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

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).
