Noetherian module implies Noetherian ring?

I know that a finitely generated $R$-module $M$ over a Noetherian ring $R$ is Noetherian. I wonder about the converse. I believe it has to be false and I am looking for counterexamples.

Also I wonder if $M$ Noetherian imply that $R$ is Noetherian is true? And if $M$ Noetherian implies $M$ finitely generated is true?

That is, do both implications fail or only one of them?

• I don't know what definition you are using for Noetherian modules, but I guarantee that an equivalent condition is that every submodule of $M$ is finitely generated. In particular, every Noetherian module is finitely generated, to answer one of your later questions. Commented Dec 13, 2012 at 19:10

I assume that $R$ is commutative in your question. As other answers have already pointed out, the answer to your question is false for trivial reasons. But here is a closely related statement which is true. (It relies on the fact that a direct sum of two Noetherian modules is again Noetherian; I leave this for you to investigate, but feel free to leave a comment if you would like more information.)

Theorem: If a ring $R$ has a faithful Noetherian module $M$, then $R$ is Noetherian.

Proof: Because $M$ is Noetherian, it is finitely generated. Say $M = \sum Rm_i$ for some finite generating set $m_1, \dots, m_n$. Consider the homomorphism $f \colon R \to M^n$ given by $f(r) = (rm_1, \dots, rm_n)$. This is injective because the annihilator of $M$ is trivial: if $f(r) = 0$ then $rm_i = 0$ for all $i$, whence $rM = r \sum Rm_i = \sum Rrm_i = 0$. Thus $R$ is isomorphic as an $R$-module to a submodule of the Noetherian module $M^n$. It follows that $R$ is Noetherian.

Corollary: If $M$ is a Noetherian $R$-module, then $R/\mathrm{ann}(M)$ is Noetherian.

What happens for noncommutative rings? The results above no longer hold. A ring may have a faithful simple (hence Noetherian) left module but fail to be either left or right Noetherian. For instance, let $V$ be an infinite dimensional vector space over a field $k$, and let $R = \mathrm{End}_k(V)$ be the ring of $k$-linear endomorphisms of $V$. Then $V$ is a simple left $R$-module, but it can be shown that $R$ is neither left nor right Noetherian.

• +1: This answer appeared just as I was finishing up mine. If you had posted it a few minutes earlier, I wouldn't have bothered. Commented Dec 13, 2012 at 19:23
• What would be a concrete example of a noetherian module over a non-noetherian ring, (which, by what you have shown, must be unfaitfhul)? Commented Dec 1, 2023 at 13:42
• Some other answers here give general constructions, but to be very explicit: Let's take $R = \prod_{n=1}^{\infty} \mathbb{Z}$ and $M = R/I \cong \mathbb{Z}$ for the ideal $I = 0 \times \mathbb{Z} \times \mathbb{Z} \times \cdots$. Commented Dec 2, 2023 at 15:50

It is not clear to me exactly what you are asking in your main question. If you are asking:

$$\bullet$$ If a ring $$R$$ admits a Noetherian module, must $$R$$ be Noetherian?

Then this is trivially false: Alex Youcis and Thomas Andrews have each shown that every commutative ring admits Noetherian modules. (As a general rule, if you are looking for a counterexample to an assertion about modules and you haven't checked the zero module, you haven't looked hard enough. Also looking at modules of the form $$R/I$$ is something to try early on.)

$$\bullet$$ If for a ring $$R$$ every finitely generated $$R$$-module is Noetherian, must $$R$$ be Noetherian?

Then this is trivially true, as $$R$$ is a finitely generated $$R$$-module.

A less trivial statement is the following:

Lemma (Kaplansky): A ring is Noetherian iff it admits a faithful Noetherian module.

Another result vaguely along these lines is:

Theorem (Eakin-Nagata) Let $$R \subset S$$ be a ring extension such that $$S$$ is finitely generated as an $$R$$-module. Then $$R$$ is Noetherian iff $$S$$ is Noetherian.

Proofs of these and other results which are (even more) vaguely related to your question can be found in $$\S 8.8$$ of my commutative algebra notes.

• +1: Of course, it was inevitable that such a remark would be simultaneously posted twice. I kept waiting for it to show as a new answer as I was typing... Commented Dec 13, 2012 at 19:30

What about taking $R$ any non-Noetherian ring and $M=\{0\}$?

Let $$R$$ be a commutative non-Noetherian ring and let $$\mathcal m$$ be a maximal ideal. Then $$R/\mathcal m$$ is finitely generated and Noetherian - it only has two sub-$$R$$-modules.

Note that, even if $$R$$ isn't Noetherian, it contains a maximal ideal, by Krull's Theorem.

• Sorry, but why is $R/m$ finitely generated? Commented Dec 13, 2012 at 19:22
• In general, if $\phi:R\to S$ is an onto homomorphism of rings, then $S$ is generated as an $R$-module by $\phi(1)$. Not just finitely generated, but singularly generated. Commented Dec 13, 2012 at 19:28
• @harajm: it is generated by the image of $1$. More generally, a quotient of a finitely generated module is finitely generated: take the image in the quotient of a finite generating set. Commented Dec 13, 2012 at 19:28