Properties of Noetherian modules

Suppose A is a commutative ring with 1. We say that A-module module M is Noetherian if every ascending chain of submodules in M eventually terminates. This is similar to how we define a ring to be Noetherian (a commutative ring A is Noetherian if every ascending chain of ideals in A eventually terminates).

In “Undergraduate Commutative Algebra” by Miles Reid, the following proposition is proved (page 53), from which some important properties of Noetherian modules can be derived.

Proposition. Let 0\longrightarrow L\overset{\alpha}{\longrightarrow} M\overset{\beta}{\longrightarrow} N\longrightarrow 0 be a short exact sequence of A-modules. Then,

\displaystyle  M\textrm{ is Noetherian } \Longleftrightarrow L \textrm{ and } N \textrm{ are Noetherian}

I have written up proof of this here. (I have filled in some details, where Miles have left to the reader).

Here are some consequences:

(1) If M_i are Noetherian modules (for i=1, 2, ..., r), then \bigoplus_{i=1}^{r} M_i is Noetherian.

(2) If A is a Noetherian ring, then A-module M is Noetherian if and only if it is finitely generated A-module.

(3) If A is a Noetherian ring, M is a finitely generated A-module, then any subdmoule N\subseteq M is again finitely-generated A-module.

Proof. (1) We recall that direct sum M=M_1\oplus M_2 can be realized as an exact sequence 0\longrightarrow M_1\overset{\alpha}{\longrightarrow} M\overset{\beta}{\longrightarrow} M_2\longrightarrow 0, where \alpha(m_1)=(m_1, 0), and \beta((m_1, m_2))=m_2. Applying the Proposition, we obtain that M= M_1\oplus M_2 is Noetherian module. By simple induction, this is extended to direct sum of any finite number of Noetherian modules.

(2) If A-module M is finitely generated, then we have a surjection A^{m}\to M for some positive integer m, where A^{m} is a free module of rank m. In other words, we have an exact sequence A^{m}\overset{\beta}{\longrightarrow} M\longrightarrow 0. From (1), we know A^{m} is Noetherian. By the one of the directions (\Rightarrow) of the Proposition, we obtain that M is also Noetherian. Conversely, if M is a Noetherian A-module, it is clear that M is finitely generated A-module, for otherwise we would obtain ascending chain of submodules

(m_1)\subsetneq (m_1, m_2)\subsetneq (m_1, m_2, m_3)\subsetneq \cdots

that does not terminate. Here (m_1, m_2, ..., m_k) is the submodule generated by elements m_1, m_2, ..., m_k.

(3) Since M is finitely generated A-module, we get from (2) that M is a Noetherian A-module. It is clear that any submodule of a Noetherian module is again Noetherian module. So a submodule N\subseteq M is Noetherian. Applying (2) again, we get that N is finitely-generated A-module, as desired.

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