Let $R$ be a Noetherian ring. Suppose all the nonzero proper ideals of $R$ have nonzero annihilators. Show that if $M$ is a maximal ideal of $R$, then $\exists$ $x \in R $ such that $M$ = $\mathrm{ann}(x)$ (ann means annihilator). Also $x \notin M$ and $x \in N $ for all other maximal ideals $N$ of $R$.
It is quite clear why $M = ann(x)$ for some $x \in R$. Actually any $x\in ann(M)$ will do the purpose.
I am facing difficulty in proving why $x\notin M$ and $x\in N$ for all other maximal ideals $N$ of $R$.
Thanks in advance!!
The first assertion is not true, consider the ring $\mathrm{Vect}_{\mathbb C}(1,x)$ generated by $\mathbb C$ and $x$ such that $x^2=0$; it is Noetherian and its unique maximal ideal is $\mathbb Cx$, $x$ is an element of $\mathrm{ann}(\mathbb Cx)$.
Let $M$ be a maximal ideal, its annihilator is not trivial, thus $x\in ann(M)$ and $Mx=0$ it follows that $M\subset ann(x)$ and $ann(x)=M$.
If $N$ is another maximal element, there exists $m\in M$, $m$ not in $N$,suppose that $x$ is not in $N$, let $p:R\rightarrow R/N$ the projection. $p(x)p(m)=0$ impossible since $R/N$ is a field and $p(x),p(m)\neq 0$ since $x,m$ are not in $N$.