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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.

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    $\begingroup$ 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. $\endgroup$ Commented Jan 29, 2022 at 1:24
  • $\begingroup$ 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. $\endgroup$
    – slowspider
    Commented Jan 29, 2022 at 7:00
  • $\begingroup$ This here math.stackexchange.com/questions/1680007/… might tell you, what not to try. $\endgroup$ Commented Jan 29, 2022 at 7:37
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    $\begingroup$ Another hint: a finitely generated projective module is always finitely presented. $\endgroup$ Commented Jan 29, 2022 at 8:19
  • $\begingroup$ Thank you all for the helpful comments/references! $\endgroup$
    – slowspider
    Commented Jan 30, 2022 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.)

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  • $\begingroup$ Nice! So we could use something like $R = \mathbb{Q}[x_1,x_2,\dots]$ and $I = \langle x_1,x_2,\dots, \rangle$. $\endgroup$
    – slowspider
    Commented Jan 29, 2022 at 20:02
  • $\begingroup$ The fact that $R/I$ is not $R$-finitely presented follows from this fact. $\endgroup$ Commented Jun 21 at 12:38

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