An element annihilates a module

Question

Let $R$ be a ring, $A$ an $R$-module, $y$ belong in $R$ such that $1+y$ annihilates $A$. Then for any ideal $I$ containing $y$, prove that $IA=A$.

Answer 1

Since $I\subset R$, we have $IA\subset RA=A$. For the other inclusion, let $a\in A$ and notice that $(1+y)a=0$ implies that $ya=-a$.