Chain Conditions in Commutative Rings

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On this page:

• Introduction
• Noetherian Rings
• Worked Exercises

Introduction

Instrumental to the study of rings is the study of factorizations. On this page, we will study the chain conditions a bit with the goal of factorization in mind.

Noetherian Rings

[definition] [deftitle]Definition (Noetherian $R$-modules).[/deftitle]

Let $R$ be a commutative ring and $M$ an $R$-module. We say that $M$ is Noetherian provided that every submodule of $M$ is f.g. In particular, $R$ is Noetherian if all of its ideals are f.g. [/definition]

There are various different equivalent ways we can express the Noetherian property, all of them having their uses. For instance, we see that if $N \le M$, then $N$ is f.g. and so we can write

where each $m_i \in M$. The fact that $M$ is Noetherian implies that, at some point, we can no longer add generators to the list $m_1, \ldots, m_t$ without getting the same module at some point. Thus, we arrive at the following proposition:

[theorem] [thmtitle]Proposition (Noetherian equivalencies).[/thmtitle]

Let $R$ be a commutative ring and $M$ an $R$-module. Then the following are equivalent:

1. $M$ is Noetherian.
2. $M$ satisfies the ascending chain condition (ACC): Every ascending chain of submodules of $M$ stabilizes.
3. Every nonempty family of submodules of $M$ has a maximal element with respect to inclusion. [/theorem]

[proof] (1) $\implies$ (2). Let ${N_i}$ be an ascending chain of submodules of $M$. Define

Since $M$ is Noetherian, $N$ is f.g. and can be written as

By definition of $N$ being the union of an ascending chain, there must be some $N_i$ that contains all of these generators. Thus,

So all of the $N_{j}$ for $j \ge i$ are all just $N$.

(2) $\implies$ (3). Let $\mathcal{F}$ be a nonempty family of submodules of $M$. By (2), it follows that any chain has a maximal element that is in $\mathcal{F}$ so Zorn’s lemma implies that $\mathcal{F}$ has a maximal element.

(3) $\implies$ (2). Let $\mathcal{F}$ be the family of f.g. submodules of $M$. Now $\mathcal{F}$ is nonempty since the zero module is contained in $\mathcal{F}$. Now, if $N$ is a submodule of $M$, say not f.g., then there is some nonzero $n_1 \in N$ for which

Thus, there is some $n_2 \in N \setminus N_1$ for which

Again, there is some $n_3 \in N \setminus (N_1 \cup N_2)$ for which

Thus, we can inductively construct an asending chain of f.g. submodules that are contained in $N$. By (3), there is some maximal element $P$ of $\mathcal{F}$. Now, we must have

by the definition of $N$. But this means that there is some f.g. submodule that is between $P$ and $N$ (by the above construction). Thus, $P = N$ as desired, contradiction. Thus, $N$ is f.g. and we are done. [/proof]

A very natural question is to ask if submodules and quotients of a Noetherian module are again Noetherian. In fact, if $N \le M$ and $M/N$ are Noetherian, then so is $M$!

[theorem] [thmtitle]Proposition (Submodules and quotients of Noetherian modules are Noetherian).[/thmtitle]

Let $M$ be a module over the commutative ring $R$ and $N$ be a submodule of $M$. Then $M$ is Noetherian if and only both $N$ and $M/N$ are Noetherian. [/theorem]

[proof] ($\implies$) If $M$ is Noetherian, then $N$ must be Noetherian because every submodule of $N$ is also a submodule of $M$ and is, thus, f.g. To see that $M/N$ is Noetherian, we use the correspondence theorem for modules: the submodules of $M/N$ of the form $P/N$ for some submodule $P$ of $M$ that contains $N$. Thus, $P$ is f.g. and so the respresentors of the generators of $P$ will generate $P/N$.

($\impliedby$) Conversely, if both $N$ and $M/N$ are Noetherian, then consider some submodule $P$ of $M$. Now $P \cap N$ is a submodule of $N$ and is f.g. by $N$ being Noetherian. The second isomorphism theorem implies that

Since $(P + N)/N$ is a submodule of $M/N$, it follows that $(P + N)/N$ is a f.g. by $M/N$ being Noetherian. Thus, $P/(P \cap N)$ is f.g.. We claim that this implies that $P$ is, thus, also f.g.. Indeed, since $P\cap N$ and $P/(P\cap N)$ are both f.g., it follows that $P$ itself is f.g. and we are done. [/proof]

Noetherian rings are well known for being nice to work with. Recall that we know that every f.g. $R$-algebra is isomorphic to $R[\mathbf{x}]/J$ where $J$ is the ideal that encodes the relations in the $R$-algebra. In the case where $R$ is Noetherian, this gives rise to the result:

This is useful because it gives us a lot of examples of Noetherian rings. We will prove this by first showing that $R[\mathbf{x}]$ itself is Noetherian then prove a more general lemma that directly implies the result.

[theorem] [thmtitle]Lemma (F.g. modules over Noetherian rings are Noetherian).[/thmtitle]

Let $R$ be a commutative Noetherian ring and $M$ a f.g. $R$-module. Then $M$ is Noetherian. [/theorem]

[proof] Since $M$ is f.g., it is isomorphic to a quotient of $R^{\oplus n}$. By the proposition above, it suffices to show that $R^{\oplus n}$ itself is f.g. as an $R$-module. We will show this by induction on $n$. The basis is obvious since $R^{\oplus 1} = R$ is Noetherian. For the inductive step, we identity $R^{\oplus (n-1)}$ with the submodule $R^{\oplus (n-1)} \oplus 0$ of $R^{\oplus n}$. Noting that

by the First Isomorphism Theorem applied to the $n$th-factor projection map. We apply the proposition above by using the inductive hypothesis to get Noetherian-ness of $R^{\oplus(n-1)}$ and noting that $R$ is Noetherian by assumption. Thus, $R^{\oplus n}$ is Noetherian. [/proof]

[theorem] [thmtitle]Theorem (Hilbert Basis Theorem).[/thmtitle]

Let $R$ be a commutative Noetherian ring. Then $R[x_1, \ldots, x_n]$ is also a commutative Noetherian ring. [/theorem]

[proof] We proceed by induction on $n$. For the base case $n = 1$, this amounts to proving that $R[x]$ is a Noetherian ring. Let $I \lhd R[x]$. Our goal is to show that $I$ is f.g. in $R[x]$. We will take the approach of Aluffi’s Algebra: Chapter 0 book. We define

Now $A \subseteq R$ and it is fairly immediate that $A$ is also an ideal of $R$. As $R$ is Noetherian, $A$ is f.g.:

This implies that there are elements $p_1(x), \ldots, p_r(x) \in I$ for which $a_i$ is the leading coefficient of $p_i(x)$.

Now define $d_i \coloneqq \deg p_i(x)$ and let $d = \max_i d_i$. Let

This is a f.g. $R$-module and so it is Noetherian by the lemma above. Thus, $M \cap I$ is a submodule of $M$ and is f.g. over $R$:

We claim that

Since all of the generators of the ideal on the right are in $I$, the $\supseteq$ direction is immediate. Let us know show $\subseteq$. Let $f(x)$ be some element of $I$. Now, if $\deg f < d$, then $f(x) \in M$ and in $M \cap I$. If $\deg f \ge d$, then we can do the following procedure: Let $a$ be the leading coefficient of $f(x)$. Then $a \in A$ and there exist $b_1, \ldots, b_r \in R$ such that

If $e = \deg f(x)$, then it follows that

has $\deg f_1(x) < \deg f$. We repeat this process on $f_1(x)$ to obtain $f_2(x)$ and so on until we obtain some

such that $\deg f_m < d$, where each $g_i(x)$ is obtain by collecting terms in the $f_i(x)$. But this means $f_m(x) \in M\cap I$ and so

where $c_i \in R$. Thus,

which implies that

as desired. Thus, $I$ is f.g..

For the inductive step, we just note that

which reduces the inductive step to the basis of induction. Thus, $R[x_1, \ldots, x_n]$ is Noetherian and we are done. [/proof]

Now that we have the Hilbert Basis Theorem, proving that quotients of $R[\mathbf{x}]$ is completely trivial.

[theorem] [thmtitle]Theorem.[/thmtitle]

Let $R$ be a Noetherian ring and $J$ be an ideal of $R[\mathbf{x}]$. Then the ring $R[\mathbf{x}]/J$ is Noetherian. [/theorem]

[proof] By the Hilbert Basis Theorem, $R[\mathbf{x}]$ is Noetherian and so it automatically follows that $J$ and $R[\mathbf{x}]/J$ are Noetherian too. [/proof]

[example] [extitle]Remark.[/extitle]

The above result is especially useful for results in classical algebraic geometry. The coordinate rings can be realized as ring quotients $k[\mathbf{x}]/J$ for an algebraically closed field $k$ — That informs us about the structures of the varieties that arise.

Aluffi also says that the quotients $R[\mathbf{x}]/J$ also include examples from number theory, although, I do not know anything about this side of the story since I know close to nothing about number theory. [/example]

Worked Exercises

[example] [extitle]Aluffi Chapter 0 Problem 5.1.1[/extitle]

Let $R$ be a commutative Noetherian ring and $I$ an ideal of $R$. Then $R/I$ is a Noetherian ring. [/example]

[proof] A commutative ring is always a module over itself and its submodules are the ideals of the ring. Furthermore, we know that given a module and one of its submodules, that module is Noetherian iff its submodule and the quotient by that submodule are. Specialing to $R$ gives precisely the desired result. [/proof]

[example] [extitle]Aluffi Chapter 0 Problem 5.1.2[/extitle]

Let $R$ be a commutative ring. Then if $R[x]$ is a Noetherian ring, so is $R$. [/example]

[proof] Let $I$ be an ideal of $R$. If we let $I[x]$ be the subset of $R[x]$ consisting of the polynomials in $x$ with coefficients in $I$, then it follows that $I[x]$ is an ideal of $R[x]$. Thus, $I[x]$ is f.g. and can be written as

We claim that

($\subseteq$) If $i \in I$, then it is also an element of $I[x]$ and so

Evaluating both sides on $0$ gives $i \in \generator{f_1(0), \ldots, f_s(0)}$.

($\supseteq$) Immediate.

Thus, we see that $I$ is f.g. and we are done. [/proof]

[example] [extitle]Aluffi Chapter 0 Problem 5.1.7[/extitle]

Let $R$ be a commutative Noetherian ring. Then the ring of power series $R[[x]]$ is also Noetherian. [/example]

Let $I$ be an ideal of $R[[x]]$. Our goal is to show that $I$ is f.g.. Define the order of the power series $\sum_{i\ge 0} a_i x^i$ to be the smallest $i$ for which $a_i \neq 0$. The dominant coefficient or order $i$ is $a_i$.

Define $A_i \subseteq R$ to be the set of dominant coefficients of power series of order $i$ in $I$ together with $0$. It is immediate that $A_i$ is an ideal. Furthermore, the $A_i$ form an AC

as any nonzero element of $A_i$ can be made of order $i+1$ by multiplying by $x$. So $A_{i+1}$ necessarily must contain the dominant coefficients of order $i$. As $R$ is Noetherian, the AC above stabilizes and each $A_i$ is f.g..

TODO: FINISH

[example] [extitle]Aluffi Chapter 0 Problem 5.1.10[/extitle]

If $R$ is a commutative Artinian ring and $I \subseteq R$ is an ideal, then $R/I$ is Artinian. [/example]

[proof] By the correspondence theorem, we know that the ideals of $R/I$ are of the form $J/I$ where $J$ is an ideal of $R$ containing $I$. Let ${J_n/I}$ be a DC of ideals of $R/I$. As $R$ is Artinian, we know that the DC of ideals ${J_n}$ stabilizes, say at $J_n$. Accordingly, it follows that

Thus, $R/I$ satisfies the DCC and we are done. [/proof]

[example] [extitle]Aluffi Chapter 0 Problem 5.1.10 (cont.)[/extitle]

If $R$ is an Artinian integral domain, then $R$ is a field. [/example]

[proof] Let $r$ be a nonzero element of $R$. Note that

is a DC of ideals of $R$. Since $R$ is Artinian, it follows that the chain stabilizes and

for some $n$. Since $R$ is an integral domain, $r^n$ and $r^{n+1}$ are associates so $r^n = ur^{n+1}$ for some unit $u \in R$. The cancellative property of integral domains implies that $1 = ur$, that is $r$ is a unit. This implies that every nonzero element of $R$ is a unit and, therefore, a field. [/proof]

[example] [extitle]Aluffi Chapter 0 Problem 5.1.10 (still cont.)[/extitle]

If $R$ is a commutative Artinian ring that is not an integral domain, then $R$ has Krull dimension $0$. [/example]

[proof] Let $P$ be prime ideal of $R$. Then $R/P$ is an integral domain which is a field. So there are no proper prime ideals $P’$ satisfying $P \subset P’ \subset R$. [/proof]