Recall that Nakayama's lemma states that
Let $R$ be a commutative ring with unity, and let $J$ be the Jacobson radical of $R$ (the intersection of all the maximal ideals of $R$). For any finitely generated $R$-module $M$, if $IM=M$ for some ideal $I\subseteq J$, then $M=0$.
I was trying to come up with a simple / illustrative example of the above situation where $M$ is non-zero and not finitely generated, but $IM=M$ for some $I\subseteq J$.
The only example I managed to come up with was
$$R=k[[\overline{x}_0,\overline{x}_1,\overline{x}_2,\ldots]]=k[[x_0,x_1,x_2,\ldots]]/(x_0^2,x_1^2-x_0,x_2^2-x_1,\ldots)$$
where $k$ is a field, and $I=M=(\overline{x}_0,\overline{x}_1,\overline{x}_2,\ldots)\neq0$. Since $k[[x_0,x_1,x_2,\ldots]]$ is a local ring with maximal ideal $(x_0,x_1,x_2,\ldots)$, we have that $R$ is a local ring with maximal ideal $I$, so that $I$ is the Jacobson radical of $R$, and $$IM=I^2=(\overline{x}_0^2,\overline{x}_1^2,\overline{x}_2^2,\ldots,\overline{x}_0\overline{x}_1,,\ldots,\overline{x}_1\overline{x}_2,\ldots)=(0,\overline{x}_0,\overline{x}_1,\ldots,\overline{x}_0\overline{x}_1,\ldots,\overline{x}_1\overline{x}_2,\ldots)\supseteq I$$
so that $IM=I^2=I=M$.
Is there an easier example – maybe with $R$ noetherian? Also, what is the geometric picture here? As I understand it, letting $M$ be non-finitely generated corresponds to looking at non-coherent sheaves, which I don't have a whole lot of intuition for – even less than my novice understanding of coherent sheaves.
Best Answer
If $(R,\mathfrak m)$ is a local domain which is not a field, then for its fraction field $K$ we have $\mathfrak m K=K$.
Amusingly, this proves that $K$ is not a finitely generated $R$-module.