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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.

egreg
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user371231
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    Hint: https://en.wikipedia.org/wiki/Schanuel%27s_lemma – Phil. Z Nov 24 '17 at 08:45
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    But can't apply Schanuel's lemma directly since don't know if $N$ is projective or not. – user371231 Nov 24 '17 at 08:48
  • By assumption there is an epimorphism $F\to N$ where $F$ is a free module of finite rank. Consider the exact sequence $$0\to N'\times_{N} F\to F\to M\to 0$$ Now Schanuel's lemma shows the module $ N'\times_{N} F$ is finitely generated. The result follows from that $ N'\times_{N} F\to N'$ is epic. – Phil. Z Nov 24 '17 at 09:10
  • What is the module $N^{'}X_{N} F$ and what is the map to $F$? – user371231 Nov 24 '17 at 09:15
  • $N'\times_N F$ is the pullback of the diagram $N'\to N\leftarrow F$. More precisely, it is the submodule of $N'\oplus F$ consisting of the elements $(x,y)\in N'\oplus F$ such that $x$ and $y$ map to the same element in $N$. – Phil. Z Nov 24 '17 at 09:22
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    See https://math.stackexchange.com/a/1201068/62967 – egreg Nov 24 '17 at 10:22

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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.

k.stm
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