I asked the following question at < https://math.stackexchange.com/questions/1508168/is-max-r-a-hausdorff-space >, but I pose it here for any help.
Recall a space is totally disconnected if the only connected subsets are singletons (one-point subsets). Now let $R $ be a commutative ring with identity such that $\operatorname{Max}(R)$ is a totally disconnected space, in the sense of the $Zariski$ topology. I want to know if $\operatorname{Max}(R)$ Hausdorff in this case?
Thanks for any help.
For a ring $R$, the space $\operatorname{MaxSpec} R$ -- the set of maximal ideals of $R$ endowed with the Zariski topology -- has two evident properties: its singleton sets are closed (i.e., the T1 separation property) and it is quasi-compact. In his thesis, M. Hochster showed that conversely any quasi-compact T1 topological space $X$ is homeomorphic to $\operatorname{MaxSpec} R$ for some commutative ring $R$.
A totally disconnected topological space is automatically T1: closures of points are connected sets! So your question reduces to one of general topology: is there a totally disconnected, quasi-compact non-Hausdorff topological space $X$? The answer is yes and indeed Eric Wofsey's answer to your earlier math.stackexchange.com question provides an example.
In summary: no, not necessarily.
Added: On the other hand, $\operatorname{Spec} R$ -- the set of prime ideals of $R$ endowed with the Zariski topology -- is T1 iff it is Hausdorff iff it is totally disconnected iff $\operatorname{Spec} R = \operatorname{MaxSpec} R$. See e.g. $\S$13.3 of these notes. Maybe this is helpful for you.
