# Question in proving Nakayama's Lemma

There was a prior question regarding intuiting Nakayama's Lemma:

Intuitive explanation of Nakayama's Lemma

I am currently studying Reid's "Undergrad. Commutative Algebra." His statement of the lemma is specifically in the context of a local ring $(A,m)$ where in his notation, $m$ is the maximal ideal, M is a finite $A$-module; then $M = mM$ implies that $M = 0$.

I feel this question is probably quite naive, so forgive a self-studier:

Can you not simply say that in this case, $m$ has no units, so the only way for

$M = mM$ is for $M$ to be $0$? Thanks.

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Suppose $M = \mathbb{Q}$ thought of as a $\mathbb{Z}$-module and $m$ is the ideal $(2)$ in $\mathbb{Z}$. Then $mM = M$ even though $m$ contains no units. (Of course several of the hypotheses of Nakayama's lemma are not satisfied here.) – user29743 Dec 21 '12 at 23:07
@countinghaus Nice example to show that the hypothesis of finite generation is necessary. – user38268 Dec 21 '12 at 23:24

Here is an example of a proper ideal $I$ such that $IM = M$. Take $R = C[0,1]$, the continuous real-valued functions on $\mathbb{R}$ and let $M = \mathbb{R}$, with the action $f*a = f(0)a$. Exercise: check that $M$ is a finitely generated $C[0,1]$-module.
Let $I$ be the ideal of functions vanishing at $1$. Then $IM = M$. But $I$ contains no units. Of course, this kind of example won't work for a local ring, since any ideal is already contained in the Jacobson radical (i.e. the maximal ideal), but it serves to show that you can't conclude the statement merely by appealing to an ideal lacking units.
Here is another (counter)example: Let $R = k[[x,x^{1/2},x^{1/4}, x^{1/8},\ldots,x^{1/2^n},\ldots]],$ the ring of formal power series over a field $k$ in powers of $2$-power roots of $x$. Then $R$ is a local ring, whose maximal ideal $m$ is the set of powers series with vanishing constant term. Note that $m^2 = m$, although $m \neq 0$. (The point is that $m$ is not fintely generated.)