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

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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]

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

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.

Depends on

Used by

Dependency tree · two levels

25 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

M. Hochster, Introduction to Commutative Algebra, Math 614, Proposition 5.11 (standard reference, not scraped)
A. Altman and S. Kleiman, A Term of Commutative Algebra, 13th ed., Exercise (16.9) (standard reference, not scraped)