Exercise with finite faithful module

Let $M$ be a faithful finite $A$-module, for a (commutative unitary) ring $A$; assume also that there are two ideals, $I\subseteq J\subset A$, such that $IM=JM$. Show that $I=J$.

Let $\{m_1,\dots ,m_n\}\subseteq M$ be a set of generators of $M$. Then for every $a\in J\setminus I$, exist $a_{i1},\dots ,a_{in}\in I$ such that $am_i=a_{i1}m_1+\dots +a_{in}m_n$, for all $1\le i\le n$. So I have a $n\times n$ matrix defined by $(a_{ij})$. What I should prove is, likely, that there is a non-zero $b\in J$ such that $b\in \operatorname {Ann} M$. How do I recover it though?

Actually you have a matrix whose $(i,i)$ entry is $a_{ii}-a$. The determinant of this matrix annihilates $M$. So $(a^n-b)M=0$ for some $b\in I$. Since $M$ is faithful we get $a^n=b\in I$. (In fact, we get that $a$ is integral over $I$.) But this doesn't mean that $a\in I$.

Counterexample. $A=K[X,Y]$, $I=(X^2,Y^2)$, $J=(X^2,XY,Y^2)$ and $M=J$.

Let $M=\mathbb{Z}/4\mathbb{Z} \oplus \mathbb{Z}/4\mathbb{Z}$, so $M$ is finite. Let $A$ be the ring of two by two matrices over $\mathbb{Z}/4\mathbb{Z}$, generated by: $$1=\left(\begin{array}{cc}1&0\\0&1\end{array}\right),\qquad s=\left(\begin{array}{cc}2&0\\0&0\end{array}\right),\qquad t=\left(\begin{array}{cc}0&2\\0&0\end{array}\right)$$

By construction, $A$ acts faithfully on $M$ (through left multiplication, where elements of $M$ are regarded as column vectors). Also $A$ is a commutative ring:$$st=ts=s^2=t^2=2t=2s=0.$$

Let $I=\langle s\rangle$ and let $J=\langle s,t\rangle$. Then $I\neq J$ but $IM=JM$ and $I\subseteq J\subsetneq A$.