During my exercise session of Commutative algebra we had the following question:

Suppose $f: M \to N$ is a morphism of $R$-modules (where $R$ is a commutative ring with 1). Suppose $N$ is Noetherian, is $\ker f$ Noetherian?

I know the answer is 'no', since if we consider the zeromap from any non-Noetherian $R$-module M to the zero module, then we have a counterexample.

However, I wondered what we could say if $f$ is not the zero map and I have the following attempt: I have tried to prove that $f^{-1}(N)$ is Noetherian.

Let $S$ be a submodule of $f^{-1}(N)$, then $f(S)$ is a submodule of $N$, hence it is finitely generated since $N$ is Noetherian. Let us denote $f(S) = \langle s_1, \ldots, s_n\rangle$. Since $s_1, \ldots, s_n$ are elements of the image of $f$, we can choose elements $m_1, \ldots m_n$ in $M$ such that $f(m_i) = s_i$. If $s \in S$, then $f(s) = r_1s_1 + \ldots r_ns_n$ and so we can write that $s = r_1m_1 + \ldots r_nm_n$. Therefore $S$ is a finitely generated submodule of $f^{-1}(N)$ and so $f^{-1}(N)$ is a Noetherian $R$-module.

Since $\ker f$ is a submodule of $f^{-1}(N)$, we have that $\ker f$ is Noetherian.

I guess the step where I write $s$ as linear combination of the $m_i$ is dodgy, so my question is: is the inverse image of a Noetherian module (under a non-zero map) Noetherian?


Take your favorite Noetherian module $N$ and consider a direct sum $M=N^{(X)}$ of copies of $N$, with $X$ an infinite set.

Now prove that the map $$ \sigma\colon N^{(X)}\to N,\qquad (n_x)_{x\to X}\mapsto \sum_{x\in X}n_x $$ has Noetherian kernel if and only if $N=0$.

Another simpler example, where $M$ is even finitely generated. Take a non Noetherian ring $R$, a maximal ideal $\mathfrak{m}$ and set $M=R$, $N=R/\mathfrak{m}$. Then $N$ is obviously Noetherian, but the kernel of the projection $R\to R/\mathfrak{m}$ is not Noetherian.

Now, let's try being more general: prove the following.

Suppose $f\colon M\to N$ is a morphism of $R$-modules. If the image $f(N)$ is Noetherian and $\ker f$ is Noetherian, then also $M$ is Noetherian.

Where does your attempt go wrong? Note that $f^{-1}(N)=M$, by definition. If $S$ is a submodule of $M$, then $f(S)$ is certainly finitely generated, but it's not true that, given that $f(m_1),\dots,f(m_n)$ generate $f(S)$, also $m_1,\dots,m_n$ generate $S$. What you can say is only that $$ m_1R+\dots+m_nR+\ker f=S $$ but then you're doomed, because you see you need $\ker f$ to be Noetherian.

  • $\begingroup$ so we can not say, in general that $f^{-1}(N)$ is Noetherian if $N$ is Noetherian. Does my 'proof' goes wrong in the step where I write $s$ as the sum of the $m_i$ or did I made mistakes elsewhere? Moreover thank you for your quick response! $\endgroup$ – Student Jan 14 '17 at 10:38
  • $\begingroup$ your more general statement can be proven by considering the short exact sequence $0 \to \ker f \to M \to f(N) \to 0$ where we used the inclusion and then the map $f$, right? $\endgroup$ – Student Jan 14 '17 at 10:44
  • $\begingroup$ @Student I added what you asked for. And yes, the exact sequence is the key for the argument. $\endgroup$ – egreg Jan 14 '17 at 10:45
  • $\begingroup$ I should have known that, thank you very much for your help and counterexamples! $\endgroup$ – Student Jan 14 '17 at 10:48

A very simple counter-example: take non-finitely generated maximal ideal of a non-noetherian ring, viz; the ideal $(X_0,X_1,\dots, X_n,\dots)$ in the polynomial ring $k[X_0,X_1,\dots, X_n,\dots]$ over a field $k$.

Then $k[X_0,X_1,\dots, X_n,\dots]/(X_0,X_1,\dots, X_n,\dots)\simeq k$ is noetherian (and artinian) since it is a field.

In the same spirit: Take a non-discrete height $1$ valuation domain $V$ and its residual field. The maximal ideal of $V$ is not finitely generated, since this would imply $V$ is a discrete-valuation domain.

  • $\begingroup$ Thank you for your quick response, that is indeed a simple counter-example! If I could have accepted two answers, I would have! $\endgroup$ – Student Jan 14 '17 at 10:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.