If $\operatorname{MSpec}(A)$ with Zariski topology is Hausdorff, is $A$ a pm-ring? May You help me with the following proof?
Thank You!
Let $A$ be a commutative ring with identity $1_A\ne 0_A$. Let $\operatorname{Spec}(A)$ be the set of all prime ideals of $A$ and let $\operatorname{MSpec}(A)$ be the set of all  maximal ideals of $A$.\
We know that if $A$ is a pm-ring (i.e. each prime ideal of $A$ is contained in only one maximal ideal of $A$) then $\operatorname{MSpec}(A)$ with Zariski topology $\mathscr{Z}_M$, inherited by  Zariski topology $\mathscr{Z}$ on $\operatorname{Spec}(A)$, is $T_2$.
We know that  $\operatorname{MSpec}(A)$ with Zariski topology $\mathscr{Z}_M$ is $T_2$ in and only if $A\left/\mathscr{J}(A)\right.$, where $\mathscr{J}(A)=\bigcap \operatorname{MSpec}(A)$ is the Jacobson radical, is a pm-ring.
1) $\forall\ \mathfrak{p}\in \operatorname{Spec}(A)\quad\forall\ \mathfrak{m}_1,\mathfrak{m}_2\in \operatorname{MSpec}(A)\qquad \mathfrak{p}\subseteq\mathfrak{m}_1\cap\mathfrak{m}_2\quad\Rightarrow\quad \mathfrak{m}_1=\mathfrak{m}_2$;
2) $\left(\operatorname{MSpec}(A),\mathscr{Z}_M\right)$ is $T_2$.
I tried to prove that $2) \Rightarrow 1)$, but I think  there is something wrong.
We suppose, by absurdum, that:
$$\exists\ \mathfrak{p}_0\in \operatorname{Spec}(A)\quad\exists\ \mathfrak{n}_1,\mathfrak{n}_2\in \operatorname{MSpec}(A)\qquad \mathfrak{p}_0\subseteq\mathfrak{n}_1\cap\mathfrak{n}_2\quad\wedge\quad \mathfrak{n}_1\ne\mathfrak{n}_2.$$
Then, there exist $W_1,W_2\in \mathscr{Z}_M$ such that $\mathfrak{n}_1\in W_1$, $\mathfrak{n}_2\in W_2$ and $W_1\cap W_2=\emptyset$.
$$\mathfrak{p}_0\subseteq\mathfrak{n}_1\:\Rightarrow\:\mathfrak{n}_1\in\overline{\left\{\mathfrak{p}_0\right\}};$$
$$\mathfrak{p}_0\subseteq\mathfrak{n}_2\:\Rightarrow\:\mathfrak{n}_2\in\overline{\left\{\mathfrak{p}_0\right\}}.$$
Let be
$$\overline{\left\{\mathfrak{p}_0\right\}}^{\, r}=\operatorname{MSpec}(A)\cap\overline{\left\{\mathfrak{p}_0\right\}}$$
the closure of $\left\{\mathfrak{p}_0\right\}$ in $\left(\operatorname{MSpec}(A),\mathscr{Z}_M\right)$. So, because $\mathfrak{n}_1\in\overline{\left\{\mathfrak{p}_0\right\}}^{\, r}$, $W_1\in\mathscr{Z}_M$ and $\mathfrak{n}_1\in W_1$, for closure's point definition, we have that:
$$W_1\cap \left\{\mathfrak{p}_0\right\}\ne\emptyset.$$
So, we deduce that $\mathfrak{p}_0\in W_1$. In the same way, we have that $\mathfrak{p}_0\in W_2$. At the end, we obtain that $\mathfrak{p}_0\in W_1\cap W_2\ne\emptyset$. The absurdum is reached.
 A: I think what you're trying to say is, if I understand it correctly, is this: 
If a prime ideal $\mathfrak{p}$ would have two distinct maximal extensions $\mathfrak{m}$ and $\mathfrak{n}$ (so not $(1)$) we would have that any two open Zariski-neighbourhoods of them in $\operatorname{Spec}(A)$ would intersect in $\mathfrak{p}$, and this is correct I believe, but this does not yet mean that these open Zariski-neighbourhoods cannot be disjoint in $\operatorname{MSpec}(A)$;the intersecting point $\mathfrak{p} \notin \operatorname{MSpec}(A)$.. 
So I see no contradiction yet with $(2)$. You'd need to find a common maximal ideal instead, and maybe that's possible with more ring-theory knowledge.
Added It seems that this implication (2) to (1) is in fact false, by   remark 4.7 in this preprint on page. This implication is from a paper from Simmons for which an erratum also was published. The aforementioned remark even gives a counterexample to the implication $(2) \to (1)$ and gives an extra condition (involving he Jacobsson radical) in proposition 4.8 to fix the implication. 
So not so strange you couldn't prove it. 
