Question about associated primes and annihilator I am trying to solve the following exercise:

Let $R$ be noetherian and $M$ a finitely genererated $R$-module. Show that $\mathrm{Ass}(R/\mathrm{Ann}(M)) \subseteq \mathrm{Ass}(M)$ and both sets have the same minimal elements. Show that in general the inclusion is not an equality.  

There is a hint that I should first prove $(\mathrm{Ann}(M):c)_R=\mathrm{Ann}(cM)$ for every $c \in R$. I have managed to do that but I have no clue where to use it. Thx for your time!

Let me demonstrate the inclusion using the equality you proved. 
Let $p \in Ass (R/ ann (M))$. Then there exists $x$ in $R$ such that $p = ann(M) :_R x$. You showed is that $ann(M) :_R x = 0 :_R xM$. Since $M$ is finitely generated, you may choose a finite generating set $\{m_1,\dots,m_n\}$ of $M$. Then
$0:_R xM = \cap\,\, 0:_R xm_i$. 
Summing up we have 
$$p = \cap\,\, 0:_R xm_i.$$
Since the intersection on the RHS is finite and $p$ is prime, $p \supset 0:_R xm_i$ for some $i$. As $Rxm_i \subset xM$, you have the other inclusion. Thus $p$ is in $ass(M)$.
