Noetherian ring

A ring $𝑅$ is called (left) Noetherian iff any of the following equivalent conditions hold:¹ #m/def/ring

every (left) ideal $𝐼 ⊴ 𝑅$ is finitely generated as a (left) $𝑅$ -module, i.e. $𝑅$ is a (left) Noetherian module;
(ascending chain condition or ACC) every increasing sequence $𝐼 1 ⊴ 𝐼 2 ⊴ \dots$ of (left) ideals of $𝑅$ has a largest element;
every non-empty set of (left) ideals of $𝑅$ contains a maximal element.

Proof

Suppose ^N1 holds, and let $𝐼 1 ⊴ 𝐼 2 ⊴ \dots$ be an increasing sequence of (left/right) ideals. Then

𝐼 = \infty ⋃ 𝑖 = 1 𝐼 𝑛

is an ideal, since if $𝑥 \in 𝐼$ then $𝑥 \in 𝐼 𝑗$ for some $𝑗$ and thus $𝑟 𝑥 \in 𝐼 𝑗$ resp. $𝑥 𝑟 \in 𝐼 𝑗$ for any $𝑟 \in 𝑅$ . Hence $𝐼$ is finitely generated, and all these generators must be exhausted by some $𝐼 𝑛$ , implying ACC.

Suppose ACC holds, and assume towards contradiction there exists a set $ℑ$ of (left/right) ideals with no maximal element. One can then form a strictly increasing sequence of $𝐼 𝑛 \in ℑ$ , contradicting ACC. Therefore ACC implies ^N3.

Suppose ^N3 holds, and let $𝐼 ⊴ 𝑅$ be an arbitrary (left/right) ideal. Letting $ℑ$ be the set of finitely generated ideals contained in $𝐼$ , which is inhabited since $0 \in ℑ$ , and thus contains a maximal element $𝐼' \in ℑ$ . Assume towards contradiction $𝐼' \neq 𝐼$ . Then we can take $𝑥 \in 𝐼 ∖ 𝐼'$ , and form $𝐼 + 𝑅 𝑥$ resp. $𝐼 + 𝑥 𝑅$ , which is a finitely generated (left/right) ideal strictly larger than $𝐼'$ , contradicting maximality. Therefore $𝐼' = 𝐼$ and $𝐼$ is finitely generated, so ^N3 implies ^N1.

Properties

Let $𝑅$ be two-sided Noetherian.

Let $𝐼 ◃ 𝑅$ be a nonzero proper ideal. Then there exist nonzero prime ideals $𝔭 1, \dots, 𝔭 𝑛 ◃ 𝑅$ such that $𝔭 1 \dots 𝔭 𝑛 \subseteq 𝐼$ .

Proof

Let $I$ be the set of all ideals for which ^P1 fails, and assume towards contradiction its maximal element is $𝐼 \in I$ , which cannot be a prime ideal, so there exist $𝑎, 𝑏 \in 𝑅 ∖ 𝐼$ such that $𝑎 𝑏 \in 𝐼$ . Let $𝔞 = (𝐼, 𝑎)$ and $𝔟 = (𝐼, 𝑏)$ , whence $𝐼 ⊊ 𝔞, 𝔟$ , so by maximality $𝔞, 𝔟 \notin I$ and thus there exist nonzero prime ideals $𝔭 1, \dots, 𝔭 𝑚, 𝔮 1, \dots 𝔮 𝑛 ◃ 𝑅$ such that

𝔞 \subseteq (𝔭 1 \dots 𝔭 𝑚), 𝔟 \subseteq (𝔮 1 \dots 𝔮 𝑛) .

Then

(𝔞 𝔟) = (𝐼 2, 𝑎 𝐼, 𝑏 𝐼, 𝑎 𝑏) \subseteq 𝐼

whence

(𝔭 1 \dots 𝔭 𝑚 𝔮 1 \dots 𝔮 𝑛) \subseteq 𝐼,

a contradiction. Therefore $I = \emptyset$ .

Other results

Finitely generated modules over a noetherian ring are noetherian (^P2)

#state/tidy | #lang/en | #SemBr

2022. Algebraic number theory course notes, §2.5, pp. 14–15 ↩

Noetherian ring

Properties

Other results

Footnotes