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Give an example of a finitely generated $R$-module $M$ (for some commutative ring $R$) that is not projective and is not finitely presented.

I was able to find an example of a finitely generated $R$-module that is not projective; if $A$ is a nonzero finite abelian group, then $A$ is not projective over $\mathbb{Z}$. However, it seems that these are finitely presented.

Note: Here I say that a module is finitely presented if and only if there exists an exact sequence $F_0 \rightarrow F_1 \rightarrow M \rightarrow 0$ where $F_0$ and $F_1$ are free with finite bases.

user26857
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slowspider
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    As an initial hint: if $R$ is Noetherian, then any finitely generated $R$-module is also finitely presented. So, to get an example, you would need to have $R$ to be some non-Noetherian commutative ring. – Daniel Schepler Jan 29 '22 at 01:24
  • I have a few examples of non-Noetherian commutative rings. The wanted to prove the projective part by using the fact that unitary modules over a PIDs are free if and only if the module is projective. However, PID's are Noetherian, so I'm still stuck. – slowspider Jan 29 '22 at 07:00
  • This here https://math.stackexchange.com/questions/1680007/polynomial-ring-in-infinitely-many-variables-over-a-noetherian-ring-is-coherent might tell you, what not to try. – Severin Schraven Jan 29 '22 at 07:37
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    Another hint: a finitely generated projective module is always finitely presented. – Jeremy Rickard Jan 29 '22 at 08:19
  • Thank you all for the helpful comments/references! – slowspider Jan 30 '22 at 20:24

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If $I\subset R$ is an ideal which is not finitely generated, then $R/I$ is not finitely presented as $R$-module, and not projective as well. (If $R/I$ is projective, then $I$ is a direct summand, so $I$ is generated by an idempotent.)

user26857
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