Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-26
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Every quotient and every localisation of a Noetherian ring is Noetherian

Statement

Let R be a Noetherian commutative ring. Then

  1. R/I is Noetherian for every ideal I of R (The quotient ring R/I with (r+I)(s+I)=rs+I), and
  2. S−1R is Noetherian for every multiplicative subset S⊆R (Multiplicative subsets and the localisation S−1R as equivalence classes of fractions).

Both cases include their degenerate instances: I=R makes R/I the zero ring, and 0∈S makes S−1R the zero ring, and the zero ring is Noetherian.

Facts & Assumptions

Given: A Noetherian commutative ring R, an ideal I of R with canonical projection π ⁣:R→R/I, and a multiplicative subset S⊆R with localisation map λS ⁣:R→S−1R.

[L1]

For I⊴R, the maps J↦J/I and K↦π−1(K) are inverse inclusion-preserving bijections between the ideals J of R containing I and the ideals K of R/I (Correspondence theorem: ideals of R/I correspond to ideals of R containing I).

[L2]

The canonical projection R→R/I is a surjective ring homomorphism with kernel I (The canonical projection R→R/I is a surjective ring homomorphism with kernel I).

[L3]

The quotient ring R/I has underlying set the additive cosets of I and multiplication (r+I)(s+I)=rs+I (The quotient ring R/I with (r+I)(s+I)=rs+I).

[L4]

In a commutative ring, (S) consists of finite sums ∑risi, and (a)=Ra; the empty sum is included and equals 0 (In a commutative ring, (S) consists of finite sums ∑risi, and (a)=Ra).

[L5]

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).

[L6]

For an ideal J of S−1R the contraction a=λS−1(J) is an ideal of R, and whenever a=(a1,…,an) one has J=(a1/1,…,an/1) (Every ideal of a localisation is generated by the images of any generating set of its contraction).

Proof

technique · direct
1.1

Every ideal of R/I is of the form J/I=π(J) for an ideal J of R containing I, by the correspondence between the ideals of R/I and the ideals of R containing I.

L1L3given
1.2

Since R is Noetherian, every ideal of R is finitely generated; in particular the ideal J of the previous sentence is, and so is the contraction of any ideal of S−1R.

L5given
2.1

Fix an ideal of R/I and write it as J/I with J=(x1,…,xn), n∈N. Then J/I is generated in R/I by π(x1),…,π(xn): an element of J/I is π(y) with y=∑icixi, so it equals ∑iπ(ci)π(xi); conversely a finite sum ∑icˉiπ(xi) has each cˉi=π(ci) by surjectivity of π, hence equals π(∑icixi)∈π(J). So every ideal of R/I is finitely generated.

L2L4step 1.1step 1.2
2.2

Fix an ideal J′ of S−1R and let a=λS−1(J′) be its contraction, an ideal of R. By step 1.2 there are a1,…,an∈R, with n∈N, such that a=(a1,…,an), and then J′=(a1/1,…,an/1). So every ideal of S−1R is finitely generated.

L6step 1.2given
3.1

A commutative ring all of whose ideals are finitely generated is Noetherian, so R/I is Noetherian by step 2.1 and S−1R is Noetherian by step 2.2.

L5step 2.1step 2.2∎

Remarks

  • The converse of neither half holds. A quotient or a localisation of a non-Noetherian ring can be Noetherian: the fraction field of a non-Noetherian integral domain is a field, and a field is Noetherian. The theorem is therefore stated in one direction only.

  • Where the Noetherian hypothesis is spent. It is used exactly twice, both times in step 1.2, to produce a finite generating list: once for an ideal of R containing I, and once for the contraction of an ideal of S−1R. Neither half needs the ascending chain condition or the maximal condition, so neither half uses a choice principle beyond 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 already records.

  • The quotient half uses only surjectivity of the projection. Step 2.1 appeals to no other property of π, so the same argument shows that the image of a finitely generated ideal under any surjective homomorphism of commutative rings is finitely generated.

Depends on

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Sources