# Finitely generated $R$-module that is not projective or finitely presented

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.

• 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. Commented Jan 29, 2022 at 1:24
• 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. Commented Jan 29, 2022 at 7:00
• This here math.stackexchange.com/questions/1680007/… might tell you, what not to try. Commented Jan 29, 2022 at 7:37
• Another hint: a finitely generated projective module is always finitely presented. Commented Jan 29, 2022 at 8:19
• Thank you all for the helpful comments/references! Commented Jan 30, 2022 at 20:24

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