Is every Noetherian module finitely generated?

Question

I was just wondering whether the following statement is correct.

Let R be a ring and M a noetherian R module. Then M is finitely generated.

Answer 1

The three standard equivalences for Noetherian are:

Theorem. Let $M$ be an $R$-module. Assuming the Axioms of Choice, the following are equivalent:

1. $M$ has ACC on submodules.
2. Every submodule of $M$ is finitely generated.
3. Every nonempty set of submodules of $M$ has maximal elements.

Proof. 1$\implies$2. (Uses dependent choice) Assume $N$ is a submodule of $M$ that is not finitely generated. We define a sequence of elements of $N$ inductively as follows: since $N$ is not finitely generated, $N\neq 0$. Let $n_1\in N$, $n_1\neq 0$. Since $N$ is not finitely generated, $\langle n_1\rangle\subsetneq N$, so there exists $n_2\in N-\langle n_1\rangle$. Assume we have chosen elements $n_1,\ldots,n_k\in N$ such that $$\langle n_1\rangle \subsetneq \langle n_1,n_2\rangle\subsetneq\cdots\subsetneq \langle n_1,\ldots,n_k\rangle.$$ Since $N$ is not finitely generated, $\langle n_1,\ldots,n_k\rangle \subseteq N$, so there exists $n_{k+1}\in N\setminus\langle n_1,\ldots,n_k\rangle$.

Thus, we have an infinite ascending chain of submodules in $M$, so $M$ does not satisfy ACC.

1$\implies 3$ (uses Zorn's Lemma): Since every ascending chain in $M$ is finite, any nonempty collection of submodules of $M$ satisfies the hypothesis of Zorn's Lemma under the partial order of inclusion (take the maximum of a chain to get an upper bound). Hence the set has maximal elements.

2$\implies$1 (Does not require AC) Let $N_1\subseteq N_2\subseteq\cdots$ be an ascending chain of submodules. Then $N=\cup N_i$ is a submodules of $N$, hence is finitely generated, $N=\langle n_1,\ldots,n_k\rangle$. For each $i$, let $m_i$ be such that $n_i\in N_{m_i}$. Let $m=\max\{m_1,\ldots,m_k\}$. Then $N\subseteq N_m\subseteq N_{m+k} \subseteq \cup N_i\subseteq N$, so $N= N_m=N_{m+k}$ for all $k$; that is, the chain stabilizes after finitely many steps.

2$\implies$3 (Uses AC) Essentially, go through 1 to show any nonempty collection of submodules satisfies Zorn's Lemma to conclude the collection has maximal elements.

3$\implies$1 (Does not require AC) Given an ascending chain of submodules, by 3 the chain has maximal (hence a maximum) element. Thus, it stabilizes after finitely many steps.

3$\implies$2 (Does not require AC) Let $N$ be a submodule of $M$. Let $S$ be the collection of all finitely generated submodules of $N$. It is not empty (it contains $0$), hence has a maximal element $\mathcal{N}$ which is a fortiori finitely generated. For every $n\in N$, $\langle \mathcal{N},n\rangle$ is finitely generated, and $\mathcal{N}\subseteq \langle \mathcal{N},n\rangle$, so maximality of $\mathcal{N}$ gives $\mathcal{N}=\langle \mathcal{N},n\rangle$. Thus, for every $n\in N$, $n\in \mathcal{N}$, so $N\subseteq \mathcal{N}\subseteq N$, proving that $N=\mathcal{N}$ and so $N$ is finitely generated. $\Box$

In particular, if $M$ is noetherian, then every submodule of $M$ is finitely generated, and in particular the module $M$ itself is finitely generated.

Answer 2

A module $M$ is Noetherian if it satisfies the ascending chain condition, or ACC. That is, there are no infinite chains of submodules $$ M_0\subset M_1\subset M_2\subset\dotsb $$ where inclusion is strict. It's a pretty easy exercise to show that a module $M$ has the ACC if and only if every submodule of $M$ is finitely generated. For the forward direction, consider an arbitrary submodule, and then keep adding elements that are not already in the submodule to build a chain. In the reverse direction, consider a chain and take the union of all the elements in that chain.

Since $M$ is a submodule of itself, $M$ must be finitely generated. To ask a similar question about Artinian rings, you need the additional concept of Krull dimension.

Definition: Let $A$ be a commutative ring and consider chains of prime ideals $$ P_0\subset P_1\subset\dotsb\subset P_s $$ in $A$. Then the Krull dimension of $A$ is the supremum of all such $s$ for all such chains.

You can show that Artinian rings have Krull dimension $0$, and once you have that, you can get the following theorem classifying Artinian rings.

Theorem: A ring $A$ is Artinian $\Leftrightarrow A$ is Noetherian and $\dim A=0$.

Answer 3

it appears to me that every answer here uses the much stronger fact that being Noetherian implies that every submodule is finitely generated. some answers state clearly that this requires the axiom of dependent choice.

however, the question itself can be shown without any choice, using the fact that any module is a direct limit of its finitely presentable (and therefore finitely generated) submodules, which is true intuitionistically. if the chosen module is also Noetherian, such a sequence must stabilise after finitely many steps, which proves that the module itself is finitely presentable (and therefore finitely generated).