This is exercise I-17 of The Geometry of Schemes by D. Eisenbud and J. Harris.
I have thought this. For $X\subset\mathrm{Spec}(R),$ an element $U$ of an arbitrary open cover $\mathscr{U}$ of $X$ is given by $$U=\mathrm{Spec}(R)-V(S)=\mathrm{Spec}(R)-\cap_{f\in S}V(f)=\cup_{f\in S}\mathrm{Spec}(R_f), $$ for some $S\subset R\ $ and with $R_f$ being the localization of $R$ at the element $f.$
So we have that $X=\cup_{U\in\mathscr{U}}U=\cup_S(\cup_{f\in S}\mathrm{Spec}(R_f)),$ and the prime ideals in $R_f$ are in one-to-one correspondence with the ideals in $R$ by $\mathfrak{p}\subset R\leftrightarrow \mathfrak{p}R_f\subset R_f.$ Therefore, $\cup_S(\cup_{f\in S}\mathrm{Spec}(R_f))$ can be seen as an ascending chain of ideals in a noetherian ring, so it must stabilize, so we conclude that $\mathscr{U}$ has a finite subcover.
Does this work? Is there any other way to do this?
Best Answer
A finite union of a quasicompacts is quasi compact. If $U$ is an open, then $U= X\setminus V(I) $ where $I$ is an ideal of $R$, there is $f_1,...f_r$ in$R$ such that $I=(f_1,...f_r)$. Then $U=D(f_1)\cup... \cup D(f_r) $ and quasi compact.