Proposition

Show that $\Spec(A)$ is irreducible if and only if the nilradical of $A$ is a prime ideal.

Solution

Let $\mathfrak{R}$ denote the nilradical of $A$. Let $a, b \in A$. Suppose that $ab \in \mathfrak{R}$. We will show that $a \in \mathfrak{R}$ or $b \in \mathfrak{R}$.

• Case 1: $V(\{ a \})^c$ is empty. Then $V(\{ a \}) = \Spec(A)$, so every prime ideal contains $a$. Thus $a \in \mathfrak{R}$.
• Case 2: $V(\{ b \})^c$ is empty. Similarly, $b \in \mathfrak{R}$.
• Case 3: Neither $V(\{ a \})^c$ nor $V(\{ b \})^c$ is empty. Then $V(\{ a \})^c \cap V(\{ b \})^c \ne \emptyset$ Thus there exists $x \in \Spec(A)$ such that $x \notin V(\{ a \})$ and $x \notin V(\{ b \})$. This implies $a \notin p_x$ and $b \notin p_x$. This is a contradiction because $ab \in \mathfrak{R} \subset p_x$. Therefore, this case is not possible.

Therefore, if $\Spec(A)$ is irreducible then the nilradical is prime.

Suppose $\mathfrak{R}$ is a prime ideal.

As shown in this post, $\{ X_f \mid f \in A \}$ forms a basis for the Zariski topology. Since every open set is the union of some basis elements [Lemma 13.1, Munkres], it suffices to show that the intersection of any two nonempty basis elements is nonempty. Let $X_{f_1}, X_{f_2}$ be two nonempty basis elements. Then $V(f_1) \ne \Spec(A)$ and $V(f_2) \ne \Spec(A)$. This implies that there exists $p_x \in \Spec(A)$ such that $f_1 \notin p_x$ Since $\mathfrak{R} \subset p_x$, $f_1 \notin \mathfrak{R}$. Thus $\mathfrak{R} \notin V(f_1)$, so $\mathfrak{R} \in X_{f_1}$. Similarly, $\mathfrak{R} \in X_{f_2}$. Thus $X_{f_1} \cap X_{f_2} \ne \emptyset$.