Ring theory MOC

Noetherian ring

A ring is called (left) Noetherian iff any of the following equivalent conditions hold:¹ ring

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

Proof

Suppose ^N1 holds, and let be an increasing sequence of (left/right) ideals. Then

is an ideal, since if then for some and thus resp. for any . Hence is finitely generated, and all these generators must be exhausted by some , implying ^N2.

Suppose ^N2 holds, and assume towards contradiction there exists a set of (left/right) ideals with no maximal element. One can then form a strictly increasing sequence of , contradicting ^N2. Therefore ^N2 implies ^N3.

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

Properties

Let be two-sided Noetherian.

1. Let be a nonzero proper ideal. Then there exist nonzero prime ideals such that .

Proof

Let be the set of all ideals for which ^P1 fails, and assume towards contradiction its maximal element is , which cannot be a prime ideal, so there exist such that . Let and , whence , so by maximality and thus there exist nonzero prime ideals such that

Then

whence

a contradiction. Therefore .

Other results

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

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Footnotes

1. 2022. Algebraic number theory course notes, §2.5, pp. 14–15 ↩