Let $M$ be a finitely presented $R$ module and $0 \to N' \to N \to M \to 0 $ be an exact sequence with $N$ a finitely generated $R$ module. Prove that $N'$ is also a finitely generated $R$ module.
By $M$ finitely presented $R$ module we mean there is a ses $0 \to K \to F \to M \to 0$ where $F$ is free of finite rank and $K$ is finitely generated $R$ module. I proved the case when $N$ is free of finite rank but can't show it in this case. Any help will be appreciated. Thanks.
Let $F'$ be free of finite rank and $F' → N$ be surjective. You get an exact sequence $$0 → K' → F' → M → 0,$$ for some kernel $K'$, where the right arrow is $F' → N → M$. Apply Schanuel to this sequence to get that $K'$ is finitely generated.
The arrow $K' → F' → N$ factorises via $N' → N$. So you get an arrow of exact sequences
$$ \require{AMScd} \begin{CD} 0 @>>> K' @>>> F' @>>> M @>>> 0 \\ @| @VVV @VVV @| @| \\ 0 @>>> N' @>>> N @>>> M @>>> 0 \\ \end{CD}, $$ upon which you can apply a four lemma.