[proofplan] We apply the [Determinant Trick](/theorems/2934) to the identity map on $M$ to find an element $a \in \mathfrak{a}$ such that $(1 - a)m = 0$ for all $m \in M$. Since $a \in \mathfrak{a} \subseteq \operatorname{Jac}(R)$, the [Characterisation of the Jacobson Radical](/theorems/2860) guarantees that $1 - a$ is a unit. Multiplying by $(1-a)^{-1}$ gives $m = 0$ for every $m \in M$. [/proofplan] [step:Apply the Determinant Trick to obtain an element $a \in \mathfrak{a}$ with $(1-a)m = 0$ for all $m \in M$] The hypotheses of the [Determinant Trick](/theorems/2934) require a finitely generated $R$-module and an ideal $\mathfrak{a}$ such that $\mathfrak{a}M = M$. Both conditions hold by assumption. Applying the Determinant Trick, there exists $a \in \mathfrak{a}$ such that $am = m$ for all $m \in M$, equivalently \begin{align*} (1 - a)m = 0 \quad \text{for all } m \in M. \end{align*} [/step] [step:Use the Jacobson radical characterisation to invert $1 - a$ and conclude $M = 0$] Since $a \in \mathfrak{a} \subseteq \operatorname{Jac}(R)$, the [Characterisation of the Jacobson Radical](/theorems/2860) (applied with $x = a$ and $y = 1$) gives that $1 - a \cdot 1 = 1 - a$ is a unit in $R$. Let $u = (1-a)^{-1} \in R$. For every $m \in M$, we have $(1-a)m = 0$. Multiplying both sides on the left by $u$: \begin{align*} m = u \cdot (1-a)m = u \cdot 0 = 0. \end{align*} Since $m$ was arbitrary, $M = 0$. [guided] The Determinant Trick produces $a \in \mathfrak{a}$ with $(1-a)m = 0$ for all $m \in M$. At this point, $(1-a)$ annihilates the entire module $M$. If we could invert $1-a$, we would immediately conclude $m = 0$ for all $m$, hence $M = 0$. Can we invert $1 - a$? Since $a \in \mathfrak{a} \subseteq \operatorname{Jac}(R)$, the [Characterisation of the Jacobson Radical](/theorems/2860) tells us that $1 - ax$ is a unit for every $x \in R$. Taking $x = 1$ gives $1 - a \in R^\times$. Let $u = (1-a)^{-1}$. For every $m \in M$: \begin{align*} m = u \cdot (1-a) m = u \cdot 0 = 0. \end{align*} Since every element of $M$ is zero, $M = 0$. This is the heart of Nakayama's Lemma: the condition $\mathfrak{a} \subseteq \operatorname{Jac}(R)$ ensures that the element $1 - a$ produced by the Determinant Trick is invertible. Without the Jacobson radical condition, we would only know that $(1-a)$ annihilates $M$, but $1-a$ could be a zero divisor, and we could not conclude $M = 0$. [/guided] [/step]