Lately I've been thinking to annihilator of modules and I've conjectured a proposition I can't prove, so I'll expose my claim.
Let $A$ be a noetherian reduced (commutative) ring and let $M$ be a faithful finitely generated module on $A$. Is it true that there exists an element $m \in M$ such that $\operatorname{Ann}(m)=0$?
Let's recall that $\operatorname{Ann}_{S^{-1}R}(S^{-1}M)=S^{-1}\operatorname{Ann}_R(M)$ for any finitely generated $R$-module $M$ and every multiplicative set $S\subset R$.
Now let $S$ be the set of all non-zerodivisors in $R$, and consider the $S^{-1}R$-module $S^{-1}M$. This is finitely generated and faithful. It is well known that $S^{-1}R$, the total ring of fractions of $R$, is isomorphic to a finite direct product of fields. Set $S^{-1}R\simeq K_1\times\cdots\times K_n$ where $K_i$ are fields. Then $S^{-1}M$ decomposes as $N_1\times\dots\times N_n$ with $N_i$ a $K_i$-module. If there is $N_i=0$ then $\operatorname{Ann}_{S^{-1}R}(S^{-1}M)\ne0$, a contradiction. Thus for each $i=1,\dots,n$ we can pick $x_i\in N_i$, $x_i\ne 0$. Setting $x=(x_1,\dots,x_n)$ we get $\operatorname{Ann}(x)=0$.
In terms of rings and modules of fractions we have found a non-zero element $\frac ms\in S^{-1}M$ with $\operatorname{Ann}_{S^{-1}R}(\frac ms)=0$. If $a\in R$ is such that $am=0$, then $\frac a1\frac ms=\frac 01$, and therefore $\frac a1=\frac 01$ hence there is $t\in S$ such that $ta=0\Rightarrow a=0$. This gives us $\operatorname{Ann}(m)=0$, and we are done.
(since $A$ is reduced we can) Assume that $0= p_1\cap ...\cap p_n$. Since $M_{p_i}\neq 0$ there exists $m_i$ such that $Ann\ m_i\subset p_i$. Now $Ann\ (m_1+...+ m_n)\subset p_1\cap ...\cap p_n=0$
