Nakayama's Lemma for local rings

Let \(A\) be a local ring with maximal ideal \(\mathfrak{m}\) and \(M\) a finitely generated \(A\)-module. Then if for the product of a module with the ideal

\begin{align*} \mathfrak{m}M = M \end{align*}

we conclude, that

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

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

\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. local ring and subtraction with 1 as unit), so we conclude

\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 \(M = 0\)