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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.

KReiser
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Aniket Bhattacharyea
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  • Here is a link related to the max-spectrum: https://math.stackexchange.com/questions/132854/why-is-better-to-work-with-the-spectrum-of-prime-ideals-than-with-the-maximal-on/3971302#3971302 – hm2020 Oct 26 '21 at 12:35

1 Answers1

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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.

KReiser
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