## 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.