Exercise 1.23

Question

Let $A$ be a Boolean ring (Exercise $\hyperref[exc:1.11]{11}$), and let $X = \operatorname{Spec}(A)$.
  \begin{enumerate}[label=\roman*)]
    \item For each $f \in A$, the set $X_f$ (Exercise 17) is both open and closed in $X$.
    \item Let $f_1,\ldots,f_n \in A$. Show that $X_{f_1} \cup \cdots \cup X_{f_n} = X_f$ for some $f \in A$.
    \item The sets $X_f$ are the only subsets of $X$ which are both open and closed.
    \item $X$ is a compact Hausdorff space.
  \end{enumerate}

Amit Awasthi · Accepted Answer

\begin{proof}[Proof of $i)$]
  $X_f$ is open by definition. Now if $\mathfrak{p} \in X$, then $f(1-f) = 0$ implies either $f \in \mathfrak{p}$ or $(1-f) \in \mathfrak{p}$, exclusively, hence $X_f = V(1-f)$ is also closed.
\end{proof}
\begin{proof}[Proof of $ii)$]
  By $i)$, $X_{f_1} \cup \cdots \cup X_{f_n} = V(\mathfrak{a})$ for some ideal $\mathfrak{a} \subseteq A$. By Exercise $11iii)$, $\mathfrak{a} = (g)$ for some $g \in A$, so letting $f = 1-g$, $X_f = V(g) = X_{f_1} \cup \cdots \cup X_{f_n}$ by $i)$.
\end{proof}
\begin{proof}[Proof of $iii)$]
  If $Y \subseteq X$ is closed, it is quasi-compact since $X$ is by Exercise $17v)$. If $Y$ is also open, $Y = X_{f_1} \cup \cdots \cup X_{f_n}$ for some $f_i$ by Exercise $ef{exc:1.17}vii)$. By $ii)$, $Y = X_f$ for some $f$.
\end{proof}
\begin{proof}[Proof of $iv)$]
  $X$ is quasi-compact by Exercise $ef{exc:1.17}v)$. Suppose $\mathfrak{p} 
e \mathfrak{q}$. By Exercise $11iii)$, $\mathfrak{p} = (f)$ for some $f \in A$, and $f 
otin \mathfrak{q}$. By the argument in $i)$, $\mathfrak{p} \in V(f) = X_{1-f}$ while $\mathfrak{q} \in X_f$, hence $X$ is Hausdorff.
\end{proof}

Exercise 1.23

Answers

Comments

Exercise 1.23

Answers

Comments

Add answer