Module theory MOC

Noetherian module

A (left) -module is (left) noetherian iff every submodule of is finitely generated as a (left) -module.¹ module

Properties

1. Let and . Then is noetherian iff both and are.
2. Let be a noetherian ring and be finitely generated. Then is noetherian.

Proof

If is Noetherian, then so is , since any submodule of is just the projection of some submodule of which is thus finitely generated, and so is , since its submodules are also submodules of .

For the converse, assume and are Noetherian. Given a , we need to prove is finitely generated. Since , this is finitely generated. By the Second isomorphism theorem,

since must be finitely generated, so must be. Therefore by ^P1, is finitely generated, proving ^P1.

^P2 is a corollary: Since there is a Module epimorphism , the First isomorphism theorem says that is isomorphic to a quotient of . By ^P1 is suffices to show is noetherian, which we do by induction.

The statement is true for since is noetherian. For , assume is noetherian, where

So by ^P1, is noetherian.

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Footnotes

1. 2009. Algebra: Chapter 0, §III.6.4, pp. 170–171 ↩