Session-authored (Fable 5 assisted)

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

✓ 34 results · all verified · 32 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 2 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

Noetherian Rings and Hilbert Basis

1 · Prerequisites

2 · Summary

This page works in commutative rings with identity, using the regular-module definition of a Noetherian ring and the module theorem identifying finite generation, the ascending chain condition, and the maximal condition. It also uses the earlier algebra pages on ideals, quotient rings, localisations, polynomial rings, evaluation, and integrality, together with the module pages on free modules, exact sequences, and chain conditions. Those inputs let ideal-theoretic statements be transported to modules and back, and they supply the basic quotient, localisation, and polynomial constructions used throughout the later arguments.

The development first rewrites Noetherianity in ideal language and derives Noetherian induction, then proves stability under quotients, localisations, retractions, and finite products. It next breaks the Hilbert basis theorem into leading-coefficient ideals, one cancellation step, and the degree argument that makes polynomial ideals finitely generated, then passes to finite-variable polynomial rings and finite-type algebras. The later items treat finite presentation, finiteness of Hom, module-finite algebras, the Artin-Tate lemma and its integral form, and finally Noether's finiteness theorem for invariants and Cohen's criterion.

3 · Logical flowchart

4 · Definitions, theorems and proofs

RemarkRemark: AI-adaptedProof: Not applicablejudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

Conventions for this development and where dependent choice and Zorn's lemma are used

Rings. Every ring on this page is commutative (Commutative ring) and carries an identity, since Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides builds the identity into the definition of a ring; the word "ring" below means that wherever a statement does not say otherwise. A statement that reads "let $R$ be a ring" with no commutativity hypothesis is proved without using commutativity and applies in particular to the commutative rings this page is about. Ring homomorphisms preserve the identity (Ring homomorphism: additive, multiplicative, and required to send $1$ to $1$ ) and subrings contain the ambient identity (Subring: a subset containing $1_{R}$ and closed under addition, additive inverses and multiplication). Nothing here requires $1 \neq 0$ : the zero ring is a ring, it is admitted by every statement that does not exclude it in as many words, and it is Noetherian, its one ideal being $0 = R$ and generated by the empty set.

Noetherian and Artinian are cited, not redefined. Left and right Noetherian rings calls $R$ left Noetherian when the left regular module $_{R} R$ is Noetherian, and Noetherian modules: every submodule is finitely generated calls a module Noetherian when every submodule of it is finitely generated. For a commutative ring the left and right conditions coincide, so the side is not written below. The descending-chain dual is Left and right Artinian rings, which calls $R$ left Artinian when $_{R} R$ is Artinian; no statement on this page assumes or concludes that condition, and no argument below uses it. The equivalence of finite generation with the ascending chain condition and with the maximal condition is Finite generation, ACC, and maximal-condition characterizations of Noetherian modules, proved for modules; the ideal-level form is obtained on this page by transporting that theorem across the identification of the ideals of $R$ with the submodules of $_{R} R$ , rather than by proving the cycle a second time.

Where a choice principle is used, and where none is. The definition of finite generation, and every argument below that produces a finite generating set from data already in hand, are theorems of ZF. Two places are different, and both are marked where they occur.

Dependent choice. The implication from the ascending chain condition to the maximal condition is not choice-free: from a nonempty family of ideals with no maximal member one builds a strictly ascending chain by choosing, at each stage, an ideal strictly containing the one already chosen, and the sequence of choices depends on the choices already made. Finite generation, ACC, and maximal-condition characterizations of Noetherian modules attributes exactly that implication to dependent choice, and the ideal-level statement on this page carries the attribution unchanged. The reverse implication, and the two implications relating finite generation to the other conditions, need no choice principle. Sources differ here: Milne §3 attributes the implication to dependent choice and Altman–Kleiman (16.4) to countable choice. The two principles are not the same, and the published module theorem's attribution is the one in force inside this library.
Zorn's lemma, hence the axiom of choice. Cohen's criterion below is proved by producing an ideal maximal among those that are not finitely generated, in a ring not yet known to be Noetherian. No chain condition is available to supply that maximal element, and it is obtained from Zorn's lemma applied to the set of non-finitely-generated ideals ordered by inclusion. This is a strictly stronger commitment than the dependent choice above, and the lemmas leading to Cohen's criterion say so in their own statements.

What "finitely generated" is measured against. For an ideal it means finite generation as an ideal of the ring, for a module finite generation as a module over the ring named, and for an algebra it means the algebra generated by finitely many elements. These are three different conditions and the ring or module against which each is taken is written out at every occurrence, because a subring of a ring $B$ can be finitely generated as an algebra over $A$ and not as an $A$ -module.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

The submodule generated by a subset consists of the finite $R$ -linear combinations of that subset

Statement

Let $R$ be a ring, let $M$ be a left $R$ -module and let $S \subseteq M$ . Then the submodule generated by $S$ (Generated submodule, cyclic and finitely generated modules, module basis and free module) is the set of finite $R$ -linear combinations of elements of $S$ :

$⟨ S ⟩_{R} = {\sum_{i = 1}^{k} r_{i} s_{i} : k \in N, r_{1}, \dots, r_{k} \in R, s_{1}, \dots, s_{k} \in S},$

where the term with $k = 0$ is the empty sum $0_{M}$ . In particular $⟨ \emptyset ⟩_{R} = 0$ . Commutativity of $R$ is not used, and the elements $s_{1}, \dots, s_{k}$ are not required to be distinct.

Facts & Assumptions

Given: A ring $R$ , a left $R$ -module $M$ and a subset $S \subseteq M$ . Write $L$ for the set displayed in the Statement.

[L1]

The submodule generated by $S$ is $⟨ S ⟩_{R} : = ⋂ {N \leq M : S \subseteq N}$ , the family being nonempty because $M \leq M$ ; thus $⟨ S ⟩_{R}$ is the smallest submodule of $M$ containing $S$ (Generated submodule, cyclic and finitely generated modules, module basis and free module).

[L2]

For a left $R$ -module $M$ , a nonempty subset $N \subseteq M$ is a submodule if and only if $r u + v \in N$ for all $r \in R$ and $u, v \in N$ (The one-step submodule criterion; intersections and sums of submodules are submodules).

[L3]

A left $R$ -module is an abelian group $(M, +, 0_{M})$ with an action satisfying $r (m + n) = r m + r n$ , $(r + s) m = r m + s m$ , $(r s) m = r (s m)$ and $1_{R} m = m$ (Unital left and right modules over a ring; unqualified module means left module).

[L4]

In a commutative ring, $(S)$ consists of finite sums $\sum r_{i} s_{i}$ , and $(a) = R a$ (In a commutative ring, $(S)$ consists of finite sums $\sum r_{i} s_{i}$ , and $(a) = R a$ ).

Proof

technique · direct

1.1

Let $L$ be the set of all sums $\sum_{i = 1}^{k} r_{i} s_{i}$ with $k \in N$ , $r_{1}, \dots, r_{k} \in R$ and $s_{1}, \dots, s_{k} \in S$ , the value at $k = 0$ being the empty sum $0_{M}$ ; in particular $0_{M} \in L$ , so $L$ is nonempty even when $S$ is empty.

L1given

2.1

Every $s \in S$ lies in $L$ , being the one-term sum $1_{R} s$ , which equals $s$ by the unitality axiom of a module; hence $S \subseteq L$ .

L3step 1.1given

2.2

$L$ is a submodule of $M$ : it is nonempty, and for $r \in R$ and $u = \sum_{i = 1}^{k} r_{i} s_{i}$ , $v = \sum_{j = 1}^{m} t_{j} u_{j}$ in $L$ the element $r u + v$ is the sum $\sum_{i = 1}^{k} (r r_{i}) s_{i} + \sum_{j = 1}^{m} t_{j} u_{j}$ , again a finite $R$ -linear combination of elements of $S$ with $k + m$ terms, so $r u + v \in L$ and the submodule criterion applies.

L2L3step 1.1algebra

2.3

Every submodule $N \leq M$ with $S \subseteq N$ contains $L$ : each $s_{i}$ lies in $N$ , so repeated use of the submodule criterion [L2] puts every $r_{i} s_{i}$ and every finite sum of those elements in $N$ ; the case $k = 0$ gives $0_{M} \in N$ .

L2step 1.1algebra

3.1

$⟨ S ⟩_{R} = L$ : by steps 2.1 and 2.2 the set $L$ is one of the submodules over which the intersection defining $⟨ S ⟩_{R}$ runs, so $⟨ S ⟩_{R} \subseteq L$ ; and $⟨ S ⟩_{R}$ is itself a submodule containing $S$ , so step 2.3 applied to it gives $L \subseteq ⟨ S ⟩_{R}$ .

L1step 2.1step 2.2step 2.3

4.1

Two consequences follow by reading the description at its extreme cases. Taking $S = \emptyset$ leaves only the empty sum, so $⟨ \emptyset ⟩_{R} = 0$ . Taking $M$ to be the regular module over a commutative $R$ , where the submodules are the ideals, the description becomes the finite-sum description of the ideal generated by $S$ , so the two agree and this lemma extends that one from ideals to modules rather than competing with it.

L4step 3.1algebra∎

Remarks

Unitality is where the containment $S \subseteq ⟨ S ⟩_{R}$ comes from. Step 2.1 writes $s = 1_{R} s$ , which is available because Unital left and right modules over a ring; unqualified module means left module builds $1_{R} m = m$ into the definition of a module. Over a non-unital action the set of finite $R$ -linear combinations need not contain $S$ , and then it is not the generated submodule.
No finiteness is assumed of $S$ . Each element of $⟨ S ⟩_{R}$ carries its own finite list of coefficients and elements; different elements may use different lists, and no bound on the length is claimed.

LemmaStatement: Literature-sourcedProof: Literature-sourcedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

Every generating set of a finitely generated module contains a finite generating subset

Statement

Let $R$ be a ring, let $M$ be a finitely generated left $R$ -module (Generated submodule, cyclic and finitely generated modules, module basis and free module) and let $S \subseteq M$ satisfy $⟨ S ⟩_{R} = M$ . Then some finite subset $S^{'} \subseteq S$ already satisfies $⟨ S^{'} ⟩_{R} = M$ .

No hypothesis is placed on $S$ itself, which may be infinite, and none on $R$ beyond being a ring.

Facts & Assumptions

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

[L1]

$M$ is finitely generated when $M = ⟨ T ⟩_{R}$ for some finite $T$ , and $⟨ T ⟩_{R}$ is the smallest submodule of $M$ containing $T$ (Generated submodule, cyclic and finitely generated modules, module basis and free module).

[L2]

For a ring $R$ , a left $R$ -module $M$ and a subset $S \subseteq M$ , 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

Fix a finite $T \subseteq M$ with $⟨ T ⟩_{R} = M$ , available because $M$ is finitely generated, and recall the hypothesis $⟨ S ⟩_{R} = M$ .

L1given

2.1

For each $t \in T$ we have $t \in M = ⟨ S ⟩_{R}$ , so $t = \sum_{i = 1}^{k} r_{i} s_{i}$ for some $k \in N$ , some $r_{1}, \dots, r_{k} \in R$ and some $s_{1}, \dots, s_{k} \in S$ ; write $S_{t} : = {s_{1}, \dots, s_{k}} \subseteq S$ , a finite set with $t \in ⟨ S_{t} ⟩_{R}$ . One such expression is selected for each of the finitely many elements of $T$ , so this is a finite sequence of selections and no choice axiom is used.

L2step 1.1given

3.1

Put $S^{'} : = ⋃_{t \in T} S_{t}$ . This is a union of finitely many finite sets, hence finite, and $S^{'} \subseteq S$ ; moreover $S_{t} \subseteq S^{'}$ gives $⟨ S_{t} ⟩_{R} \subseteq ⟨ S^{'} ⟩_{R}$ , so $t \in ⟨ S^{'} ⟩_{R}$ for every $t \in T$ , that is $T \subseteq ⟨ S^{'} ⟩_{R}$ .

step 2.1algebra

4.1

$⟨ S^{'} ⟩_{R}$ is a submodule of $M$ containing $T$ , so it contains the smallest such submodule, namely $⟨ T ⟩_{R} = M$ ; and $⟨ S^{'} ⟩_{R} \subseteq M$ always. Hence $⟨ S^{'} ⟩_{R} = M$ with $S^{'} \subseteq S$ finite.

L1step 1.1step 3.1∎

Remarks

The finite generating subset depends on the chosen $T$ , and no smallest one is claimed. Different finite generating sets $T$ produce different subsets $S^{'}$ , and the lemma asserts only that some finite subset of $S$ generates. It says nothing about the least possible size of such a subset.
The hypothesis that $M$ is finitely generated cannot be dropped. Without it the conclusion is the assertion that every generating set of every module has a finite generating subset, which would make every module finitely generated, since a module always generates itself.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

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.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

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.

LemmaStatement: Literature-sourcedProof: Literature-sourcedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

A subring that admits a module retraction from a Noetherian ring is Noetherian

Statement

Let $R^{'}$ be a Noetherian commutative ring and let $R \subseteq R^{'}$ be a subring (Subring: a subset containing $1_{R}$ and closed under addition, additive inverses and multiplication), so that $R^{'}$ becomes an $R$ -algebra through the inclusion and in particular an $R$ -module (Algebras over a commutative ring, central structure maps, and algebra homomorphisms). Suppose there is a map

$ρ ⁣ : R^{'} ⟶ R$

that is $R$ -linear (Module homomorphism and isomorphism, kernel, image and cokernel) and restricts to the identity on $R$ , that is $ρ (x) = x$ for every $x \in R$ . Then $R$ is Noetherian.

The map $ρ$ is not assumed to be a ring homomorphism; additivity and $ρ (a x) = a ρ (x)$ for $a \in R$ are all that is used.

Facts & Assumptions

Given: A Noetherian commutative ring $R^{'}$ , a subring $R \subseteq R^{'}$ , and an $R$ -linear $ρ : R^{'} \to R$ with $ρ (x) = x$ for $x \in R$ . For an ideal $a$ of $R$ write $a R^{'}$ for the ideal of $R^{'}$ generated by the subset $a$ .

[L1]

A subset $S \subseteq R$ is a subring of $R$ when (T1) $1_{R} \in S$ ; (T2) $x, y \in S$ implies $x + y \in S$ ; (T3) $x \in S$ implies $- x \in S$ ; (T4) $x, y \in S$ implies $x y \in S$ (Subring: a subset containing $1_{R}$ and closed under addition, additive inverses and multiplication).

[L2]

An $R$ -algebra is a unital ring $A$ together with a unital ring homomorphism $η_{A} : R \to A$ whose image is central; the induced scalar action is $r a : = η_{A} (r) a$ , making $A$ an $R$ -module (Algebras over a commutative ring, central structure maps, and algebra homomorphisms).

[L3]

A function $f : M \to N$ between left $R$ -modules is an $R$ -module homomorphism if $f (m + m^{'}) = f (m) + f (m^{'})$ and $f (r m) = r f (m)$ for all $m, m^{'} \in M$ and $r \in R$ (Module homomorphism and isomorphism, kernel, image and cokernel).

[L4]

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).

[L5]

If a left module $M$ over a ring is finitely generated and $S \subseteq M$ satisfies $⟨ S ⟩_{R} = M$ , then some finite subset of $S$ already generates $M$ (Every generating set of a finitely generated module contains a finite generating subset).

[L6]

[L7]

Proof

technique · direct

1.1

Fix an ideal $a$ of $R$ . Because $R$ is a subring of $R^{'}$ the two rings share the identity, so $a \subseteq a R^{'}$ , each $a \in a$ being $a \cdot 1_{R^{'}}$ ; and $ρ$ is additive with $ρ (a x) = a ρ (x)$ for $a \in R$ and $x \in R^{'}$ , since $R^{'}$ carries the $R$ -action $a \cdot x = a x$ coming from the inclusion.

L1L2L3given

2.1

The ideal $a R^{'}$ of $R^{'}$ is exactly the set of finite sums $\sum_{i} a_{i} x_{i}$ with $a_{i} \in a$ and $x_{i} \in R^{'}$ , which is also the $R^{'}$ -submodule of $R^{'}$ generated by the subset $a$ ; and $a R^{'}$ is finitely generated because $R^{'}$ is Noetherian. Applying the finite-subset lemma to the generating set $a$ of that module produces finitely many elements $a_{1}, \dots, a_{n} \in a$ , with $n \in N$ , generating $a R^{'}$ .

L4L5L6L7step 1.1given

3.1

Let $a \in a$ . By step 2.1 and the description of a generated ideal, $a = \sum_{i = 1}^{n} x_{i} a_{i}$ for some $x_{1}, \dots, x_{n} \in R^{'}$ . Applying $ρ$ and using $ρ (a) = a$ together with $R$ -linearity, and noting $a_{i} \in R$ , gives $a = ρ (a) = \sum_{i = 1}^{n} a_{i} ρ (x_{i})$ with every $ρ (x_{i}) \in R$ .

L3L6step 1.1step 2.1

4.1

Hence $a \subseteq (a_{1}, \dots, a_{n}) R$ , and the reverse inclusion holds because each $a_{i}$ lies in $a$ ; so $a = (a_{1}, \dots, a_{n}) R$ is finitely generated. As $a$ was arbitrary, every ideal of $R$ is finitely generated and $R$ is Noetherian.

L4L6step 3.1∎

Remarks

Why a ring retraction is not asked for. The only properties of $ρ$ used are additivity and $R$ -homogeneity, and both are used only in step 3.1, to push the relation $a = \sum x_{i} a_{i}$ down into $R$ . Requiring $ρ$ to be multiplicative would exclude the averaging maps that are the standard source of such retractions.
The subring hypothesis is what makes $a \subseteq a R^{'}$ available. A subring contains $1_{R^{'}}$ by (T1) of Subring: a subset containing $1_{R}$ and closed under addition, additive inverses and multiplication, so each $a \in a$ is visibly a member of the ideal it generates in $R^{'}$ . Without a shared identity the inclusion can fail and the retraction would have nothing to act on.
Every ideal of $R$ needs its own finite list. The list $a_{1}, \dots, a_{n}$ produced in step 2.1 depends on $a$ , and no bound uniform in $a$ is claimed or available.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

Every ideal of a localisation is generated by the images of any generating set of its contraction

Statement

Let $R$ be a commutative ring, let $S \subseteq R$ be multiplicative and let $λ_{S} ⁣ : R \to S^{- 1} R$ , $r \mapsto r / 1$ , be the localisation map (Multiplicative subsets and the localisation $S^{- 1} R$ as equivalence classes of fractions). Let $J$ be an ideal of $S^{- 1} R$ and let

$a : = λ_{S}^{- 1} (J) = {r \in R : r / 1 \in J}$

be its contraction. Then $a$ is an ideal of $R$ , and for every subset $T \subseteq R$ with $a = (T)$ the ideal $J$ is generated in $S^{- 1} R$ by the image $λ_{S} (T) = {t / 1 : t \in T}$ . In particular, if $a = (a_{1}, \dots, a_{n})$ with $n \in N$ then $J = (a_{1} / 1, \dots, a_{n} / 1)$ .

Only the $S$ -saturated half of the ideal correspondence is used. Nothing here says that extension and contraction are mutually inverse on all ideals of $R$ ; that is false in general, and it is the contraction of an ideal of $S^{- 1} R$ , not an arbitrary ideal of $R$ , that this lemma starts from.

Facts & Assumptions

Given: A commutative ring $R$ , a multiplicative subset $S \subseteq R$ , an ideal $J$ of $S^{- 1} R$ , and a subset $T \subseteq R$ generating the contraction $a = λ_{S}^{- 1} (J)$ .

[L1]

A subset $S \subseteq R$ of a commutative ring is multiplicative if $1 \in S$ and $s, t \in S$ implies $s t \in S$ ; the localisation $S^{- 1} R$ has elements $r / s$ with the displayed arithmetic, and the localisation map is the ring homomorphism $λ_{S} : R \to S^{- 1} R$ , $r \mapsto r / 1$ (Multiplicative subsets and the localisation $S^{- 1} R$ as equivalence classes of fractions).

[L2]

For a commutative ring $R$ and a multiplicative $S \subseteq R$ , extension $I \mapsto S^{- 1} I = {r / s : r \in I, s \in S}$ and contraction $J \mapsto λ_{S}^{- 1} (J)$ give inverse inclusion-preserving bijections between $S$ -saturated ideals of $R$ and ideals of $S^{- 1} R$ (Ideals of $S^{- 1} R$ correspond to $S$ -saturated ideals of $R$ , and prime ideals correspond to primes disjoint from $S$ ).

[L3]

[L4]

For a subset $T$ of a ring, $(T)$ is the intersection of all two-sided ideals of that ring containing $T$ (The ideal generated by a subset and principal ideals).

Proof

technique · direct

1.1

Fix the data of the Given line. The set $a = λ_{S}^{- 1} (J)$ is the preimage of an ideal under the ring homomorphism $λ_{S}$ .

L1given

2.1

The contraction of an ideal of $S^{- 1} R$ is one of the two directions of the ideal correspondence, so $a$ is an $S$ -saturated ideal of $R$ , and because extension and contraction are mutually inverse on that class of ideals, its extension recovers $J$ : $J = S^{- 1} a = {r / u : r \in a, u \in S}$ .

L2step 1.1

3.1

Every element of $J$ lies in the ideal generated by the image of $T$ . Indeed such an element is $r / u$ with $r \in a = (T)$ and $u \in S$ , and $r$ is then a finite sum $r = \sum_{i = 1}^{k} c_{i} t_{i}$ with $c_{i} \in R$ and $t_{i} \in T$ ; dividing by $u$ gives $r / u = \sum_{i = 1}^{k} (c_{i} / u) (t_{i} / 1)$ , a finite $S^{- 1} R$ -linear combination of elements of the image of $T$ . The empty sum $k = 0$ gives $r = 0$ and $r / u = 0$ , which lies in every ideal.

L3L4step 2.1algebra

3.2

Conversely each $t / 1$ with $t \in T$ lies in $J$ , because $t \in a = λ_{S}^{- 1} (J)$ says exactly that $λ_{S} (t) = t / 1$ belongs to $J$ ; and $J$ is an ideal, so it contains the ideal generated by that image.

L4step 2.1

4.1

The two inclusions of steps 3.1 and 3.2 give $J = (λ_{S} (T))$ . Taking $T = {a_{1}, \dots, a_{n}}$ finite yields $J = (a_{1} / 1, \dots, a_{n} / 1)$ .

L4step 3.1step 3.2∎

Remarks

What the cited correspondence does and does not give. It is a bijection between the $S$ -saturated ideals of $R$ and all ideals of $S^{- 1} R$ (Ideals of $S^{- 1} R$ correspond to $S$ -saturated ideals of $R$ , and prime ideals correspond to primes disjoint from $S$ ). An arbitrary ideal $b$ of $R$ need not be $S$ -saturated, and then $λ_{S}^{- 1} (S^{- 1} b)$ is strictly larger than $b$ . The proof above never applies the correspondence to such a $b$ : it starts from an ideal of $S^{- 1} R$ and contracts.
The generating set is not required to be finite. The argument uses only that each element of $a$ is a finite $R$ -linear combination of elements of $T$ ; the finite case is stated separately because it is the one the Noetherian application needs.
The degenerate localisation is covered. If $0 \in S$ then $S^{- 1} R$ is the zero ring, its only ideal is $0$ , and the contraction is all of $R$ ; the conclusion holds because every subset of the zero ring generates its only ideal.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-26Open item page →

Every quotient and every localisation of a Noetherian ring is Noetherian

Statement

Let $R$ be a Noetherian commutative ring. Then

$R / I$ is Noetherian for every ideal $I$ of $R$ (The quotient ring $R / I$ with $(r + I) (s + I) = r s + I$ ), and
$S^{- 1} R$ is Noetherian for every multiplicative subset $S \subseteq R$ (Multiplicative subsets and the localisation $S^{- 1} R$ as equivalence classes of fractions).

Both cases include their degenerate instances: $I = R$ makes $R / I$ the zero ring, and $0 \in S$ makes $S^{- 1} R$ 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 $π ⁣ : R \to R / I$ , and a multiplicative subset $S \subseteq R$ with localisation map $λ_{S} ⁣ : R \to S^{- 1} R$ .

[L1]

For $I ⊴ R$ , the maps $J \mapsto J / I$ and $K \mapsto π^{- 1} (K)$ 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 $R \to R / I$ is a surjective ring homomorphism with kernel $I$ (The canonical projection $R \to R / I$ is a surjective ring homomorphism with kernel $I$ ).

[L3]

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

[L4]

[L5]

[L6]

For an ideal $J$ of $S^{- 1} R$ the contraction $a = λ_{S}^{- 1} (J)$ is an ideal of $R$ , and whenever $a = (a_{1}, \dots, a_{n})$ one has $J = (a_{1} / 1, \dots, a_{n} / 1)$ (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 $S^{- 1} R$ .

L5given

2.1

Fix an ideal of $R / I$ and write it as $J / I$ with $J = (x_{1}, \dots, x_{n})$ , $n \in N$ . Then $J / I$ is generated in $R / I$ by $π (x_{1}), \dots, π (x_{n})$ : an element of $J / I$ is $π (y)$ with $y = \sum_{i} c_{i} x_{i}$ , so it equals $\sum_{i} π (c_{i}) π (x_{i})$ ; conversely a finite sum $\sum_{i} {\overset{ˉ}{c}}_{i} π (x_{i})$ has each ${\overset{ˉ}{c}}_{i} = π (c_{i})$ by surjectivity of $π$ , hence equals $π (\sum_{i} c_{i} x_{i}) \in π (J)$ . So every ideal of $R / I$ is finitely generated.

L2L4step 1.1step 1.2

2.2

Fix an ideal $J^{'}$ of $S^{- 1} R$ and let $a = λ_{S}^{- 1} (J^{'})$ be its contraction, an ideal of $R$ . By step 1.2 there are $a_{1}, \dots, a_{n} \in R$ , with $n \in N$ , such that $a = (a_{1}, \dots, a_{n})$ , and then $J^{'} = (a_{1} / 1, \dots, a_{n} / 1)$ . So every ideal of $S^{- 1} R$ 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 $S^{- 1} R$ 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 $S^{- 1} R$ . 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.

CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

A ring with finitely many ideals of zero intersection whose quotients are Noetherian rings is Noetherian

Statement

Let $R$ be a commutative ring and let $a_{1}, \dots, a_{r}$ be ideals of $R$ , with $r \in N$ and $r \geq 1$ , such that

$⋂_{i = 1}^{r} a_{i} = 0$

and every quotient ring $R / a_{i}$ (The quotient ring $R / I$ with $(r + I) (s + I) = r s + I$ ) is Noetherian. Then $R$ is Noetherian.

The restriction $r \geq 1$ avoids the vacuous endpoint. If $r = 0$ , the empty intersection is $R$ itself (Ideal criteria and intersections of ideals), so the displayed condition forces $R = 0$ ; the conclusion is then still true because the zero ring is Noetherian.

Facts & Assumptions

Given: A commutative ring $R$ , ideals $a_{1}, \dots, a_{r}$ with $r \geq 1$ and zero intersection, and Noetherian quotient rings $R / a_{i}$ with canonical projections $π_{i} ⁣ : R \to R / a_{i}$ .

[L1]

A finite direct sum is Noetherian if and only if every summand is Noetherian, and it is Artinian if and only if every summand is Artinian. The empty direct sum is included (Finite direct sums preserve and reflect Noetherian and Artinian conditions).

[L2]

For a unital ring $R$ and a family $(M_{i})_{i \in I}$ of left $R$ -modules, the direct sum $⨁_{i \in I} M_{i}$ is the submodule of the coordinatewise product consisting of the families of finite support (The direct sum of an indexed family of modules).

[L3]

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

[L4]

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

[L5]

An $R$ -algebra is a unital ring $A$ with a unital ring homomorphism $η_{A} : R \to A$ of central image; the induced scalar action $r a : = η_{A} (r) a$ makes $A$ an $R$ -module (Algebras over a commutative ring, central structure maps, and algebra homomorphisms).

[L6]

[L7]

Proof

technique · direct

1.1

Each $R / a_{i}$ is a Noetherian $R$ -module for the action $r \cdot \overset{ˉ}{x} : = π_{i} (r) \overset{ˉ}{x}$ , which is the algebra action along the surjective projection $π_{i}$ . A subset of $R / a_{i}$ closed under that action is closed under multiplication by every element of $R / a_{i}$ , because $π_{i}$ is onto, so the $R$ -submodules of $R / a_{i}$ are exactly its ideals; each such ideal is generated over $R / a_{i}$ by a finite list, since $R / a_{i}$ is a Noetherian ring, and the same list generates it over $R$ because every coefficient in $R / a_{i}$ is $π_{i}$ of a coefficient in $R$ .

L3L4L5L7given

1.2

The map $ϕ ⁣ : R \to ⨁_{i = 1}^{r} R / a_{i}$ , $ϕ (x) = (π_{1} (x), \dots, π_{r} (x))$ , is $R$ -linear, and $ϕ (x) = 0$ says $x \in a_{i}$ for every $i$ , so $\ker ϕ = ⋂_{i = 1}^{r} a_{i} = 0$ and $ϕ$ is injective.

L2L6given

2.1

The direct sum $⨁_{i = 1}^{r} R / a_{i}$ has finitely many summands, each Noetherian as an $R$ -module, so it is a Noetherian $R$ -module.

L1L2step 1.1

3.1

Let $b$ be an ideal of $R$ . Its image $ϕ (b)$ is an $R$ -submodule of the direct sum, being the image of a submodule under an $R$ -linear map, so it is generated by finitely many of its own elements $ϕ (b_{1}), \dots, ϕ (b_{k})$ with $b_{1}, \dots, b_{k} \in b$ and $k \in N$ . For $b \in b$ write $ϕ (b) = \sum_{j = 1}^{k} c_{j} ϕ (b_{j}) = ϕ (\sum_{j = 1}^{k} c_{j} b_{j})$ with $c_{j} \in R$ ; injectivity of $ϕ$ gives $b = \sum_{j = 1}^{k} c_{j} b_{j}$ , so $b = (b_{1}, \dots, b_{k})$ .

L3L6step 1.2step 2.1

4.1

Every ideal of $R$ is therefore finitely generated, and a commutative ring with that property is Noetherian.

L7step 3.1∎

Remarks

Neither hypothesis can be dropped. Without the zero intersection the map $ϕ$ of step 1.2 has a kernel and step 3.1 cannot pull generators back; taking $r = 1$ and $a_{1} = R$ shows what goes wrong, since the zero ring is Noetherian while $R$ need not be. Without finiteness of the list, the direct sum in step 2.1 need not be Noetherian.
The ideals are not assumed distinct, comparable or proper. Repetitions and the values $0$ and $R$ are all admitted; only the intersection and the Noetherian quotients are used.
The intersection condition says $R$ embeds in the product of its quotients. That is the entire content of step 1.2, and it is why the corollary is about an embedding rather than about a decomposition: no claim is made that $ϕ$ is surjective.

CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

A product of two Noetherian rings is Noetherian

Statement

Let $R$ and $S$ be Noetherian commutative rings. Then the product ring $R \times S$ (The product ring $R \times S$ with componentwise operations, its identity $(1_{R}, 1_{S})$ and its units $R^{\times} \times S^{\times}$ ) is Noetherian.

Facts & Assumptions

Given: Noetherian commutative rings $R$ and $S$ , and their product ring $R \times S$ with componentwise operations.

[L1]

The product ring $R \times S$ is the cartesian product of the underlying sets with the componentwise operations $(a, b) + (a^{'}, b^{'}) = (a + a^{'}, b + b^{'})$ and $(a, b) (a^{'}, b^{'}) = (a a^{'}, b b^{'})$ , zero $(0_{R}, 0_{S})$ and identity $(1_{R}, 1_{S})$ (The product ring $R \times S$ with componentwise operations, its identity $(1_{R}, 1_{S})$ and its units $R^{\times} \times S^{\times}$ ).

[L2]

For a ring homomorphism $f ⁣ : R \to S$ there is a ring isomorphism $R / \ker f ≅ im f$ (First isomorphism theorem for rings: $R / \ker f ≅ im f$ ).

[L3]

If a commutative ring $T$ has ideals $a_{1}, \dots, a_{r}$ with $r \geq 1$ and zero intersection, and every quotient ring $T / a_{i}$ is Noetherian, then $T$ is Noetherian (A ring with finitely many ideals of zero intersection whose quotients are Noetherian rings is Noetherian).

Proof

technique · direct

1.1

Put $a_{1} = {0_{R}} \times S$ and $a_{2} = R \times {0_{S}}$ . Each is an additive subgroup of $R \times S$ closed under multiplication by an arbitrary element, since $(a, b) (0_{R}, t) = (0_{R}, b t)$ and $(a, b) (u, 0_{S}) = (a u, 0_{S})$ , so each is an ideal; and $a_{1} \cap a_{2} = {(0_{R}, 0_{S})} = 0$ . The ring $R \times S$ is commutative because $R$ and $S$ are and the operations are componentwise.

L1givenalgebra

2.1

The coordinate maps $p_{1} (a, b) = a$ and $p_{2} (a, b) = b$ are ring homomorphisms, since the operations are componentwise and $p_{1} (1_{R}, 1_{S}) = 1_{R}$ , $p_{2} (1_{R}, 1_{S}) = 1_{S}$ ; they are surjective, with $\ker p_{1} = a_{1}$ and $\ker p_{2} = a_{2}$ . The first isomorphism theorem therefore gives ring isomorphisms $(R \times S) / a_{1} ≅ R$ and $(R \times S) / a_{2} ≅ S$ .

L1L2step 1.1algebra

3.1

A ring isomorphism carries ideals to ideals and finite generating lists to finite generating lists, so a ring isomorphic to a Noetherian ring is Noetherian; hence both quotients in step 2.1 are Noetherian. With the zero intersection of step 1.1 this is the hypothesis of the preceding corollary at $r = 2$ , and it gives that $R \times S$ is Noetherian.

L3step 1.1step 2.1algebra∎

Remarks

Every finite product of Noetherian rings is Noetherian, by induction on the number of factors using $R_{1} \times \dots \times R_{n + 1} ≅ (R_{1} \times \dots \times R_{n}) \times R_{n + 1}$ ; the base case of one factor is tautological. A product indexed by an infinite set is not defined by The product ring $R \times S$ with componentwise operations, its identity $(1_{R}, 1_{S})$ and its units $R^{\times} \times S^{\times}$ , which introduces the product of two rings only, so no claim is made about one here.
The converse holds as well, though it is not what is claimed above: step 2.1 exhibits each factor as isomorphic to a quotient of $R \times S$ , and Every quotient and every localisation of a Noetherian ring is Noetherian makes every quotient of a Noetherian ring Noetherian. So $R \times S$ is Noetherian exactly when both factors are.
When both factors are nonzero, the two coordinate ideals are incomparable, and their intersection is zero while their sum is the unit ideal. If one factor is the zero ring, one coordinate ideal is the whole product and the other is zero, so they are comparable; the proof uses only their zero intersection and covers that degenerate case as well.

LemmaStatement: Literature-sourcedProof: Literature-sourcedprecheck passaudited 2026-08-26Open item page →

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 $n$

Statement

Let $R$ be a commutative ring, let $a$ be an ideal of the polynomial ring $R [x]$ (The polynomial ring over a commutative ring as finitely supported coefficient sequences with convolution), and for $n \in N$ set

$a_{n} : = {0} \cup {lc (f) : f \in a, f \neq 0, \deg f = n} \subseteq R .$

Then $a_{n}$ is an ideal of $R$ for every $n \in N$ , and

$a_{0} \subseteq a_{1} \subseteq a_{2} \subseteq \dots .$

The index runs over $N$ , so the chain begins at $a_{0}$ , whose members are $0$ together with the nonzero constant polynomials that lie in $a$ .

Adjoining $0$ is not cosmetic. The zero polynomial has no degree and no leading coefficient (Degree, leading coefficient and monic polynomial, with the zero polynomial having no degree), so it contributes no element, and without the adjunction the set would be empty whenever $a$ contains no element of degree exactly $n$ .

Facts & Assumptions

Given: A commutative ring $R$ , an ideal $a$ of $R [x]$ , and $n \in N$ . A nonzero $f \in a$ of degree $n$ with $lc (f) = c$ is said to realise $c$ at stage $n$ .

[L1]

$R [x]$ is the set of finitely supported functions $a ⁣ : N \to R$ , with $(a + b)_{i} = a_{i} + b_{i}$ and $(a b)_{i} = \sum_{j + k = i} a_{j} b_{k}$ ; the constant $r$ is the sequence supported at $0$ with value $r$ , and $x$ is the sequence with coefficient $1_{R}$ at index $1$ and zero elsewhere (The polynomial ring over a commutative ring as finitely supported coefficient sequences with convolution).

[L2]

For $0 \neq f = \sum_{i} a_{i} x^{i} \in R [x]$ the degree is $\deg f = \max {i \in N : a_{i} \neq 0}$ and the leading coefficient is $lc (f) = a_{\deg f}$ ; the zero polynomial has no degree and no leading coefficient (Degree, leading coefficient and monic polynomial, with the zero polynomial having no degree).

[L3]

A nonempty subset $I \subseteq R$ is a two-sided ideal exactly when it is closed under $x - y$ and under $r x, x r$ for all $r \in R$ , $x, y \in I$ (Ideal criteria and intersections of ideals).

[L4]

For nonzero $f, g \in R [x]$ over a commutative ring: if $f + g \neq 0$ then $\deg (f + g) \leq \max {\deg f, \deg g}$ ; the coefficient of $x^{\deg f + \deg g}$ in $f g$ is $lc (f) lc (g)$ , and if $f g \neq 0$ then $\deg (f g) \leq \deg f + \deg g$ (Degree inequalities for sums and products over a commutative ring).

[L5]

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 in a commutative ring the left, right and two-sided notions agree (Left, right and two-sided ideals).

Proof

technique · direct

1.1

Fix $n \in N$ and read the displayed definition: $a_{n}$ consists of $0$ together with the leading coefficients of those elements of $a$ that are nonzero of degree exactly $n$ . In particular $0 \in a_{n}$ , so $a_{n}$ is a nonempty subset of $R$ ; and no element of $a_{n}$ other than the adjoined $0$ is $0$ , since a leading coefficient is nonzero by definition.

L1L2given

2.1

$a_{n}$ is closed under differences. Let $c, c^{'} \in a_{n}$ . If $c^{'} = 0$ then $c - c^{'} = c \in a_{n}$ . If $c = 0$ and $c^{'} \neq 0$ , take $g \in a$ realising $c^{'}$ at stage $n$ ; negation is coefficientwise, so $- g \in a$ is nonzero of degree $n$ with $lc (- g) = - c^{'}$ , and $c - c^{'} = - c^{'} \in a_{n}$ . If both are nonzero, take realisers $f, g \in a$ at stage $n$ ; then $- g$ is nonzero of degree $n$ , the coefficient of $x^{n}$ in $f - g$ is $c - c^{'}$ , and the coefficients of $f - g$ above index $n$ all vanish. Should $c - c^{'} = 0$ , the difference lies in $a_{n}$ as the adjoined element; otherwise $f - g \neq 0$ , the degree law gives $\deg (f - g) \leq \max {n, n} = n$ , and the nonvanishing coefficient at $x^{n}$ forces $\deg (f - g) = n$ with $lc (f - g) = c - c^{'}$ , so $c - c^{'} \in a_{n}$ .

L1L2L4step 1.1algebra

2.2

$a_{n}$ is closed under multiplication by elements of $R$ . Let $c \in a_{n}$ and $r \in R$ . If $r c = 0$ the product lies in $a_{n}$ as the adjoined element, and this covers $c = 0$ and $r = 0$ . Otherwise $c \neq 0$ and $r \neq 0$ ; take $f \in a$ realising $c$ at stage $n$ and read $r$ as a constant polynomial, which is nonzero of degree $0$ with leading coefficient $r$ . The coefficient of $x^{0 + n}$ in the product $r f$ is $r c \neq 0$ , so $r f \neq 0$ ; the degree law then gives $\deg (r f) \leq 0 + n$ , and the nonvanishing coefficient at $x^{n}$ forces $\deg (r f) = n$ with $lc (r f) = r c$ . Since $a$ is an ideal of $R [x]$ we have $r f \in a$ , so $r c \in a_{n}$ .

L1L2L4step 1.1algebra

2.3

The stages ascend. Let $c \in a_{n}$ with $c \neq 0$ and let $f \in a$ realise it at stage $n$ . Multiplying by $x$ shifts coefficients: by the convolution rule the coefficient of $x^{i}$ in $x f$ is the coefficient of $x^{i - 1}$ in $f$ for $i \geq 1$ and is $0$ at $i = 0$ . So $x f$ has zero coefficients above index $n + 1$ and coefficient $c \neq 0$ at index $n + 1$ , whence $x f$ is a nonzero element of $a$ of degree $n + 1$ with $lc (x f) = c$ . Thus $c \in a_{n + 1}$ , and $0 \in a_{n + 1}$ as well, so $a_{n} \subseteq a_{n + 1}$ .

L1L2step 1.1

3.1

By the ideal criterion, a nonempty subset of a commutative ring closed under differences and under multiplication by arbitrary ring elements is an ideal; steps 2.1 and 2.2 supply exactly those closures, so $a_{n}$ is an ideal of $R$ .

L3L5step 2.1step 2.2

4.1

Step 3.1 holds for every $n \in N$ and step 2.3 gives $a_{n} \subseteq a_{n + 1}$ for every $n \in N$ , so the stages form an ascending chain of ideals of $R$ indexed by $N$ and beginning at $a_{0}$ . A nonzero element of $a$ has degree $0$ exactly when it is a nonzero constant, and its leading coefficient is then that constant, so $a_{0}$ is the set of constants lying in $a$ , the zero constant included.

step 2.3step 3.1∎

Remarks

Exact degree, not degree at most $n$ . Defining the stage by "degree at most $n$ " gives the same ideals, but then the ideal property itself needs the shifting argument of step 2.3 rather than only the ascent. With exact degree the two facts separate cleanly, which is what the Hilbert basis argument uses: it needs a generator of a prescribed degree, not merely of bounded degree.
The chain need not be strictly increasing, and it need not stabilise. Nothing above assumes $R$ Noetherian. Stabilisation is exactly what the Noetherian hypothesis will buy, and it is not available here.
Why $a_{n}$ is closed under multiplication even where degrees drop. Over a ring with zero divisors $r f$ can have degree below $n$ , and then $r c = 0$ ; step 2.2 records that case separately and sends it to the adjoined $0$ rather than pretending the degree is preserved.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

A single cancellation step lowers the degree of a polynomial in an ideal once its leading coefficient lies in a realised stage

Statement

Let $R$ be a commutative ring, let $a$ be an ideal of $R [x]$ , let $n \in N$ , and let $a_{n}$ be the stage ideal of 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 $n$ . Suppose $m \in N$ and $g_{1}, \dots, g_{m} \in a$ are nonzero polynomials of degree $n$ whose leading coefficients $c_{j} = lc (g_{j})$ generate $a_{n}$ , so that $a_{n} = (c_{1}, \dots, c_{m})$ .

Let $f \in a$ be nonzero of degree $d$ with $d \geq n$ and $lc (f) \in a_{n}$ . Then there is a polynomial $h$ in the ideal of $R [x]$ generated by $g_{1}, \dots, g_{m}$ such that

$f - h = 0 or f - h \neq 0 and \deg (f - h) < d .$

At $d = 0$ the second alternative is impossible, so there $f = h$ . At $m = 0$ the hypothesis on $f$ cannot be met: $a_{n}$ is then the zero ideal while a leading coefficient is nonzero, so the statement is not vacuously producing a degree drop out of nothing.

Facts & Assumptions

Given: A commutative ring $R$ , an ideal $a$ of $R [x]$ , an index $n \in N$ , polynomials $g_{1}, \dots, g_{m} \in a$ nonzero of degree $n$ with leading coefficients $c_{1}, \dots, c_{m}$ generating $a_{n}$ , and a nonzero $f \in a$ of degree $d \geq n$ with $lc (f) \in a_{n}$ .

[L1]

$R [x]$ is the set of finitely supported functions $a ⁣ : N \to R$ , with $(a + b)_{i} = a_{i} + b_{i}$ and $(a b)_{i} = \sum_{j + k = i} a_{j} b_{k}$ ; $x$ is the sequence with coefficient $1_{R}$ at index $1$ and zero elsewhere (The polynomial ring over a commutative ring as finitely supported coefficient sequences with convolution).

[L2]

[L3]

[L4]

For an ideal $a$ of $R [x]$ and $n \in N$ , the set $a_{n}$ of leading coefficients of the nonzero degree- $n$ elements of $a$ , together with $0$ , is an ideal of $R$ , and $a_{n} \subseteq a_{n + 1}$ (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 $n$ ).

Proof

technique · direct

1.1

Fix the data of the Given line and write $c = lc (f)$ , a nonzero element of $R$ lying in $a_{n}$ ; the exponent $d - n$ is a natural number because $d \geq n$ .

L1L2given

2.1

Since $a_{n}$ is the ideal generated by $c_{1}, \dots, c_{m}$ and $c$ lies in it, the finite-sum description gives $r_{1}, \dots, r_{m} \in R$ with $c = \sum_{j = 1}^{m} r_{j} c_{j}$ . Any one such list may be taken; the construction below uses no property of it beyond this equation.

L3L4step 1.1

3.1

Put $h : = \sum_{j = 1}^{m} r_{j} x^{d - n} g_{j}$ , which lies in the ideal of $R [x]$ generated by $g_{1}, \dots, g_{m}$ because each coefficient $r_{j} x^{d - n}$ is an element of $R [x]$ . By the convolution rule the coefficient of $x^{i}$ in $r_{j} x^{d - n} g_{j}$ is $r_{j}$ times the coefficient of $x^{i - (d - n)}$ in $g_{j}$ , read as $0$ when $i < d - n$ . For $i > d$ that index exceeds $n$ , so the coefficient vanishes; at $i = d$ that index is exactly $n$ , so the coefficient is $r_{j} c_{j}$ . Summing over $j$ , the polynomial $h$ has zero coefficients at every index above $d$ and coefficient $\sum_{j = 1}^{m} r_{j} c_{j} = c$ at index $d$ .

L1L2L3step 2.1

4.1

The polynomial $f - h$ lies in $a$ and has zero coefficient at every index $i \geq d$ : above $d$ both $f$ and $h$ vanish, and at $d$ the two coefficients are $c$ and $c$ . So either $f - h = 0$ , or $f - h$ is nonzero and every index carrying a nonzero coefficient is below $d$ , which by the definition of degree means $\deg (f - h) < d$ .

L2step 3.1

5.1

Two extreme cases are worth recording. When $d = n$ the exponent $d - n$ is $0$ and $x^{0} = 1$ , so $h = \sum_{j} r_{j} g_{j}$ and no shift occurs. When $d = 0$ there is no natural number below $d$ , so the second alternative of step 4.1 cannot occur and the conclusion is the equality $f = h$ . When $m = 0$ the sum defining $a_{n}$ is empty, $a_{n} = 0$ , and no nonzero leading coefficient lies in it, so no $f$ satisfies the hypothesis.

L2L3step 3.1step 4.1∎

Remarks

Nothing is divided and no leading coefficient is inverted. The correction $h$ is built by multiplying the given $g_{j}$ by ring elements and a power of $x$ , so the argument runs over an arbitrary commutative ring rather than only over a field or a domain. That is exactly why the stage ideals are needed: over a field one generator per stage would do.
The $g_{j}$ must have degree exactly $n$ . If some $g_{j}$ had degree below $n$ the coefficient computation in step 3.1 would place $r_{j} c_{j}$ at an index below $d$ and the cancellation at $x^{d}$ would fail. This is what makes the exact-degree definition of the stage ideal the usable one.
The lemma produces one step, not a terminating procedure. Iterating it requires knowing that the leading coefficient of $f - h$ again lies in a realised stage, which is what the ascending chain condition supplies in the finite-generation lemma that uses this one.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

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

Statement

Let $R$ be a Noetherian commutative ring and let $a$ be an ideal of $R [x]$ . Let $a_{0} \subseteq a_{1} \subseteq \dots$ be the stage ideals of 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 let $N \in N$ be an index at which that chain stabilises, so $a_{n} = a_{N}$ for every $n \geq N$ . For each $n \leq N$ choose finitely many nonzero elements $c_{n, 1}, \dots, c_{n, m_{n}}$ of $a_{n}$ generating it, and for each of them a polynomial $g_{n, j} \in a$ , nonzero of degree $n$ , with $lc (g_{n, j}) = c_{n, j}$ .

Then the finitely many polynomials $g_{n, j}$ , for $n \leq N$ and $1 \leq j \leq m_{n}$ , generate $a$ as an ideal of $R [x]$ . In particular every ideal of $R [x]$ is finitely generated.

The selections are possible: each $a_{n}$ is an ideal of the Noetherian ring $R$ , hence has a finite generating set, from which the zero element may be discarded without loss, and every nonzero element of $a_{n}$ is by definition the leading coefficient of some nonzero degree- $n$ element of $a$ . Only finitely many selections are made, so no choice axiom is used.

Facts & Assumptions

Given: A Noetherian commutative ring $R$ , an ideal $a$ of $R [x]$ , and the stage ideals $a_{n}$ for $n \in N$ .

[L1]

For a commutative ring, being Noetherian is equivalent to every ideal being finitely generated, and to every ascending chain of ideals indexed by $N$ stabilising (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]

[L3]

[L4]

With $a_{n} = (c_{1}, \dots, c_{m})$ realised at stage $n$ by $g_{1}, \dots, g_{m} \in a$ , every nonzero $f \in a$ of degree $d \geq n$ with $lc (f) \in a_{n}$ admits an $h$ in the ideal generated by $g_{1}, \dots, g_{m}$ with $f - h = 0$ or $\deg (f - h) < d$ (A single cancellation step lowers the degree of a polynomial in an ideal once its leading coefficient lies in a realised stage).

[L5]

For $0 \neq f \in R [x]$ the degree is the largest index carrying a nonzero coefficient and the leading coefficient is the coefficient there; the zero polynomial has no degree (Degree, leading coefficient and monic polynomial, with the zero polynomial having no degree).

[L6]

Every nonempty subset $S \subseteq N$ has a least element: there is $ℓ \in S$ with $ℓ \leq s$ for all $s \in S$ (The well-ordering principle).

Proof

technique · contradiction

1.1

The stage ideals form an ascending chain of ideals of $R$ indexed by $N$ , and $R$ is Noetherian, so that chain stabilises: fix $N \in N$ with $a_{n} = a_{N}$ for every $n \geq N$ .

L1L2given

2.1

For each $n \leq N$ the ideal $a_{n}$ of $R$ is finitely generated; discard the zero element from a finite generating set, which changes nothing it generates, and realise each remaining generator $c_{n, j}$ by a nonzero $g_{n, j} \in a$ of degree $n$ . This is a selection over the finitely many pairs $(n, j)$ with $n \leq N$ , so it is a finite selection. Let $b$ be the ideal of $R [x]$ generated by all the $g_{n, j}$ ; since every $g_{n, j}$ lies in $a$ , we have $b \subseteq a$ .

L1L2L3step 1.1

3.1

Suppose $b \neq a$ , so that $a ∖ b$ is nonempty. The zero polynomial lies in $b$ , so every element of $a ∖ b$ is nonzero and therefore has a degree; the set of those degrees is a nonempty subset of $N$ and so has a least element $d$ . Fix $f \in a ∖ b$ with $\deg f = d$ .

assume-contraL3L5L6step 2.1

4.1

Put $n = \min (d, N)$ , so $n \leq N$ and $d \geq n$ . If $d \leq N$ then $n = d$ and $lc (f)$ lies in $a_{d} = a_{n}$ by the definition of the stage; if $d > N$ then $n = N$ and $lc (f) \in a_{d} = a_{N} = a_{n}$ by the stabilisation of step 1.1. The polynomials $g_{n, 1}, \dots, g_{n, m_{n}}$ realise generators of $a_{n}$ at stage $n$ , so the cancellation lemma applies and yields $h$ in the ideal generated by them, hence $h \in b$ , with $f - h = 0$ or $\deg (f - h) < d$ .

L2L4L5step 2.1step 3.1

5.1

Both alternatives are impossible. If $f - h = 0$ then $f = h \in b$ , contradicting $f \in a ∖ b$ . If $f - h \neq 0$ with $\deg (f - h) < d$ , then $f - h$ lies in $a$ and not in $b$ , since $h \in b$ and $f \notin b$ , so its degree belongs to the set whose least element is $d$ , contradicting $\deg (f - h) < d$ . Therefore $a = b$ , and $a$ is generated by the finitely many $g_{n, j}$ .

step 3.1step 4.1discharge-contradiction∎

Remarks

The generating list is not canonical. It depends on the stabilisation index $N$ , on the finite generating sets chosen for the stages, and on the realisers; a larger $N$ or a larger generating set produces a longer list that generates the same ideal. Nothing above claims minimality.
Where the Noetherian hypothesis is used, and where it is not. It is used twice, in step 1.1 for the stabilisation of the chain and in step 2.1 for finite generation of each stage. 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 A single cancellation step lowers the degree of a polynomial in an ideal once its leading coefficient lies in a realised stage hold over any commutative ring; only the passage from "one cancellation step" to "a finite generating list" needs the chain condition.
The minimal-degree argument replaces an induction that would not terminate on its own. A single cancellation step lowers the degree, but iterating it from an arbitrary element gives no bound; choosing a counterexample of least degree turns one step into a contradiction.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

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]

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.

CorollaryStatement: Literature-sourcedProof: Literature-sourcedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

If $R$ is Noetherian then $R [x_{1}, \dots, x_{n}]$ is Noetherian for every $n \in N$

Statement

Let $R$ be a Noetherian commutative ring. Then the iterated polynomial ring $R [x_{1}, \dots, x_{n}]$ of Polynomial rings in finitely many commuting indeterminates by iteration is Noetherian for every $n \in N$ .

The index starts at $0$ , where the published definition sets $R [x_{1}, \dots, x_{0}] = R$ and the assertion is the hypothesis itself.

Facts & Assumptions

Given: A Noetherian commutative ring $R$ .

[L1]

Polynomial rings in finitely many commuting indeterminates are defined recursively by $R [x_{1}, \dots, x_{0}] : = R$ and $R [x_{1}, \dots, x_{n + 1}] : = R [x_{1}, \dots, x_{n}] [x_{n + 1}]$ (Polynomial rings in finitely many commuting indeterminates by iteration).

[L2]

If $R$ is a Noetherian commutative ring then $R [x]$ is a Noetherian commutative ring (Hilbert basis theorem: if $R$ is Noetherian then $R [x]$ is Noetherian).

Proof

technique · induction

1.1

At $n = 0$ the recursive definition gives $R [x_{1}, \dots, x_{0}] = R$ , which is Noetherian by hypothesis; this is the base of the induction and is not skipped.

baseL1given

1.2

Let $n \in N$ and assume $R [x_{1}, \dots, x_{n}]$ is Noetherian.

2.1

The recursive definition gives $R [x_{1}, \dots, x_{n + 1}] = R [x_{1}, \dots, x_{n}] [x_{n + 1}]$ , a polynomial ring in one indeterminate over the ring assumed Noetherian in step 1.2; the Hilbert basis theorem applied to that ring makes $R [x_{1}, \dots, x_{n + 1}]$ Noetherian.

L1L2step 1.2

3.1

The base case of step 1.1 and the passage of step 2.1 give, by induction on $n$ , that $R [x_{1}, \dots, x_{n}]$ is Noetherian for every $n \in N$ .

step 1.1step 2.1discharge-induction∎

Remarks

Finitely many indeterminates is essential. The induction produces a proof for each $n \in N$ separately and says nothing about a ring of polynomials in infinitely many indeterminates; the companion examples page carries a witness that the conclusion fails there.
The converse holds too, by iterating $R [x]$ is Noetherian if and only if $R$ is Noetherian down the tower of coefficient rings.

CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

$R [x]$ is Noetherian if and only if $R$ is Noetherian

Statement

Let $R$ be a commutative ring. Then $R [x]$ is Noetherian if and only if $R$ is Noetherian.

Facts & Assumptions

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

[L1]

If $R$ is a Noetherian commutative ring then $R [x]$ is a Noetherian commutative ring (Hilbert basis theorem: if $R$ is Noetherian then $R [x]$ is Noetherian).

[L2]

For commutative rings $R, S$ , a unital ring homomorphism $φ ⁣ : R \to S$ and $s \in S$ , there is a unique unital ring homomorphism ${ev}_{φ, s} ⁣ : R [x] \to S$ that extends $φ$ on constant polynomials and sends $x$ to $s$ (Universal property of $R [x]$ : a coefficient homomorphism and the image of $x$ determine a unique ring homomorphism).

[L3]

For a unital ring homomorphism $φ ⁣ : R \to S$ between commutative rings, $s \in S$ and $f = \sum_{i} a_{i} x^{i} \in R [x]$ , the value of $f$ at $s$ along $φ$ is $f_{φ} (s) = \sum_{i} φ (a_{i}) s^{i}$ (Evaluation and roots of a polynomial in a commutative target ring).

[L4]

For a ring homomorphism $f ⁣ : R \to S$ there is a ring isomorphism $R / \ker f ≅ im f$ (First isomorphism theorem for rings: $R / \ker f ≅ im f$ ).

[L5]

Every quotient $R / I$ of a Noetherian commutative ring by an ideal is Noetherian (Every quotient and every localisation of a Noetherian ring is Noetherian).

Proof

technique · direct

1.1

For the direction from $R$ to $R [x]$ , the Hilbert basis theorem applied to $R$ gives at once that $R [x]$ is Noetherian.

L1given

1.2

For the converse direction, take $φ$ to be the identity of $R$ and $s = 0_{R}$ in the universal property: there is a unital ring homomorphism $ε ⁣ : R [x] \to R$ , evaluation at $0$ , which is the identity on constant polynomials and sends $x$ to $0$ . Being the identity on constants makes $ε$ surjective, so $im ε = R$ .

L2L3given

2.1

Still for the converse direction, the first isomorphism theorem applied to $ε$ gives a ring isomorphism $R [x] / \ker ε ≅ R$ .

L4step 1.2

3.1

Still for the converse direction, assume $R [x]$ Noetherian. Its quotient $R [x] / \ker ε$ is then Noetherian, and a ring isomorphism carries ideals to ideals and finite generating lists to finite generating lists, so the isomorphic ring $R$ is Noetherian.

L5step 2.1algebra

4.1

Step 1.1 is one implication and step 3.1 is the other, so $R [x]$ is Noetherian exactly when $R$ is.

step 1.1step 3.1∎

Remarks

Which half is the theorem. The direction from $R$ to $R [x]$ is the Hilbert basis theorem and carries all the work; the converse is three citations, since $R$ is a quotient of $R [x]$ and quotients of Noetherian rings are Noetherian.
Evaluation at any element of $R$ would do. The argument uses only that $ε$ is a surjective ring homomorphism $R [x] \to R$ ; evaluation at $0$ is chosen because it is the one whose kernel, the ideal of polynomials with zero constant term, is the easiest to name.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

Subalgebra generated by a subset, algebras of finite type, and module-finite algebras

Definition

Let $R$ be a commutative ring and let $A$ be a commutative $R$ -algebra with structure map $η_{A} ⁣ : R \to A$ (Algebras over a commutative ring, central structure maps, and algebra homomorphisms).

The subalgebra generated by finitely many elements. Let $n \in N$ and $a_{1}, \dots, a_{n} \in A$ . Iterating the universal property of a polynomial ring (Universal property of $R [x]$ : a coefficient homomorphism and the image of $x$ determine a unique ring homomorphism) along the recursion $R [x_{1}, \dots, x_{k + 1}] = R [x_{1}, \dots, x_{k}] [x_{k + 1}]$ of Polynomial rings in finitely many commuting indeterminates by iteration gives a unique unital ring homomorphism

$ev ⁣ : R [x_{1}, \dots, x_{n}] ⟶ A$

that agrees with $η_{A}$ on constants and sends $x_{i}$ to $a_{i}$ for each $i$ ; at each step of the recursion the one-variable universal property supplies existence and uniqueness of the extension, and at $n = 0$ the map is $η_{A}$ itself. Its image is written $R [a_{1}, \dots, a_{n}]$ and is called the $R$ -subalgebra of $A$ generated by $a_{1}, \dots, a_{n}$ . It is a subring of $A$ containing $η_{A} (R)$ , and it is the smallest such subring containing $a_{1}, \dots, a_{n}$ , since any subring with those properties is closed under the sums and products that make up a polynomial expression. At $n = 0$ it is $η_{A} (R)$ , the image of $R$ in $A$ .

Finite type. $A$ is of finite type over $R$ , equivalently a finitely generated $R$ -algebra, when $A = R [a_{1}, \dots, a_{n}]$ for some $n \in N$ and some $a_{1}, \dots, a_{n} \in A$ . Equivalently, $A$ is isomorphic as an $R$ -algebra to a quotient $R [x_{1}, \dots, x_{n}] / a$ for some $n \in N$ and some ideal $a$ : the map $ev$ above is then surjective, and First isomorphism theorem for rings: $R / \ker f ≅ im f$ identifies $A$ with $R [x_{1}, \dots, x_{n}] / \ker ev$ ; conversely the composite of the canonical projection with the inclusion of the indeterminates exhibits any such quotient as generated by the residues of $x_{1}, \dots, x_{n}$ .

Module-finite. $A$ is module-finite over $R$ , equivalently a finite $R$ -algebra, when $A$ is finitely generated as an $R$ -module (Generated submodule, cyclic and finitely generated modules, module basis and free module) for the action $r \cdot a = η_{A} (r) a$ .

Module-finite implies finite type. If $b_{1}, \dots, b_{n}$ generate $A$ as an $R$ -module then $R [b_{1}, \dots, b_{n}]$ , being a subring of $A$ that contains $η_{A} (R)$ and every $b_{i}$ , contains every $R$ -linear combination $\sum_{i} η_{A} (r_{i}) b_{i}$ and hence all of $A$ ; so $A = R [b_{1}, \dots, b_{n}]$ . The converse fails, and the companion examples page carries a witness.

Remarks

Three conditions, three rings. "Finitely generated" is ambiguous on its own: an ideal may be finitely generated as an ideal, a module as a module, and an algebra as an algebra, and the three are different requirements. This page writes "of finite type" for the algebra condition and "module-finite" for the module condition, and always names the ring over which the condition is taken. Sources differ in vocabulary: Totaro and Milne write "finite algebra" for what is called module-finite here, and Altman–Kleiman write "module finite" and "algebra finite".
The generators need not be algebraically independent. Nothing above asks $ev$ to be injective. When it is, $A$ is a polynomial ring and the ideal $a$ is zero; that is a special case, not the definition.
The empty list is allowed and is not the same as $A = R$ . At $n = 0$ the subalgebra generated is $η_{A} (R)$ , which is a quotient of $R$ rather than a copy of it unless $η_{A}$ is injective.

CorollaryStatement: Literature-sourcedProof: Literature-sourcedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

Every algebra of finite type over a Noetherian ring is a Noetherian ring

Statement

Let $R$ be a Noetherian commutative ring and let $A$ be a commutative $R$ -algebra of finite type (Subalgebra generated by a subset, algebras of finite type, and module-finite algebras). Then $A$ is a Noetherian ring.

Facts & Assumptions

Given: A Noetherian commutative ring $R$ and a commutative $R$ -algebra $A$ of finite type, with structure map $η_{A} ⁣ : R \to A$ .

[L1]

An $R$ -algebra $A$ is of finite type over $R$ when $A = R [a_{1}, \dots, a_{n}]$ for some $n \in N$ and $a_{1}, \dots, a_{n} \in A$ , where $R [a_{1}, \dots, a_{n}]$ is the image of the unital ring homomorphism $R [x_{1}, \dots, x_{n}] \to A$ agreeing with $η_{A}$ on constants and sending $x_{i}$ to $a_{i}$ (Subalgebra generated by a subset, algebras of finite type, and module-finite algebras).

[L2]

For a ring homomorphism $f ⁣ : R \to S$ there is a ring isomorphism $R / \ker f ≅ im f$ (First isomorphism theorem for rings: $R / \ker f ≅ im f$ ).

[L3]

If $R$ is a Noetherian commutative ring then $R [x_{1}, \dots, x_{n}]$ is Noetherian for every $n \in N$ (If $R$ is Noetherian then $R [x_{1}, \dots, x_{n}]$ is Noetherian for every $n \in N$ ).

[L4]

Every quotient $R / I$ of a Noetherian commutative ring by an ideal is Noetherian (Every quotient and every localisation of a Noetherian ring is Noetherian).

Proof

technique · direct

1.1

By the finite-type hypothesis there are $n \in N$ and $a_{1}, \dots, a_{n} \in A$ with $A = R [a_{1}, \dots, a_{n}]$ , so the evaluation homomorphism $ev ⁣ : R [x_{1}, \dots, x_{n}] \to A$ sending $x_{i}$ to $a_{i}$ has image all of $A$ and is therefore surjective.

L1given

2.1

The first isomorphism theorem applied to $ev$ gives a ring isomorphism $R [x_{1}, \dots, x_{n}] / \ker ev ≅ A$ .

L2step 1.1

3.1

The ring $R [x_{1}, \dots, x_{n}]$ is Noetherian because $R$ is, so its quotient by the ideal $\ker ev$ is Noetherian; a ring isomorphism carries ideals to ideals and finite generating lists to finite generating lists, so $A$ is Noetherian.

L3L4step 2.1algebra∎

Remarks

This is the form of the Hilbert basis theorem later pages use. A ring presented by finitely many generators and any relations at all is Noetherian, with no hypothesis on the relations; the number of generators is what matters, not their independence.
The converse is false and is not claimed. A Noetherian ring need not be of finite type over a Noetherian subring: a field extension generated by infinitely many algebraic elements is a field, hence Noetherian, and is not of finite type over the base field as an algebra.

CorollaryStatement: Literature-sourcedProof: Literature-sourcedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

Every algebra of finite type over a principal ideal domain is a Noetherian ring

Statement

Let $R$ be a principal ideal domain (Principal ideal domain) and let $A$ be a commutative $R$ -algebra of finite type (Subalgebra generated by a subset, algebras of finite type, and module-finite algebras). Then $A$ is a Noetherian ring.

Facts & Assumptions

Given: A principal ideal domain $R$ and a commutative $R$ -algebra $A$ of finite type. A field is a principal ideal domain under the definition in force here: it is an integral domain, and its only ideals are $(0)$ and $(1)$ , both principal. So the field case falls under this statement rather than outside it.

[F1]

An integral domain $R$ is a principal ideal domain (PID) if every ideal $I ⊴ R$ is principal: there is an $a \in R$ with $I = (a)$ (Principal ideal domain).

[L1]

Every principal ideal domain is Noetherian (Every principal ideal domain is Noetherian).

[L2]

Every commutative algebra of finite type over a Noetherian commutative ring is a Noetherian ring (Every algebra of finite type over a Noetherian ring is a Noetherian ring).

Proof

technique · direct

1.1

The base ring $R$ is a principal ideal domain, hence an integral domain and in particular a commutative ring, and hence a Noetherian ring.

F1L1given

2.1

The algebra $A$ is of finite type over the Noetherian commutative ring $R$ , so it is a Noetherian ring.

L2step 1.1∎

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

Finitely presented modules and finitely presented algebras

Definition

Let $R$ be a commutative ring. For $n \in N$ write $R^{n}$ for the free $R$ -module $R^{({1, \dots, n})}$ on an $n$ -element set, with standard basis $e_{1}, \dots, e_{n}$ (The free module on a set and its standard basis); at $n = 0$ the index set is empty and $R^{0} = 0$ .

Finitely presented modules. An $R$ -module $M$ is finitely presented when there are $m, n \in N$ and module homomorphisms making

$R^{m} \overset{α}{\to} R^{n} \overset{β}{\to} M ⟶ 0$

an exact sequence (Exact sequences and short exact sequences of modules): that is, $im α = \ker β$ and $β$ is surjective (Module homomorphism and isomorphism, kernel, image and cokernel).

The equivalent quotient form. $M$ is finitely presented exactly when $M ≅ R^{n} / K$ for some $n \in N$ and some finitely generated submodule $K \leq R^{n}$ (Generated submodule, cyclic and finitely generated modules, module basis and free module). Given a presentation, put $K = \ker β = im α$ , which is generated by $α (e_{1}), \dots, α (e_{m})$ , and First isomorphism theorem for modules: $M / \ker f ≅ im f$ gives $M ≅ R^{n} / K$ . Conversely, given $M ≅ R^{n} / K$ with $K$ generated by $k_{1}, \dots, k_{m}$ , the universal property of the free module (Universal property of the free module on a set) supplies $α ⁣ : R^{m} \to R^{n}$ with $α (e_{j}) = k_{j}$ , whose image is $K$ , and composing the canonical projection with the isomorphism gives $β$ ; the displayed sequence is then exact.

Finitely presented algebras. A commutative $R$ -algebra $A$ is finitely presented when there are $n \in N$ and a finitely generated ideal $a \subseteq R [x_{1}, \dots, x_{n}]$ (The ideal generated by a subset and principal ideals) with $A ≅ R [x_{1}, \dots, x_{n}] / a$ as an $R$ -algebra.

Finitely presented implies finitely generated, in both senses. A finitely presented module is generated by $β (e_{1}), \dots, β (e_{n})$ , since $β$ is surjective. A finitely presented algebra is of finite type (Subalgebra generated by a subset, algebras of finite type, and module-finite algebras), being a quotient of $R [x_{1}, \dots, x_{n}]$ .

The boundary values are admitted. Taking $m = n = 0$ presents the zero module, and taking $m = 0$ with $n$ arbitrary presents the free module $R^{n}$ ; taking $a = 0$ presents the polynomial algebra itself.

Remarks

What finite presentation adds to finite generation is a bound on the relations. Finite generation says $M$ is a quotient of some $R^{n}$ ; finite presentation says the kernel of that quotient map is itself finitely generated. Over an arbitrary commutative ring the second is strictly stronger.
The two notions carry the same name and are not the same condition. A finitely presented algebra is finitely presented as an algebra, which constrains a defining ideal in a polynomial ring, and says nothing on its own about the underlying module.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

Over a Noetherian ring a module is Noetherian exactly when it is finitely generated, exactly when it is finitely presented

Statement

Let $R$ be a Noetherian commutative ring and let $M$ be an $R$ -module. The following are equivalent.

$M$ is a Noetherian module (Noetherian modules: every submodule is finitely generated).
$M$ is finitely generated (Generated submodule, cyclic and finitely generated modules, module basis and free module).
$M$ is finitely presented (Finitely presented modules and finitely presented algebras).

The proof below shows exactly where the Noetherian hypothesis is used to make conditions 2 and 3 agree.

Facts & Assumptions

Given: A Noetherian commutative ring $R$ and an $R$ -module $M$ . For $n \in N$ , $R^{n}$ denotes the free module on an $n$ -element set with standard basis $e_{1}, \dots, e_{n}$ .

[L1]

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

[L2]

$M$ is finitely generated when $M = ⟨ S ⟩_{R}$ for some finite $S \subseteq M$ , and $⟨ S ⟩_{R}$ is the smallest submodule of $M$ containing $S$ (Generated submodule, cyclic and finitely generated modules, module basis and free module).

[L3]

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

[L4]

In the free module $R^{(X)}$ every element has a unique expression $\sum_{x \in F} r_{x} e_{x}$ with $F \subseteq X$ finite, where $e_{x}$ has coordinate $1_{R}$ at $x$ and zero elsewhere; for $X = \emptyset$ the module is $0$ (The free module on a set and its standard basis).

[L5]

Every set map $u ⁣ : X \to M$ extends uniquely to an $R$ -module homomorphism $\overset{ˉ}{u} ⁣ : R^{(X)} \to M$ with $\overset{ˉ}{u} (e_{x}) = u (x)$ , given by $\overset{ˉ}{u} (\sum_{x \in F} r_{x} e_{x}) = \sum_{x \in F} r_{x} u (x)$ (Universal property of the free module on a set).

[L6]

[L7]

A sequence of modules and homomorphisms is exact at a module where two arrows meet when the image of the incoming map equals the kernel of the outgoing one (Exact sequences and short exact sequences of modules).

[L8]

An $R$ -module $M$ is finitely presented when there are $m, n \in N$ and an exact sequence $R^{m} \to R^{n} \to M \to 0$ (Finitely presented modules and finitely presented algebras).

Proof

technique · direct

1.1

Condition 1 implies condition 2, because $M$ is a submodule of itself and a Noetherian module has all of its submodules finitely generated.

L1L2given

1.2

Condition 2 implies condition 1, since $R$ is a Noetherian ring and a finitely generated module over such a ring is Noetherian.

L3given

1.3

Condition 3 implies condition 2: an exact sequence $R^{m} \to R^{n} \to M \to 0$ has its right-hand map $β$ surjective, because exactness at $M$ says the image of $β$ is the kernel of the zero map $M \to 0$ , which is all of $M$ ; every element of $R^{n}$ is $\sum_{i = 1}^{n} r_{i} e_{i}$ , so every element of $M$ is $\sum_{i = 1}^{n} r_{i} β (e_{i})$ and $M$ is generated by the finite set ${β (e_{1}), \dots, β (e_{n})}$ .

L2L4L6L7L8given

1.4

Assume condition 2 and fix a finite generating set $m_{1}, \dots, m_{n}$ of $M$ , with $n \in N$ . The universal property of the free module gives an $R$ -module homomorphism $β ⁣ : R^{n} \to M$ with $β (e_{i}) = m_{i}$ , and its image is the set of all finite sums $\sum_{i} r_{i} m_{i}$ , which is $⟨ m_{1}, \dots, m_{n} ⟩_{R} = M$ ; so $β$ is surjective.

L2L4L5L6given

2.1

Still assuming condition 2, the module $R^{n}$ is generated by the finite set ${e_{1}, \dots, e_{n}}$ , hence is Noetherian over the Noetherian ring $R$ ; therefore its submodule $K : = \ker β$ is finitely generated, say by $k_{1}, \dots, k_{m}$ with $m \in N$ .

L1L2L3L4step 1.4

3.1

Still assuming condition 2, the universal property gives $α ⁣ : R^{m} \to R^{n}$ with $α (e_{j}) = k_{j}$ , whose image is $⟨ k_{1}, \dots, k_{m} ⟩_{R} = K = \ker β$ ; together with the surjectivity of $β$ this makes $R^{m} \to R^{n} \to M \to 0$ exact, so $M$ is finitely presented and condition 2 implies condition 3.

L5L6L7L8step 1.4step 2.1

4.1

Steps 1.1, 1.2, 1.3 and 3.1 close the cycle: condition 1 gives condition 2, condition 2 gives condition 1 and condition 3, and condition 3 gives condition 2. The three conditions are therefore equivalent.

step 1.1step 1.2step 1.3step 3.1∎

Remarks

Where the Noetherian hypothesis is spent. Only in step 1.2, through Finitely generated modules over a left Noetherian ring are Noetherian, and in step 2.1, to make $\ker β$ finitely generated. Step 1.3 and step 1.4 hold over any commutative ring.
The presentation is not canonical. It depends on the chosen generating set of $M$ and on the chosen generating set of $\ker β$ ; different choices give different $m$ and $n$ , and nothing above claims either is minimal.

CorollaryStatement: Literature-sourcedProof: Literature-sourcedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

Every algebra of finite type over a Noetherian ring is finitely presented

Statement

Facts & Assumptions

Given: A Noetherian commutative ring $R$ and a commutative $R$ -algebra $A$ of finite type.

[L1]

An $R$ -algebra is of finite type over $R$ exactly when it is isomorphic as an $R$ -algebra to a quotient $R [x_{1}, \dots, x_{n}] / a$ for some $n \in N$ and some ideal $a$ (Subalgebra generated by a subset, algebras of finite type, and module-finite algebras).

[L2]

If $R$ is a Noetherian commutative ring then $R [x_{1}, \dots, x_{n}]$ is Noetherian for every $n \in N$ (If $R$ is Noetherian then $R [x_{1}, \dots, x_{n}]$ is Noetherian for every $n \in N$ ).

[L3]

[L4]

A commutative $R$ -algebra is finitely presented when it is isomorphic as an $R$ -algebra to $R [x_{1}, \dots, x_{n}] / a$ for some $n \in N$ and some finitely generated ideal $a$ (Finitely presented modules and finitely presented algebras).

Proof

technique · direct

1.1

Being of finite type, $A$ is isomorphic as an $R$ -algebra to $R [x_{1}, \dots, x_{n}] / a$ for some $n \in N$ and some ideal $a$ of $R [x_{1}, \dots, x_{n}]$ .

L1given

2.1

The ring $R [x_{1}, \dots, x_{n}]$ is Noetherian because $R$ is, so its ideal $a$ is finitely generated.

L2L3step 1.1

3.1

A presentation by a polynomial ring in finitely many variables modulo a finitely generated ideal is exactly what finite presentation as an algebra asks for, so $A$ is finitely presented over $R$ .

L4step 1.1step 2.1∎

Remarks

Over a Noetherian base the two algebra conditions coincide. Finite presentation always implies finite type; this corollary supplies the converse when the base ring is Noetherian, so no relation-finiteness hypothesis needs to be carried in that setting.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

Over a commutative ring the homomorphism group ${Hom}_{R} (M, N)$ is an $R$ -module

Statement

Let $R$ be a commutative ring and let $M, N$ be $R$ -modules. Then the abelian group ${Hom}_{R} (M, N)$ of The abelian group ${Hom}_{R} (M, N)$ and maps induced by pre- and postcomposition becomes an $R$ -module under the pointwise scalar action

$(r f) (m) : = r f (m) (r \in R, f \in {Hom}_{R} (M, N), m \in M) .$

The underlying additive group is the published one, unchanged: this extends the abelian-group structure rather than replacing it. Commutativity of $R$ is used, and is used only to see that $r f$ is again $R$ -linear.

Facts & Assumptions

Given: A commutative ring $R$ and $R$ -modules $M, N$ .

[L1]

For left $R$ -modules $M, N$ , the set ${Hom}_{R} (M, N)$ of module homomorphisms is an abelian group under pointwise addition, with zero the zero homomorphism and inverse $(- f) (m) = - f (m)$ (The abelian group ${Hom}_{R} (M, N)$ and maps induced by pre- and postcomposition).

[L2]

A ring $R$ is commutative when its multiplication is commutative, $x y = y x$ for all $x, y \in R$ (Commutative ring).

[L3]

[L4]

Proof

technique · direct

1.1

The set ${Hom}_{R} (M, N)$ already carries the pointwise addition making it an abelian group, with the zero homomorphism as neutral element and $(- f) (m) = - f (m)$ as inverse. Nothing below alters that addition.

L1given

2.1

For $r \in R$ and $f \in {Hom}_{R} (M, N)$ the function $r f$ defined by $(r f) (m) = r f (m)$ is again an $R$ -module homomorphism. Additivity is pointwise: $(r f) (m + m^{'}) = r (f (m) + f (m^{'})) = r f (m) + r f (m^{'})$ . Homogeneity is where commutativity enters: for $s \in R$ , $(r f) (s m) = r f (s m) = r (s f (m)) = (r s) f (m) = (s r) f (m) = s (r f (m)) = s (r f) (m)$ .

L2L3L4step 1.1algebra

3.1

The module axioms hold pointwise, each being an identity in $N$ evaluated at an arbitrary $m \in M$ : $((r + r^{'}) f) (m) = (r + r^{'}) f (m) = r f (m) + r^{'} f (m)$ , $(r (f + g)) (m) = r (f (m) + g (m)) = r f (m) + r g (m)$ , $((r r^{'}) f) (m) = (r r^{'}) f (m) = r (r^{'} f (m))$ and $(1_{R} f) (m) = 1_{R} f (m) = f (m)$ .

L1L3step 2.1algebra

4.1

So ${Hom}_{R} (M, N)$ with the addition of step 1.1 and the action of step 2.1 satisfies the definition of an $R$ -module, and its additive group is the published abelian group unchanged.

step 1.1step 3.1∎

Remarks

Commutativity is not decoration here. Over a noncommutative $R$ the function $r f$ need not be $R$ -linear, since the computation in step 2.1 turns on $r s = s r$ . That is why The abelian group ${Hom}_{R} (M, N)$ and maps induced by pre- and postcomposition gives only an abelian group in general, and why the module structure is recorded separately rather than being read into the definition.
The induced maps are $R$ -linear for this structure. For $u ⁣ : M \to M^{'}$ and $v ⁣ : N \to N^{'}$ the maps $u^{*}$ and $v_{*}$ of The abelian group ${Hom}_{R} (M, N)$ and maps induced by pre- and postcomposition satisfy $u^{*} (r g) = r u^{*} (g)$ and $v_{*} (r f) = r v_{*} (f)$ , both by evaluating at a point, so nothing about the published functoriality has to be revisited.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

For a commutative ring, ${Hom}_{R} (R^{n}, N) ≅ N^{n}$

Statement

Let $R$ be a commutative ring, let $n \in N$ , let $R^{n}$ be the free $R$ -module on an $n$ -element set with standard basis $e_{1}, \dots, e_{n}$ (The free module on a set and its standard basis), and let $N$ be an $R$ -module. Write $N^{n}$ for the direct sum of $n$ copies of $N$ (The direct sum of an indexed family of modules), which for a finite index set is the coordinatewise product. Then

$Φ ⁣ : {Hom}_{R} (R^{n}, N) ⟶ N^{n}, Φ (f) = (f (e_{1}), \dots, f (e_{n})),$

is an isomorphism of $R$ -modules, the source carrying the module structure of Over a commutative ring the homomorphism group ${Hom}_{R} (M, N)$ is an $R$ -module.

At $n = 0$ both sides are the zero module.

Facts & Assumptions

Given: A commutative ring $R$ , a natural number $n$ , the free module $R^{n}$ with standard basis $e_{1}, \dots, e_{n}$ , and an $R$ -module $N$ .

[L1]

[L2]

In $R^{(X)}$ the standard basis vector $e_{x}$ has coordinate $1_{R}$ at $x$ and zero elsewhere, and every element has a unique expression $\sum_{x \in F} r_{x} e_{x}$ with $F \subseteq X$ finite; for $X = \emptyset$ the module is $0$ (The free module on a set and its standard basis).

[L3]

For a family $(M_{i})_{i \in I}$ of left $R$ -modules the direct sum is the submodule of the coordinatewise product consisting of the families of finite support; for $I = \emptyset$ both product and direct sum are the zero module (The direct sum of an indexed family of modules).

[L4]

For a commutative ring $R$ and $R$ -modules $M, N$ , the abelian group ${Hom}_{R} (M, N)$ is an $R$ -module under $(r f) (m) = r f (m)$ , with the published addition unchanged (Over a commutative ring the homomorphism group ${Hom}_{R} (M, N)$ is an $R$ -module).

Proof

technique · direct

1.1

$Φ$ is a bijection. It is injective: two homomorphisms $R^{n} \to N$ agreeing on $e_{1}, \dots, e_{n}$ are the unique extension of the same set map on the index set, hence equal. It is surjective: given $(y_{1}, \dots, y_{n}) \in N^{n}$ , the set map $i \mapsto y_{i}$ extends to a homomorphism $f ⁣ : R^{n} \to N$ with $f (e_{i}) = y_{i}$ , so $Φ (f) = (y_{1}, \dots, y_{n})$ .

L1L2given

2.1

$Φ$ is $R$ -linear. Addition in ${Hom}_{R} (R^{n}, N)$ and in $N^{n}$ is pointwise and coordinatewise respectively, so $Φ (f + g) = ((f + g) (e_{1}), \dots) = Φ (f) + Φ (g)$ ; and the scalar action on the source is pointwise, so $Φ (r f) = (r f (e_{1}), \dots, r f (e_{n})) = r Φ (f)$ .

L3L4step 1.1algebra

3.1

A bijective $R$ -module homomorphism is an isomorphism of $R$ -modules, so $Φ$ is one. At $n = 0$ the index set is empty: $R^{0} = 0$ , the only homomorphism $0 \to N$ is the zero map, and $N^{0}$ is the zero module, so both sides are zero and $Φ$ is the unique map between them.

L2L3step 1.1step 2.1∎

Remarks

The isomorphism depends on the chosen basis. A different ordered basis of $R^{n}$ gives a different $Φ$ ; what is canonical is that ${Hom}_{R} (R^{n}, N)$ is isomorphic to $N^{n}$ , not any particular isomorphism.
Finiteness of the index set is what makes the target a direct sum. For an infinite index set $X$ the same argument identifies ${Hom}_{R} (R^{(X)}, N)$ with the coordinatewise product of copies of $N$ , not with the direct sum, because a homomorphism may be nonzero on infinitely many basis vectors.

CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

Over a Noetherian ring the homomorphism module between two finitely generated modules is finitely generated

Statement

Let $R$ be a Noetherian commutative ring and let $M, N$ be finitely generated $R$ -modules. Then ${Hom}_{R} (M, N)$ , with the $R$ -module structure of Over a commutative ring the homomorphism group ${Hom}_{R} (M, N)$ is an $R$ -module, is a finitely generated $R$ -module.

The proof exhibits ${Hom}_{R} (M, N)$ as isomorphic to a submodule of $N^{n}$ for a suitable $n \in N$ . It does not assert that the embedding is onto, and in general it is not.

Facts & Assumptions

Given: A Noetherian commutative ring $R$ and finitely generated $R$ -modules $M$ and $N$ .

[L1]

A module is finitely generated when it equals $⟨ S ⟩_{R}$ for some finite subset $S$ (Generated submodule, cyclic and finitely generated modules, module basis and free module).

[L2]

Every set map $u ⁣ : X \to M$ extends uniquely to an $R$ -module homomorphism $\overset{ˉ}{u} ⁣ : R^{(X)} \to M$ with $\overset{ˉ}{u} (e_{x}) = u (x)$ (Universal property of the free module on a set).

[L3]

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 submodule generated by a subset consists of the finite $R$ -linear combinations of that subset).

[L4]

For a homomorphism $u ⁣ : M \to N$ and a module $X$ , precomposition gives a map $u^{*} ⁣ : {Hom}_{R} (N, X) \to {Hom}_{R} (M, X)$ , $g \mapsto g \circ u$ (The abelian group ${Hom}_{R} (M, N)$ and maps induced by pre- and postcomposition).

[L5]

For a commutative ring $R$ and $R$ -modules $M, N$ , the abelian group ${Hom}_{R} (M, N)$ is an $R$ -module under $(r f) (m) = r f (m)$ (Over a commutative ring the homomorphism group ${Hom}_{R} (M, N)$ is an $R$ -module).

[L6]

For a commutative ring $R$ , $n \in N$ and an $R$ -module $N$ , the map $f \mapsto (f (e_{1}), \dots, f (e_{n}))$ is an isomorphism of $R$ -modules ${Hom}_{R} (R^{n}, N) \to N^{n}$ (For a commutative ring, ${Hom}_{R} (R^{n}, N) ≅ N^{n}$ ).

[L7]

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

[L8]

A finite direct sum is Noetherian if and only if every summand is Noetherian (Finite direct sums preserve and reflect Noetherian and Artinian conditions).

[L9]

In $R^{(X)}$ the standard basis vector $e_{x}$ has coordinate $1_{R}$ at $x$ and zero elsewhere, and every element is uniquely a finite $R$ -linear combination of them (The free module on a set and its standard basis).

[L10]

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

Proof

technique · direct

1.1

Fix a finite generating set $m_{1}, \dots, m_{n}$ of $M$ , with $n \in N$ . The universal property of the free module gives $π ⁣ : R^{n} \to M$ with $π (e_{i}) = m_{i}$ , and its image is the set of finite sums $\sum_{i} r_{i} m_{i}$ , that is $⟨ m_{1}, \dots, m_{n} ⟩_{R} = M$ ; so $π$ is surjective.

L1L2L3L9given

2.1

Precomposition with $π$ gives $π^{*} ⁣ : {Hom}_{R} (M, N) \to {Hom}_{R} (R^{n}, N)$ , $g \mapsto g \circ π$ . It is $R$ -linear for the module structures above, since $(g + g^{'}) \circ π = g \circ π + g^{'} \circ π$ and $(r g) \circ π = r (g \circ π)$ , both by evaluating at a point of $R^{n}$ . It is injective: if $g \circ π = 0$ then $g$ vanishes on $im π = M$ , so $g = 0$ .

L4L5step 1.1

2.2

The module ${Hom}_{R} (R^{n}, N)$ is Noetherian. Indeed $N$ is finitely generated over the Noetherian ring $R$ , hence a Noetherian module; the finite direct sum $N^{n}$ of copies of $N$ is then Noetherian; and ${Hom}_{R} (R^{n}, N)$ is isomorphic to $N^{n}$ , while an isomorphism of modules carries submodules to submodules and finite generating sets to finite generating sets, so the isomorphic module is Noetherian too.

L6L7L8L10step 1.1

3.1

The image $π^{*} ({Hom}_{R} (M, N))$ is a submodule of the Noetherian module ${Hom}_{R} (R^{n}, N)$ , hence finitely generated; and $π^{*}$ is injective and $R$ -linear, so ${Hom}_{R} (M, N)$ is isomorphic to that image and is therefore finitely generated as well.

L10step 2.1step 2.2∎

Remarks

The injectivity of $π^{*}$ is an instance of left exactness. Covariant and contravariant $Hom$ are left exact gives it for any exact $A \to B \to C \to 0$ ; step 2.1 writes out the case needed here, which uses only that $π$ is surjective and so does not require assembling the exact sequence first.
No claim of surjectivity. A homomorphism $R^{n} \to N$ descends to $M$ exactly when it kills $\ker π$ , and most do not; the corollary needs only the embedding.
Both hypotheses of finite generation are used, and for different reasons. Finite generation of $M$ produces the free cover in step 1.1; finite generation of $N$ makes the target Noetherian in step 2.2.

LemmaStatement: Literature-sourcedProof: Literature-sourcedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

Module finiteness is transitive along a tower of algebras

Statement

Let $φ ⁣ : A \to B$ be a homomorphism of commutative rings, so that $B$ is an $A$ -algebra (Algebras over a commutative ring, central structure maps, and algebra homomorphisms) and every $B$ -module becomes an $A$ -module through $a \cdot z : = φ (a) z$ . Suppose $B$ is generated as an $A$ -module by $b_{1}, \dots, b_{m}$ with $m \in N$ , and let $M$ be a $B$ -module generated as a $B$ -module by $u_{1}, \dots, u_{n}$ with $n \in N$ . Then the $m n$ products $b_{i} u_{j}$ generate $M$ as an $A$ -module.

In particular, if $ψ ⁣ : B \to C$ is a homomorphism of commutative rings making $C$ module-finite over $B$ , and $B$ is module-finite over $A$ , then $C$ is module-finite over $A$ (Subalgebra generated by a subset, algebras of finite type, and module-finite algebras): the products $ψ (b_{i}) c_{j}$ of a finite $A$ -generating list of $B$ with a finite $B$ -generating list of $C$ generate $C$ over $A$ .

Facts & Assumptions

Given: Commutative rings $A$ and $B$ , a ring homomorphism $φ ⁣ : A \to B$ , a finite $A$ -module generating list $b_{1}, \dots, b_{m}$ of $B$ , a $B$ -module $M$ and a finite $B$ -module generating list $u_{1}, \dots, u_{n}$ of $M$ .

[L1]

An $R$ -algebra is a unital ring $A$ with a unital ring homomorphism $η_{A} ⁣ : R \to A$ of central image, and the induced scalar action $r a : = η_{A} (r) a$ makes $A$ an $R$ -module (Algebras over a commutative ring, central structure maps, and algebra homomorphisms).

[L2]

$A$ is module-finite over $R$ when it is finitely generated as an $R$ -module (Subalgebra generated by a subset, algebras of finite type, and module-finite algebras).

[L3]

A left $R$ -module is an abelian group with an action satisfying $r (m + n) = r m + r n$ , $(r + s) m = r m + s m$ , $(r s) m = r (s m)$ and $1_{R} m = m$ (Unital left and right modules over a ring; unqualified module means left module).

[L4]

Proof

technique · direct

1.1

The $A$ -action on $M$ is $a \cdot z = φ (a) z$ , computed in the $B$ -module $M$ ; it satisfies the module axioms because $φ$ is a ring homomorphism and $M$ is a $B$ -module. Each product $b_{i} u_{j}$ is an element of $M$ .

L1L3given

2.1

Let $z \in M$ . Since $u_{1}, \dots, u_{n}$ generate $M$ over $B$ , the finite-sum description gives $β_{1}, \dots, β_{n} \in B$ with $z = \sum_{j = 1}^{n} β_{j} u_{j}$ . Since $b_{1}, \dots, b_{m}$ generate $B$ over $A$ , the same description gives, for each $j$ , elements $α_{1 j}, \dots, α_{m j} \in A$ with $β_{j} = \sum_{i = 1}^{m} φ (α_{i j}) b_{i}$ .

L4step 1.1

3.1

Substituting and using the $B$ -module axioms, $z = \sum_{j = 1}^{n} (\sum_{i = 1}^{m} φ (α_{i j}) b_{i}) u_{j} = \sum_{i = 1}^{m} \sum_{j = 1}^{n} φ (α_{i j}) (b_{i} u_{j}) = \sum_{i, j} α_{i j} \cdot (b_{i} u_{j})$ , an $A$ -linear combination of the $m n$ products. As $z$ was arbitrary, those products generate $M$ as an $A$ -module. Taking $M = C$ with its $B$ -module structure gives the transitivity statement.

L2L3L4step 2.1algebra∎

Remarks

Both degenerate cases collapse rather than fail. If $m = 0$ then $B$ is generated over $A$ by the empty list, so $B = 0$ and $1_{B} = 0_{B}$ ; every $B$ -module then satisfies $z = 1_{B} z = 0$ , so $M = 0$ and the empty set of products generates it. If $n = 0$ then $M = 0$ directly. The count $m n$ is $0$ in both cases, which is the correct answer.
The products need not be distinct or independent. The list $b_{i} u_{j}$ may repeat entries and may be far from minimal; only the finiteness of the count is used.

TheoremStatement: Literature-sourcedProof: Literature-sourcedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

A module-finite algebra over a Noetherian ring is a Noetherian ring, and so is every ring between the two

Statement

Let $A$ be a Noetherian commutative ring and let $B$ be a commutative $A$ -algebra that is module-finite over $A$ (Subalgebra generated by a subset, algebras of finite type, and module-finite algebras). Then:

$B$ is a Noetherian ring;
if in addition the $A$ -algebra structure map is the inclusion of $A$ as a subring of $B$ (Subring: a subset containing $1_{R}$ and closed under addition, additive inverses and multiplication), then every subring $C$ of $B$ with $A \subseteq C \subseteq B$ is module-finite over $A$ and is a Noetherian ring;
every finitely generated $B$ -module is finitely generated as an $A$ -module, and hence is a Noetherian $A$ -module.

Facts & Assumptions

Given: A Noetherian commutative ring $A$ , a commutative $A$ -algebra $B$ with structure map $η_{B} ⁣ : A \to B$ that is module-finite over $A$ , and, for the second clause, the additional hypotheses that $η_{B}$ is the inclusion of $A$ as a subring of $B$ and that $C$ is a subring of $B$ with $A \subseteq C \subseteq B$ (Subring: a subset containing $1_{R}$ and closed under addition, additive inverses and multiplication).

[L1]

$B$ is module-finite over $A$ when it is finitely generated as an $A$ -module (Subalgebra generated by a subset, algebras of finite type, and module-finite algebras).

[L2]

An $A$ -algebra is a unital ring $B$ with a unital ring homomorphism $η_{B} ⁣ : A \to B$ of central image, and the induced scalar action $a b : = η_{B} (a) b$ makes $B$ an $A$ -module (Algebras over a commutative ring, central structure maps, and algebra homomorphisms).

[L3]

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

[L4]

In a commutative ring, $(S)$ consists of finite sums $\sum r_{i} s_{i}$ , and $(a) = R a$ (In a commutative ring, $(S)$ consists of finite sums $\sum r_{i} s_{i}$ , and $(a) = R a$ ).

[L5]

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

[L6]

[L7]

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

[L8]

If $B$ is generated as an $A$ -module by $b_{1}, \dots, b_{m}$ and a $B$ -module $M$ is generated as a $B$ -module by $u_{1}, \dots, u_{n}$ , then the $m n$ products $b_{i} u_{j}$ generate $M$ as an $A$ -module (Module finiteness is transitive along a tower of algebras).

Proof

technique · direct

1.1

Let $B^{'}$ be any commutative $A$ -algebra that is module-finite over $A$ . Then $B^{'}$ is a finitely generated module over the Noetherian ring $A$ , hence a Noetherian $A$ -module: every $A$ -submodule of $B^{'}$ is finitely generated over $A$ . This applies in particular to $B^{'} = B$ .

L1L2L3L5given

2.1

First clause. Let $b$ be an ideal of $B^{'}$ . It is an additive subgroup of $B^{'}$ closed under the $A$ -action, since $a \cdot x = η_{B^{'}} (a) x \in b$ for $x \in b$ , so it is an $A$ -submodule and therefore generated over $A$ by finitely many $x_{1}, \dots, x_{k} \in b$ . Every element of $b$ is then $\sum_{i} η_{B^{'}} (a_{i}) x_{i}$ , which lies in the ideal $(x_{1}, \dots, x_{k})$ of $B^{'}$ ; and that ideal is contained in $b$ because each $x_{i}$ lies in $b$ . So $b = (x_{1}, \dots, x_{k})$ is finitely generated as an ideal, and $B^{'}$ is a Noetherian ring. Taking $B^{'} = B$ gives the first clause.

L4L5L6L7step 1.1

2.2

Second clause, first half. Suppose the $A$ -algebra structure map is the inclusion $A \subseteq B$ and $C$ is a subring of $B$ with $A \subseteq C \subseteq B$ . Then $C$ is an additive subgroup of $B$ closed under the $A$ -action, because $a \cdot x = a x$ is a product of two elements of $C$ ; so $C$ is an $A$ -submodule of the Noetherian $A$ -module $B$ and is therefore finitely generated as an $A$ -module, that is, module-finite over $A$ .

L5L7step 1.1given

2.3

Third clause. Let $M$ be a finitely generated $B$ -module, say generated over $B$ by $u_{1}, \dots, u_{n}$ , and let $b_{1}, \dots, b_{m}$ generate $B$ over $A$ . The $m n$ products $b_{i} u_{j}$ generate $M$ as an $A$ -module, so $M$ is a finitely generated $A$ -module and hence a Noetherian $A$ -module.

L1L3L8step 1.1

3.1

Step 2.2 makes $C$ a commutative $A$ -algebra, through the inclusion $A \subseteq C$ , that is module-finite over $A$ ; steps 1.1 and 2.1 were proved for an arbitrary such algebra, so applying them with $B^{'} = C$ makes $C$ a Noetherian ring. With steps 2.1 and 2.3 this establishes all three clauses.

step 2.1step 2.2step 2.3∎

Remarks

The intermediate-ring clause is the one that gets used. It says nothing about $C$ being of finite type over $A$ as an algebra, only that it is module-finite, which is stronger; the Artin–Tate lemma is what handles the situation where only $C$ sits between $A$ and a finite-type algebra without $B$ being module-finite over $A$ .
Module-finite is strictly stronger than finite type here. A finite-type algebra over a Noetherian ring is Noetherian as well, but its ideals need not be finitely generated over the base ring, and the argument of step 2.1 would not run.
The hypothesis that $A$ is a subring is used only in the second clause. Clauses 1 and 3 need no injectivity of $η_{B}$ .

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

A subalgebra generated by finitely many integral elements is module-finite

Statement

Let $A \subseteq B$ be commutative rings, $A$ a subring of $B$ (Subring: a subset containing $1_{R}$ and closed under addition, additive inverses and multiplication), let $n \in N$ and let $b_{1}, \dots, b_{n} \in B$ be integral over $A$ (Integral elements over a commutative ring and algebraic integers). Then the $A$ -subalgebra $A [b_{1}, \dots, b_{n}]$ of $B$ is module-finite over $A$ (Subalgebra generated by a subset, algebras of finite type, and module-finite algebras).

The zero ring is not excluded: if $A = 0$ then $B = 0$ and the conclusion holds with the empty generating list.

Facts & Assumptions

Given: Commutative rings $A \subseteq B$ with $A$ a subring of $B$ , a natural number $n$ , and elements $b_{1}, \dots, b_{n} \in B$ integral over $A$ .

[L1]

A subring of $B$ contains $1_{B}$ and has the same zero and identity as $B$ (Subring: a subset containing $1_{R}$ and closed under addition, additive inverses and multiplication).

[L2]

$R [a_{1}, \dots, a_{n}]$ is the image of the unital ring homomorphism $R [x_{1}, \dots, x_{n}] \to A$ agreeing with the structure map on constants and sending $x_{i}$ to $a_{i}$ ; it is the smallest subring of $A$ containing the image of $R$ and $a_{1}, \dots, a_{n}$ . An algebra is module-finite over $R$ when it is finitely generated as an $R$ -module (Subalgebra generated by a subset, algebras of finite type, and module-finite algebras).

[L3]

For a homomorphism of commutative rings $A \to B$ , an element $b \in B$ is integral over $A$ when it is a root of a monic polynomial in $A [X]$ (Integral elements over a commutative ring and algebraic integers).

[L4]

Let $A \subseteq B$ be commutative rings with $A \neq 0$ and let $b \in B$ . Then $b$ is integral over $A$ if and only if $A [b]$ is finitely generated as an $A$ -module (Integrality and finite-module characterizations for one element).

[L5]

If $B$ is generated as an $A$ -module by finitely many elements and a $B$ -module $M$ is generated as a $B$ -module by finitely many elements, then the products of the two lists generate $M$ as an $A$ -module; module finiteness is transitive along a tower of commutative algebras (Module finiteness is transitive along a tower of algebras).

Proof

technique · induction

1.1

Dispose of the zero base ring first. If $A = 0$ then $1_{A} = 0_{A}$ , and a subring shares the identity and zero of the ambient ring, so $1_{B} = 0_{B}$ and $B = 0$ ; every subalgebra of $B$ is then the zero module over $A$ , generated by the empty list. For the rest of the argument assume $A \neq 0$ , so that $1_{B} \neq 0_{B}$ and every subring of $B$ containing $A$ is a nonzero commutative ring.

L1L2given

1.2

The case $n = 0$ : the subalgebra $A []$ generated by the empty list is the smallest subring of $B$ containing $A$ , which is $A$ itself, and $A$ is generated as an $A$ -module by $1_{A}$ .

baseL2given

1.3

Let $k \in N$ and suppose $A_{k} : = A [b_{1}, \dots, b_{k}]$ is module-finite over $A$ .

2.1

Write $A_{k + 1} : = A [b_{1}, \dots, b_{k + 1}]$ . Both $A_{k + 1}$ and $A_{k} [b_{k + 1}]$ are the smallest subring of $B$ containing $A$ and $b_{1}, \dots, b_{k + 1}$ , so they are equal. The element $b_{k + 1}$ is a root of some monic $f \in A [X]$ ; the coefficients of $f$ lie in $A \subseteq A_{k}$ , so $f$ is a monic polynomial in $A_{k} [X]$ with the same coefficients, and evaluating it at $b_{k + 1}$ gives the same element of $B$ , namely $0$ . Hence $b_{k + 1}$ is integral over $A_{k}$ . Since $A_{k}$ is a nonzero commutative subring of $B$ by step 1.1, the integrality criterion gives that $A_{k} [b_{k + 1}] = A_{k + 1}$ is a finitely generated $A_{k}$ -module; with the assumption of step 1.3 that $A_{k}$ is a finitely generated $A$ -module, transitivity makes $A_{k + 1}$ a finitely generated $A$ -module.

L2L3L4L5step 1.1step 1.3

3.1

The base case of step 1.2 and the passage of step 2.1 give, by induction on $n$ , that $A [b_{1}, \dots, b_{n}]$ is module-finite over $A$ for every $n \in N$ .

step 1.2step 2.1discharge-induction∎

Remarks

The nonzero hypothesis of the cited integrality theorem is why the zero ring is disposed of first. Integrality and finite-module characterizations for one element assumes $A \neq 0$ ; step 1.1 removes that case by hand rather than leaving the citation standing over a ring the source excludes.
Integrality over the enlarged ring is inherited, not re-proved. The same monic polynomial serves at every stage, which is what keeps the induction from needing a new integrality hypothesis at each step.
Finitely many elements is essential. The subalgebra generated by an infinite set of integral elements is integral over $A$ but need not be module-finite; each finite subfamily is, and the union of an increasing chain of finite modules need not be finite.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

The Artin–Tate coefficient subalgebra is a Noetherian algebra of finite type

Statement

$C = A [t_{1}, \dots, t_{r}]$ is of finite type over $A$ , with $r \in N$ and $t_{1}, \dots, t_{r} \in C$ (Subalgebra generated by a subset, algebras of finite type, and module-finite algebras);
$y_{1}, \dots, y_{n} \in C$ generate $C$ as a $B$ -module, with $n \geq 1$ and $y_{1} = 1_{C}$ .

Choose elements $z_{i j} \in B$ for $1 \leq i \leq r$ and $1 \leq j \leq n$ , and $z_{i j k} \in B$ for $1 \leq i, j, k \leq n$ , such that

$t_{i} = \sum_{j = 1}^{n} z_{i j} y_{j}, y_{i} y_{j} = \sum_{k = 1}^{n} z_{i j k} y_{k} .$

Let $A^{'}$ be the $A$ -subalgebra of $B$ generated by all the $z_{i j}$ and all the $z_{i j k}$ . Then $A \subseteq A^{'} \subseteq B$ , the algebra $A^{'}$ is of finite type over $A$ , and $A^{'}$ is a Noetherian ring.

Such coefficients exist, because the $y_{j}$ generate $C$ as a $B$ -module and $t_{i}$ and $y_{i} y_{j}$ are elements of $C$ . They are chosen, not canonical: only their existence and their finite number are used, and a different choice gives a possibly different $A^{'}$ with the same properties.

Facts & Assumptions

Given: Commutative rings $A \subseteq B \subseteq C$ , each a subring of the next, with $A$ Noetherian; a finite list $t_{1}, \dots, t_{r}$ with $C = A [t_{1}, \dots, t_{r}]$ ; a list $y_{1}, \dots, y_{n}$ generating $C$ as a $B$ -module with $y_{1} = 1_{C}$ ; and coefficients $z_{i j}, z_{i j k} \in B$ as displayed.

[L1]

$R [a_{1}, \dots, a_{n}]$ is the image of the unital ring homomorphism $R [x_{1}, \dots, x_{n}] \to A$ agreeing with the structure map on constants and sending $x_{i}$ to $a_{i}$ , and it is the smallest subring of $A$ containing the image of $R$ and $a_{1}, \dots, a_{n}$ ; an algebra is of finite type over $R$ when it equals $R [a_{1}, \dots, a_{n}]$ for some finite list (Subalgebra generated by a subset, algebras of finite type, and module-finite algebras).

[L2]

Every commutative algebra of finite type over a Noetherian commutative ring is a Noetherian ring (Every algebra of finite type over a Noetherian ring is a Noetherian ring).

[L3]

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 submodule generated by a subset consists of the finite $R$ -linear combinations of that subset).

[L4]

A subring contains the identity of the ambient ring and shares its zero, identity and additive inverses (Subring: a subset containing $1_{R}$ and closed under addition, additive inverses and multiplication).

Proof

technique · direct

1.1

The coefficients exist. Each $t_{i}$ and each product $y_{i} y_{j}$ is an element of $C$ , and $C$ is generated as a $B$ -module by $y_{1}, \dots, y_{n}$ , so by the finite-sum description each of them is a $B$ -linear combination of $y_{1}, \dots, y_{n}$ ; fix one such expression for each, which is a selection over a finite index set.

L3L4given

2.1

The elements $z_{i j}$ and $z_{i j k}$ form a finite list of elements of $B$ , indexed by the finitely many pairs $(i, j)$ with $i \leq r$ , $j \leq n$ and the finitely many triples $(i, j, k)$ with $i, j, k \leq n$ . Let $A^{'}$ be the $A$ -subalgebra of $B$ they generate, that is, the smallest subring of $B$ containing $A$ and all of them. Then $A \subseteq A^{'} \subseteq B$ , and $A^{'}$ is of finite type over $A$ by definition, being generated as an $A$ -algebra by a finite list.

L1L4step 1.1

3.1

The base ring $A$ is Noetherian and $A^{'}$ is a commutative $A$ -algebra of finite type, so $A^{'}$ is a Noetherian ring.

L2step 2.1∎

Remarks

Nothing here uses that $B$ is Noetherian, and nothing may. Whether $B$ is Noetherian is not among the hypotheses of the Artin–Tate lemma; the point of passing to $A^{'}$ is to obtain a Noetherian ring inside $B$ without assuming one.
The normalisation $y_{1} = 1_{C}$ costs nothing. A finite $B$ -module generating list of $C$ stays finite and generating when $1_{C}$ is adjoined to it, so it may always be arranged; it is used where the relations are read back, not here.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

In the Artin–Tate setup the intermediate ring is module-finite over the coefficient subalgebra

Statement

Keep the data and the notation of The Artin–Tate coefficient subalgebra is a Noetherian algebra of finite type: commutative rings $A \subseteq B \subseteq C$ , each a subring of the next, with $A$ Noetherian, $C = A [t_{1}, \dots, t_{r}]$ , a $B$ -module generating list $y_{1}, \dots, y_{n}$ of $C$ with $y_{1} = 1_{C}$ , coefficients $z_{i j}, z_{i j k} \in B$ as displayed there, and $A^{'}$ the $A$ -subalgebra of $B$ they generate.

Then $C$ is generated as an $A^{'}$ -module by $y_{1}, \dots, y_{n}$ , so $C$ is module-finite over $A^{'}$ ; and $B$ is module-finite over $A^{'}$ as well (Subalgebra generated by a subset, algebras of finite type, and module-finite algebras).

Facts & Assumptions

Given: The data of The Artin–Tate coefficient subalgebra is a Noetherian algebra of finite type, and the set $M : = \sum_{j = 1}^{n} A^{'} y_{j} = {\sum_{j = 1}^{n} a_{j} y_{j} : a_{j} \in A^{'}} \subseteq C$ .

[L1]

In that setup $A \subseteq A^{'} \subseteq B$ , the algebra $A^{'}$ is of finite type over $A$ , and $A^{'}$ is a Noetherian ring (The Artin–Tate coefficient subalgebra is a Noetherian algebra of finite type).

[L2]

$R [a_{1}, \dots, a_{n}]$ is the smallest subring of $A$ containing the image of $R$ and $a_{1}, \dots, a_{n}$ ; an algebra is module-finite over $R$ when it is finitely generated as an $R$ -module (Subalgebra generated by a subset, algebras of finite type, and module-finite algebras).

[L3]

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 submodule generated by a subset consists of the finite $R$ -linear combinations of that subset).

[L4]

Let $A$ be a Noetherian commutative ring and $B$ a commutative $A$ -algebra module-finite over $A$ with $A$ a subring of $B$ ; then every ring $C$ with $A \subseteq C \subseteq B$ is module-finite over $A$ and is a Noetherian ring (A module-finite algebra over a Noetherian ring is a Noetherian ring, and so is every ring between the two).

[L5]

A subring contains the identity of the ambient ring and is closed under sums, additive inverses and products (Subring: a subset containing $1_{R}$ and closed under addition, additive inverses and multiplication).

Proof

technique · direct

1.1

The set $M = \sum_{j = 1}^{n} A^{'} y_{j}$ is the $A^{'}$ -submodule of $C$ generated by $y_{1}, \dots, y_{n}$ , by the finite-sum description of a generated submodule; it is in particular an additive subgroup of $C$ closed under multiplication by elements of $A^{'}$ . Since $y_{1} = 1_{C}$ we have $A^{'} = A^{'} y_{1} \subseteq M$ , and therefore $A \subseteq M$ and $1_{C} \in M$ .

L1L3L5given

2.1

$M$ is closed under multiplication, hence is a subring of $C$ . It suffices to multiply two generators with coefficients: for $a, a^{'} \in A^{'}$ , $(a y_{i}) (a^{'} y_{j}) = a a^{'} (y_{i} y_{j}) = a a^{'} \sum_{k = 1}^{n} z_{i j k} y_{k} = \sum_{k = 1}^{n} (a a^{'} z_{i j k}) y_{k}$ , and each $a a^{'} z_{i j k}$ lies in $A^{'}$ because $A^{'}$ is a subring of $B$ containing every $z_{i j k}$ . A product of two elements of $M$ expands by distributivity into a finite sum of such terms, so it lies in $M$ ; with the additive subgroup property and $1_{C} \in M$ from step 1.1, $M$ is a subring of $C$ .

L1L3L5step 1.1

3.1

$M$ contains each $t_{i}$ , since $t_{i} = \sum_{j = 1}^{n} z_{i j} y_{j}$ and every $z_{i j}$ lies in $A^{'}$ . So $M$ is a subring of $C$ containing $A$ and $t_{1}, \dots, t_{r}$ . But $C = A [t_{1}, \dots, t_{r}]$ is the smallest such subring, so $C \subseteq M$ ; and $M \subseteq C$ by construction. Hence $C = M$ , and $y_{1}, \dots, y_{n}$ generate $C$ as an $A^{'}$ -module, so $C$ is module-finite over $A^{'}$ .

L2L3step 1.1step 2.1

4.1

Now $A^{'}$ is a Noetherian commutative ring, $A^{'}$ is a subring of $C$ , and $C$ is module-finite over $A^{'}$ ; the intermediate-ring clause applied to $A^{'} \subseteq B \subseteq C$ gives that $B$ is module-finite over $A^{'}$ .

L1L4step 3.1∎

Remarks

No induction on total degree is needed. Showing directly that $\sum_{j} A^{'} y_{j}$ is a subring containing $A$ and the algebra generators, and then invoking minimality of $A [t_{1}, \dots, t_{r}]$ , replaces the degreewise reduction of a polynomial expression; the multiplication table $y_{i} y_{j} = \sum_{k} z_{i j k} y_{k}$ is exactly what makes the closure argument work in one line.
$y_{1} = 1_{C}$ is used, and used here. It is what puts $A^{'}$ itself inside $M$ , without which $M$ need not contain $A$ and the minimality argument would not apply.
Only $A^{'}$ is known to be Noetherian. The final step is applied with $A^{'}$ as the Noetherian base, never with $B$ , which is exactly the point of the Artin–Tate argument.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

Artin–Tate lemma: an intermediate ring over which a finite-type algebra is module-finite is itself of finite type

Statement

Let $A \subseteq B \subseteq C$ be commutative rings, each a subring of the next (Subring: a subset containing $1_{R}$ and closed under addition, additive inverses and multiplication), with $A$ Noetherian. Suppose $C$ is of finite type over $A$ and module-finite over $B$ (Subalgebra generated by a subset, algebras of finite type, and module-finite algebras). Then $B$ is of finite type over $A$ .

No Noetherian hypothesis is placed on $B$ ; that is what the argument has to do without, and it is why a Noetherian subring of $B$ is manufactured on the way.

Facts & Assumptions

Given: Commutative rings $A \subseteq B \subseteq C$ , each a subring of the next, with $A$ Noetherian, $C$ of finite type over $A$ and $C$ module-finite over $B$ .

[L1]

$R [a_{1}, \dots, a_{n}]$ is the smallest subring of $A$ containing the image of $R$ and $a_{1}, \dots, a_{n}$ ; an algebra is of finite type over $R$ when it equals $R [a_{1}, \dots, a_{n}]$ for some finite list, and module-finite over $R$ when it is finitely generated as an $R$ -module (Subalgebra generated by a subset, algebras of finite type, and module-finite algebras).

[L2]

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 submodule generated by a subset consists of the finite $R$ -linear combinations of that subset).

[L3]

With $A \subseteq B \subseteq C$ as above, $C = A [t_{1}, \dots, t_{r}]$ , a $B$ -module generating list $y_{1}, \dots, y_{n}$ of $C$ with $y_{1} = 1_{C}$ and coefficients $z_{i j}, z_{i j k} \in B$ satisfying $t_{i} = \sum_{j} z_{i j} y_{j}$ and $y_{i} y_{j} = \sum_{k} z_{i j k} y_{k}$ : the $A$ -subalgebra $A^{'} \subseteq B$ generated by those coefficients satisfies $A \subseteq A^{'} \subseteq B$ , is of finite type over $A$ , and is a Noetherian ring (The Artin–Tate coefficient subalgebra is a Noetherian algebra of finite type).

[L4]

In that same setup $C$ is module-finite over $A^{'}$ , and $B$ is module-finite over $A^{'}$ (In the Artin–Tate setup the intermediate ring is module-finite over the coefficient subalgebra).

[L5]

Proof

technique · direct

1.1

Fix $r \in N$ and $t_{1}, \dots, t_{r} \in C$ with $C = A [t_{1}, \dots, t_{r}]$ , available because $C$ is of finite type over $A$ . Fix a finite list generating $C$ as a $B$ -module and adjoin $1_{C}$ to it; the result $y_{1}, \dots, y_{n}$ is still finite and still generates, and it may be indexed so that $n \geq 1$ and $y_{1} = 1_{C}$ .

L1L2given

2.1

Each $t_{i}$ and each product $y_{i} y_{j}$ lies in $C$ , so each is a $B$ -linear combination of $y_{1}, \dots, y_{n}$ ; fix coefficients $z_{i j}, z_{i j k} \in B$ with $t_{i} = \sum_{j} z_{i j} y_{j}$ and $y_{i} y_{j} = \sum_{k} z_{i j k} y_{k}$ , and let $A^{'}$ be the $A$ -subalgebra of $B$ they generate. Then $A \subseteq A^{'} \subseteq B$ , $A^{'}$ is of finite type over $A$ , and $A^{'}$ is a Noetherian ring.

L2L3step 1.1

3.1

In this setup $B$ is module-finite over $A^{'}$ : fix $s \in N$ and $w_{1}, \dots, w_{s} \in B$ generating $B$ as an $A^{'}$ -module, so that every element of $B$ is $\sum_{l = 1}^{s} a_{l} w_{l}$ with $a_{l} \in A^{'}$ .

L2L4step 2.1

4.1

Let $D$ be the smallest subring of $B$ containing $A$ , all the coefficients $z_{i j}$ and $z_{i j k}$ , and $w_{1}, \dots, w_{s}$ ; that is, $D$ is the $A$ -subalgebra of $B$ generated by that finite list. Then $D$ contains $A$ and the coefficients, hence contains the smallest subring of $B$ containing them, which is $A^{'}$ ; and $D$ contains each $w_{l}$ and is closed under products and sums, so $D$ contains every $\sum_{l} a_{l} w_{l}$ with $a_{l} \in A^{'}$ , that is all of $B$ . Since also $D \subseteq B$ , we get $B = D$ , so $B$ is generated as an $A$ -algebra by a finite list and is of finite type over $A$ .

L1L2L5step 2.1step 3.1∎

Remarks

The Noetherian ring in play is $A^{'}$ , never $B$ . The hypothesis is that $A$ is Noetherian, and $A^{'}$ inherits that as a finite-type algebra over $A$ ; the conclusion about $B$ is then read off from $B$ being a finite $A^{'}$ -module.
The generating list of $B$ is explicit. It consists of the structure coefficients $z_{i j}$ and $z_{i j k}$ together with the $A^{'}$ -module generators $w_{l}$ of $B$ , all of which depend on the choices made in steps 1.1 to 3.1. No minimality is claimed.

CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

The Artin–Tate lemma with integrality in place of module finiteness

Statement

Let $A \subseteq B \subseteq C$ be commutative rings, each a subring of the next (Subring: a subset containing $1_{R}$ and closed under addition, additive inverses and multiplication), with $A$ Noetherian. Suppose $C$ is of finite type over $A$ (Subalgebra generated by a subset, algebras of finite type, and module-finite algebras) and every element of $C$ is integral over $B$ (Integral elements over a commutative ring and algebraic integers). Then $C$ is module-finite over $B$ , and $B$ is of finite type over $A$ .

Facts & Assumptions

Given: Commutative rings $A \subseteq B \subseteq C$ , each a subring of the next, with $A$ Noetherian, $C$ of finite type over $A$ , and every element of $C$ integral over $B$ .

[L1]

[L2]

[L3]

For commutative rings $A \subseteq B$ with $A$ a subring of $B$ and $b_{1}, \dots, b_{n} \in B$ integral over $A$ , the subalgebra $A [b_{1}, \dots, b_{n}]$ is module-finite over $A$ (A subalgebra generated by finitely many integral elements is module-finite).

[L4]

For commutative rings $A \subseteq B \subseteq C$ , each a subring of the next, with $A$ Noetherian, $C$ of finite type over $A$ and $C$ module-finite over $B$ , the ring $B$ is of finite type over $A$ (Artin–Tate lemma: an intermediate ring over which a finite-type algebra is module-finite is itself of finite type).

[L5]

Proof

technique · direct

1.1

Fix $r \in N$ and $t_{1}, \dots, t_{r} \in C$ with $C = A [t_{1}, \dots, t_{r}]$ . The subring $B [t_{1}, \dots, t_{r}]$ of $C$ contains $B$ , hence contains $A$ , and contains every $t_{i}$ ; since $A [t_{1}, \dots, t_{r}]$ is the smallest subring of $C$ with those two properties, $C = A [t_{1}, \dots, t_{r}] \subseteq B [t_{1}, \dots, t_{r}] \subseteq C$ . So $C = B [t_{1}, \dots, t_{r}]$ is of finite type over $B$ , generated by the same elements.

L1L5given

2.1

Every element of $C$ is integral over $B$ , so in particular each $t_{i}$ is; $B$ is a subring of $C$ ; hence $B [t_{1}, \dots, t_{r}]$ is module-finite over $B$ . By step 1.1 that ring is $C$ , so $C$ is module-finite over $B$ .

L2L3step 1.1

3.1

The hypotheses of the Artin–Tate lemma are now all in place for $A \subseteq B \subseteq C$ : $A$ is Noetherian, $C$ is of finite type over $A$ , and $C$ is module-finite over $B$ . Therefore $B$ is of finite type over $A$ .

L4step 2.1∎

Remarks

Finite type over $A$ gives finite type over $B$ for free. Step 1.1 uses only that $A \subseteq B$ : the same algebra generators work over the larger base ring. That is what lets the integrality hypothesis be applied to the finitely many $t_{i}$ rather than to all of $C$ at once.
Integrality of every element is more than integrality of the generators, and only the generators are used. The hypothesis as stated is the usual one and is what the applications supply; the proof needs it only at $t_{1}, \dots, t_{r}$ .

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

A group acting on a ring by automorphisms and its invariant subring

Definition

Let $G$ be a group (Group and abelian group) and $C$ a commutative ring. An action of $G$ on $C$ by ring automorphisms is a left action $G \times C \to C$ (Left group actions, transitive actions, and faithful actions), written $(g, c) \mapsto g \cdot c$ , such that for every $g \in G$ the map $c \mapsto g \cdot c$ is a ring homomorphism (Ring homomorphism: additive, multiplicative, and required to send $1$ to $1$ ). Each such map is then automatically bijective, with inverse the map given by $g^{- 1}$ , since $g^{- 1} \cdot (g \cdot c) = (g^{- 1} g) \cdot c = c$ and likewise in the other order; so each $g$ acts as a ring automorphism, and in particular $g \cdot 1_{C} = 1_{C}$ and $g \cdot 0_{C} = 0_{C}$ .

The invariant subring is

$C^{G} : = {c \in C : g \cdot c = c for every g \in G} .$

It is a subring of $C$ (Subring: a subset containing $1_{R}$ and closed under addition, additive inverses and multiplication). It contains $1_{C}$ , because every $g$ acts as a unital ring homomorphism; and for $c, c^{'} \in C^{G}$ and $g \in G$ one has $g \cdot (c + c^{'}) = g \cdot c + g \cdot c^{'} = c + c^{'}$ , $g \cdot (- c) = - (g \cdot c) = - c$ and $g \cdot (c c^{'}) = (g \cdot c) (g \cdot c^{'}) = c c^{'}$ , so $C^{G}$ satisfies (T1) to (T4).

Actions by algebra automorphisms. When $C$ is a commutative $A$ -algebra with structure map $η_{C} ⁣ : A \to C$ (Algebras over a commutative ring, central structure maps, and algebra homomorphisms), an action by $A$ -algebra automorphisms is one in which every $g$ fixes $η_{C} (A)$ pointwise: $g \cdot η_{C} (a) = η_{C} (a)$ for all $g \in G$ and $a \in A$ . Then $η_{C} (A) \subseteq C^{G}$ , so $C^{G}$ is an $A$ -subalgebra of $C$ and $A \subseteq C^{G} \subseteq C$ whenever $A$ is a subring of $C$ .

The trivial group. If $G = {e}$ then $e \cdot c = c$ for every $c$ , so $C^{G} = C$ .

Remarks

This agrees with the notation already in use for symmetric polynomials, and does not compete with it. Symmetric polynomials as the invariants of variable permutations lets $σ \in {Sym}_{n}$ act on $R [x_{1}, \dots, x_{n}]$ by $σ \cdot f (x_{1}, \dots, x_{n}) = f (x_{σ (1)}, \dots, x_{σ (n)})$ and writes $R [x_{1}, \dots, x_{n}]^{{Sym}_{n}}$ for the fixed subset. That is exactly $C^{G}$ for $C = R [x_{1}, \dots, x_{n}]$ and $G = {Sym}_{n}$ : the same set, under the same notation, and the definition here is the general form of it.
Only invariance is asked for, not any finiteness. $G$ may be infinite, and $C^{G}$ may then be small; the finiteness of $G$ is a hypothesis of the results about $C^{G}$ , not part of this definition.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

For a finite group of ring automorphisms the orbit polynomial is monic over the invariant subring, so the ring is integral over its invariants

Statement

Let $G$ be a finite group acting by ring automorphisms on a nonzero commutative ring $C$ , and let $C^{G}$ be the invariant subring (A group acting on a ring by automorphisms and its invariant subring). Extend the action to the polynomial ring $C [T]$ coefficientwise, fixing $T$ . For $x \in C$ put

$P_{x} (T) : = \prod_{g \in G} (T - g \cdot x) \in C [T] .$

Then $P_{x}$ is monic of degree $∣ G ∣$ , every coefficient of $P_{x}$ lies in $C^{G}$ , and $P_{x} (x) = 0$ . Consequently every element of $C$ is integral over $C^{G}$ (Integral elements over a commutative ring and algebraic integers).

Finiteness of $G$ is a hypothesis, not a convenience: the product is over the index set $G$ and is a polynomial only because that set is finite. The hypothesis $C \neq 0$ is what makes $1_{C} \neq 0_{C}$ , so that a monic polynomial exists at all.

Facts & Assumptions

Given: A finite group $G$ acting by ring automorphisms on a nonzero commutative ring $C$ , an element $x \in C$ , and the polynomial ring $C [T]$ .

[L1]

For an action of a group $G$ on a commutative ring $C$ by ring automorphisms, $C^{G} = {c \in C : g \cdot c = c$ for every $g \in G}$ is a subring of $C$ , and each $g$ acts as a ring automorphism, so $g \cdot 1_{C} = 1_{C}$ (A group acting on a ring by automorphisms and its invariant subring).

[L2]

In a group every element $h$ has a two-sided inverse $h^{- 1}$ , and multiplication is associative (Group and abelian group).

[L3]

A left action satisfies $e \cdot c = c$ and $(g h) \cdot c = g \cdot (h \cdot c)$ for all $g, h \in G$ and $c \in C$ (Left group actions, transitive actions, and faithful actions).

[L4]

$C [T]$ is the set of finitely supported functions $N \to C$ with $(a + b)_{i} = a_{i} + b_{i}$ and $(a b)_{i} = \sum_{j + k = i} a_{j} b_{k}$ ; the constant $c$ is supported at $0$ and $T$ has coefficient $1_{C}$ at index $1$ (The polynomial ring over a commutative ring as finitely supported coefficient sequences with convolution).

[L5]

For $0 \neq f \in C [T]$ the degree is the largest index carrying a nonzero coefficient, the leading coefficient is the coefficient there, and $f$ is monic when that coefficient is $1_{C}$ (Degree, leading coefficient and monic polynomial, with the zero polynomial having no degree).

[L6]

For nonzero $f, g \in C [T]$ : the coefficient of $T^{\deg f + \deg g}$ in $f g$ is $lc (f) lc (g)$ , and if $f g \neq 0$ then $\deg (f g) \leq \deg f + \deg g$ (Degree inequalities for sums and products over a commutative ring).

[L7]

For commutative rings $C, S$ , a unital ring homomorphism $φ ⁣ : C \to S$ and $s \in S$ , there is a unique unital ring homomorphism $C [T] \to S$ extending $φ$ on constants and sending $T$ to $s$ (Universal property of $R [x]$ : a coefficient homomorphism and the image of $x$ determine a unique ring homomorphism).

[L8]

The value of $f = \sum_{i} a_{i} T^{i}$ at $s$ along $φ$ is $f_{φ} (s) = \sum_{i} φ (a_{i}) s^{i}$ , and $s$ is a root of $f$ when that value is $0$ (Evaluation and roots of a polynomial in a commutative target ring).

[L9]

[L10]

For a commutative ring $C$ , the polynomial ring $C [T]$ is a commutative ring (Polynomial convolution makes $R [x]$ a commutative ring containing $R$ as its constant subring).

Proof

technique · direct

1.1

For $g \in G$ let $g$ act on $C [T]$ by $g \cdot \sum_{i} a_{i} T^{i} : = \sum_{i} (g \cdot a_{i}) T^{i}$ . This is additive because addition of polynomials is coefficientwise, multiplicative because $g \cdot \sum_{j + k = i} a_{j} b_{k} = \sum_{j + k = i} (g \cdot a_{j}) (g \cdot b_{k})$ by the ring-homomorphism property of $g$ on $C$ , and it sends $1$ to $1$ and $T$ to $T$ ; the action axioms are inherited coefficientwise, so this is again an action of $G$ by ring automorphisms, restricting to the given one on constants.

L1L2L3L4given

2.1

$P_{x} = \prod_{g \in G} (T - g \cdot x)$ is monic of degree $∣ G ∣$ . Each factor $T - g \cdot x$ is nonzero of degree $1$ with leading coefficient $1_{C} \neq 0_{C}$ , because $C \neq 0$ . Multiplying the factors one at a time: if $f$ is monic of degree $d$ then the coefficient of $T^{d + 1}$ in $f \cdot (T - g \cdot x)$ is $1_{C} \cdot 1_{C} = 1_{C} \neq 0_{C}$ , so that product is nonzero, its degree is at most $d + 1$ , and the nonvanishing coefficient at $T^{d + 1}$ forces the degree to be exactly $d + 1$ with leading coefficient $1_{C}$ . Since $G$ is finite and nonempty, the product over all of $G$ is monic of degree $∣ G ∣$ .

L4L5L6step 1.1

2.2

$P_{x}$ is fixed by the action. For $h \in G$ , applying $h$ coefficientwise to a product is applying it to each factor, so $h \cdot P_{x} = \prod_{g \in G} (T - (h g) \cdot x)$ ; the map $g \mapsto h g$ is a bijection of $G$ onto itself, with inverse $g \mapsto h^{- 1} g$ , so it merely permutes the factors of a product in the commutative ring $C [T]$ and $h \cdot P_{x} = P_{x}$ . Since the action on $C [T]$ is coefficientwise, every coefficient of $P_{x}$ is fixed by every $h \in G$ , that is, lies in $C^{G}$ .

L1L2L3L10step 1.1

3.1

$P_{x} (x) = 0$ . Evaluation at $x$ along the identity of $C$ is the unique unital ring homomorphism $C [T] \to C$ fixing constants and sending $T$ to $x$ , so it carries the product $\prod_{g} (T - g \cdot x)$ to $\prod_{g} (x - g \cdot x)$ . The factor indexed by the identity $e$ of $G$ is $x - e \cdot x = x - x = 0$ , and a product in a commutative ring with a zero factor is zero.

L3L7L8step 2.1

4.1

By step 2.2 all coefficients of $P_{x}$ lie in the subring $C^{G}$ , so $P_{x}$ is a polynomial in $C^{G} [T]$ ; its leading coefficient is $1_{C} = 1_{C^{G}}$ , so it is monic there as well, and by step 3.1 the element $x$ is a root of it. Hence $x$ is integral over $C^{G}$ , and since $x \in C$ was arbitrary, $C$ is integral over $C^{G}$ .

L1L5L9step 2.1step 2.2step 3.1∎

Remarks

The product is over the whole group, not over the orbit. Repeated factors are tolerated and are what keeps the degree equal to $∣ G ∣$ independently of the stabiliser of $x$ ; taking the product over the orbit would give a polynomial of varying degree and would need the orbit to be a set of distinct elements.
Nothing is said about $C^{G}$ being large. For a faithful action with many invariants the polynomial $P_{x}$ is informative; for the trivial group it is $(T - x)^{1}$ and the statement is the tautology that every element of $C$ is integral over $C$ .
Infinite $G$ gives nothing here. The construction produces no polynomial at all, since an infinite product of linear factors is not an element of $C [T]$ , and the conclusion can fail.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

Noether's finiteness theorem: the invariants of a finite group acting on a finite-type algebra over a Noetherian ring form an algebra of finite type

Statement

Let $A$ be a Noetherian commutative ring, let $C$ be a commutative $A$ -algebra of finite type (Subalgebra generated by a subset, algebras of finite type, and module-finite algebras) with $A$ a subring of $C$ (Subring: a subset containing $1_{R}$ and closed under addition, additive inverses and multiplication), and let $G$ be a finite group acting on $C$ by $A$ -algebra automorphisms (A group acting on a ring by automorphisms and its invariant subring). Then the invariant subring $C^{G}$ is of finite type over $A$ .

Facts & Assumptions

Given: A Noetherian commutative ring $A$ , a commutative $A$ -algebra $C$ of finite type with $A$ a subring of $C$ , and a finite group $G$ acting on $C$ by $A$ -algebra automorphisms.

[L1]

For an action by ring automorphisms, $C^{G} = {c \in C : g \cdot c = c$ for every $g \in G}$ is a subring of $C$ ; when the action is by $A$ -algebra automorphisms and $A$ is a subring of $C$ , one has $A \subseteq C^{G} \subseteq C$ (A group acting on a ring by automorphisms and its invariant subring).

[L2]

For a finite group acting by ring automorphisms on a nonzero commutative ring $C$ , every element of $C$ is integral over $C^{G}$ (For a finite group of ring automorphisms the orbit polynomial is monic over the invariant subring, so the ring is integral over its invariants).

[L3]

For commutative rings $A \subseteq B \subseteq C$ , each a subring of the next, with $A$ Noetherian, $C$ of finite type over $A$ and every element of $C$ integral over $B$ , the ring $B$ is of finite type over $A$ (The Artin–Tate lemma with integrality in place of module finiteness).

[L4]

An algebra is of finite type over $R$ when it equals $R [a_{1}, \dots, a_{n}]$ for some finite list, and $R []$ is the image of $R$ (Subalgebra generated by a subset, algebras of finite type, and module-finite algebras).

[L5]

A subring contains the identity of the ambient ring and shares its zero and identity (Subring: a subset containing $1_{R}$ and closed under addition, additive inverses and multiplication).

Proof

technique · direct

1.1

Dispose of the zero ring. If $C = 0$ then $1_{C} = 0_{C}$ , and since $A$ is a subring of $C$ it has the same zero and identity, so $A = 0$ ; also $C^{G} = 0$ , which is the image of $A$ in $C^{G}$ and hence equals $A []$ , an algebra of finite type over $A$ . For the rest of the argument assume $C \neq 0$ .

L1L4L5given

1.2

The invariant subring sits between the two: $C^{G}$ is a subring of $C$ , and every $g \in G$ fixes $A$ pointwise because the action is by $A$ -algebra automorphisms and $A$ is a subring of $C$ , so $A \subseteq C^{G} \subseteq C$ , each a subring of the next.

L1L5given

2.1

Since $G$ is finite and $C$ is nonzero, every element of $C$ is integral over $C^{G}$ .

L2step 1.1

3.1

The three rings $A \subseteq C^{G} \subseteq C$ satisfy the hypotheses of the integral form of the Artin–Tate lemma: $A$ is Noetherian, $C$ is of finite type over $A$ , and every element of $C$ is integral over $C^{G}$ . Therefore $C^{G}$ is of finite type over $A$ .

L3L4step 1.2step 2.1∎

Remarks

The theorem says finite type, and no more. It produces finitely many algebra generators of $C^{G}$ over $A$ and identifies none of them; for the symmetric group acting on a polynomial ring the companion examples page compares this with the classical description by elementary symmetric polynomials, which is strictly more information.
Finiteness of $G$ is used only through For a finite group of ring automorphisms the orbit polynomial is monic over the invariant subring, so the ring is integral over its invariants, and there it is essential: it is what makes the orbit polynomial a polynomial.
No hypothesis on the characteristic, and no invertibility of $∣ G ∣$ . The route through integrality and the Artin–Tate lemma avoids averaging entirely, which is why nothing here breaks when the order of $G$ is not invertible in $C$ .

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

If some ideal is not finitely generated, there is one maximal among the ideals that are not

Statement

Let $R$ be a commutative ring and suppose at least one ideal of $R$ is not finitely generated. Let

$Σ : = {a ⊴ R : a is not finitely generated}$

be ordered by inclusion. Then $Σ$ has a maximal element (Maximal element and greatest element): an ideal that is not finitely generated and that no ideal of $Σ$ strictly contains.

This uses Zorn's lemma, hence the axiom of choice (Zorn's lemma). No Noetherian hypothesis is available: by 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, $Σ$ is nonempty exactly because $R$ is not Noetherian, so the maximal element cannot come from a maximal condition. Maximal here means maximal in $Σ$ , not maximal among the proper ideals of $R$ .

Facts & Assumptions

Given: A commutative ring $R$ with at least one ideal that is not finitely generated, and the set $Σ$ of its non-finitely-generated ideals, ordered by inclusion.

[L1]

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

[L2]

For $S \subseteq R$ , $(S)$ is the intersection of all two-sided ideals of $R$ containing $S$ ; in particular $S \subseteq (S)$ , and $({a})$ is written $(a)$ (The ideal generated by a subset and principal ideals).

[L3]

[L4]

A nonempty subset $I \subseteq R$ is a two-sided ideal exactly when it is closed under $x - y$ and under $r x, x r$ for all $r \in R$ , $x, y \in I$ (Ideal criteria and intersections of ideals).

[L5]

A subset $C$ of a poset is a chain when any two of its elements are comparable; the empty set is a chain (Chain in a poset).

[L6]

An element $m$ of a poset is maximal when no element is strictly above it (Maximal element and greatest element).

[L7]

Assuming the Axiom of Choice, a nonempty poset in which every chain has an upper bound has a maximal element (Zorn's lemma).

Proof

technique · direct

1.1

Inclusion partially orders $Σ$ , and $Σ$ is nonempty by hypothesis. The empty chain is bounded in $Σ$ : its upper bounds are all the elements of $Σ$ , and there is at least one.

L1L2L6given

2.1

Let $C \subseteq Σ$ be a nonempty chain and put $U : = ⋃ C$ . Then $U$ is an ideal of $R$ . It is nonempty, since some $I \in C$ contains $0$ . For $x, y \in U$ pick $I, J \in C$ with $x \in I$ and $y \in J$ ; the two are comparable, so both lie in the larger one, whose being an ideal gives $x - y$ there and hence in $U$ . For $r \in R$ and $x \in U$ , choosing $I \in C$ with $x \in I$ gives $r x \in I \subseteq U$ . The ideal criterion applies.

L1L4L5step 1.1

3.1

$U$ is not finitely generated, so $U \in Σ$ and $U$ is an upper bound of $C$ . Suppose instead $U = (u_{1}, \dots, u_{k})$ with $k \in N$ ; each $u_{i}$ lies in $U$ , hence in some member of $C$ . A nonempty finite subset of a chain has a greatest member, by induction on its size using comparability of any two elements, so there is $I \in C$ containing every $u_{i}$ ; when $k = 0$ take any $I \in C$ , which exists because $C$ is nonempty. Then $U = (u_{1}, \dots, u_{k}) \subseteq I \subseteq U$ , so $I = U$ is finitely generated, contradicting $I \in Σ$ .

L2L3L5step 2.1

4.1

Every chain in $Σ$ , empty or not, therefore has an upper bound in $Σ$ , and $Σ$ is nonempty; Zorn's lemma gives a maximal element of $Σ$ . This is the one place the argument leaves ZF, and it uses the full axiom of choice rather than a countable or dependent form.

L6L7step 1.1step 3.1∎

Remarks

Maximal in $Σ$ , not maximal in $R$ . The ideal produced is maximal among those that fail to be finitely generated. A maximal ideal of $R$ in the usual sense may perfectly well be finitely generated, and nothing here says the two notions meet.
No chain condition is used or available. The hypothesis is the opposite of a chain condition, so the maximal element has to be bought with Zorn's lemma; that is the whole reason this lemma is separated from the criterion it serves.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

An ideal maximal among the non-finitely-generated ideals is prime

Statement

Let $R$ be a commutative ring and let $p$ be an ideal of $R$ that is maximal in the set $Σ$ of non-finitely-generated ideals of $R$ (If some ideal is not finitely generated, there is one maximal among the ideals that are not): that is, $p$ is not finitely generated, and every ideal of $R$ strictly containing $p$ is finitely generated. Then $p$ is a prime ideal (Prime ideals and maximal ideals in a commutative ring).

Facts & Assumptions

Given: A commutative ring $R$ and an ideal $p$ maximal in the set $Σ$ of ideals of $R$ that are not finitely generated (If some ideal is not finitely generated, there is one maximal among the ideals that are not). For an ideal $a$ and $a \in R$ , write $(a : a) : = {r \in R : r a \in a}$ and $a a : = {a x : x \in a}$ .

[L1]

A proper ideal $P ⊊ R$ of a commutative ring is prime when $a b \in P$ implies $a \in P$ or $b \in P$ (Prime ideals and maximal ideals in a commutative ring).

[L2]

[L3]

For $S \subseteq R$ , $(S)$ is the intersection of all two-sided ideals containing $S$ , so $S \subseteq (S)$ ; $({a})$ is written $(a)$ (The ideal generated by a subset and principal ideals).

[L4]

For ideals $I, J$ of $R$ , the sum is $I + J = {i + j : i \in I, j \in J}$ (The sum $I + J$ and product $I J$ of two-sided ideals).

[L5]

A nonempty subset $I \subseteq R$ is a two-sided ideal exactly when it is closed under $x - y$ and under $r x, x r$ for all $r \in R$ , $x, y \in I$ (Ideal criteria and intersections of ideals).

Proof

technique · contradiction

1.1

$p$ is a proper ideal: the unit ideal $R = (1)$ is generated by one element, so $R \notin Σ$ , whereas $p \in Σ$ .

L2L3given

1.2

Suppose $p$ is not prime. By the previous step it is proper, so there are $a, b \in R$ with $a b \in p$ , $a \notin p$ and $b \notin p$ .

assume-contraL1given

2.1

The ideal $p + (a)$ strictly contains $p$ , since $a$ lies in it and not in $p$ , so by maximality it is finitely generated. Every element of $p + (a)$ has the form $x + w a$ with $x \in p$ and $w \in R$ , because $(a) = R a$ ; so a finite generating list may be written $x_{1} + w_{1} a, \dots, x_{n} + w_{n} a$ with $x_{i} \in p$ , $w_{i} \in R$ and $n \in N$ .

L2L3L4step 1.2

2.2

The set $q : = (p : a)$ is an ideal of $R$ : it contains $0$ , and if $r a, r^{'} a \in p$ then $(r - r^{'}) a = r a - r^{'} a \in p$ and $(s r) a = s (r a) \in p$ for every $s \in R$ . It contains $p$ , since $x a \in p$ for $x \in p$ , and it contains $b$ , since $b a = a b \in p$ ; as $b \notin p$ the containment $p \subseteq q$ is strict, so by maximality $q$ is finitely generated, say $q = (q_{1}, \dots, q_{m})$ with $m \in N$ .

L2L5step 1.2

3.1

The set $a q$ is an ideal, being closed under differences and under multiplication by $R$ because $q$ is, and it is generated by $a q_{1}, \dots, a q_{m}$ : any $q \in q$ is $\sum_{j} r_{j} q_{j}$ , so $a q = \sum_{j} r_{j} (a q_{j})$ . Moreover $a q \subseteq p$ by the definition of $q$ .

L2L5step 2.2

3.2

$p = (x_{1}, \dots, x_{n}) + a q$ . The inclusion from right to left holds because each $x_{i}$ lies in $p$ and $a q \subseteq p$ . For the other inclusion take $z \in p \subseteq p + (a)$ and write $z = \sum_{i = 1}^{n} c_{i} (x_{i} + w_{i} a) = \sum_{i = 1}^{n} c_{i} x_{i} + y a$ with $c_{i} \in R$ and $y = \sum_{i = 1}^{n} c_{i} w_{i}$ ; then $y a = z - \sum_{i} c_{i} x_{i}$ lies in $p$ , so $y \in q$ and $y a \in a q$ , whence $z \in (x_{1}, \dots, x_{n}) + a q$ .

L2L4step 2.1step 2.2

4.1

Both summands are finitely generated, so $p = (x_{1}, \dots, x_{n}, a q_{1}, \dots, a q_{m})$ is finitely generated, contradicting $p \in Σ$ . The supposition of step 1.2 is therefore untenable: whenever $a b \in p$ , either $a \in p$ or $b \in p$ , and with step 1.1 this makes $p$ prime.

L1L2L4step 1.1step 3.1step 3.2discharge-contradiction∎

Remarks

Where each maximality use goes. Maximality of $p$ in $Σ$ is used exactly twice, in step 2.1 on $p + (a)$ and in step 2.2 on the colon ideal $(p : a)$ ; both are strictly larger than $p$ precisely because $a \notin p$ and $b \notin p$ .
The generators of $p + (a)$ are normalised, not merely chosen. Writing them as $x_{i} + w_{i} a$ with $x_{i} \in p$ is what lets step 3.2 separate the part of $z$ lying in $(x_{1}, \dots, x_{n})$ from the multiple of $a$ ; an unnormalised list would not split that way.
No Noetherian hypothesis anywhere. The lemma is used inside a proof whose conclusion is that the ring is Noetherian, so assuming a chain condition here would be circular.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

Cohen's criterion: a commutative ring in which every prime ideal is finitely generated is Noetherian

Statement

Let $R$ be a commutative ring in which every prime ideal (Prime ideals and maximal ideals in a commutative ring) is finitely generated. Then $R$ is Noetherian.

The proof uses Zorn's lemma through If some ideal is not finitely generated, there is one maximal among the ideals that are not, and therefore the axiom of choice. It does not use, and could not use, a maximal condition on the ideals of $R$ : that condition is part of what is being proved.

Facts & Assumptions

Given: A commutative ring $R$ in which every prime ideal is finitely generated.

[L1]

A proper ideal $P ⊊ R$ of a commutative ring is prime when $a b \in P$ implies $a \in P$ or $b \in P$ (Prime ideals and maximal ideals in a commutative ring).

[L2]

If at least one ideal of a commutative ring is not finitely generated, then the set $Σ$ of its non-finitely-generated ideals, ordered by inclusion, has a maximal element; the proof uses Zorn's lemma (If some ideal is not finitely generated, there is one maximal among the ideals that are not).

[L3]

An ideal maximal in $Σ$ is prime (An ideal maximal among the non-finitely-generated ideals is prime).

[L4]

Proof

technique · contradiction

1.1

Suppose $R$ is not Noetherian. By the ideal-level characterisation, some ideal of $R$ is then not finitely generated, so the set $Σ$ of non-finitely-generated ideals of $R$ is nonempty.

assume-contraL4given

2.1

Since $Σ$ is nonempty, it has a maximal element $p$ : an ideal that is not finitely generated and that no non-finitely-generated ideal strictly contains. This is where Zorn's lemma is used; no chain condition on $R$ is available, and none is invoked.

L2step 1.1

3.1

That maximal element is a prime ideal of $R$ .

L3step 2.1

4.1

By hypothesis every prime ideal of $R$ is finitely generated, so $p$ is finitely generated, contradicting $p \in Σ$ . The supposition of step 1.1 fails: every ideal of $R$ is finitely generated, and $R$ is Noetherian.

L1L4step 1.1step 3.1discharge-contradiction∎

Remarks

The hypothesis is about primes only, and that is the whole point. Checking finite generation of every ideal is the definition; checking it on primes is a strictly smaller task, and Cohen's criterion says the smaller task suffices.
The maximal element does not come from Noetherian induction. Noetherian induction: a property that passes to an ideal whenever it holds for every strictly larger ideal holds for every ideal and 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 both presuppose the conclusion here. The element is produced by Zorn's lemma applied to a poset of ideals in a ring assumed not to satisfy any chain condition.
A prime ideal of the ring, not a maximal one. An ideal maximal among the non-finitely-generated ideals is prime produces primeness, not maximality among proper ideals, and the hypothesis is applied to primes accordingly.

RemarkRemark: AI-adaptedProof: Not applicablejudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

Which constructions preserve the Noetherian condition, and which do not

What is proved above. Starting from a Noetherian commutative ring $R$ , each of the following is again Noetherian:

every quotient $R / I$ and every localisation $S^{- 1} R$ (Every quotient and every localisation of a Noetherian ring is Noetherian);
the polynomial ring $R [x]$ (Hilbert basis theorem: if $R$ is Noetherian then $R [x]$ is Noetherian) and, by induction, $R [x_{1}, \dots, x_{n}]$ for each $n \in N$ (If $R$ is Noetherian then $R [x_{1}, \dots, x_{n}]$ is Noetherian for every $n \in N$ );
every commutative $R$ -algebra of finite type (Every algebra of finite type over a Noetherian ring is a Noetherian ring);
every module-finite commutative $R$ -algebra, and every ring between $R$ and such an algebra (A module-finite algebra over a Noetherian ring is a Noetherian ring, and so is every ring between the two);
the product $R \times S$ with $S$ Noetherian (A product of two Noetherian rings is Noetherian);
a subring $R_{0} \subseteq R$ that admits an $R_{0}$ -linear retraction $R \to R_{0}$ fixing $R_{0}$ pointwise (A subring that admits a module retraction from a Noetherian ring is Noetherian).

What fails, and why it is worth saying. Every entry above supplies some map or some finiteness relating the new ring to $R$ . Two natural-looking weakenings supply neither, and both fail.

An arbitrary subring. Being an additive subgroup closed under multiplication gives no way to pull a generating set of an ideal of the subring back from the larger ring, and the conclusion is false: a subring of a Noetherian ring need not be Noetherian. The companion examples page of this pair works a witness inside a polynomial ring in two variables, where the failing ideal is visibly not finitely generated. The retraction hypothesis is exactly what is missing: it is a map back, and with it the argument runs.

Infinitely many indeterminates. The polynomial-ring statement is proved by an induction on the number of indeterminates, and that induction has no limit stage: it covers a finite list and no more. The companion examples page of this pair carries the witness, a ring of polynomials in countably many indeterminates in which the ideals generated by initial segments of the variables form a strictly ascending chain. The same remark applies to the product statement, which is proved for two factors and extends by iteration to a finite list; a product indexed by an infinite set is not among the constructions A product of two Noetherian rings is Noetherian speaks about.

A hypothesis that is not needed anywhere above. No result above assumes that $R$ is an integral domain, that $R$ is nonzero, or that its ideals are principal. The zero ring is Noetherian and is admitted throughout, and rings with zero divisors are admitted in the Hilbert basis argument in particular, which never multiplies two leading coefficients together.

5 · Examples, counterexamples and false statements

None yet.

Noetherian Rings and Hilbert Basis

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

Conventions for this development and where dependent choice and Zorn's lemma are used

The submodule generated by a subset consists of the finite RR-linear combinations of that subset

Statement

Facts & Assumptions

Proof

Remarks

Every generating set of a finitely generated module contains a finite generating subset

Statement

Facts & Assumptions

Proof

Remarks

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

Facts & Assumptions

Proof

Remarks

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

Statement

Facts & Assumptions

Proof

Remarks

A subring that admits a module retraction from a Noetherian ring is Noetherian

Statement

Facts & Assumptions

Proof

Remarks

Every ideal of a localisation is generated by the images of any generating set of its contraction

Statement

Facts & Assumptions

Proof

Remarks

Every quotient and every localisation of a Noetherian ring is Noetherian

Statement

Facts & Assumptions

Proof

Remarks

A ring with finitely many ideals of zero intersection whose quotients are Noetherian rings is Noetherian

Statement

Facts & Assumptions

Proof

Remarks

A product of two Noetherian rings is Noetherian

Statement

Facts & Assumptions

Proof

Remarks

The leading coefficients of the degree-nn elements of an ideal of R[x]R[x], together with 00, form an ideal of RR, and these ideals ascend with nn

Statement

Facts & Assumptions

Proof

Remarks

A single cancellation step lowers the degree of a polynomial in an ideal once its leading coefficient lies in a realised stage

Statement

Facts & Assumptions

Proof

Remarks

Over a Noetherian ring, an ideal of R[x]R[x] is generated by finitely many polynomials realising generators of its stages up to the stabilisation degree

Statement

Facts & Assumptions

Proof

Remarks

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

Statement

Facts & Assumptions

Proof

Remarks

If RR is Noetherian then R[x1,…,xn]R[x1​,…,xn​] is Noetherian for every n∈Nn∈N

Statement

Facts & Assumptions

Proof

Remarks

R[x]R[x] is Noetherian if and only if RR is Noetherian

Statement

Facts & Assumptions

Proof

The submodule generated by a subset consists of the finite $R$ -linear combinations of that subset

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 $n$

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

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

If $R$ is Noetherian then $R [x_{1}, \dots, x_{n}]$ is Noetherian for every $n \in N$

$R [x]$ is Noetherian if and only if $R$ is Noetherian

Over a commutative ring the homomorphism group ${Hom}_{R} (M, N)$ is an $R$ -module

For a commutative ring, ${Hom}_{R} (R^{n}, N) ≅ N^{n}$