Commutative Algebra (VI): Chain Conditions

14 Feb 2019

• commutative algebra

We have previously considred finite generated and finite presented modules. In this post, we look at other finiteness properties of modules based on chain conditions.

Chain conditions

Proposition: Let $\Gamma$ be a partially ordered set. Then the following conditions are equivalent:

1. Any nonempty $S\subseteq\Gamma$ has a maximal element (ie. an element $x\in S$ such that $x\nless x’$ for all $x’\in S$);
2. Any ascending chain $x_1< x_2<\cdots$ of elements of $\Gamma$ must stop after finitely many steps (ie. there does not exist an infinite ascending chain);
3. Any nondecreasing chain $x_1\leq x_2\leq\cdots$ of elements in $\Gamma$ must be eventually constant (ie. there exists $N\geq1$ such that $x_n=x_N$ for all $n\geq N$).

Proof: ($1\Rightarrow2$) The contrapositive is clear: if $x_1< x_2<\cdots$ is an infinite ascending chain, then $S=\{x_n\,:\,n\geq1\}$ does not have a maximal element.

($2\Rightarrow3$) By removing repeated terms from a nondecreasing chain, we get an ascending chain, which only has finitely many terms by hypothesis; hence all terms after the first appearance of the largest term are equal to the largest term.

($3\Rightarrow1$) We prove the contrapositive. Take $S\subseteq\Gamma$ nonempty such that $S$ has no maximal element. By the axiom of choice, there exists a function $\varphi:S\to S$ such that $\varphi(x)>x$ for all $x\in S$. Pick any $x_1\in S$, and inductively define $x_{n+1}=\varphi(x_n)$. Then $x_1< x_2<\cdots$ is an infinite ascending chain, which in particular is not eventually constant. $\qed$

The equivalent conditions above are called the ascending chain condition (a.c.c.) for a partially ordered set $\Gamma$. By reversing the order, we define the descending chain condition (d.c.c.) analogously.

Proposition: If $\Gamma$ satisfies a.c.c. (resp. d.c.c.) and $\Gamma’\subseteq\Gamma$, then $\Gamma’$ also satisfies a.c.c. (resp. d.c.c.).

Proof: Every nonempty $S\subseteq\Gamma$ has a maximal (resp. minimal) element; in particular, so does every nonempty $S\subseteq\Gamma’$. $\qed$

Noetherian and Artinian modules

Let $A$ be a (commutative unital) ring, and $M$ an $A$-module. Note that the set of submodules of $M$ form a partially ordered set under inclusion. We say that $M$ is a Noetherian module (resp. Artinian module) if its set of submodules satisfies a.c.c. (resp. d.c.c.).

For instance, consider $\bb Z$ as a $\bb Z$-module. Its submodules are $\{0\}$ and $n\bb Z$ for integers $n\geq1$. Hence ascending chains of (nonzero) submodules correspond to decreasing sequences of positive integers, which must terminate in finitely many steps by the well-ordering property of $\bb Z_{>0}$. Thus $\bb Z$ is Noetherian. However, we have the infinite descending sequence

$$\bb Z\supsetneq2\bb Z\supsetneq2^2\bb Z\supsetneq\cdots,$$

so $\bb Z$ is not Artinian.

For another example, let $W$ be the abelian group of dyadic rationals,

$$W=\left\{\frac a{2^k}\,:\,a\in\bb Z,\,k\geq0\right\}.$$

Now consider the $\bb Z$-module $M=W/\bb Z$. For any submodule $N\subseteq M$, note that if $\frac a{2^k}\in N$ and $a$ is odd, then $a$ is invertible mod $2^k$ implies that $\frac1{2^k}\in N$. Hence we must have one of the following:

$$\begin{aligned} N&=0\\ \text{or }N&=\frac1{2^k}\bb Z\qquad\text{for some }k\geq0\\ \text{or }N&=\bigcup_{k\geq0}\frac1{2^k}\bb Z=M. \end{aligned}$$

Hence descending chains of submodules of $M$ correspond to decreasing sequences of $k$, which must terminate after finitely many steps. Thus $M$ is Artinian. However, we have the infinite ascending sequence

$$\{0\}\subsetneq\frac12\bb Z\subsetneq\frac1{2^2}\bb Z\subsetneq\cdots,$$

so $M$ is not Noetherian.

The Noetherian and Artinian properties carry over to submodules and quotient modules:

Proposition: Let $M$ be a Noetherian (resp. Artinian) $A$-module, and let $N\subseteq M$ be a submodule. Then $N$ and $M/N$ are also Noetherian (resp. Artinian).

Proof: Note that the set of submodules of $N$ is a subset of the set of submodules of $M$. Also, by the lattice isomorphism theorem, the set of submodules of $M/N$ is order-isomorphic to a subset of the set of submodules of $M$ (namely, those which contain $N$). The result now follows from the previous proposition. $\qed$.

In the other direction, we have the following result:

Proposition: Let $M$ be an $A$-module, and let $N\subseteq M$ be a submodule. If $N$ and $M/N$ are both Noetherian (resp. Artinian), then $M$ is also Noetherian (resp. Artinian).

Proof: We prove the statement for Noetherian modules; the proof for Artinian modules is analogous.

Let $M_1\subseteq M_2\subseteq\cdots$ be a nondecreasing chain of submodules of $M$. By hypothesis, the nondecreasing chains

$$\begin{aligned} M_1\cap N&\subseteq M_2\cap N\subseteq\cdots,\\ \frac{M_1+N}N&\subseteq\frac{M_2+N}N\subseteq\cdots \end{aligned}$$

are both eventually constant. Hence there exists $n_0$ such that for all $n\geq n_0$, we have

$$\begin{aligned} M_n\cap N&=M_{n_0}\cap N,\\ \frac{M_n}{M_n\cap N}\cong\frac{M_n+N}N&=\frac{M_{n_0}+N}N\cong\frac{M_{n_0}}{M_{n_0}\cap N}. \end{aligned}$$

But $M_{n_0}\subseteq M_n$, so $M_n=M_{n_0}$ for all $n\geq n_0$.

Hence the nondecreasing chain $M_1\subseteq M_2\subseteq\cdots$ is eventually constant, so $M$ is Noetherian. $\qed$

Recall that a composition series of $M$ is a chain of submodules

$$M=M_0\supseteq M_1\supseteq\cdots\supseteq M_r=0,$$

such that each $M_i/M_{i+1}$ is simple. We have proven in the previous post that any two composition series of $M$ have the same length, known as the (composition) length $\ell(M)$.

Proposition: Let $M$ be an $A$-module. Then $M$ has a composition series if and only if it is both Noetherian and Artinian.

Proof: ($\Rightarrow$) We proved that for any submodules $M_1\subsetneq M_2$ of $M$, we have $\ell(M_1)<\ell(M_2)$. Hence any ascending or descending chain of submodules of $M$ has at most $\ell(M)+1$ elements, so $M$ is Noetherian and Artinian.

($\Leftarrow$) Let $M_0=\{0\}$, and inductively define $M_{k+1}$ to be a minimal submodule strictly containing $M_k$. Since $M$ is Artinian, this always exists unless $M_k=M$. Note that by minimality, $M_{k+1}/M_k$ has no proper submodules except $0$, ie. $M_{k+1}/M_k$ is simple.

Now we have an ascending chain $M_0\subsetneq M_1\subsetneq\cdots$ of submodules of $M$. Since $M$ is Noetherian, this chain must stop after finitely many steps, so $M_n=M$ for some $n\geq0$. Then

$$0=M_0\subseteq M_1\subseteq\cdots\subseteq M_n=M$$

is a composition series for $M$. $\qed$

Hence being both Noetherian and Artinian is equivalent to the finiteness of composition length.

Noetherian and Artinian rings

We say that $A$ is a Noetherian ring (resp. Artinian ring) if it is Noetherian (resp. Artinian) as an $A$-module; equivalently, if its set of ideals satisfies a.c.c. (resp. d.c.c.). The above results for Noetherian and Artinian modules yield the following:

Proposition: Let $A$ be a Noetherian (resp. Artinian) ring, and let $I\subseteq A$ be an ideal. Then $A/I$ is also Noetherian (resp. Artinian).

Proof: We proved that $A/I$ is a Noetherian (resp. Artinian) $A$-module. Now note that the $A$-submodules of $A/I$ are precisely the $A/I$-submodules, namely the ideals of $A/I$. $\qed$

Hence the Noetherian and Artinian properties carry over to quotient rings, but not subrings in general (since subrings of $A$ are not $A$-modules).

Proposition: Let $A$ be a Noetherian (resp. Artinian) ring. Then any finite $A$-module is also Noetherian (resp. Artinian).

Proof: First we induct on $n$ to show that the free module $A^n$ is Noetherian (resp. Artinian). This is given for $n=1$. If this is true for $n$, then the $A$-modules $A^n$ and $A^{n+1}/A^n\cong A$ are Noetherian (resp. Artinian), hence by the previous result, so is $A^{n+1}$.

Now any finite $A$-module is a quotient of some free module $A^n$, and hence is also Noetherian (resp. Artinian). $\qed$

Corollary: If $A$ is a Noetherian (resp. Artinian) ring and $B\supseteq A$ is a ring which is finitely generated as an $A$-module, then $B$ is also a Noetherian (resp. Artinian) ring.

Proof: By the previous proposition, $B$ is a Noetherian (resp. Artinian) $A$-module. Since every $B$-submodule of $B$ is also an $A$-submodule, $B$ is also a Noetherian (resp. Artinian) $B$-module. $\qed$

A warning to the reader: note that, for instance, $A[X]$ is not finitely generated as an $A$-module, even though it is finitely generated as an $A$-algebra (by $X$). For Noetherian rings, the extension of the above result to finitely generated algebras uses the Hilbert basis theorem, which we will cover in the next post.

Artinian rings

We have seen that the Noetherian and Artinian conditions for modules are independent. However, for rings we have the following surprising result:

Akizuki-Hopkins-Levitzki Theorem: Any Artinian ring is Noetherian.

We proceed in several steps.

Proposition: Let $A$ be an Artinian ring. Then $A$ has finitely many maximal ideals.

Proof: Suppose otherwise; then let $\mf m_1,\mf m_2,\ldots$ be pairwise distinct maximal ideals. For any $k\geq2$, pick $x_i\in\mf m_i\backslash\mf m_k$ for $1\leq i\leq k-1$. Since $\mf m_k$ is prime, we have

$$x_1\cdots x_{k-1}\in\mf m_1\cdots\mf m_{k-1}\backslash\mf m_k.$$

Then we have the infinite descending chain of ideals

$$\mf m_1\supsetneq\mf m_1\mf m_2\supsetneq\mf m_1\mf m_2\mf m_3\supsetneq\cdots,$$

which is the desired contradiction. $\qed$

Proposition: Let $A$ be an Artinian ring. Then the Jacobson radical $\rad(A)$ is nilpotent.

Proof: Write $I=\rad(A)$, and consider the nonincreasing chain

$$I\supseteq I^2\supseteq\cdots,$$

which must be eventually constant; hence there exists $s\geq1$ with $I^s=I^{s+1}$.

Assume for sake of contradiction that $I^s\neq(0)$. Consider the family of ideals $J\subseteq A$ such that $JI^s\neq(0)$; since $II^s=I^{s+1}=I^s\neq(0)$, this family is nonempty, so it has a minimal element $J$.

Now there exists $x\in J$ such that $I^sx\neq0$, so $I^s(Ax)\neq(0)$ and $Ax\subseteq J$ implies $J=Ax$ by minimality of $J$; in particular, $J$ is finitely generated as an $A$-module.

Also, note that

$$(JI^s)I^s=JI^{2s}=JI^s\neq(0),$$

so $JI^s\subseteq J$ implies $JI^s=J$ by minimality of $J$. By Nakayama’s lemma, we have $J=(0)$, so $JI^s=(0)$, which is the desired contradiction. $\qed$

Corollary: Let $A$ be an Artinian ring. Then every prime ideal of $A$ is maximal.

Proof: Let the maximal ideals of $A$ be $\mf m_1,\ldots,\mf m_r$, and let $\mf p$ be any prime ideal of $A$. Since the Jacobson radical $\rad(A)=\mf m_1\mf m_2\cdots\mf m_r$ is nilpotent, we have

$$(\mf m_1\cdots\mf m_r)^s=(0)\subseteq\mf p$$

for some $s\geq1$. Hence we must have $\mf m_k\subseteq\mf p$ for some $k$, so $\mf p=\mf m_k$. $\qed$

Proof of Akizuki-Hopkins-Levitzki Theorem: Let $A$ be an Artinian ring, with maximal ideals $\mf m_1,\ldots,\mf m_r$, and Jacobson radical $I=\rad(A)=\mf m_1\mf m_2\cdots\mf m_r$ satisfying $I^s=(0)$. Consider the chain of ideals

$$\begin{aligned} A&\supseteq\mf m_1\supseteq\mf m_1\mf m_2\supseteq\cdots\supseteq I\\ &\supseteq I\mf m_1\supseteq I\mf m_1\mf m_2\supseteq\cdots\supseteq I^2\\ &\supseteq\cdots\\ &\supseteq I^s=(0). \end{aligned}$$

For any two consecutive terms in this sequence, say $M$ and $M\mc m_i$, note that $M/M\mf m_i$ is a $(A/\mf m_i)$-vector space, whose $(A/\mf m_i)$-vector subspaces are precisely its $A$-submodules. Hence

$$\begin{aligned} A&\text{ is an Artinian }A\text{-module}\\ \implies M&\text{ is an Artinian }A\text{-module}\\ \implies M/M\mf m_i&\text{ is an Artinian }A\text{-module}\\ \implies M/M\mf m_i&\text{ is an Artinian }(A/\mf m_i)\text{-vector space}, \end{aligned}$$

so $M/M\mf m_i$ is finite-dimensional over $A/\mf m_i$. Thus $\ell(M/M\mf m_i)<\infty$. By additivity of composition length, we get $\ell(A)<\infty$, so $A$ has a composition series implies $A$ is a Noetherian ring. $\qed$