Nakayama's lemma $:$
Let $M$ be a finitely generated $R$-module and $I$ be an ideal of $R$ such that $IM=M,$ then there exists $r \in R$ such that $r \equiv 1\ (\text{mod}\ I)$ and $rM = 0.$
In order to prove this lemma we need the following theorem.
Theorem $:$
Let $M$ be a finitely generated $R$-module. Let $\varphi : M \longrightarrow M$ be an $R$-module homomorphism and $I$ be an ideal of $R$ such that $\varphi (M) \subseteq IM.$ Then there is a relation of the form
$${\varphi}^n + a_1 {\varphi}^{n-1} + \cdots + a_n = 0,\ a_i \in I.$$
$\text{for some}\ n.$
Our instructor gave a proof of this theorem and I have understood it on my own. After the proof of this theorem it has been stated in the lecture notes that Nakayama's lemma directly follows from the above theorem by taking $\varphi = \text{id}_M.$ But I find difficulty to understand this. Also I find difficulty to understand the second assertion made in the lemma which is $rM=0.$ The problem is that since $I$ is an ideal of $R$ then $rM = r(IM) = (rI)M = IM = M = 0.$ So $rM = 0 \implies M=0.$ Where have I done mistakes? Would anybody please help overcoming me from this situation?
Thank you very much.
For the first question note $\varphi^n = \varphi = \text{id}$ in the special case. Thus $$\varphi\cdot(1+a_1+\cdots+a_n)=0$$ so $r= 1 + \sum a_i$ works.
As for your second question, $rI \subset I$ is not necessarily equality so you can only deduce $rM \subset M$ which is of course true.