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How does one show that the prime spectrum of a domain is irreducible in the Zariski topology?

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$0$ is the only minimal prime of $R$. Suppose, $R=V(I)\cup V(J)$. If $I,J$ are nonzero, then the zero ideal is in neither closed set. So, one of them must be zero, in which case, the corresponding closed set is the entire spectrum. – Dev Bappa Feb 6 '11 at 20:04
A little typo: it's $\text{Spec } R=V(I)\cup V(J)$. – Poirot Feb 7 '11 at 2:32
The set $\{x\}$ is closed (we say that $x$ is a "closed point") in $\mathrm{Spec} (A) \Leftrightarrow $x is maximal. – user155434 Jun 8 '14 at 11:31

A topological space is irreducible if and only if it is not empty and any two nonempty open subsets have nontrivial intersection. Equivalently, if and only if the space is not empty, and cannot be written as a union of two proper closed subsets.

(To see the equivalence: suppose any two nonempty open subsets have nontrival intersection, and let $C_1$ and $C_2$ be closed subsets such that $C_1\cup C_2 = X$. Let $O_1=X-C_1$, $O_2=X-C_2$; since $O_1\cap O_2 = (X-C_1)\cap(X-C_2) = X-(C_1\cup C_2) = \emptyset$, either $O_1$ is empty or $O_2$ is empty, hence either $C_1 =X$ or $C_2=X$. Conversely, suppose that $X$ is not the union of two proper closed subsets, and let $O_1$ and $O_2$ be nonempty open sets. Then $C_1=X-O_1$ and $C_2=X-O_2$ are proper closed subsets, so $\emptyset\neq X-(C_1\cup C_2) = (X-C_1)\cap(X-C_2) = O_1\cap O_2$.)

For a ring $R$, $\mathrm{Spec}(R)$ as a set consists of all prime ideals of $R$, and the closed sets are the sets of the form $V(E)$, where $E\subseteq R$ is a subset, and $V(E) = \{\mathfrak{p}\in\mathrm{Spec}(R) \mid E\subseteq \mathfrak{p}\}$.

So, suppose $R$ is a domain. Is $\mathrm{Spec}(R)$ empty? Hint: Remember that if $S$ is a commutative ring with identity, then $\mathfrak{I}$ is a prime ideal if and only if $S/\mathfrak{I}$ is an integral domain.

Now suppose that $E$ and $F$ are subsets of $R$ such that $\mathrm{Spec}(R) = V(E) \cup V(F)$. That means: every prime ideal of $R$ contains either $E$ or $F$ (or both).

In particular, there is a very special prime ideal that must contain either $E$ or $F$, which tells you that $E$ or $F$ is actually equal to blank. And therefore, either $V(E)$ or $V(F)$ is blah. Which means that $\mathrm{Spec}(R)$ is blankety-blank.

For an encore, try Exercise I.19 of Atiyah-MacDonald, which says that in general, $\mathrm{Spec}(A)$ is irreducible if and only if the nilradical of $A$ is a prime ideal (and notice that your proposition is just one implication of a special case of this).

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It's missing an "open" here: "any two nonempty subsets", and in other places too. Sorry, I can't edit. – Poirot Feb 7 '11 at 2:36
@John: Thank you. – Arturo Magidin Feb 7 '11 at 2:40

A basic fact about the topological space $\operatorname{Spec} R$ is that the irreducible components of $\operatorname{Spec} R$ correspond naturally to the minimal prime ideals of $R$. For a proof of that see Section 13.4 of my commutative algebra notes. (I don't know why I haven't stated this in the form "Theorem: ..." It just appears in the text. I'll fix that eventually...)

If you admit this fact, then your question becomes "What are the minimal primes in a domain?" which is very easy to answer.

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If anyone has any suggestions for improving this answer, I will gladly take them. – Pete L. Clark Dec 27 '13 at 1:22

It can be shown that $Spec(R)$ is irreducible iff nilradical of $R$ is a prime ideal. Using this and the fact that nilradical of a domain is just $0$ which is prime shows that prime spectrum of domain is irreducible.

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