I am trying to prove (or disprove) the following:

For a non-Noetherian ring $R$ and a non-finitely generated ideal $I$ the $R$-module $R/I$ is not finitely presented.

I don't think that it is enough to consider s.e.s. $0\rightarrow I\rightarrow R\rightarrow R/I \rightarrow 0$ and argue that $I$ is not finitely generated so we are done. When I consider a general surjection $0\rightarrow \ker(f)\rightarrow R^n\rightarrow R/I\rightarrow 0$ I don't know how to proceed. I could show that $\ker f$ contains $I$ as a submodule, that's all.

If this is not true in general, I wonder if it holds for the case where $R=k[x_1,...]$ with infinite generators and $I=(x_1,...)$ the maximal ideal.


1 Answer 1


It's generally true over any ring that

if a module $M$ is finitely presented, $L$ is finitely generated and $f\colon L\to M$ is a surjective homomorphism, then $\ker f$ is finitely generated.

The trick is to consider a surjective homomorphism $g\colon R^n\to L$ and consider the diagram with exact rows $$\require{AMScd} \begin{CD} 0 @>>> \ker f\circ g @>>> R^n @>f\circ g>> M @>>> 0 \\ @. @VVV @VgVV @| @. \\ 0 @>>> \ker f @>>> L @>f>> M @>>> 0 \end{CD} $$ It is easily seen that $\ker f\circ g\to \ker f$ is surjective as well. Thus we are reduced to prove the special case when $L=R^n$ is finitely generated and free.

By assumption, there exists a surjective homomorphism $h\colon R^m\to M$ such that $\ker h$ is finitely generated. Take the pull-back $N$ of $f$ and $h$ along $M$: $$\begin{CD} {} @. {} @. 0 @. 0 \\ @. @. @VVV @VVV \\ {} @. {} @. \ker f @= \ker f \\ @. @. @VVV @VVV \\ 0 @>>> \ker h @>>> N @>>> R^n @>>> 0 \\ @. @| @VVV @VfVV \\ 0 @>>> \ker h @>>> R^m @>h>> M @>>> 0 \\ @. @. @VVV @VVV \\ {} @. {} @. 0 @. 0 \\ \end{CD}$$ Now note that $N\cong R^n\oplus\ker h\cong R^m\oplus\ker f$, so $\ker f$ is finitely generated.

  • $\begingroup$ Thank you! By the way, there might be a minor misprint where you say that $N=\ker h$. Everything else is clear. $\endgroup$
    – definition
    Jul 11, 2019 at 0:01

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