a / s \in S^{- 1} R$ with $u \notin S^{- 1} p$ . Then $a \notin p$ , so multiplication by $a$ on $N$ is injective by [L1]. Because $s / 1$ is a unit, multiplication by $u = (a / 1) (s / 1)^{- 1}$ on $S^{- 1} N$ is also injective. The module $S^{- 1} M$ is finitely generated over the Noetherian ring $S^{- 1} R$ : if $m_{1}, \dots, m_{t}$ generate $M$ , then $m_{1} / 1, \dots, m_{t} / 1$ generate $S^{- 1} M$ . Hence [L1], applied over $S^{- 1} R$ , shows via [L3] that $S^{- 1} Q$ is $S^{- 1} p$ -primary in $S^{- 1} M$ .

L1L3L4step 1.2algebra

3.1

Steps 1.1 and 2.1 prove the two localization alternatives.

step 1.1step 2.1∎

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A primary component is recovered by contracting its localization away from the radical

Statement

Let $R$ be a Noetherian commutative ring, let $M$ be a finitely generated left $R$ -module, let $Q ⊊ M$ be a $p$ -primary submodule, and let the radical $p$ be prime. Let $S \subseteq R$ be multiplicative with $S \cap p = \emptyset$ . Then

Q = {m \in M : m / 1 \in S^{- 1} Q} .

Facts & Assumptions

Given: A Noetherian commutative ring $R$ , a finitely generated left $R$ -module $M$ , a $p$ -primary submodule $Q ⊊ M$ for a prime ideal $p$ , and a multiplicative subset $S \subseteq R$ disjoint from $p$ .

[L1]

For the quotient $N = M / Q$ , every $a \notin p$ acts injectively on $N$ (Primary submodules of finite modules are characterized by a singleton associated-prime set).

[L2]

Localisation commutes with quotient modules, so $S^{- 1} (M / Q) ≅ (S^{- 1} M) / (S^{- 1} Q)$ (Localisation commutes with quotient modules and arbitrary direct sums).

Proof

technique · direct

1.1

The inclusion $Q \subseteq {m \in M : m / 1 \in S^{- 1} Q}$ is immediate, because every element of $Q$ localizes into $S^{- 1} Q$ .

given

1.2

Conversely, let $m \in M$ with $m / 1 \in S^{- 1} Q$ . Under the identification of [L2], the class of $m + Q$ in $S^{- 1} (M / Q)$ is zero. Hence some $s \in S$ satisfies $s (m + Q) = 0$ in $M / Q$ , so $s m \in Q$ . Since $s \notin p$ , [L1] makes multiplication by $s$ injective on $M / Q$ , and the equality $s (m + Q) = 0$ forces $m + Q = 0$ . Therefore $m \in Q$ .

L1L2choosealgebra

2.1

Steps 1.1 and 1.2 prove that $Q$ is exactly the contraction of $S^{- 1} Q$ .

step 1.1step 1.2∎

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-28Open item page →

Isolated primary components are recovered by localization and contraction

Statement

Assume the Axiom of Choice.

Let $R$ be a Noetherian commutative ring, let $M$ be a finitely generated left $R$ -module, and let

N = Q_{1} \cap \dots \cap Q_{r}

be a minimal primary decomposition with each $Q_{i}$ $p_{i}$ -primary for a prime ideal $p_{i}$ . If $p_{i}$ is isolated, then

Q_{i} = M \cap N_{p_{i}}

inside $M_{p_{i}}$ . In particular, each isolated primary component is uniquely determined by $N \subseteq M$ .

Facts & Assumptions

Given: The Axiom of Choice, a Noetherian commutative ring $R$ , a finitely generated left $R$ -module $M$ , and a minimal primary decomposition $N = Q_{1} \cap \dots \cap Q_{r}$ with each $Q_{i}$ $p_{i}$ -primary for a prime ideal $p_{i}$ .

[L1]

In a minimal primary decomposition, isolated means minimal under inclusion among the component radicals, and the radicals are pairwise distinct (Primary decompositions, minimality, and isolated components).

[L2]

Assuming the Axiom of Choice, in the Noetherian finite-module setting, localizing a $p$ -primary component at its prime radical keeps it primary, while localizing at a multiplicative set meeting its radical gives the whole localized module (Localisation of a primary submodule either stays primary or becomes the whole module).

[L3]

In the Noetherian finite-module setting, a $p$ -primary component with prime radical is recovered by contracting its localization at a multiplicative set disjoint from $p$ (A primary component is recovered by contracting its localization away from the radical).

[L4]

Localisation commutes with finite intersections of submodules (Localisation commutes with finite intersections of submodules).

[L5]

In the Noetherian finite-module setting, the prime component radicals of a minimal primary decomposition are intrinsic and equal the associated primes of the quotient (The radicals in a minimal primary decomposition are intrinsic).

Proof

technique · direct

1.1

Fix an isolated component $Q_{i}$ . By [L1], if $j \neq i$ then $p_{j} ⊈ p_{i}$ , since otherwise minimality of $p_{i}$ would force $p_{j} = p_{i}$ , contradicting the distinct-radical part of [L1]. Hence for each $j \neq i$ we may choose $a_{j} \in p_{j} ∖ p_{i}$ . Localizing at $R ∖ p_{i}$ , fact [L2] gives $(Q_{j})_{p_{i}} = M_{p_{i}}$ for $j \neq i$ , while $(Q_{i})_{p_{i}}$ remains a proper primary submodule. Using [L4], $N_{p_{i}} = ⋂_{j = 1}^{r} (Q_{j})_{p_{i}} = (Q_{i})_{p_{i}} .$

L1L2L4choosealgebra

2.1

Fact [L3] recovers $Q_{i}$ by contracting $(Q_{i})_{p_{i}}$ back to $M$ . Since step 1.1 identifies $(Q_{i})_{p_{i}}$ with $N_{p_{i}}$ , this gives $Q_{i} = M \cap N_{p_{i}} .$ By [L5], the prime $p_{i}$ depends only on $M / N$ and occurs in every minimal primary decomposition. Applying the same localization-and-contraction argument to any such decomposition recovers its $p_{i}$ -component from the same right-hand side. Hence the isolated component is unique.

L3L5step 1.1

3.1

Therefore every isolated component is recovered by localization and contraction.

step 2.1∎

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An ideal contained in a finite union of prime ideals lies in one of them

Statement

Let $R$ be a commutative ring, let $I ⊴ R$ be an ideal, and let $p_{1}, \dots, p_{n}$ be prime ideals with $n \geq 1$ . If

I \subseteq p_{1} \cup \dots \cup p_{n},

then $I \subseteq p_{i}$ for some $i$ .

Facts & Assumptions

Given: A commutative ring $R$ , an ideal $I ⊴ R$ , a positive integer $n$ , and prime ideals $p_{1}, \dots, p_{n}$ with $I \subseteq p_{1} \cup \dots \cup p_{n}$ .

[L1]

A prime ideal is proper and contains one factor whenever it contains a product (Prime ideals and maximal ideals in a commutative ring).

Proof

technique · direct

1.1

We argue by induction on $n$ . The case $n = 1$ is immediate. Assume $n \geq 2$ and that the claim is known for smaller families. Suppose, for contradiction, that $I ⊈ p_{i}$ for every $i$ . Then for each $k$ , the ideal $I$ is not contained in $⋃_{i \neq k} p_{i}$ either, because the induction hypothesis would then force $I \subseteq p_{i}$ for some $i \neq k$ . Hence for each $k$ we may choose $x_{k} \in I ∖ ⋃_{i \neq k} p_{i}$ . Since $I \subseteq ⋃_{i} p_{i}$ , each $x_{k}$ must lie in $p_{k}$ .

L1choosealgebra

2.1

If $n = 2$ , then $x_{1} + x_{2} \in I$ . It is not in $p_{1}$ , because $x_{1} \in p_{1}$ would then force $x_{2} = (x_{1} + x_{2}) - x_{1} \in p_{1}$ , contradicting the choice of $x_{2}$ ; the same argument shows $x_{1} + x_{2} \notin p_{2}$ . This contradicts $I \subseteq p_{1} \cup p_{2}$ .

step 1.1algebra

2.2

If $n \geq 3$ , put $z = x_{n} + x_{1} x_{2} \dots x_{n - 1} \in I .$ For $i < n$ , the product term lies in $p_{i}$ because $x_{i} \in p_{i}$ , while $x_{n} \notin p_{i}$ by the choice in step 1.1; hence $z \notin p_{i}$ . Also each $x_{j}$ with $j < n$ lies outside $p_{n}$ , so [L1] implies $x_{1} \dots x_{n - 1} \notin p_{n}$ ; since $x_{n} \in p_{n}$ , one gets $z \notin p_{n}$ as well. This again contradicts $I \subseteq p_{1} \cup \dots \cup p_{n}$ .

L1step 1.1algebra

3.1

The contradictions in steps 2.1 and 2.2 show that the assumption $I ⊈ p_{i}$ for every $i$ is impossible. Therefore $I \subseteq p_{i}$ for some $i$ .

step 1.1step 2.1step 2.2∎

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Downward-closed intersections of primary components are intrinsic

Statement

Assume the Axiom of Choice.

Let $R$ be a Noetherian commutative ring, let $M$ be a finitely generated left $R$ -module, and let

N = Q_{1} \cap \dots \cap Q_{r}

be a minimal primary decomposition with each $Q_{i}$ $p_{i}$ -primary for a prime ideal $p_{i}$ . Let $Σ \subseteq {Ass}_{R} (M / N)$ be downward-closed under inclusion. Then

⋂_{p_{i} \in Σ} Q_{i}

depends only on $N \subseteq M$ and on $Σ$ , not on the chosen minimal decomposition. When $Σ = \emptyset$ , the empty intersection is interpreted as $M$ .

Facts & Assumptions

Given: The Axiom of Choice, a Noetherian commutative ring $R$ , a finitely generated left $R$ -module $M$ , a submodule $N \subseteq M$ , and a minimal primary decomposition $N = Q_{1} \cap \dots \cap Q_{r}$ with each $Q_{i}$ $p_{i}$ -primary for a prime ideal $p_{i}$ .

[L1]

In the Noetherian finite-module setting, a minimal primary decomposition whose component radicals are prime has radical set ${p_{1}, \dots, p_{r}} = {Ass}_{R} (M / N)$ (The radicals in a minimal primary decomposition are intrinsic).

[L2]

Assuming the Axiom of Choice, in the Noetherian finite-module setting, localizing a primary component away from its prime radical keeps it, while localizing at a set meeting its radical turns it into the whole localized module (Localisation of a primary submodule either stays primary or becomes the whole module).

[L3]

In the Noetherian finite-module setting, a primary component with prime radical is recovered by contracting its localization at a multiplicative set disjoint from that radical (A primary component is recovered by contracting its localization away from the radical).

[L4]

Localisation commutes with finite intersections of submodules (Localisation commutes with finite intersections of submodules).

[L5]

If an ideal is contained in a finite union of prime ideals, then it lies in one of them (An ideal contained in a finite union of prime ideals lies in one of them).

[L6]

For a multiplicative subset $S$ , the canonical localization map is $λ_{M} : M \to S^{- 1} M$ , $m \mapsto m / 1$ (Localisation of a module at a multiplicative subset).

Proof

technique · direct

1.1

Let $Σ \subseteq {Ass}_{R} (M / N)$ be downward-closed, and set $S = R ∖ ⋃_{p \in Σ} p .$ Because the members of $Σ$ are prime, $S$ is multiplicative. If $p_{i} \in Σ$ , then $S \cap p_{i} = \emptyset$ by definition. If $p_{i} \notin Σ$ and $S \cap p_{i} = \emptyset$ , then $p_{i} \subseteq ⋃_{p \in Σ} p$ , so [L5] forces $p_{i} \subseteq p$ for some $p \in Σ$ . Since $p_{i} \in {Ass}_{R} (M / N)$ by [L1] and $Σ$ is downward-closed, this would imply $p_{i} \in Σ$ , a contradiction. Hence $S \cap p_{i} = \emptyset ⟺ p_{i} \in Σ .$

L1L5givenalgebra

2.1

By [L4], $S^{- 1} N = ⋂_{i = 1}^{r} S^{- 1} Q_{i} .$ For $p_{i} \in Σ$ , step 1.1 and [L2] say that $S^{- 1} Q_{i}$ remains a proper localized primary component. For $p_{i} \notin Σ$ , step 1.1 and [L2] give $S^{- 1} Q_{i} = S^{- 1} M$ . Therefore $S^{- 1} N = ⋂_{p_{i} \in Σ} S^{- 1} Q_{i} .$ When $Σ = \emptyset$ , the right-hand side is the empty intersection $S^{- 1} M$ , which matches the fact that $0 \in S$ and hence $S^{- 1} N = S^{- 1} M = 0$ .

L2L4step 1.1algebra

3.1

Let $λ_{M} : M \to S^{- 1} M$ be the canonical map of [L6]. Contracting the equality of step 2.1 back to $M$ gives $λ_{M}^{- 1} (S^{- 1} N) = ⋂_{p_{i} \in Σ} λ_{M}^{- 1} (S^{- 1} Q_{i}),$ because inverse image commutes with intersections. For every $p_{i} \in Σ$ , step 1.1 and [L3] identify $λ_{M}^{- 1} (S^{- 1} Q_{i})$ with $Q_{i}$ . Hence $λ_{M}^{- 1} (S^{- 1} N) = ⋂_{p_{i} \in Σ} Q_{i} .$

L3L6step 1.1step 2.1algebra

4.1

The left-hand side of step 3.1 depends only on $N \subseteq M$ and the set $S$ , hence only on $N \subseteq M$ and on $Σ$ . By [L1], any other minimal primary decomposition has the same associated-prime set, so it yields the same $S$ and therefore the same intersection.

L1step 3.1

5.1

Thus the intersection of the primary components indexed by any downward-closed subset of ${Ass}_{R} (M / N)$ is intrinsic.

step 4.1∎

CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-28Open item page →

A radical ideal in a Noetherian ring is the intersection of its minimal primes

Statement

Assume Dependent Choice.

Let $R$ be a Noetherian commutative ring and let $I ⊴ R$ be a radical ideal. Then there exist finitely many prime ideals $p_{1}, \dots, p_{m}$ minimal over $I$ such that

I = p_{1} \cap \dots \cap p_{m} .

When $I = R$ , this is the empty intersection.

Facts & Assumptions

Given: Dependent Choice, a Noetherian commutative ring $R$ , and a radical ideal $I ⊴ R$ .

[L1]

Assuming Dependent Choice, every ideal of a Noetherian ring has a minimal primary decomposition (Every submodule of a finite module over a Noetherian ring has a minimal primary decomposition).

[L2]

The radical of a primary ideal is prime (The radical of a primary ideal is prime).

[L3]

The radical of an intersection of $n \geq 1$ ideals is the intersection of their radicals (The radical of a finite intersection).

Proof

technique · direct

1.1

If $I = R$ , then no prime ideal contains $I$ , and the empty intersection is $R$ . This is exactly the stated boundary case.

given

1.2

Assume $I \neq R$ . By [L1], choose a minimal primary decomposition $I = Q_{1} \cap \dots \cap Q_{r} .$ Since the empty intersection is $R$ , one has $r \geq 1$ . For each $i$ , put $p_{i} = \sqrt{Q_{i}}$ . Fact [L2] makes every $p_{i}$ prime. Since $I$ is radical, $I = \sqrt{I} = \sqrt{Q_{1} \cap \dots \cap Q_{r}} = \sqrt{Q_{1}} \cap \dots \cap \sqrt{Q_{r}} = p_{1} \cap \dots \cap p_{r}$ by [L3].

L1L2L3choosealgebra

2.1

Let $M$ be the set of inclusion-minimal members among the finite family ${p_{1}, \dots, p_{r}}$ . If one prime in the family contains another, removing the larger one does not change the intersection, so the intersection over $M$ is still $I$ . Every prime $q$ minimal over $I$ contains the product $p_{1} \dots p_{r}$ , which is contained in $I$ , so primality of $q$ forces $p_{i} \subseteq q$ for some $i$ . Since $I \subseteq p_{i}$ and $q$ is minimal among primes containing $I$ , this gives $q = p_{i}$ . Thus the members of $M$ are exactly the minimal prime ideals over $I$ .

step 1.2algebra

3.1

Steps 1.1, 1.2, and 2.1 prove that a radical ideal is the finite intersection of its minimal primes.

step 1.1step 1.2step 2.1∎

5 · Examples, counterexamples and false statements

None yet.