# Finitely generated but not finitely presented module

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.

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.
• Thank you! By the way, there might be a minor misprint where you say that $N=\ker h$. Everything else is clear. – definition Jul 11 '19 at 0:01