# Non-finitely generated R-module

Currently studying for qualifying exams and came across the following problem:

Give an explicit example of a ring $$R$$ (commutative with identity) and a surjection $$\psi: M \rightarrow N$$ of finitely generated $$R$$-modules such that $$\ker(\psi)$$ is not a finitely generated $$R$$-module.

The given hint is to let $$R = \mathbb{Z}[x_1, x_2, ... ]$$, however I am confused how such a construction would yield finitely generated submodules? It seems clear that we want to invoke the general strategy of picking an ideal $$I$$ of $$R$$ and defining the natural surjection $$\phi : R \rightarrow R/I$$ If we let $$I = \langle x_1, x_2, ...., \rangle$$ then $$\ker(\phi) \cong I$$ and $$I$$ is not finitely generated, however $$R$$ is also not finitely generated. Thus, we do not satisfy the problem statement? Any help on this with the given hint would be greatly appreciated!

• $R$ is certainly finitely generated over itself, no? Dec 20, 2021 at 16:13
• That is it. I was definitely confusing finitely generated ring and finitely generated R-modules. Thank you! Dec 20, 2021 at 16:15

A (commutative with identity) ring $$R$$ is Noetherian if and only if every ideal of $$R$$ is finitely generated (as a module over $$R$$ or as an ideal, it's the same).
A ring $$R$$ is Noetherian if and only if every submodule of every finitely generated $$R$$-module is finitely generated.
Thus your example requires a ring that's not Noetherian and the simplest example is $$R=\mathbb{Z}[x_1,x_2,\dotsc]$$, the ring of polynomials on countably many indeterminates over $$\mathbb{Z}$$, because the ideal $$I$$ generated by the indeterminates isn't finitely generated.
Indeed there is the strictly increasing sequence of ideal $$(0)\subset (x_1)\subset (x_1,x_2)\subset\dotsb$$ and the union of these ideals is $$I$$. Why is this sequence strictly increasing? Because if $$x_{n+1}=x_1f_1+x_2f_2+\dots+x_nf_n \qquad(f_1,f_2,\dots,f_n\in R)$$ evaluating at $$x_1=x_2=\dots=x_n=0$$ and $$x_{n+1}=1$$ leads to a contradiction. But if $$I$$ is finitely generated, its generators belong to one of the ideals in the chain: again a contradiction.