2
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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 '19 at 0:01

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.