2.3 Noetherian Rings and Modules

Definition

Let $I$ be an indexing set, and let $m_i\in M$. For $i\in I$:

• say $\{m_i{\}}_{i\in I}$ generate $M$ if for any $m\in M$, $m={\sum }_{i\in I}{a}_i{m}_i$, and say $\{m_i{\}}_{i\in I}$ is a set of generator;
• say $M$ is a free $A$-module, if we can choose some set if generator $\{m_i{\}}_{i\in I}$ such that for any $m\in M$ can be uniquely expressed as $m\in {\sum }_{i\in I}{a}_i{m}_i$. (Equivalently, ${\sum }_{i\in I}{a}_i{m}_i=0$ iff $a_i=0$ for all $i\in I$.)

Furthermore, we say $M$ is finitely generated or finite free if $|I|<\infty$. In this case, we can write $M=A^{\oplus d}={\oplus }_{i=1}^d{A}_i$.

Theorem

The followings are equivalent:

• Every non-empty family of submodule of $M$ has a maximal element. Namely, given $P=\{M_j{\}}_{j\in J}$, there exists $N\in P$ such that if $N\subseteq M_j$, then $M_j=N$.
• Every increasing sequence of submodule $\{M_i{\}}_{i=1}^{\infty }$ is stationary. Namely, there exists $N$ such that $M_i=M_N$ for all $i⩾N$.
• Every submodule of $M$ is finitely generated.

Non-example. Consider $\mathbb{Z}$-module $\mathbb{Q}$. It is not finitely generated. Let $M_n=⟨\frac{1}{2^n}⟩$, then $\{M_n\}$ is not stationary.

Example. A vector space over a field.

\begin{proof} i)→iii) Let $E\subseteq M$ be a submodule. Define $\Phi =\{G\subseteq E:G\text{ is finitely generated}\}$. Note that $(0)\in \Phi$. By i), there exists a maximal element $F\in \Phi$. Assume that $F\ne E$, then there exists $x\in F\setminus E$ such that $⟨F,x⟩\in \Phi$, which is a contradiction. Therefore, $F=E$ and so $E$ is finitely generated. By the arbitrary of $E$, we finish the proof of iii).

iii)→ii) Consider the increasing sequence $\{M_i{\}}_{i=1}^{\infty }$ and define $E={\cup }_{i=1}^{\infty }{M}_i$, then $E$ is still a submodule of $M$ (for any $x,y\in M$, verify $ax+by\in M$). By iii), $E$ is finitely generated. Suppose that $\{e_1,\cdots ,e_d\}$ is a set of generators of $E$. Assume that $e_i\in M_{n_i}$, then $\{e_1,\cdots ,e_d\}\in M_N$, where $N=\max \{n_1,\cdots ,n_d\}$. It follows that $E=M_N$ and $M_n=M_N$ for all $n⩾N$.

ii)→i) Consider $P=\{M_i{\}}_{i\in I}$. Assume that there does not exist maximal element, then there is a strictly increasing sequence

$$N_1⊊N_2⊊N_3\cdots$$

which contradicts with ii). \end{proof}

Definition

If an $A$-module satisfies conditions in ^4af1b0, then we say $M$ is a Noetherian $A$-module.

If the $B$-module $B$ satisfies condition in ^4af1b0, then we say $B$ is a Noetherian ring.

Remark. $R$ is a Noetherian ring iff any ideal is finitely generated.

Examples.

• $\mathbb{Z}$ is a Noetherian ring, as each ideal is finitely generated.
• $R:=\mathbb{Z}[x_1,x_2,x_3,\cdots ]\supseteq (x_1,x_2,\cdots )$ is not finitely generated and so $\mathbb{Z}[x_1,x_2,\cdots ]$ is not a Noetherian ring.
• Consider $R\twoheadrightarrow \mathbb{Z}[x_1,x_2,\cdots ]/(x_1,x_2,\cdots )\simeq \mathbb{Z}\twoheadrightarrow \mathbb{Z}/p\mathbb{Z}$, then $\mathbb{Z}/p\mathbb{Z}$ is an $\mathbb{R}$-module and it is a Noetherian module.
• $\mathbb{Z}$ is a Noetherian ring, but $\mathbb{Q}$ is not a Noetherian $\mathbb{Z}$-module.
• $\mathbb{Q}$ is a Noetherian $\mathbb{Q}$-module, i.e., $\mathbb{Q}$ is a Noetherian ring.

Proposition

Assume that $E$ is an $A$-module. If $E^{\prime }\subseteq E$ is a submodule, then $E$ is a Noetherian $A$-module iff $E^{\prime }$ and $E/E^{\prime }$ are both Noetherian $A$-modules.

\begin{proof} Assume that $E$ is Noetherian. For any submodule $F\subseteq E$, $F$ is finitely generated by ^4af1b0. Then it is easy to show $E^{\prime }$ is Noetherian. For any $G\subseteq E/E^{\prime }$, $G$ can be written as $G=H/E^{\prime }$ for some $E\supseteq H\supseteq E^{\prime }$ and so $G$ is finitely generated. It follows that $E/E^{\prime }$ is Noetherian.

On the other hand, assume that $E^{\prime }$ and $E/E^{\prime }$ are both Noetherian $A$-modules. Let $G\subseteq E$ is a submodule, then we aim to show $G$ is finitely generated.

Idea:

$$\begin{array}{ccccccccc}0& \to & G\cap E^{\prime }& \to & G& \stackrel{\pi }{\to }& G/G\cap E^{\prime }\simeq (G+E^{\prime })/E^{\prime }& \to & 0\\ & & \downarrow & & \downarrow & & \downarrow & & \\ 0& \to & E^{\prime }& \to & E& \to & E/E^{\prime }& \to & 0\end{array}$$

Since $E^{\prime }$ is Noetherian, $G\cap E^{\prime }$ is finitely generated. Choose a set of generator $\{f_1,\cdots ,f_m\}$. Since $E/E^{\prime }$ is Noetherian, $G/G\cap E^{\prime }$ is finitely generated. Choose a set of generator $\{e_1,\cdots ,e_n\}$, and choose some pre-image under $\pi$, that is, $\{{\hat{e}}_1,\cdots {\hat{e}}_n\}\in G$. Claim that $\{f_1,\cdots ,f_m,{\hat{e}}_1,\cdots ,{\hat{e}}_n\}$ generates $G$. Let $g\in G$, then $\pi (g)={\sum }_{i=1}^n{a}_i{e}_i$ for some $a_i\in A$. Consider $\tilde{g}=g-({\sum }_{i=1}^n{a}_i{\hat{e}}_i)$, then $\pi (\tilde{g})=0$ and so $\tilde{g}\in \ker \pi =G\cap E^{\prime }$. Thus $\tilde{g}={\sum }_{j=1}^m{b}_j{f}_j$ and so $g={\sum }_{i=1}^n{a}_i{e}_i+{\sum }_{i=1}^m{b}_j{f}_j$. \end{proof}

Corollary

If $M_1,\cdots ,M_n$ are Noetherian $A$-module, then so for ${\prod }_{i=1}^n{M}_i$.

\begin{proof} When $n=2$, consider $0\to M_1\to M_1\times M_2\to M_2\to 0$ and by ^0af43c $M_1\times M_2$ is Noetherian. By induction we finish the proof. \end{proof}

Corollary

If $A$ is Noetherian, then any finitely generated $A$-module is a Noetherian $A$-module.

\begin{proof} Note that $M$ is a finitely generated $A$-module iff there exists a surjective $A^{\oplus n}\twoheadrightarrow M$. Note that $A$ is a Noetherian $A$-module iff $A^{\oplus n}$ is a Noetherian $A$-module. By ^2f2111, $A^{\oplus n}$ is Noetherian $A$-module. By ^0af43c, $M$ is Noetherian. \end{proof}

Hilbert basis theorem

If $A$ is Noetherian ring, then $A[x]$ is also a Noetherian ring.

\begin{proof} Suppose $\mathfrak{a}\subseteq A[X]$ is a non-finitely generated left ideal. Then by recursion (using the axiom of dependent choice) there is a sequence of polynomials $\left\{f_0,f_1,\dots \right\}$ such that if ${\mathfrak{b}}_n$ is the left ideal generated by $f_0,\dots ,f_{n-1}$ then $f_n\in \mathfrak{a}\backslash {\mathfrak{b}}_n$ is of minimal degree. By construction, $\left\{\mathrm{deg}\left(f_0\right),\mathrm{deg}\left(f_1\right),\dots \right\}$ is a non-decreasing sequence of natural numbers. Let $a_n$ be the leading coefficient of $f_n$ and let $\mathfrak{b}$ be the left ideal in $R$ generated by $a_0,a_1,\dots$. Since $A$ is Noetherian, the chain of ideals

$$\left(a_0\right)\subset \left(a_0,a_1\right)\subset \left(a_0,a_1,a_2\right)\subset \cdots$$

must terminate. Thus $\mathfrak{b}=\left(a_0,\dots ,a_{N-1}\right)$ for some integer $N$. So in particular,

$$a_N=\sum _{i<N}{u}_i{a}_i,u_i\in A.$$

Now consider $g={\sum }_{i<N}{u}_i{X}^{\mathrm{deg}\left(f_N\right)-\mathrm{deg}\left(f_i\right)}{f}_i$, whose leading term is equal to that of $f_N$; moreover, $g\in {\mathfrak{b}}_N$. However, $f_N\notin {\mathfrak{b}}_N$, which means that $f_N-g\in \mathfrak{a}\backslash {\mathfrak{b}}_N$ has degree less than $f_N$, contradicting the minimality. \end{proof}