Determinant trick

Remarks

Apply the Cayley–Hamilton theorem for modules to $\varphi = \id_M$, which satisfies $\varphi(M) = M = IM$. It gives $a_1, \ldots, a_n \in I$ with $(1 + a_1 + \cdots + a_n) \id_M = 0$, so $r = 1 + a_1 + \cdots + a_n$ works. The name comes from the determinant of the matrix $I_n - (a_{ij})$ used in that proof. If $I$ lies in the Jacobson radical of $R$, then $r$ is a unit, so $M = 0$; this is Nakayama's lemma. The hypothesis that $M$ is finitely generated is needed: for $R = \Z$, $I = 2\Z$ and $M = \Q$ one has $IM = M$, but the only $r$ with $r\Q = 0$ is $r = 0$, and $0 - 1 \notin 2\Z$.

Going deeper

For a student

This is the most flexible form of Nakayama's lemma. It gives Nakayama's lemma for local rings and for the Jacobson radical at once. It also shows that a finitely generated ideal $I$ with $I^2 = I$ is generated by an idempotent. Indeed, there is $r \in 1 + I$ with $rI = 0$. Then $e = 1 - r \in I$ satisfies $e x = x$ for $x \in I$, so $e^2 = e$ and $I = eR$.