Let $X = \{x_1, x_2, \dots, x_n\}$ be a minimal set of generators for $M$ in the sense that $M$ is not generated by any proper subset of $X$

Elements of $\mathfrak{a}M$ can be written as linear combinations $\sum a_i x_i$ where $a_i \in \mathfrak{a}$

Suppose that $|X| > 0$ Since $M = \mathfrak{a}M$ we can express $x_1$ as a such a linear combination: $$x_1 = \sum a_i x_i.$$ Moving the term involving $a_1$ to the left, we have $$(1 - a_1)x_1 = \sum_{i > 1} a_i x_i.$$ But $a_1 \in J(R)$ so $1-a_1$ is invertible, say with inverse $b$ Therefore, $$x_1 = \sum_{i > 1} b a_i x_i.$$ But this means that $x_1$ is redundant as a generator of $M$ and so $M$ is generated by the subset $\{x_2, x_3, \dots, x_n\}$ This contradicts the minimality of $X$

We conclude that $|X| = 0$ and therefore $M = 0$