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I have trouble proving the following question. Suppose we have an exact sequence $L\to M \to N \to 0$ of $R$-modules, with $M $ finitely presented and $L$ finitely generated. Show that $N$ is finitely presented.

Previous result: Suppose we have maps of $R$-modules $f:L\to M$ and $g:M\to N$. Then we can construct an exact sequence $0 → ker(f) → ker(g ◦ f) → ker(g) → coker(f) → coker(g ◦ f) → coker(g) → 0.$

And the hint is to use this previous result on a map $R^n\to M\to N$, this can be constructed by using the finite generation of $L$. But then I am stuck.

Joost van Geffen
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1 Answers1

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since M is finite presented,so we have the following commutative diagram and $Ker \epsilon$ is finite generated. enter image description here The morphism $\alpha$ and $\beta$ are induced.

By 5 lemma we can show that $\alpha$ is epimorphism.by snake lemma we have $\beta $ is an isomorphism.

since L is finite generated,$Im f$ is finite generated.And we get $\alpha$ is epimorphism and $Ker \alpha$ is finite generated before.So $Ker g\epsilon$ is finite generated.

So N if finite presented.

Jian
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