Let $R$ be a Noetherian local ring and $M$ be a finitely generated $R$-module. Suppose that two ideals $I$ and $(x)$ are consisting of zerodiviors on $M$. Given a nonzerodivisor $a\in (I,x)$ on $M$, is it true that the ideal $(I,x)$ consists of zerodivisors on $M/aM$?
I want to show $(I,x)$ is contained in some prime in $\operatorname{Ass}(M/aM)$, but found it not so easy. For example, we have $\sqrt{\operatorname{Ann}(M/aM)}=\sqrt{\operatorname{Ann}M+(a)}$, but how can I proceed?
Set $J=(x)$. We know that $\mathrm{grade}(I,M)=\mathrm{grade}(J,M)=0$. We also know that $\mathrm{grade}(I\cap J,M)=0$ (see Bruns and Herzog, Proposition 1.2.10(c)). Now use the short exact sequence $$0\to R/I\cap J\to R/I\oplus R/J\to R/(I+J)\to 0$$ and the long exact homology sequence for Ext in order to conclude that $\mathrm{grade}(I+J,M)\le1$. By hypothesis $\mathrm{grade}(I+J,M)\ge1$, so $\mathrm{grade}(I+J,M)=1$. Then $$\mathrm{grade}(I+J,M/aM)=\mathrm{grade}(I+J,M)-1=0,$$ and we are done.