Counterexamples to Nakayama's Lemma if $M$ is not finitely generated

Question

One of the most famous forms of Nakayama's lemma says:

Let $I$ be an ideal in $R$ and $M$ a finitely-generated $R$ module. If $IM = M$, then there exists an $r \in R$ with $r ≡ 1 \pmod I$, such that $rM = 0$.

Someone knows counterexamples to the assert IN THIS FORM if $M$ is not finitely generated?

To be more precise, I'm looking to an $R$-molude $M$ such than doesn't exists an element $r \in R$ with the properties above (and, possibly, an explicit proof of this "non existence").

Thank you.

Answer 1

Let $M = \mathbb{Q}$ thought of as a $\mathbb{Z}$-module and let $I = (2)$. Then $IM = M$, and odd numbers don't kill $\mathbb{Q}$.

Answer 2

A corollary is: If $R$ is a commutative local ring and $M$ is a finitely generated $R$-module such that $M=\mathfrak{m}M$, then $M=0$.

This doesn't hold when $M$ is not finitely generated: Take any DVR $R$ (for example $R=\mathbb{Z}_{(p)}$ or $R=\mathbb{Z}_p$) with uniformizer $p$ and field of fractions $K$. Then $K$ is an $R$-module with $K = pK$, but $K \neq 0$.

Another corollary is: If $R$ is a commutative ring, $M$ is a finitely generated $R$-module, then every surjective endomorphism of $M$ is an isomorphism (apply Nakayama to the $R[x]$-module $M$ where $x$ acts as the given endomorphism). But for $R \neq 0$ there are lots of $R$-modules (not finitely generated) which have a surjective endomorphism, which is not injective, for example the shift $R^{\mathbb{N}} \to R^{\mathbb{N}}, (a_0,a_1,\dotsc) \mapsto (a_1,a_2,\dotsc)$.

Answer 3

Extending on hunter's answer (for this answer, I only work over a domain $R$): an $R$-module $M$ is called divisible if $M = rM$ for all $0 \ne r \in R$. By definition, Nakayama's Lemma utterly fails for divisible modules. Any injective module is divisible, so divisible modules always exist (and over a PID, such as $\mathbb{Z}$, divisible implies injective). A direct consequence of this is that nonzero injective modules are never finitely generated (if $R$ is not a field).

As divisibility is fairly straightforward to check, we easily get examples over $\mathbb{Z}$: the abelian groups $\mathbb{Q}$, $\mathbb{Q}/\mathbb{Z}$, $\mathbb{Z}_{p^\infty}$ (for $p$ prime) are all divisible (and hence not finitely generated). This exhausts all divisible abelian groups, in the sense that every divisible abelian group is a direct sum of copies of these (in fact, just the first and third are needed).