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.

  • $\begingroup$ @user26857 Where $Z = \mathbb{Z}$ :-) $\endgroup$ – user144221 Oct 17 '16 at 7:15

You're making things complex where it's simple.

Any ideal $I$ in $R$, is a sub-$\mathbf Z$-module of the finitely generated $\mathbf Z$-module $R$, hence it's finitely generated as a $\mathbf Z$-module. A fortiori, it is finitely generated as a $R$-module (He who can do the more can do the less…)

  • $\begingroup$ How do we know submodules of a finitely generated $\mathbb{Z}$-module must be finitely generated? Maybe I'm just missing something obvious, but in my mind I can only right now envision how to prove that for $F$-modules, where $F$ is a field. (i.e. vector spaces) $\endgroup$ – ASKASK Oct 17 '16 at 0:29
  • $\begingroup$ Nevermind, the statement is clearly obvious. $\endgroup$ – ASKASK Oct 17 '16 at 0:31
  • $\begingroup$ I was imagining the statement that submodules of a finitely generated free z-module must be also be free $\endgroup$ – ASKASK Oct 17 '16 at 0:32
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    $\begingroup$ Finitely generated modules over a noetherian ring are noetherian modules, i.e. all their submodules are finitely generated. $\endgroup$ – Bernard Oct 17 '16 at 0:32
  • $\begingroup$ Submodules of a finitely generated free module $L$ over a P.I.D. $R$ are indeed free, and a theorem asserts we can find an adapted basis, i.e. a basis $(e_1,\dots,e_n) $ of $L$ and elements $a_1, \dots,a_r\in R$ $(r\le n)$ such that 1) $a_1 \mid \dots\mid a_r$ and 2) $(a_1e_1,\dots,e_re_r)$ is a basis of the submodule. $\endgroup$ – Bernard Oct 17 '16 at 0:37

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