6 Chain Condition

Proposition

Let $\Sigma$ be a poset. The followings are equivalent.

• Any increasing sequence $x_1⩽\cdots ⩽x_n⩽\cdots$ in $\Sigma$ is stationary.
• Any non-empty subset has a maximal element.

\begin{proof} Easy. \end{proof}

Remark. Let $M$ be an $A$-module, and let $\Sigma =\{\text{mbox}submoduleofM\}$. If $(\Sigma ,\subseteq )$ is the poset in ^1ab5b8, then it is denoted as a.c.c. If $(\Sigma ,\supseteq )$ is the poset in ^1ab5b8, then it is denoted as d.c.c.

Definition

If $M$ satisfies a.c.c., then we say $M$ is a Noetherian $A$-module.

If $M$ satisfies d.c.c., then we say $M$ is a Artinian $A$-module.

Say the ring $A$ is a Noetherian ring (rep. Artinian ring) if $A$ is a Noetherian (rep. Artinian) $A$-module.

Examples.

• Any finite abelian group is both Noetherian and Artinian $\mathbb{Z}$-module.
• $\mathbb{Z}$ as a $\mathbb{Z}$-module is a.c.c., not d.c.c.
• Let $G=(\mathbb{Q}/\mathbb{Z})[p^{\infty }]=\{x\in \mathbb{Q}/\mathbb{Z}:p^{n}x=0\text{mbox}forsomen⩾1\}$, that is, $G$ is a $\mathbb{Z}$-module generated by $\{1/p,1/p^2,\cdots \}$. It is not a.c.c., but it is d.c.c. In fact, since $p^m$ has primitive root, we can prove that all non-trivial submodules are $\mathbb{Z}\cdot \frac{1}{p^n}/\mathbb{Z}$.
  • This example gives a non-Noetherian module over a Noetherian ring, and it gives a Artinian module over a non-Artinian ring.
• $\mathbb{C}[x_1,\cdots ,x_n,\cdots ]$ is non-Noetherian and non-Artinian.

Remark.

• Later in ^myy86x, Artinian ring $⟺$ Noetherian ring with $\dim =0$. Namely, Artinian rings are the most basic Noetherian rings.
• However, Artinian modules in general behave poorly.

Proposition

$M$ is a Noetherian $A$-module iff any submodule is finitely generated.

\begin{proof} See ^4af1b0. \end{proof}

Proposition

Let $0\to M^{\prime }\stackrel{\alpha }{\to }M\stackrel{\beta }{\to }{M}^{\prime \prime }\to 0$ be a short exact sequence of $A$-module, then

• $M$ is Noetherian iff $M^{\prime }$ and $M^{\prime \prime }$ are Noetherian;
• $M$ is Artinian iff $M^{\prime }$ and $M^{\prime \prime }$ are Artinian.

\begin{proof} i) See ^0af43c.

ii) Assume that $M$ is Artinian. For any descending chains $M^{\prime }=M_0^{\prime }⊋M_1^{\prime }⊋\cdots ⊋M_n^{\prime }⊋\cdots$ and $M^{\prime \prime }=M_0^{\prime \prime }⊋M_1^{\prime }⊋\cdots ⊋M_n^{\prime \prime }⊋\cdots$, they induce descending chains of $M$

$$\begin{array}{rl}& M={\beta }^{-1}(M_0^{\prime \prime })⊋{\beta }^{-1}(M_1^{\prime \prime })⊋\cdots ⊋{\beta }^{-1}(M_n^{\prime \prime })⊋\cdots ,\\ & M=\alpha (M_0^{\prime })⊋\alpha (M_1^{\prime })⊋\cdots ⊋\alpha (M_n^{\prime })⊋\cdots .\end{array}$$

Since $M$ is Artinian, there exist $s$ and $t$ such that ${\beta }^{-1}(M_k^{\prime \prime })={\beta }^{-1}(M_{k+1}^{\prime \prime })$ and $\alpha (M_t^{\prime })=\alpha (M_{t+1}^{\prime })$. Since $\beta$ is surjective and $\alpha$ is injective, it deduces that $M_k^{\prime \prime }=M_{k+1}^{\prime \prime }$ and $M_t^{\prime }=M_{t+1}^{\prime }$. Hence $M^{\prime }$ and $M^{\prime \prime }$ are Artinian.

Conversely, assume that $M^{\prime }$ and $M^{\prime \prime }$ are Artinian. If $M$ has a descending chain which is not stationary

$$M=M_0⊋M_1⊋\cdots ⊋M_n⊋\cdots$$

then we have infinitely many short exact sequences $0\to {\alpha }^{-1}(M_i)\to M_i\to \beta (M_i)\to 0$. Since $M_i⊋M_{i+1}$, at least one of $\{{\alpha }^{-1}(M_i),{\alpha }^{-1}(M_{i+1})\}$ and $\{{\beta }^{-1}(M_i),{\beta }^{-1}(M_{i+1})\}$ contain two distinct elements.

Consider the following chains of $M^{\prime }$ and $M^{\prime \prime }$

$$\begin{array}{rl}& M^{\prime }={\alpha }^{-1}(M_0)\supseteq {\alpha }^{-1}(M_1)\supseteq \cdots \supseteq {\alpha }^{-1}(M_n)\supseteq \cdots ,\\ & M^{\prime \prime }=\beta (M_0)\supseteq \beta (M_1)\supseteq \cdots \supseteq \beta (M_n)\supseteq \cdots .\end{array}$$

where some terms may repeat. Since $M^{\prime }$ and $M^{\prime \prime }$ are Artinian, these two chains must contain infinitely many same terms. That is, there exists $n_1$ and $n_2$ such that ${\alpha }^{-1}(M_n)={\alpha }^{-1}(M_{n+1})$ for all $n⩾n_1$ and $\beta (M_n)=\beta (M_{n+1})$ for all $n⩾n_2$. Take $n=n_1+n_2$, then ${\alpha }^{-1}(M_n)={\alpha }^{-1}(M_{n+1})$ and $\beta (M_n)=\beta (M_{n+1})$ hold, which is impossible. \end{proof}

Remark. Similarly to ^4af1b0., Artinian also has equivalent definition, see here.

Corollary

If $M_1,\cdots ,M_n$ are Noetherian (Artinian) $A$-module, then so is ${\oplus }_{i=1}^n{M}_i$.

Proposition

Let $A$ be a Noetherian/Artinian ring, and let $M$ be a finitely generated $A$-module. Then $M$ is a Noetherian/Artinian $A$-module.

\begin{proof} See ^gmjk45. \end{proof}

Proposition

Let $A$ be a Noetherian (rep. Artinian) ring, and let $\alpha \subseteq A$ be an ideal. Then $A/\alpha$ is a Noetherian (rep. Artinian) ring.

\begin{proof} Recall that ideals of quotient ring are quotient of ideals. \end{proof}

Composition Series

Definition

A chain of submodule of $M$ is a (finite) sequence of submodule

$$M=M_0⊋M_1⊋\cdots ⊋M_n=0.$$

Say its length is $n$. Say a chain is a composition series if no extra submodule can be inserted, that is, $M_{i-1}/M_i$ is a simple module for all $1⩽i⩽n$.

Proposition

Suppose $M$ has a composition series of length $n$. Then every composition series of $M$ has length $n$, and every chain in $M$ can be extended to a composition series.

\begin{proof} Define $\ell (M)$ be the minimal length of all possible composition series.

Step 1. Claim that if $N\subseteq M$, then $\ell (N)⩽\ell (M)$. And if $\ell (N)=\ell (M)$, then $M=N$.

Let $M=M_0\supset M_1\supset \cdots \supset M_{\ell (M)}=0$ be the composition series with the minimal length. Define $N_i=M_i\cap N$, then we have

$$N=N_0\supseteq N_1\supseteq \cdots \supseteq N_{\ell (M)}=0.\text{\hspace{1em}(*)}$$

But $N_{i-1}/N_i=N\cap M_{i-1}/N\cap M_i$ and there is a injective map $N_{i-1}/N_i\to M_{i-1}/M_i$. As $M_{i-1}/M_i$ is simple, $N_{i-1}/N_i\in \{0,M_{i-1}/M_i\}$. After removing repeated items, $(\ast )$ becomes a composition series of $N$. Hence $\ell (N)⩽\ell (M)$. Also if $\ell (N)=\ell (M)$, then $N_i=M_i$ and so $M=N$.

Step 2. Claim that any chain has length $⩽\ell (M)$.

Indeed, if $M=M_0⊋M_1⊋\cdots ⊋M_n=0$, then use step 1, $\ell (M)=\ell (M_0)>\ell (M_1)>\cdots >\ell (M_n)=0$ and so $\ell (M)⩾n$.

Step 3. By Step 2, any composition series has length $\ell (M)$.

Step 4. Claim that any chain can be refined to be a composition series.

Indeed, if $M_{i-1}/M_i$ is not simple, then we can strictly add a module $M_{i-1}⊋N⊋M_i$ to make its length strictly bigger. This process stops after finite times by Step 2. \end{proof}

Proposition

$M$ has a finite length composition series iff $M$ is Noetherian and Artinian.

\begin{proof} "→" Both acc and dcc are satisfied because the length of a chain is bounded by Step 2 above.

"←" $M$ is Noetherian yields $M$ has a proper maximal submodule $M_1$. Then $M_1$ is Noetherian by ^a5f817 and it has a proper maximal submodule $M_2$. The process stops by $M$ Artinian. Since $M_i$ is a maximal submodule of $M_{i-1}$, $M_{i-1}/M_i$ is simple. Thus, $M$ has a finite length composition series. \end{proof}

Remark. For $\mathbb{Z}$-module $\mathbb{Z}$, $\mathbb{Z}\supseteq (2)\supseteq (4)\supseteq (8)\supseteq \cdots$ has a infinite length composition series, and it is not an Artinian ring.

Definition

If $M$ satisfies ^a89d3d, then call it a module of finite length. Denote $\ell (M)$ as the length of any composition series.

Fact (Jordan-Holder theorem for modules of finite length). if ${\left(M_i\right)}_{0<1<\pi }$ and ${\left(M_i^{\prime }\right)}_{0<1<n}$ are any two composition series of $M$, there is a one-to-one correspondence between the set of quotients ${\left(M_{i-1}/M_i\right)}_{1<1<n}$ and the set of quotients ${\left(M_{i-1}^{\prime }/M_i^{\prime }\right)}_{1<1<n}$, such that corresponding quotients are isomorphic. The proof is the same as for finite groups.

Proposition

Let $0\to M^{\prime }\to M\to M^{\prime \prime }\to 0$ be a short exact sequence. Then

• $M$ is of finite length iff both $M^{\prime }$ and $M^{\prime \prime }$ are of finite length;
• in this case, $\ell (M)=\ell (M^{\prime })+\ell (M^{\prime \prime })$.

\begin{proof} i) By ^a5f817 and ^a89d3d.

ii) Let $M^{\prime }=M_0^{\prime }\supset M_1^{\prime }\supset \cdots \supset M_{\ell (M^{\prime })}^{\prime }=0$ and $M^{\prime \prime }=M_0^{\prime \prime }\supset M_1^{\prime \prime }\supset \cdots \supset M_{\ell (M^{\prime \prime })}^{\prime \prime }=0$ be composition series. Then it is easy to show

$$M=M_0={\beta }^{-1}(M_0^{\prime \prime })\supset {\beta }^{-1}(M_1^{\prime \prime })\supset \cdots \supset {\beta }^{-1}(M_{\ell (M^{\prime \prime })}^{\prime \prime })={\beta }^{-1}(0)=\alpha (M^{\prime })\supset \cdots \supset \alpha (M_{\ell (M^{\prime })}^{\prime })=0$$

is a composition series. \end{proof}

Examples. Here are several examples to compute the length of composition series.

• ${\ell }_{\mathbb{C}}(\mathbb{C}[x]/(x^2))=2$, because $\mathbb{C}[x]/(x)\cong \mathbb{C}$ and $\mathbb{C}[x]/(x^2)/\mathbb{C}[x]/(x)\simeq (x)/(x^2)\simeq \mathbb{C}$ are simple $\mathbb{C}$-modules.
• $\ell (\mathbb{Z}/4\mathbb{Z})=2$, because $0\to \mathbb{Z}/2\mathbb{Z}\stackrel{\times 2}{\to }\mathbb{Z}/4\mathbb{Z}\stackrel{\mathrm{mod}2}{\to }\mathbb{Z}/2\mathbb{Z}\to 0$ is exact
• ${\ell }_{\mathbb{Z}}(\mathbb{Z}[x]/(p^2,x^3))=6$, because $0\to p\mathbb{Z}[x]/(p^2,x^3)\to \mathbb{Z}[x]/(p^2,x^3)\to \mathbb{Z}[x]/(p,x^3)\to 0$ is exact, where $p\mathbb{Z}[x]/(p^2,x^3)\simeq \mathbb{Z}[x]/(p,x^3)\simeq {\mathbb{F}}_p[x]/(x^3)$.
  • ${\ell }_p({\mathbb{F}}_p[x]/(x^3))=3$, because $0\to {\mathbb{F}}_p[x]/(x)\to {\mathbb{F}}_p[x]/(x^3)\to {\mathbb{F}}_p[x]/(x^2)\to 0$ is exact and ${\ell }_p({\mathbb{F}}_p[x]/(x^2))=2$.
  • It deduces that ${\ell }_{\mathbb{Z}}(\mathbb{Z}[x]/(p^2,x^3))=2\times 3=6$. See here.

Proposition

Let $V$ be a vector space over field $k$. Then the followings are equivalent:

• finite dimension;
• finite length;
• a.c.c.;
• d.c.c.

\begin{proof} Note that i)→ii)→iii) and i)→ii)→iv) are easy. Then it remains to show iii)→i) and iv)→i).

iii)→i). Otherwise dimension is not finite, then $0\supset k{e}_1\supset k{e}_1\oplus k{e}_2\supset \cdots \supset {\oplus }_{i=1}^{n}k{e}_i\supset \cdots$ is not acc.

iv)→i). Take $V\supset {\oplus }_{i=1}^{\infty }k{e}_i$, then it gives $V\supset {\oplus }_{i=2}^{\infty }k{e}_i\supset \cdots \supset {\oplus }_{i=n}^{\infty }k{e}_i\supset \cdots$ and dcc fails. \end{proof}

Proposition

Let $A$ be a ring such that $(0)=m_1\cdot m_2\cdots m_n$ is a product of finitely many maximal ideals. Then $A$ is Noetherian iff $A$ is Artinian.

\begin{proof} Consider the chain

$$A⊋m_1\supseteq m_1{m}_2\supseteq \cdots \supseteq m_1\cdots m_n=0,$$

where each factor $m_1\cdots m_{i-1}/m_1\cdots m_i$ is a $A/m_i$-module as $m_i$ acts on $m_1\cdots m_{i-1}/m_1\cdots m_i$ trivially. Remark that $A/m_i:=k_i$ is a field. By ^e2295f, for each factor we have “acc iff dcc”. By ^a5f817, $A$ also satisfies “acc iff dcc”. \end{proof}

Text-Ex. 6.5

We say a topological space $X$ is Noetherian, if its open subsets satisfy a.c.c. If $X$ is Noetherian, then:

• it is quasi-compact; and
• any subspace is also Noetherian.

\begin{proof} i) Given any open cover $X={\cup }_{i\in I}{U}_i$, remark that $U_1\subseteq U_1\cup U_2\subseteq \cdots \subseteq {\cup }_{i\in I}{U}_i$ has to stop by a.c.c.

ii) Let $Y\subseteq X$. Consider open cover $Y={\cup }_{i\in I}{V}_i$. We aim to get $U_i$ with $U_i\cap Y=V_i$ and $U_i⊊U_{i+1}$. Take any $U_1$. For any $U_2$, take $U_2^{\prime }:=U_2\cup U_1$. That is, change the sequence $\{U_i{\}}_{i\in I}$ to $\{U_i^{\prime }={\cup }_{j⩽i}{U}_i{\}}_i$. Then $\{U_i^{\prime }\}$ stabilizes and so $\{V_i\}$ stabilizes. \end{proof}