Nakayama Lemma for Jacobson Radical

Suppose \(m_1,...,m_n\) is a generator of \(M\) with minimal \(n\). if \(n = 0\), then \(\langle \emptyset\rangle = 0\). Otherwise, for \(m_1 \in M = \mathfrak{a}M\) we get by Assumption a linear combination

\begin{align*} m_1 = \sum_{i=1}^{n} a_im_i && a_i \in \mathfrak{m} \end{align*}

But since \(a_1 \in \mathfrak{m}\), we conclude, that \(a_1 - 1\) is a unit (cf. Jacobson radical and unit)

\begin{align*} m_1 =& \sum_{i=1}^{n} a_im_i && \vert a_i \in \mathfrak{m} \\ m_1 (1 - a_1) =& \sum_{i=2}^{n} a_i m_i && \vert \cdot (1 - a_1)^{-1} \\ m_1 =& (1 - a_1)^{-1} \sum_{i=2}^{n} a_i m_i \end{align*}

Hence \(m_2,...,m_n\) is a smaller generator of \(M\). Thus by induction, \(\emptyset\) is a generator and hence \(M = 0\)

corollary of

1. general Nakayama's lemma
2. Jacobson radical and unit

where by assumption there exists an \(a \in \mathfrak{a}\) such that

\begin{align*} (1 - a) \cdot M = 0 \end{align*}

and \(1 - a \in A^{\times}\) is a unit showing

\begin{align*} 0 =& (1 - a) \cdot m \\ =& m \forall m \in M \end{align*}