Question: I've found the following claim.

If $R$ is a commutative ring, and every prime ideal is maximal, then $\mathrm{Spec}(R)$ with the Zariski topology is Hausdorff and totally disconnected.

Is it true ? Why ?

Answer: 
To say that every prime ideal is maximal is to say that each point of Spec $R$ is closed. One can then prove purely topologically that $X$ is Hausdorff (using
the basic topological properties of Spec's). (See [this question and its answers](https://math.stackexchange.com/questions/108892/direct-way-to-show-operatornamespeca-is-t-1-rightarrow-operatorna).)
if every prime ideal is maximal, then the krull dimension of R is 0.
Then choose M,N in Spec(R). AS they are maximal, M+N=R. then m+n=1, and D(m) is a basic open set in zariski's topology which contains M, and D(n) is a basic open set which contains N. D(m) and D(n) are disjoint and therefore Spec(R) is Hausdorff. (M different from N )