 Solution

Let $X=\mathrm{Spec}R$ be an affine scheme, and let $(U_{\lambda }{)}_{\lambda \in \Lambda }$ be an open cover. Write each complement as a Zariski closed subset $X\setminus U_{\lambda }=V(I_{\lambda })$ for an ideal $I_{\lambda }\subseteq R$. The intersection rule for Zariski closed subsets givesA proper ideal lies in a maximal ideal, which is a prime ideal; consequently $V(J)=\varnothing$ forces $J=R$. Thus $1\in {\sum }_{\lambda }{I}_{\lambda }$. By the definition of a sum of ideals, this membership is witnessed by finitely many elements:Hence ${\sum }_{j=1}^r{I}_{{\lambda }_j}=R$, so ${\bigcap }_{j=1}^{r}V(I_{{\lambda }_j})=\varnothing$, and therefore $X={\bigcup }_{j=1}^r{U}_{{\lambda }_j}$. If $R$ is the zero ring, $X$ is empty and the empty subcover suffices. This provesThe proof uses only complements, the intersection rule for Zariski closed subsets, and the finite expression of an element in a sum of ideals.