[proofplan]
We reduce to the standard form of [Nakayama's Lemma](/theorems/2935) by passing to the quotient module $M/N$. The hypothesis $\mathfrak{a}M + N = M$ translates to $\mathfrak{a}(M/N) = M/N$ in the quotient. Since $M/N$ is finitely generated (as a quotient of the finitely generated module $M$) and $\mathfrak{a} \subseteq \operatorname{Jac}(R)$, Nakayama's Lemma gives $M/N = 0$, which means $N = M$.
[/proofplan]
[step:Pass to the quotient $M/N$ and verify that $\mathfrak{a}(M/N) = M/N$]
Let $\pi: M \to M/N$ denote the canonical surjection. We claim that $\mathfrak{a}(M/N) = M/N$.
Since $\mathfrak{a}M + N = M$, every element $m \in M$ can be written as $m = \sum_i a_i m_i + n$ for some $a_i \in \mathfrak{a}$, $m_i \in M$, and $n \in N$. Applying $\pi$:
\begin{align*}
\pi(m) = \sum_i a_i \pi(m_i) + \pi(n) = \sum_i a_i \pi(m_i) \in \mathfrak{a}(M/N),
\end{align*}
since $\pi(n) = 0$. As $m$ was arbitrary, $M/N \subseteq \mathfrak{a}(M/N)$. The reverse inclusion $\mathfrak{a}(M/N) \subseteq M/N$ holds because $\mathfrak{a}(M/N)$ is a submodule of $M/N$. Therefore $\mathfrak{a}(M/N) = M/N$.
[guided]
The idea is to eliminate $N$ by quotienting it out. Let $\pi: M \to M/N$ denote the canonical surjection, where $\pi(m) = m + N$.
We need to verify that the hypothesis $\mathfrak{a}M + N = M$ translates to $\mathfrak{a}(M/N) = M/N$ in the quotient. Take any element $\bar{m} = m + N \in M/N$. Since $\mathfrak{a}M + N = M$, we can write $m = \sum_i a_i m_i + n$ with $a_i \in \mathfrak{a}$, $m_i \in M$, and $n \in N$. Then
\begin{align*}
\bar{m} = \pi(m) = \pi\Bigl(\sum_i a_i m_i + n\Bigr) = \sum_i a_i \pi(m_i) + \underbrace{\pi(n)}_{= 0} = \sum_i a_i \pi(m_i).
\end{align*}
This shows $\bar{m} \in \mathfrak{a}(M/N)$. Since $\bar{m}$ was arbitrary, $M/N = \mathfrak{a}(M/N)$.
Equivalently, one can express this at the level of ideals: $\mathfrak{a}(M/N) = (\mathfrak{a}M + N)/N$. The hypothesis gives $\mathfrak{a}M + N = M$, so $(\mathfrak{a}M + N)/N = M/N$.
[/guided]
[/step]
[step:Apply Nakayama's Lemma to $M/N$ to conclude $N = M$]
The module $M/N$ is finitely generated as an $R$-module: since $M$ is finitely generated with generators $m_1, \ldots, m_k$, the quotient $M/N$ is generated by $\pi(m_1), \ldots, \pi(m_k)$. We have verified $\mathfrak{a}(M/N) = M/N$, and the hypothesis gives $\mathfrak{a} \subseteq \operatorname{Jac}(R)$.
All hypotheses of [Nakayama's Lemma](/theorems/2935) are satisfied for the module $M/N$ and the ideal $\mathfrak{a}$. Applying it yields $M/N = 0$, which means $N = M$.
[guided]
We now have all the ingredients to apply the standard form of Nakayama. The module $M/N$ is finitely generated over $R$ because $M$ is finitely generated: if $M = Rm_1 + \cdots + Rm_k$, then $M/N = R\pi(m_1) + \cdots + R\pi(m_k)$, so $M/N$ has at most $k$ generators. We also have $\mathfrak{a}(M/N) = M/N$ (established in the previous step) and $\mathfrak{a} \subseteq \operatorname{Jac}(R)$ (given by hypothesis).
By [Nakayama's Lemma](/theorems/2935), $M/N = 0$.
The equation $M/N = 0$ means every coset $m + N$ equals $0 + N = N$, i.e., every $m \in M$ belongs to $N$. Hence $M \subseteq N$. Since $N \subseteq M$ by definition, we conclude $N = M$.
Why is this reduction useful? The submodule form of Nakayama is the version most commonly applied in practice. It says, roughly, that if a submodule $N$ is "close to $M$" in the sense that the gap $M/N$ is killed by the Jacobson radical, then $N$ already equals $M$. A typical application: if $M$ is finitely generated over a local ring $(R, \mathfrak{m})$ and $m_1, \ldots, m_k \in M$ generate $M/\mathfrak{m}M$ as an $R/\mathfrak{m}$-vector space, then setting $N = Rm_1 + \cdots + Rm_k$ gives $\mathfrak{m}M + N = M$, so $N = M$ -- the lifts of a basis for $M/\mathfrak{m}M$ generate $M$.
[/guided]
[/step]
