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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]

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).

[L7]

For a commutative ring, being Noetherian is equivalent to every ideal being finitely generated (A commutative ring is Noetherian exactly when every ideal is finitely generated, exactly when its ideals satisfy the ascending chain condition, and exactly when every nonempty set of ideals has a maximal member).

Proof

technique · direct

1.1

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.

Depends on

Used by

A product of two Noetherian rings is Noetherian Corollary

Dependency tree · two levels

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

Sources

A. Altman and S. Kleiman, A Term of Commutative Algebra, 13th ed., Exercise (16.18) (standard reference, not scraped)
J. S. Milne, A Primer of Commutative Algebra, v4.03, §3 (standard reference, not scraped)