proof that $\operatorname{Spec}(R)$ is quasi-compactProofThatOperatornameSpecRIsQuasiCompact2006-07-30 03:15:392007-05-30 23:44:14Proofprime spectrum0\usepackage{amssymb}
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Note that most of the notation used here is defined in the entry prime spectrum.
The following is a proof that $\operatorname{Spec}(R)$ is quasi-compact.
\begin{proof}
Let $\Lambda$ be an indexing set and $\displaystyle \left\{ U_{\lambda} \right\}_{\lambda \in \Lambda}$ be an open cover for $\operatorname{Spec}(R)$. For every $\lambda \in \Lambda$, let $I_{\lambda}$ be an ideal of $R$ with $\displaystyle U_{\lambda}=\operatorname{Spec}(R) \setminus V\left( I_{\lambda} \right)$. Since
\begin{center}
$\begin{array}{ll}
\operatorname{Spec}(R) & \displaystyle =\bigcup_{\lambda \in \Lambda} U_{\lambda} \\
& \displaystyle =\bigcup_{\lambda \in \Lambda} \bigg( \operatorname{Spec}(R) \setminus V\left( I_{\lambda} \right) \bigg) \\
& \displaystyle =\operatorname{Spec}(R) \setminus \bigcap_{\lambda \in \Lambda} V\left( I_{\lambda} \right) \\
& \displaystyle =\operatorname{Spec}(R) \setminus V\left( \sum_{\lambda \in \Lambda} I_{\lambda} \right), \end{array}$
\end{center}
$\displaystyle V\left( \sum_{\lambda \in \Lambda} I_{\lambda} \right)=\emptyset$. Thus, by \PMlinkname{this theorem}{VIemptysetImpliesIR}, $\displaystyle \sum_{\lambda \in \Lambda} I_{\lambda} =R$. Since $\displaystyle 1_R \in R=\sum_{\lambda \in \Lambda} I_{\lambda}$, there exists a finite subset $L$ of $\Lambda$ such that, for every $\ell \in L$, there exists an $i_{\ell} \in I_{\ell}$ with $\displaystyle 1_R=\sum_{\ell \in L} i_{\ell}$.
Let $r \in R$. Then $\displaystyle r=r \cdot 1_R=r\sum_{\ell \in L} i_{\ell}=\sum_{\ell \in L} r \cdot i_{\ell} \in \sum_{\ell \in L} I_{\ell}$. Thus, $\displaystyle \sum_{\ell \in L} I_{\ell}=R$. Therefore, $\displaystyle V\left( \sum_{\ell \in L} I_{\ell} \right)=\emptyset$. Since
\begin{center}
$\begin{array}{ll}
\operatorname{Spec}(R) & \displaystyle =\operatorname{Spec}(R) \setminus V\left( \sum_{\ell \in L} I_{\ell} \right) \\
& \displaystyle =\operatorname{Spec}(R) \setminus \bigcap_{\ell \in L} V\left( I_{\ell} \right) \\
& \displaystyle =\bigcup_{\ell \in L} \bigg( \operatorname{Spec}(R) \setminus V\left( I_{\ell} \right) \bigg) \\
& \displaystyle =\bigcup_{\ell \in L} U_{\ell}, \end{array}$
\end{center}
$\displaystyle \left\{ U_{\lambda} \right\}_{\lambda \in \Lambda}$ restricts to a finite subcover. It follows that $\operatorname{Spec}(R)$ is quasi-compact.
\end{proof}