TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26

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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

Statement

Let $R$ be a commutative ring. The following are equivalent.

$R$ is Noetherian (Left and right Noetherian rings).
Every ideal of $R$ is finitely generated: for every ideal $a$ there are finitely many $a_{1}, \dots, a_{n} \in a$ , with $n \in N$ , such that $a = (a_{1}, \dots, a_{n})$ .
Ascending chain condition. Every chain of ideals $a_{0} \subseteq a_{1} \subseteq a_{2} \subseteq \dots$ indexed by $N$ stabilises: there is $N \in N$ with $a_{n} = a_{N}$ for every $n \geq N$ .
Maximal condition. Every nonempty set of ideals of $R$ has a maximal member with respect to inclusion.

The implication from the ascending chain condition to the maximal condition uses dependent choice; the remaining implications are choice-free. The same attribution is carried by Finite generation, ACC, and maximal-condition characterizations of Noetherian modules, from which this statement is obtained.

Facts & Assumptions

Given: A commutative ring $R$ . Write $_{R} R$ for the additive group of $R$ carrying the scalar action $r \cdot x : = r x$ ; the ring axioms are exactly the four module axioms for this action, so $_{R} R$ is a left $R$ -module (Unital left and right modules over a ring; unqualified module means left module), and it is the left regular module named by Left and right Noetherian rings. For $S \subseteq R$ write $(S)$ for the ideal generated by $S$ (The ideal generated by a subset and principal ideals), and $(a_{1}, \dots, a_{n})$ for $({a_{1}, \dots, a_{n}})$ .

[L1]

A unital ring $R$ is left Noetherian when its left regular module $_{R} R$ is Noetherian, and right Noetherian when the right regular module $R_{R}$ is Noetherian (Left and right Noetherian rings).

[L2]

A left $R$ -module $M$ is Noetherian when every submodule of $M$ is finitely generated (Noetherian modules: every submodule is finitely generated).

[L3]

For a left $R$ -module $M$ , the following are equivalent: every submodule is finitely generated; every ascending chain of submodules stabilizes; and every nonempty family of submodules has a maximal member. The implication from ACC to the maximal condition uses dependent choice; the other displayed implications are choice-free (Finite generation, ACC, and maximal-condition characterizations of Noetherian modules).

[L4]

An additive subgroup $I \leq (R, +)$ is a left ideal when $r i \in I$ for every $r \in R$ and $i \in I$ , and a right ideal when $i r \in I$ for every such $r, i$ ; a two-sided ideal is both, and in a commutative ring these three notions agree (Left, right and two-sided ideals).

[L5]

A subset $N \subseteq M$ of a left $R$ -module $M$ is a submodule when it is a subgroup of the additive group of $M$ and is closed under scalars, $r n \in N$ for $r \in R$ and $n \in N$ (Submodule of a module).

[L6]

In a commutative ring, $(S)$ consists of finite sums $\sum r_{i} s_{i}$ , and $(a) = R a$ ; the empty sum is included and equals $0$ (In a commutative ring, $(S)$ consists of finite sums $\sum r_{i} s_{i}$ , and $(a) = R a$ ).

[L7]

For a ring $R$ , a left $R$ -module $M$ and $S \subseteq M$ , the submodule $⟨ S ⟩_{R}$ is the set of finite sums $\sum_{i = 1}^{k} r_{i} s_{i}$ with $k \in N$ , $r_{i} \in R$ and $s_{i} \in S$ , the term with $k = 0$ being $0_{M}$ (The submodule generated by a subset consists of the finite $R$ -linear combinations of that subset).

Proof

technique · direct

1.1

Unfolding the two definitions in turn, $R$ is Noetherian exactly when the left regular module $_{R} R$ is a Noetherian module, and that holds exactly when every submodule of $_{R} R$ is finitely generated as an $R$ -module. No choice principle enters here: the two definitions are being read, not compared.

L1L2given

2.1

A subset $I \subseteq R$ is a submodule of $_{R} R$ exactly when it is a subgroup of $(R, +)$ closed under the action, that is, when $r x \in I$ for all $r \in R$ and $x \in I$ ; and that is word for word the condition defining a left ideal of $R$ , which in a commutative ring is the same thing as an ideal. So the submodules of $_{R} R$ and the ideals of $R$ are the same subsets of $R$ , and since the correspondence is the identity on subsets it preserves and reflects inclusion.

L4L5step 1.1algebra

3.1

Finite generation means the same on both sides of that identification. For $S \subseteq R$ the submodule $⟨ S ⟩_{R}$ of $_{R} R$ is the set of finite sums $\sum r_{i} s_{i}$ with $r_{i} \in R$ and $s_{i} \in S$ , and the ideal $(S)$ is that same set of finite sums; so $⟨ S ⟩_{R} = (S)$ , and an ideal is generated as a module by a finite subset exactly when it is generated as an ideal by that subset.

L6L7step 2.1

4.1

Apply the module theorem to $M =_{R} R$ and rewrite each of its three conditions through the identifications just made. "Every submodule of $_{R} R$ is finitely generated" becomes condition 2; "every ascending chain of submodules of $_{R} R$ stabilises" becomes condition 3, an ascending chain of ideals being an ascending chain of submodules and conversely; "every nonempty family of submodules of $_{R} R$ has a maximal member" becomes condition 4. With step 1.1 identifying condition 1 with the first of these, conditions 1 to 4 are equivalent.

L3step 2.1step 3.1

5.1

The choice accounting transfers with the statement. The module theorem attributes exactly one of its implications, from the ascending chain condition to the maximal condition, to dependent choice, and the rewriting in step 4.1 is a change of vocabulary that uses no selection at all; so among conditions 1 to 4 the same single implication carries dependent choice and the rest are choice-free.

L1L3step 4.1∎

Remarks

Why the ideal-level form is proved rather than assumed. Left and right Noetherian rings fixes the Noetherian condition through the regular module, so the sentence "every ideal is finitely generated" is not the definition in force here but a consequence of it. Everything below cites this theorem for the ideal-level form, and does not unfold the regular module again.
The maximal condition is about a nonempty set of ideals. Dropping nonemptiness makes condition 4 false in every ring, since the empty set has no member at all, maximal or otherwise. The hypothesis is exactly the one carried by Finite generation, ACC, and maximal-condition characterizations of Noetherian modules.
Maximal, not greatest. A maximal member of a set of ideals has no member of that set strictly above it; it need not contain the others. In the set of all proper ideals of a ring with more than one maximal ideal there is no greatest element, and condition 4 does not claim one.
The chain in condition 3 is indexed from $0$ . Nothing changes if it is indexed from $1$ , since a chain indexed from $1$ extends to one indexed from $0$ by repeating its first term, but the index set is written out so that the stabilisation index $N$ is unambiguous.

Depends on

Used by

Dependency tree · two levels

16 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., (16.3)-(16.5) and (16.13) (standard reference, not scraped)
J. S. Milne, A Primer of Commutative Algebra, v4.03, §3 (3.1) (standard reference, not scraped)
B. Totaro, Commutative Algebra (Michaelmas 2011), notes by Z. Norwood, §8 (8.1) (standard reference, not scraped)