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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.

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up vote 14 down vote accepted

My favorite example is $R = \mathbf Z_{(p)}$ [which is very Noetherian] and $M = \mathbf Q$. Of course, you then think of localizing $R = k[x]$.

On that note, there was a thread on MO about understanding the lemma and Roy Smith's answer there is pretty geometric: "It's sort of like the inverse function theorem".

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+1 for the speed with which you gave the best possible example and the link to Roy's answer. You remind me of the gun slingers in the films I loved to watch in my youth :-) – Georges Elencwajg May 31 '12 at 19:51
@GeorgesElencwajg Thanks for your kind words, as usual. And I love the proof that, e.g., $\mathbf Q$ is not finitely generated! On a serious note, the man with no name didn't have to write a thesis, so I should probably start cutting down :-/ – Dylan Moreland May 31 '12 at 19:52

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.

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