Noetherian module

A (left) $𝑅$ -module $𝑀$ is (left) noetherian iff every submodule of $𝑀$ is finitely generated as a (left) $𝑅$ -module.¹ #m/def/module

Properties

Let $𝑀 \in 𝑅 𝖬 𝗈 𝖽$ and $𝑁 \leq 𝑅 𝑀$ . Then $𝑀$ is noetherian iff both $𝑁$ and $𝑀 / 𝑁$ are.
Let $𝑅$ be a noetherian ring and $𝑀 \in 𝑅 𝖬 𝗈 𝖽$ be finitely generated. Then $𝑀$ is noetherian.

Proof

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

For the converse, assume $𝑁$ and $𝑀 / 𝑁$ are Noetherian. Given a $𝑃 \leq 𝑅 𝑀$ , we need to prove $𝑃$ is finitely generated. Since $𝑃 \cap 𝑁 \leq 𝑅 𝑁$ , this is finitely generated. By the Second isomorphism theorem,

𝑃 𝑃 \cap 𝑁 ≅ 𝑅 𝑃 + 𝑁 𝑁 \leq 𝑅 𝑀 𝑁

since $𝑃 + 𝑁 𝑁$ must be finitely generated, so must $𝑃 𝑃 \cap 𝑁$ be. Therefore by ^P1, $𝑃$ is finitely generated, proving ^P1.

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

The statement is true for $𝑛 = 1$ since $𝑅$ is noetherian. For $𝑛 > 1$ , assume $𝑅 (𝑛 - 1) \leq 𝑅 𝑅 (𝑛)$ is noetherian, where

𝑅 (𝑛) 𝑅 (𝑛 - 1) ≅ 𝑅 𝑅 .

So by ^P1, $𝑅 (𝑛)$ is noetherian.

#state/develop | #lang/en | #SemBr

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

Noetherian module

Properties

Footnotes