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

$\displaystyle V\left( \sum_{\lambda \in \Lambda} I_{\lambda} \right)=\emptyset$ Thus, by this theorem, $\displaystyle \sum_{\lambda \in \Lambda} I_{\lambda} =R$ 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

$\displaystyle \left\{ U_{\lambda} \right\}_{\lambda \in \Lambda}$ restricts to a finite subcover. It follows that $\operatorname{Spec}(R)$ is quasi-compact.