Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
add comment

2 Answers

up vote 12 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".

share|improve this answer
    
+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
1  
@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
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.