# Noetherian Modules

Through this article, A is a fixed ring. For the first two sections, all modules are over A.

Recall that a submodule of a finitely generated module is not finitely generated in general. This will not happen if we constrain ourselves to a better behaved class of modules.

Definition.

Let M be an A-module. Consider the set $\Sigma$ of submodules of M, ordered by reverse inclusion, i.e. $N \le N'$ if and only if $N'\subseteq N$. We say M is noetherian if $\Sigma$ is a noetherian poset.

By proposition 1 hereM is noetherian if either of the following equivalent conditions holds.

• Every non-empty collection of submodules of M has a maximal element.
• If $N_0 \subseteq N_1 \subseteq N_2 \subseteq \ldots$ is a sequence of submodules of M, then $N_k = N_{k+1} = \ldots$ for some $k\ge 0$.

Philosophically, noetherian is a type of “finiteness” condition on modules as the results below show.

Proposition 1.

Given an exact sequence of A-modules: $0 \longrightarrow N \longrightarrow M \longrightarrow P \longrightarrow 0$,

M is noetherian if and only if N and P are.

Proof

Without loss of generality, we assume $N\subseteq M$ and $P = M/N$.

(⇒) Suppose M is noetherian. N is noetherian since submodules of N are submodules of M. Similarly, P is noetherian since submodules of P correspond bijectively to submodules of M containing N.

(⇐) Let $M_0 \subseteq M_1 \subseteq \ldots$ be a sequence of submodules of M. We get sequences of submodules $M_0 \cap N \subseteq M_1 \cap N \subseteq \ldots \subseteq N, \quad (M_0 + N)/N \subseteq (M_1 + N)/N \subseteq \ldots \subseteq M/N.$

Since N and M/N are noetherian, there is a k such that $M_k \cap N = M_{k+1} \cap N = \ldots$ and $M_k + N = M_{k+1} + N = \ldots$. Now apply the following to finish to job. ♦

Exercise A

Prove that if $Q \subseteq Q'$ and N are submodules of M such that $Q\cap N = Q'\cap N$ and $Q+N = Q'+N$, then $Q = Q'$.

Corollary 1.

• If M and N are noetherian modules, so is $M\oplus N$.
• If N and N’ are noetherian submodules of M, so is N + N’ (because it is a quotient of $N\oplus N'$). # Classification of Noetherian Modules

Lemma 1.

A noetherian module M is finitely generated.

Proof

Take the set of all finitely generated submodules of M. Since M is noetherian it has a maximal element $M_0$. If $M_0 \ne M$ pick any $x\in M - M_0$. Then $M_0 + (x)$ is a finitely generated submodule of M which properly contains $M_0$, which contradicts maximality of $M_0$. ♦

Proposition 2.

A module M is noetherian if and only if every submodule is finitely generated.

Proof

(⇒) Any submodule of a noetherian module is noetherian, and hence finitely generated.

(⇐) Pick any sequence of submodules $N_0 \subseteq N_1 \subseteq \ldots$ of M. Let $N = \cup_i N_i$. Show that this is a submodule of M. By the given condition it is finitely generated by, say, $x_1, \ldots, x_k$. Then for some n we have $x_1, \ldots, x_k \in N_n$. It follows that $N_n = N_{n+1} = \ldots = N$. ♦

Definition.

The ring A is noetherian if it is noetherian as a module over itself. By proposition 2, A is noetherian if and only if all its ideals are finitely generated.

In particular, a noetherian ring A satisfies a.c.c. on the set of its principal ideals. Hence in a noetherian integral domain, every non-zero element $a\in A$ can be factored as a product of irreducibles.

Proposition 3.

If the ring A is noetherian, an A-module M is noetherian if and only if it is finitely generated.

Proof

(⇒) Apply lemma 1. (⇐) Since A is noetherian as an A-module, so is any finite direct sum $A^n$. And since $M$ is finitely generated, it is a quotient of some $A^n$; hence M is noetherian. ♦

In summary, the class of finitely generated modules over noetherian rings is well-behaved: it is closed under taking submodules, quotient modules, finite direct sums and finite sums of submodules.

Exercise B

Let MN be modules; prove the following.

• If M and N are finitely generated, so is $M\otimes_A N$.
• If M is noetherian and N is finitely generated, then $M\otimes_A N$ is noetherian. # Constructing Noetherian Rings

Thus the question now stands: which rings are noetherian? Obviously fields are noetherian, having only two ideals. Next we have the following easy constructions.

Lemma 2.

Any principal ideal domain is noetherian.

Proof

Indeed we showed earlier that A satisfies a.c.c. on principal ideals. But all ideals of A are principal. ♦

Lemma 3.

If A is a noetherian ring, so is any localization $S^{-1}A$.

Proof

We showed that any ideal of $S^{-1}A$ is of the form $\mathfrak a (S^{-1}A)$ for some ideal $\mathfrak a \subseteq A$. If $\mathfrak a$ is generated by $x_1, \ldots, x_n$ then $\mathfrak a(S^{-1}A)$ is generated by $\frac {x_1} 1, \ldots, \frac{x_n}1$. ♦

Lemma 4.

If A and B are noetherian rings, so is $A\times B$.

Proof

We saw that any ideal of $A\times B$ is of the form $\mathfrak a \times \mathfrak b$ where $\mathfrak a, \mathfrak b$ are ideals of AB respectively. ♦

Lemma 5.

If A is noetherian, so is any quotient $A/\mathfrak a$.

Proof

Any ideal of $A/\mathfrak a$ is of the form $\mathfrak b/\mathfrak a$ where $\mathfrak b \subseteq A$ is an ideal, so it is finitely generated. ♦ # Hilbert Basis Theorem

Hilbert Basis Theorem.

If A is a noetherian ring, so is $A[X]$.

Proof

Suppose $\mathfrak b \subseteq A[X]$ is not finitely generated; we pick $f_0 \in \mathfrak b - \{0\}$ of minimum degree. Recursively, pick $f_n \in \mathfrak b - (f_0, f_1, \ldots, f_{n-1})$ of minimum degree. Let $d_i = \deg f_i$ so that $d_0 \le d_1 \le \ldots$. Suppose the leading term of $f_i$ is $a_i X^{d_i}$, where $a_i \in A$.

The sequence $(a_0) \subseteq (a_0, a_1) \subseteq \ldots$ of ideals of A eventually terminates so $a_n \in (a_0, a_1, \ldots, a_{n-1})$ for some n. We write $a_n = r_0 a_0 + \ldots + r_{n-1} a_{n-1}$ for some $r_0, \ldots, r_{n-1} \in A$. Now \begin{aligned} r_0 X^{d_n - d_0} f_0 &= r_0 a_0 X^{d_n} + (\text{smaller terms}), \\ r_1 X^{d_n - d_1} f_1 &= r_1 a_1 X^{d_n} + (\text{smaller terms}), \\ &\vdots \\ r_{n-1} X^{d_n - d_{n-1}} f_{n-1} &= r_{n-1} a_{n-1} X^{d_n} + (\text{smaller terms}).\end{aligned}

Let g be the sum of all these polynomials; then $g\in \mathfrak b$ has degree $d_n$ and leading coefficient $\sum_{i=0}^{n-1} r_i a_i = a_n$. Hence $f_n - g \in \mathfrak b$ has degree $< d_n$ and lies outside $(f_0, \ldots, f_{n-1})$, which contradicts minimality of $\deg f_n$. ♦

Immediately we have:

Corollary 2.

If A is a noetherian ring, so is any finitely generated A-algebra.

Proof

Such an algebra must be a (ring) quotient of $A[X_1, \ldots, X_n]$ for some n. ♦

Examples

Any finitely generated algebra over a field or $\mathbb Z$ is noetherian; more generally any localization of such a ring is noetherian.

Exercise C

Prove the following for A-algebras B and C.

• If B and C are of finite type over A, so is $B\otimes_A C$.
• If B is of finite type over A, and C is noetherian, then $B\otimes_A C$ is noetherian.

Note

In most cases of interest (e.g. algebraic number theory and algebraic geometry), our rings and modules will be noetherian. Most A-algebras we look at are of finite type over A but not finite over A. In other words, they are finitely generated as A-algebras but not as A-modules. Thus, if we look at such an algebra as an A-module, it is no longer noetherian and we have to be careful in applying the above results.

Next, by proposition 1 here, if M is a finitely generated module over a noetherian ring, it is flat if and only if it is projective. But that does not mean we can safely ignore their difference when dealing with concrete examples, for in the case of flatness, we are usually more interested in flat algebras rather than modules.

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