Let $R$ be a local ring with maximal ideal $I$. $M$ is a finitely generated module over $R$ generated by $a_1, \ldots, a_n$ and the generators are chosen such that their quotients in $M/IM$ form a basis. Then there is a surjective homomorphism $f: R^{n} \to M$. Suppose that $R^{m} = M \oplus \ker f$. How to show that $\ker f = I\ker f$?
Suppose $(p_1, \ldots, p_n) \in \ker f$, then $p_1a_1+\ldots+p_na_n = 0$. It then implies that $p_1a_1+\cdots+p_na_n = 0$ in $M/IM$. So $p_i \in I, \forall i$. But this is not enough to conclude that $\ker f = I\ker f$.