Let $R$ be a Noetherian ring, $M$ a finitely generated $R$-module and $p\in \operatorname{Ass}(M)$. Suppose $x$ is an $M$-regular element and $q$ is a minimal prime over $I=(p,x)$. How can we show that $q\in \operatorname{Ass}(M/xM)$?
Note: Since $q$ is minimal prime over $(p,x)$, we know that in $R_q$, $I_q$ is $qR_q$-primary.
On
I dont feel comfortable answering my own question but i am sketching a solution as I was asked to do so above. As seen in the solution by Youngsu and by YACP, we may assume that $(R,m)$ is a Noetherian local ring and $m$ is minimal over $(p,x)$. As $p\in \operatorname{Ass}M$, $p=(0:y)$ for some $0\neq y\in M$. By Krull's Intersection theorem there exists a $r$ such that $y\in x^rM\setminus x^{r+1}M$. Thus $y=x^ra$ for some $a\in M\setminus xM$. Since $x$ is $M$ regular, it follows that $p=(0:a)$. Since $p\in \operatorname{Ass}M$, we have the following natural composite map, say $f$, $Ra\cong R/p\subseteq M\to M/xM$. Note that $Im(f)\neq 0$. Observe that $Im(f)\cong R/(p,x)$ which is of finite length. Thus $m\in Ass(Im(f))\subseteq Ass(M/xM)$.
Assume to the contrary suppose that $q$ is not associated to $M/xM$. We localize at $q$ to assume that $(R,q)$ is local and $\sqrt{(p,x)} = q$. Since $q$ is not associated to $M/xM$, there exists $y \in q$ such that $x,y$ is an $M$-regular sequence. Observe that $y^l$ is $M/xM$-regular for $l \ge 1$. Therefore, we may assume that $y \in (p,x)$. Write $y = w + ax$ for some $w \in p$ and $a \in R$ and $I = (x,y) = (w,x)$. We show that $w,x$ is an $M$-sequence. Since $\operatorname{depth}(I,M) = 2$, by the statement below we conclude that $H_i(w,x;M) = 0$ for $i > 0$. This implies that $w \in p$ is $M$-regular. This contradicts the assumption $p$ associated to $M$.