Irreducibility of $\operatorname{Spec}$ $\iff$ primality of nilradical

The following is an exercise (Problem 1.19) in Atiyah and Macdonald's Introduction to Commutative Algebra text :

A topological space $$X$$ is said to be irreducible if $$X \neq \emptyset$$ and if every pair of non-empty open sets in $$X$$ intersect, or equivalently if every non-empty open set is dense in $$X$$. Show that $$\operatorname{Spec}(A)$$ is irreducible if and only if the nilradical of $$A$$ is a prime ideal.

I have a specific question about the solution to this problem found here.

Let $$X_f$$ be the set of prime ideals of $$A$$ that do not contain $$f \in A$$, and let $$X_g$$ be the set of prime ideals of $$A$$ that do not contain $$g \in A$$. Suppose $$\operatorname{Spec}(A)$$ is irreducible. Then the author says that $$X_f \cap X_g = \emptyset \Rightarrow X_f$$ is empty or $$X_g$$ is empty. Why does this implication hold ? Is this just a general fact about two prime ideals -- that if their intersection is empty, then one of them must be empty ?

I tried to show this implication by the contrapositive. Suppose that $$X_f$$ and $$X_g$$ are both nonempty. Then there is at least one prime ideal of $$A$$ that does not contain $$f$$, and there is at least one prime ideal of $$A$$ that does not contain $$g$$. However, I don't know how this implies that $$X_f \cap X_g \neq \emptyset$$ (that is, that there is at least one prime ideal that doesn't contain $$f$$ nor $$g$$).

Thanks !

2 Answers

A topological space being irreducible means that any two nonempty open sets intersect non-trivially.

Let $$nil(A) \subseteq A$$ be the nilradical and let $$\pi: A \rightarrow A/nil(A)$$ be the canonical map. The map $$\pi$$ induce a homeomorphism of topological spaces

$$H1.\text{ } Spec(A) \cong Spec(A/nil(A)).$$

Hence $$T:=Spec(A)$$ is irreduible iff $$S:=Spec(B):=Spec(A/nil(A))$$ is irreducible.

Assume $$B$$ is a domain and assume $$D(f) \cap D(g)=D(fg)=\emptyset$$. It follows $$fg$$ is nilpotent, and since $$B$$ is reduced it follows $$fg=0$$. Since $$B$$ is a domain it follows $$f$$ or $$g=0$$. Hence $$D(f)=\emptyset$$ or $$D(g)=\emptyset$$. Hence $$S$$ is irreducible.

Conversely if $$B$$ is reduced and $$S:=Spec(B)$$ is irreducible. Let $$f,g \neq 0 \in B$$ be elements with $$fg=0$$. It follows $$V(fg)=S$$ hence $$D(fg)=D(f) \cap D(g)=\emptyset$$. Since $$S$$ is irreducible we must have $$D(f)=\emptyset$$ hence $$f$$ is nilpotent. It follows $$f\in nil(B)=(0)$$ hence $$B$$ is a domain.

Note: I use the notion "irreducibility" from Hartshorne, Proposition II.3.1 and AM. Ex.1.19:

Definition: A topological $$X$$ space is "irreducible" iff every pair of nonempty open sets in $$X$$ intersect.

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