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