Let \(R\) be a commutative ring, let \(I\subseteq R\) be an ideal, and let \(M\) be a finitely generated \(R\)-module with \(IM=M\). Then there is \(i\in I\) with \(im=m\) for every \(m\in M\). In particular, if \(I\) is contained in the Jacobson radical \(J(R)\), then \(M=0\).
Let \((R,\mfm)\) be a local ring, let \(M\) and \(N\) be \(R\)-modules with \(N\) finitely generated, and let \(\varphi\colon M\to N\) be \(R\)-linear. Then \(\varphi\) is surjective if and only if the induced map \(M/\mfm M\to N/\mfm N\) of \(R/\mfm\)-vector spaces is surjective.
Let \(m_1,\ldots,m_n\) generate \(M\). Since \(M=IM\), there are \(a_{jk}\in I\) with \(m_j=\sum_ka_{jk}m_k\), so the matrix \(\mathrm{I}_n-A\), with \(A=(a_{jk})\), kills the column vector \((m_1,\ldots,m_n)^t\). Multiplying by the adjugate of \(\mathrm{I}_n-A\) gives \(\det(\mathrm{I}_n-A)\,m_j=0\) for every \(j\). Expanding the determinant, \(\det(\mathrm{I}_n-A)=1-i\) with \(i\in I\), so \(m=im\) for every \(m\in M\). If \(I\subseteq J(R)\), then \(1-i\) is a unit by FF-QWCKR, so \(M=0\).
For the local statement, only the converse needs proof. If \(M/\mfm M\to N/\mfm N\) is surjective, then \(N=\varphi(M)+\mfm N\), so the finitely generated module \(C=N/\varphi(M)\) satisfies \(\mfm C=C\), and \(C=0\) by the first part with \(I=\mfm=J(R)\).