Assume $R$ is a commutative Noetherian ring and $M$ is a nonzero finitely generated $R$-module and $I\subseteq J\subseteq R$ are ideals so that $JM\neq M$.
I'm trying to get a grasp of the notion of grade of an ideal so I conjectured that
$$ \operatorname{grade}_M I=\operatorname{grade}_M J\Leftrightarrow J\subseteq\operatorname{Zdv}_R(M/IM) $$
and tried to prove it without success.
My attempt on the direction $\Rightarrow$ is that pick a maximal $M$-sequence $(a_1,\ldots,a_n)$ in $I$ and suppose that there is $x\in J\setminus\operatorname{Zdv}_R(M/IM)$ so that $(a_1,\ldots,a_n,x)$ is an $M$-sequence in $J$ which is a contradiction. However I think $x\notin\operatorname{Zdv}_R(M/IM)$ does not imply $x\notin\operatorname{Zdv}_R(M/(a_1,\ldots,a_n)M)$ so that we actually can't deduce that $(a_1,\ldots,a_n,x)$ is an $M$-sequence.
My attempt on the direction $\Leftarrow$ is that suppose $\operatorname{grade}_M I<\operatorname{grade}_M J$ and pick a maximal $M$-sequence $(a_1,\ldots,a_n)$ in $I$. Then it is not maximal in $J$ so there is $x\in J\setminus\operatorname{Zdv}_R(M/(a_1,\ldots,a_n)M)$. Again, I would like to deduce that $x\notin\operatorname{Zdv}_R(M/IM)$ to arrive in a contradiction but I think it can't be done.
My question is: does this conjecture hold even in one of the directions and if it does not then what would be a counterexample?
The crux of the problem seems to be the relation between $\operatorname{Zdv}_R(M/IM)$ and $\operatorname{Zdv}_R(M/(a_1,\ldots,a_n)M)$ where $(a_1,\ldots,a_n)$ is a maximal $M$-sequence in $I$.
If $R=K[X,Y]/(X^2,XY)$, $M=R$, $I=(x)$ and $J=(x,y)$, then $\operatorname{grade}I=\operatorname{grade}J=0$ and $J\not\subseteq Z_R(R/I)$ for $y\notin Z_R(R/I)$.
If $R=K[X,Y]$, $M=R$, $I=(X^2,XY)$ and $J=(X,Y)$, then $1=\operatorname{grade}I<\operatorname{grade}J=2$ and $J\subseteq Z_R(R/I)$.