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Commutative Algebra (VI): Chain Conditions

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 Γ\Gamma be a partially ordered set. Then the following conditions are equivalent:

  1. Any nonempty SΓS\subseteq\Gamma has a maximal element (ie. an element xSx\in S such that xxx\nless x’ for all xSx’\in S);
  2. Any ascending chain x1<x2<x_1< x_2<\cdots of elements of Γ\Gamma must stop after finitely many steps (ie. there does not exist an infinite ascending chain);
  3. Any nondecreasing chain x1x2x_1\leq x_2\leq\cdots of elements in Γ\Gamma must be eventually constant (ie. there exists N1N\geq1 such that xn=xNx_n=x_N for all nNn\geq N).

Proof: (121\Rightarrow2) The contrapositive is clear: if x1<x2<x_1< x_2<\cdots is an infinite ascending chain, then S={xn:n1}S=\{x_n\,:\,n\geq1\} does not have a maximal element.

(232\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.

(313\Rightarrow1) We prove the contrapositive. Take SΓS\subseteq\Gamma nonempty such that SS has no maximal element. By the axiom of choice, there exists a function φ:SS\varphi:S\to S such that φ(x)>x\varphi(x)>x for all xSx\in S. Pick any x1Sx_1\in S, and inductively define xn+1=φ(xn)x_{n+1}=\varphi(x_n). Then x1<x2<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 Γ\Gamma. By reversing the order, we define the descending chain condition (d.c.c.) analogously.

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

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

Noetherian and Artinian modules

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

For instance, consider Z\bb Z as a Z\bb Z-module. Its submodules are {0}\{0\} and nZn\bb Z for integers n1n\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 Z>0\bb Z_{>0}. Thus Z\bb Z is Noetherian. However, we have the infinite descending sequence

Z2Z22Z,\bb Z\supsetneq2\bb Z\supsetneq2^2\bb Z\supsetneq\cdots,

so Z\bb Z is not Artinian.

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

W={a2k:aZ,k0}.W=\left\{\frac a{2^k}\,:\,a\in\bb Z,\,k\geq0\right\}.

Now consider the Z\bb Z-module M=W/ZM=W/\bb Z. For any submodule NMN\subseteq M, note that if a2kN\frac a{2^k}\in N and aa is odd, then aa is invertible mod 2k2^k implies that 12kN\frac1{2^k}\in N. Hence we must have one of the following:

N=0or N=12kZfor some k0or N=k012kZ=M. \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 MM correspond to decreasing sequences of kk, which must terminate after finitely many steps. Thus MM is Artinian. However, we have the infinite ascending sequence

{0}12Z122Z,\{0\}\subsetneq\frac12\bb Z\subsetneq\frac1{2^2}\bb Z\subsetneq\cdots,

so MM is not Noetherian.

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

Proposition: Let MM be a Noetherian (resp. Artinian) AA-module, and let NMN\subseteq M be a submodule. Then NN and M/NM/N are also Noetherian (resp. Artinian).

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

In the other direction, we have the following result:

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

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

Let M1M2M_1\subseteq M_2\subseteq\cdots be a nondecreasing chain of submodules of MM. By hypothesis, the nondecreasing chains

M1NM2N,M1+NNM2+NN \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 n0n_0 such that for all nn0n\geq n_0, we have

MnN=Mn0N,MnMnNMn+NN=Mn0+NNMn0Mn0N. \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 Mn0MnM_{n_0}\subseteq M_n, so Mn=Mn0M_n=M_{n_0} for all nn0n\geq n_0.

Hence the nondecreasing chain M1M2M_1\subseteq M_2\subseteq\cdots is eventually constant, so MM is Noetherian.  \qed

Recall that a composition series of MM is a chain of submodules

M=M0M1Mr=0,M=M_0\supseteq M_1\supseteq\cdots\supseteq M_r=0,

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

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

Proof: (\Rightarrow) We proved that for any submodules M1M2M_1\subsetneq M_2 of MM, we have (M1)<(M2)\ell(M_1)<\ell(M_2). Hence any ascending or descending chain of submodules of MM has at most (M)+1\ell(M)+1 elements, so MM is Noetherian and Artinian.

(\Leftarrow) Let M0={0}M_0=\{0\}, and inductively define Mk+1M_{k+1} to be a minimal submodule strictly containing MkM_k. Since MM is Artinian, this always exists unless Mk=MM_k=M. Note that by minimality, Mk+1/MkM_{k+1}/M_k has no proper submodules except 00, ie. Mk+1/MkM_{k+1}/M_k is simple.

Now we have an ascending chain M0M1M_0\subsetneq M_1\subsetneq\cdots of submodules of MM. Since MM is Noetherian, this chain must stop after finitely many steps, so Mn=MM_n=M for some n0n\geq0. Then

0=M0M1Mn=M0=M_0\subseteq M_1\subseteq\cdots\subseteq M_n=M

is a composition series for MM.  \qed

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

Noetherian and Artinian rings

We say that AA is a Noetherian ring (resp. Artinian ring) if it is Noetherian (resp. Artinian) as an AA-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 AA be a Noetherian (resp. Artinian) ring, and let IAI\subseteq A be an ideal. Then A/IA/I is also Noetherian (resp. Artinian).

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

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

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

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

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

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

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

A warning to the reader: note that, for instance, A[X]A[X] is not finitely generated as an AA-module, even though it is finitely generated as an AA-algebra (by XX). 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 AA be an Artinian ring. Then AA has finitely many maximal ideals.

Proof: Suppose otherwise; then let m1,m2,\mf m_1,\mf m_2,\ldots be pairwise distinct maximal ideals. For any k2k\geq2, pick ximi\mkx_i\in\mf m_i\backslash\mf m_k for 1ik11\leq i\leq k-1. Since mk\mf m_k is prime, we have

x1xk1m1mk1\mk.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

m1m1m2m1m2m3,\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 AA be an Artinian ring. Then the Jacobson radical rad(A)\rad(A) is nilpotent.

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

II2,I\supseteq I^2\supseteq\cdots,

which must be eventually constant; hence there exists s1s\geq1 with Is=Is+1I^s=I^{s+1}.

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

Now there exists xJx\in J such that Isx0I^sx\neq0, so Is(Ax)(0)I^s(Ax)\neq(0) and AxJAx\subseteq J implies J=AxJ=Ax by minimality of JJ; in particular, JJ is finitely generated as an AA-module.

Also, note that

(JIs)Is=JI2s=JIs(0),(JI^s)I^s=JI^{2s}=JI^s\neq(0),

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

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

Proof: Let the maximal ideals of AA be m1,,mr\mf m_1,\ldots,\mf m_r, and let p\mf p be any prime ideal of AA. Since the Jacobson radical rad(A)=m1m2mr\rad(A)=\mf m_1\mf m_2\cdots\mf m_r is nilpotent, we have

(m1mr)s=(0)p(\mf m_1\cdots\mf m_r)^s=(0)\subseteq\mf p

for some s1s\geq1. Hence we must have mkp\mf m_k\subseteq\mf p for some kk, so p=mk\mf p=\mf m_k.  \qed

Proof of Akizuki-Hopkins-Levitzki Theorem: Let AA be an Artinian ring, with maximal ideals m1,,mr\mf m_1,\ldots,\mf m_r, and Jacobson radical I=rad(A)=m1m2mrI=\rad(A)=\mf m_1\mf m_2\cdots\mf m_r satisfying Is=(0)I^s=(0). Consider the chain of ideals

Am1m1m2IIm1Im1m2I2Is=(0). \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 MM and MmiM\mc m_i, note that M/MmiM/M\mf m_i is a (A/mi)(A/\mf m_i)-vector space, whose (A/mi)(A/\mf m_i)-vector subspaces are precisely its AA-submodules. Hence

A is an Artinian A-module    M is an Artinian A-module    M/Mmi is an Artinian A-module    M/Mmi is an Artinian (A/mi)-vector space, \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/MmiM/M\mf m_i is finite-dimensional over A/miA/\mf m_i. Thus (M/Mmi)<\ell(M/M\mf m_i)<\infty. By additivity of composition length, we get (A)<\ell(A)<\infty, so AA has a composition series implies AA is a Noetherian ring.  \qed

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