Inverse image of a Noetherian module under a module morphism

Question

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?

Answer 1

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.

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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.

Answer 2

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.