Let $R$ be a commutative ring with unity. Show that the maximal spectrum of $R$ is quasi-compact under Zariski topology.
I tried by taking an open cover $\{ D(I_i) | I \in \Lambda \}$ of $\operatorname{Max} R$. Then $\operatorname{Max} R \subseteq \{ D(I_i) \mid I \in \Lambda \} \implies \operatorname{Max} R \subseteq \operatorname{Spec}(R) \setminus V(\sum I_i)$
But I don't know how to proceed from here.
Best Answer
You can reduce to the claim that $\operatorname{Spec} R$ is quasi-compact.
Suppose $\{D(I_i)\}$ covers $\operatorname{MaxSpec} R$. Then $V(\sum I_i)\subset \operatorname{Spec} R$ is a closed subscheme of $\operatorname{Spec} R$, so it is quasi-compact. Therefore if it's non-empty, it must contain a closed point. But such a closed point is a point of $\operatorname{MaxSpec} R$, and $\{D(I_i)\}$ were assumed to cover $\operatorname{MaxSpec} R$, so it must be the case that $\{D(I_i)\}$ cover $\operatorname{Spec} R$ as well, so we're down to just showing that $\operatorname{Spec} R$ is quasi-compact. This is well-covered on MSE (1 2) and elsewhere, like Stacks 00E8.