# Embedded associated prime

$\underline{\textbf{Embedded associated prime}}$

I am reading the book "Joins and Intersections". In the proof of Rees theorem I have some doubt.

Let $\mathbf M$ be a finitely generated $\mathbf A$-module and $\mathbf N$ be $A$-submodule of $\mathbf M$ generated by all such $m\in{\mathbf M}$ such that dim ${{\mathbf A}m}<{dim{\mathbf M}}$ .

Why $\frac{\mathbf M}{\mathbf N}$ does not have any embedded prime?

-

I think roughly speaking this is the reason: show for every associated prime $\mathfrak{p}$ of $M/N$, $\dim A/\mathfrak{p}=\dim M$. Then $M/N$ cannot have an embedded prime. To show the claim, suppose $\mathfrak{p}$ is an associated prime of $M/N$. Then this means $\mathfrak{p}$ is the annihilator of some element $x\in M/N$. Since $x\not\in N$, we have $\dim Ax=\dim M$. But $\dim Ax =\dim A/\mathrm{Ann}\; x=\dim A/\mathfrak{p}$.
Actually, I may have cheated a little bit here, identifying annihilator of $x$ as an element of $M/N$ with annihilator of $x$ as an element of $M$. But perhaps you can still make the idea work. –  Mahdi Majidi-Zolbanin May 18 '13 at 15:59