Is every Noetherian module finitely generated? 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.
 A: 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 $\dim A=0$. 
A: The three standard equivalences for Noetherian are:

Theorem. Let $M$ be an $R$-module. Assuming the Axioms of Choice, the following are equivalent:
  
  
*
  
*$M$ has ACC on submodules.
  
*Every submodule of $M$ is finitely generated.
  
*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.
