second Nakayamas Lemma for Jacobson Radical

Let \(A\) be a ring with Jacobson radical \(J(A)\), an ideal \(\mathfrak{a} \subseteq J(A)\) and \(M\) a finitely generated \(A\)-module and \(N \subseteq M\) a submodule such that

\begin{align*} \mathfrak{a}M + N = M \end{align*}

Then it follows that

\begin{align*} M = 0 \end{align*}

It suffices to show that the quotient module

\begin{align*} M/N = 0 \end{align*}

(cf. correspondence theorem for modules)

Note that by definition

\begin{align*} \mathfrak{a} ( M/N) \subseteq M/N \end{align*}

Lte \(m + N \in M/N\).
Since \(M = \mathfrak{a}M + N\) it follows that

\begin{align*} m = am' + n \end{align*}

for an \(a \in A, n \in N\).

This shosw

\begin{align*} m + N =& am' + n + N \\ =& am' + N \in& \mathfrak{a} (M /N) \end{align*}

therefore showing

\begin{align*} \mathfrak{a}(M/N) = M/N \end{align*}

thus it follows from the Nakayama Lemma for Jacobson Radical