I am having trouble with the following exercise.
Let $R = F[x]$, where F is some field. Let $I = (x^2)$, and let M be an R/I-module (induced action, so assuming $IM = 0$). Then I want to show that the following statements are equivalent:
i) M is finitely generated as an R-module ii) M is finitely generated as an R/I-module iii) M is finite-dimensional as a vector space over F.
i) implies ii): since a generating set for M as an R-module gives a generating set for M as an R/I-module.
ii) implies iii): Can I just expand a generating set $\{a_1, a_2, \dots , a_n\}$ for M as an R/I-module to a generating set $\{a_1, a_2, \dots , a_n, xa_1, xa_2, \dots , xa_n\}$ as an F-module?
iii) implies i) This one should be obvious, as we can use the generating set for M as an R-module for M as an R-module.
My question is whether my arguments above are correct. Is there a better way to see things? Maybe from a general point of view? Are there any rings which are particularly nice in this regard? (I am aware of the existence of the Nakayama Lemma, and R/I is a local ring, so I might be able to apply it somehow)
Your idea for i$\Rightarrow$ii will work.
ii$\Rightarrow$iii is not so generic and depends on $I$. We know that $R/I$ itself has a basis $\{1,x\}$ over $F$ hence is a finite-dimensional $F$-vector space. A finitely generated $R/I$ module $M$ is a quotient of the free $R/I$ module on some finite generating set, and since this free module is essentially a finite direct sum of copies of $R/I$ it is finite-dimensional, hence so is $M$.
Your idea given in the post will also work if you want to work it out. This may be a good exercise.