Exercise 1.17

Question

For each $f \in A$, let $X_f$ denote the complement of $V(f)$ in $X = \operatorname{Spec}(A)$. The sets $X_f$ are open. Show that they form a basis of open sets for the Zariski topology, and that
  \begin{enumerate}[label=\roman*)]
    \item $X_f \cap X_g = X_{fg}$;
    \item $X_f = \emptyset \Leftrightarrow f$ is nilpotent;
    \item $X_f = X \Leftrightarrow f$ is a unit;
    \item $X_f = X_g \Leftrightarrow r\left( (f) \right) = r\left( (g) \right)$;
    \item $X$ is quasi-compact (that is, every open covering of $X$ has a finite sub-covering).
    \item More generally, each $X_f$ is quasi-compact.
    \item An open subset of $X$ is quasi-compact if and only if it is a finite union of sets $X_f$.
  \end{enumerate}
  \par The sets $X_f$ are called \emph{basic open sets} of $X = \operatorname{Spec}(A)$.

Amit Awasthi · Accepted Answer

\begin{proof}[Proof of basis]
  Let $X \setminus V(E)$ be open; $\{X_f\}$ is a basis since by Exercise $15iii)$,
  \begin{equation*}
    X \setminus V(E) = X \setminus \bigcap_{f \in E} V(f) = \bigcup X_f.\qedhere
  \end{equation*}
\end{proof}
\begin{proof}[Proof of $i)$]
  $X_f \cap X_g = X \setminus (V(f) \cup V(g)) = X \setminus V(fg) = X_{fg}$ by Exercise $15iv)$.
\end{proof}
\begin{proof}[Proof of $ii)$]
  $X_f = \emptyset$ if and only if $f$ is contained in every prime ideal of $A$. But this holds if and only if $f \in \mathfrak{N}$ by Prop.~1.8, i.e., $f$ is nilpotent.
\end{proof}
\begin{proof}[Proof of $iii)$]
  $X_f = X$ if and only if $f$ is not contained in any prime ideal of $A$, which holds if and only if $f$ is a unit by Cor.~1.5.
\end{proof}
\begin{proof}[Proof of $iv)$]
  $\Leftarrow$. $X_f = X \setminus V(r(f)) = X \setminus V(r(g)) = X_g$ by Exercise $ef{exc:1.15}i)$.
  \par $\Rightarrow$. $X_f = X_g$ implies $\mathfrak{p} 
i f$ if and only $\mathfrak{p} 
i g$, so $r(f) = r(g)$ by Prop.~1.14.
\end{proof}
\begin{proof}[Proof of $v)$]
  Since the $\{X_f\}$ form a basis by the above, any open cover of $X$ has a refinement of the form $\bigcup X_{f_\alpha}$. Thus, by Exercises $ef{exc:1.15}ii)$ and $ef{exc:1.15}iii)$,
  \begin{align*}
    X = \bigcup_{\alpha \in A} X_{f_\alpha} &\Leftrightarrow \emptyset = X \setminus \bigcup_{\alpha \in A} X_{f_\alpha} = \bigcap_{\alpha \in A} V(f_\alpha) = V\left( (f_\alpha)_{\alpha\in A} ight)\
    &\Leftrightarrow 1 \in (f_\alpha)_{\alpha \in A}\
    &\Leftrightarrow 1 = g_1f_1 + \cdots + g_mf_m\ 	ext{for some}\ g_i \in A, \{f_1,\ldots,f_m\} \subseteq \{f_\alpha\}\
    &\Leftrightarrow \emptyset = V\left( (f_1,\ldots,f_{m}) ight) = \bigcap_{i=1}^m V(f_i)\
    &\Leftrightarrow X = X_{f_1} \cup \cdots \cup X_{f_m}.\qedhere
  \end{align*}
\end{proof}
\begin{proof}[Proof of $vi)$]
  As in $v)$, any open cover of $X_f$ has a refinement of the form $\bigcup X_{f_\alpha}$. Now again, by Exercise 15iii),
  \begin{equation*}
    X_f \subseteq \bigcup_{\alpha \in A} X_{f_\alpha} \Leftrightarrow V(f) \supseteq \bigcap_{\alpha \in A} V(f_\alpha) = V\left( (f_\alpha)_{\alpha \in A} ight),
  \end{equation*}
  and so $f \in r\left( (f_\alpha)_{\alpha \in A} ight)$ by Prop.~1.14, i.e., $f^n = g_1f_1 + \cdots + g_mf_m$ for some $g_i \in A, \{f_1,\ldots,f_m\} \subseteq \{f_\alpha\}$. Then, since $V(f) = V(f^n)$ by Exercise 15i),
  \begin{equation*}
    V(f) \supseteq V\left( (f_1,\ldots,f_m) ight) = \bigcap_{i=1}^m V(f_i) \Leftrightarrow X_f \subseteq X_{f_1} \cup \cdots \cup X_{f_m}.\qedhere
  \end{equation*}
\end{proof}
\begin{proof}[Proof of $vii)$]
  $\Leftarrow$. Trivial since each $X_f$ is quasi-compact and open.
  \par $\Rightarrow$. $X$ can be written as a union of $X_f$ since $\{X_f\}$ is a basis, and we can find a finite subcover since $X$ is quasi-compact.
\end{proof}

Exercise 1.17

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Exercise 1.17

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