Let $R$ be a commutative Noetherian local ring, $M$ a finitely generated $R$-module. If $\dim_R M=\dim_R R/\mathfrak{q}$ for all $\mathfrak{q}\in \operatorname{Min}_R M$ and $x\in R$ is $M$-regular, then $\dim_R M/xM=\dim_R R/\mathfrak{q}$ for all $\mathfrak{q}\in \operatorname{Min}_R M/xM$? (Here $\operatorname{Min}_R M$ means minimal prime ideals in $\operatorname{Supp}_R M$.)
I know that $\dim_R M/xM=\dim_R M-1$. Now, let $\mathfrak{q}\in \operatorname{Min}_R M/xM$, then $\mathfrak{q}\in \operatorname{Supp}_R M$. Clearly, $\mathfrak{q}\notin \operatorname{Min}_R M$. Then there exists $\mathfrak{p}\in \operatorname{Min}_R M$ such that $\mathfrak{p}\subsetneq \mathfrak{q}$. Note that $\dim R/\mathfrak{p}> \dim R/\mathfrak{q}$. The result can be obtained if one shows that $\dim R/\mathfrak{p}= \dim R/\mathfrak{q}+1$. I do not know how to do it. Can someone help me?