Reference: Atiyah and Macdonald, Introduction to Commutative Algebra, page 46.
Let $A$ be a commutative ring with 1 and $M$ a $A$-module. Then we define $\mathrm{Supp}(M)$ to be the set of prime ideals $\mathfrak{p}$ of $A$ such that $M_{\mathfrak{p}}\neq 0$.
It is known that in every exact sequence $0\longrightarrow N \longrightarrow M \longrightarrow M/N\longrightarrow0$ of $A$-modules one have:
$\mathrm{Supp}(N) \subseteq \mathrm{Supp}(M)=\mathrm{Supp}(N)\bigcup \mathrm{Supp}(M/N)$.
When in such an exact sequence as above, $\mathrm{Supp}(N) \subseteq \mathrm{Supp}(M/N)$ and hence $\mathrm{Supp}(M)=\mathrm{Supp}(M/N)?$
I want to prove if $\dim M=d$ and $N$ is Cohen-Macaulay of dimension $d-1$ and $M/N$ is Cohen-Macaulay of dimension $d$ then $\mathrm{Supp}(M)=\mathrm{Supp}(M/N)$.