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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Hilbert basis theorem: if $R$ is Noetherian then $R [x]$ is Noetherian

Statement

Let $R$ be a Noetherian commutative ring. Then the polynomial ring $R [x]$ is a Noetherian commutative ring.

No hypothesis beyond Noetherianity is placed on $R$ : it may have zero divisors, and it may be the zero ring.

Facts & Assumptions

Given: A Noetherian commutative ring $R$ and its polynomial ring $R [x]$ .

[L1]

For every commutative ring $R$ , the coefficientwise addition and convolution multiplication make $R [x]$ a commutative ring, and the constant-polynomial map $R \to R [x]$ is an injective unital ring homomorphism (Polynomial convolution makes $R [x]$ a commutative ring containing $R$ as its constant subring).

[L2]

Over a Noetherian commutative ring $R$ , every ideal of $R [x]$ is generated by finitely many polynomials realising generators of its stage ideals up to a stabilisation degree; in particular every ideal of $R [x]$ is finitely generated (Over a Noetherian ring, an ideal of $R [x]$ is generated by finitely many polynomials realising generators of its stages up to the stabilisation degree).

[L3]

For a commutative ring, being Noetherian is equivalent to every ideal being 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).

Proof

technique · direct

1.1

$R [x]$ is a commutative ring, so the ideal-level characterisation of the Noetherian condition applies to it.

L1given

2.1

Every ideal of $R [x]$ is finitely generated, by the finite-generation lemma applied to the Noetherian ring $R$ .

L2step 1.1

3.1

A commutative ring all of whose ideals are finitely generated is Noetherian, so $R [x]$ is Noetherian.

L3step 2.1∎

Remarks

The work is in the lemmas, and it is of three separate kinds. The stage ideals and their ascent are The leading coefficients of the degree- $n$ elements of an ideal of $R [x]$ , together with $0$ , form an ideal of $R$ , and these ideals ascend with $n$ and use no chain condition; the one degree-lowering step is A single cancellation step lowers the degree of a polynomial in an ideal once its leading coefficient lies in a realised stage and uses no chain condition either; the passage to a finite generating list is Over a Noetherian ring, an ideal of $R [x]$ is generated by finitely many polynomials realising generators of its stages up to the stabilisation degree and is where the Noetherian hypothesis is spent. Keeping them apart is what makes each checkable on its own.
No leading coefficient is inverted anywhere. A reader who has seen the argument only over a field may expect a division step; over a general commutative ring the stage ideals carry that role, which is why they are ideals of $R$ rather than single elements.

Depends on

Used by

Dependency tree · two levels

20 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., (16.11) and (16.12) (standard reference, not scraped)
J. S. Milne, A Primer of Commutative Algebra, v4.03, §3 Theorem 3.7 (standard reference, not scraped)
B. Totaro, Commutative Algebra (Michaelmas 2011), notes by Z. Norwood, §8 Theorem 8.3 (standard reference, not scraped)
M. Hochster, Introduction to Commutative Algebra, Math 614, Theorem 5.6 (standard reference, not scraped)

Hilbert basis theorem: if RR is Noetherian then R[x]R[x] is Noetherian