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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Noetherian induction: a property that passes to an ideal whenever it holds for every strictly larger ideal holds for every ideal

Statement

Let $R$ be a Noetherian commutative ring and let $P$ be a set of ideals of $R$ with the following hereditary property: an ideal $I$ of $R$ belongs to $P$ whenever every ideal $J$ of $R$ with $I ⊊ J$ belongs to $P$ . Then $P$ contains every ideal of $R$ .

Read $P$ as the ideals satisfying a property $P$ : if $P (J)$ holds for every ideal $J$ strictly containing $I$ , and this for every $I$ , then $P$ holds for every ideal. The induction runs downward from the unit ideal, not upward from $0$ : the hypothesis applied to $I = R$ has empty content on the left, since no ideal strictly contains $R$ , so it asserts $R \in P$ outright.

The proof uses the maximal condition of 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 and therefore carries the same dependent-choice cost as that condition.

Facts & Assumptions

Given: A Noetherian commutative ring $R$ and a set $P$ of ideals of $R$ with the hereditary property of the Statement. A member $I$ of a set $Σ$ of ideals is called maximal in $Σ$ when no member of $Σ$ strictly contains $I$ .

[L1]

For a Noetherian commutative ring, every nonempty set of ideals of $R$ has a maximal member with respect to inclusion (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]

The implication from the ascending chain condition to the maximal condition uses dependent choice; the remaining implications are choice-free (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).

Proof

technique · contradiction

1.1

Suppose the conclusion fails, and let $Σ$ be the set of ideals of $R$ that do not belong to $P$ ; the supposition says exactly that $Σ \neq \emptyset$ .

assume-contragiven

2.1

Since $R$ is Noetherian and $Σ$ is a nonempty set of ideals, it has a maximal member: fix an ideal $I \in Σ$ such that no member of $Σ$ strictly contains $I$ .

L1step 1.1

3.1

Let $J$ be any ideal of $R$ with $I ⊊ J$ . Then $J \notin Σ$ , for otherwise $J$ would be a member of $Σ$ strictly containing $I$ , against the maximality fixed in the previous step. So $J \in P$ , and this holds for every ideal strictly containing $I$ .

step 2.1

4.1

The hereditary property, applied to the ideal $I$ , therefore gives $I \in P$ . But $I \in Σ$ means $I \notin P$ .

step 3.1given

5.1

The supposition of step 1.1 is untenable, so $Σ = \emptyset$ and $P$ contains every ideal of $R$ . The only non-constructive input is the maximal element produced in step 2.1, whose dependent-choice cost is the one recorded by the cited characterisation; nothing else in the argument selects anything.

L2step 2.1step 4.1discharge-contradiction∎

Remarks

Where the induction starts. There is no base case to verify separately. The hereditary hypothesis at $I = R$ quantifies over an empty collection of ideals, so it holds vacuously on the left and delivers $R \in P$ ; the principle then works downward. An attempt to run the same scheme upward from the zero ideal would need a descending chain condition, which a Noetherian ring need not satisfy.
A property failing for every ideal is not a counterexample. If $P = \emptyset$ then $Σ$ is the set of all ideals, step 2.1 produces the maximal member $R$ , and the hereditary hypothesis fails at $R$ ; so such a $P$ never satisfies the hypothesis in the first place.
Maximal, not greatest, is what the argument needs. Step 3.1 uses only that nothing in $Σ$ lies strictly above $I$ . It never compares $I$ with an arbitrary member of $Σ$ , which is what a greatest element would supply and what 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 does not provide.

Depends on

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

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

10 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

M. Hochster, Introduction to Commutative Algebra, Math 614, §2.2 Noetherian induction (standard reference, not scraped)
A. Altman and S. Kleiman, A Term of Commutative Algebra, 13th ed., §16 (standard reference, not scraped)