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.
What you claim is false; a finitely generated module need not be Noetherian. As a counterexample, consider the polynomial ring $\mathbb{Q}[x_0,x_1,\dots]$ in infinitely many variables as a module over itself. It is finitely generated (by 1), but its submodule $\langle x_0,x_1,\dots\rangle$ isn't, so the module itself isn' Noetherian.
What is true is that a module over a Noetherian ring is Noetherian iff it is finitely generated (this might require the ring to be commutative with unity).
Best Answer
Is the following the proof in Atiyah & Macdonald? I find it constructive:
The $A$-module $M$ is the quotient of $A^n$ for some natural number $n$. There exists a morphism $\Phi: A^n \to A^n$ covering $\phi: M \to M$. Since $A$ is a commutative ring, determinant makes sense and one can define the characteristic polynomial $p_\Phi(t) := \det(\Phi-t\,\mathrm{Id})$. By the Cayley–Hamilton Theorem, $p_\Phi(\Phi)=0: A^n \to A^n$. Observe that this morphism covers $p_\Phi(\phi): M \to M$. Since the morphism $A^n \to M$ is surjective, it follows that $p_\Phi(\phi)=0$ also.
For a given $\phi$, I don't know anything about the uniqueness the polynomial $p_\Phi(t)$.