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Assume that M',M'' are finitely generated R-modules (here R is a commutative unitary ring) and M is another R-module. The initial problem is that if there exists a s.e.s. $0\rightarrow M'\xrightarrow{f} M\xrightarrow{g}M''\rightarrow 0 \text{ }(1)$ then we can deduce that M is also finitely generated and an answer to this can be found here Finitely generated modules in exact sequence. The answer was linear algebraish and I was wondering if we could somehow prove that the sequence spits so we get an isomorphism $M\cong M'\dot{\oplus} M''$ which would also imply that M IS Finitely generated. So the question is: Can we somehow prove that (1) splits assuming that M',M'' are finitely generated and if not is there a counterexample (this could happen if for example M'' is projective).

Arturo Magidin
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Bigalos
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    "To wander": to move about without a definite destination or purpose. "To wonder": To be filled with curiosity or doubt. – Arturo Magidin Feb 10 '24 at 02:20
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    $0 \to 2\Bbb Z \hookrightarrow \Bbb Z \twoheadrightarrow \Bbb Z/2\Bbb Z \to 0$ does not split, although $2\Bbb Z$ and $\Bbb Z/2\Bbb Z$ are finitely generated. – azif00 Feb 10 '24 at 02:26
  • The example you gave works because $\mathbb{Z}_2$ is not a projective $\mathbb{Z}$ module and now I realized that (1) splits if and only if M'' ($\mathbb{Z}_2$ in the example) is projective so the idea was to find a non-projective finitely generated module. Am I correct ? – Bigalos Feb 10 '24 at 19:19
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    @BillIconomou Certainly (1) splits if $M''$ is projective, but it may split otherwise. Indeed, if $M$ and $N$ are any $R$-modules, then there is the natural split exact sequence $0 \to M \to M \oplus N \to N \to 0$. Projectivity of $M$ is equivalent to every short exact sequence $0 \to A \to B \to M \to 0$ being split. – metalspringpro Feb 10 '24 at 22:30

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