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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 ?

user26857
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    Here's the answer. In particular, every such $Spec(R)$ is finite and discrete, and discrete spaces are Hausdorff and totally connected. – Fredrik Meyer May 08 '14 at 15:44
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    Dear @FredrikMeyer There can be infinitely many maximal ideals, so it is not always finite. Take $\prod F_2$, for example, as many copies of $F_2$ as you want. (Sorry if I misunderstood the use of finite... maybe you mean something else.) – rschwieb May 08 '14 at 18:45
  • @rschwieb Thank you for the clarification. If we assume $R$ is Noetherian (as is common in algebraic geometry), then $R$ is Artinian (by a comment on the linked answer). – Fredrik Meyer May 08 '14 at 19:39
  • Dear @FredrikMeyer : Yes, that's certainly true. Then $R/Nil(R)$ would be a product of $n$ fields and there would be $n$ maximal ideals. – rschwieb May 08 '14 at 20:19
  • Related: https://math.stackexchange.com/questions/201661 – Watson Dec 06 '16 at 08:16

3 Answers3

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If every prime is maximal, then $R/\mathord{\operatorname{nil}(R)}$ is von Neumann regular. Since the spectrum of $R$ is homeomorphic to that of $R/\mathord{\operatorname{nil}(R)}$, without loss of generality we can assume $R$ is von Neumann regular.

The thing about von Neumann reuglar rings is that they're full of idempotents, and idempotents disconnect the spectrum. Let $e$ be an idempotent. Notice that given a prime ideal, exactly one of $e$ or $1-e$ is in the prime ideal. This allows you to demonstrate that $V(eR)\cup V((1-e)R)=\operatorname{Spec}(R)$, and $V(eR)\cap V((1-e)R)=\emptyset$. Thus every set of the form $V(eR)$ is clopen.

Now suppose $X\subseteq \operatorname{Spec}(R)$ is a subset with more than one element. Take two of these elements, say $M_1$ and $M_2$. Take $a\in M_1\setminus M_2$. Since $R$ is VNR, $aR=eR$ where $e$ is an idempotent, and we now have $e\in M_1\setminus M_2$. Thus $M_1\in V(eR)$ and $M_2\in V((1-e)R)$. Thus these two clopen sets of $\operatorname{Spec}(R)$ disconnect the set $X$.

(Of course, this simultaneously establishes that any two points can be separated with open sets.)

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rschwieb
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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.)

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Matt E
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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 )

Avdlk
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