I was working on proving that $\mathbb{Z}[\sqrt{-5}]$ was a noetherian ring without resorting to Hilbert Basis Theorem, and I was wondering if this generalized easily to when the ring $R$ is a $\mathbb{Z}$-module.
My proof goes:
Suppose $R$ is a commutative ring which is finitely generated as a $Z$-module. Suppose the generators are $r_1,\ldots,r_n$.
Then we may write $R = \mathbb{Z}r_1 + \cdots + \mathbb{Z}r_n$.
Furthermore, let $I$ be an ideal of $R$. Since $r_1,\ldots,r_n$ generate $R$, we must have that every $x \in I$ may be written in the form $x = a_1 r_1 + \cdots a_n r_n$ for some $a_i \in \mathbb{Z}$.
Now, let $I_i$ denote the set of all coefficients of $r_i$ that may be obtained from writing an element $x \in I$ in the given form above. Then, by definition $I_i \subseteq \mathbb{Z}$, and it is clearly an additive subgroup of $\mathbb{Z}$, so it must be an ideal in $\mathbb{Z}$. Since $\mathbb{Z}$ is a PID, we may write $I_i = \mathbb{Z}b_i$ for some $b_i \in \mathbb{Z}$. Then, the elements $b_1r_1, \ldots, b_nr_n$ clearly generates $I$ as a $\mathbb{Z}$-module, so $I$ is finitely generated as a $Z$-module, thus it must be a finitely generated ideal.
My one concern about this proof is that the expansion $x=a_1r_1 + \cdots a_nr_n$ need not be unique, so I'm not sure if the ideal $I_i$ is well defined. Do I need to throw on the condition that $R$ is a free $\mathbb{Z}$-module? That would certainly fix the issue, but I'm wondering if it is necessary.
If a ring is noetherian as a $Z$-module, then it is noetherian as a ring, for its ideals are in particular $Z$-submodules.