Exercise 1.19

Question

A topological space $X$ is said to be \emph{irreducible} if $X 
e \emptyset$ 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.

Amit Awasthi · Accepted Answer

\begin{proof}
  $\Rightarrow$. Suppose $fg \in \mathfrak{N}$ but $f,g 
otin \mathfrak{N}$. By Exercises $17i)$ and $17ii)$, $X_f,X_g$ are non-empty even though $X_f \cap X_g = X_{fg} = \emptyset$, hence $\operatorname{Spec}(A)$ is not irreducible.
  \par $\Leftarrow$. Suppose $\operatorname{Spec}(A)$ is not irreducible, and $U,V$ are non-empty disjoint open sets. Since $\{X_f\}$ is a basis by Exercise 17, there exist $X_f \subseteq U,X_g \subseteq V$ such that $X_{fg} = X_f \cap X_g = \emptyset$ and so $fg \in \mathfrak{N}$ while $f,g 
otin \mathfrak{N}$ by Exercises $17i)$ and $17ii)$. Thus, $\mathfrak{N}$ is not prime.
\end{proof}

Exercise 1.19

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Exercise 1.19

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