Lemma 10.17.9. Let $R$ be a ring. The topology on $X = \mathop{\mathrm{Spec}}(R)$ has the following properties:

1. $X$ is quasi-compact,

2. $X$ has a basis for the topology consisting of quasi-compact opens, and

3. the intersection of any two quasi-compact opens is quasi-compact.

Proof. The spectrum of a ring is quasi-compact, see Lemma 10.17.8. It has a basis for the topology consisting of the standard opens $D(f) = \mathop{\mathrm{Spec}}(R_ f)$ (Lemma 10.17.6) which are quasi-compact by the first remark. The intersection of two standard opens is quasi-compact as $D(f) \cap D(g) = D(fg)$. Given any two quasi-compact opens $U, V \subset X$ we may write $U = D(f_1) \cup \ldots \cup D(f_ n)$ and $V = D(g_1) \cup \ldots \cup D(g_ m)$. Then $U \cap V = \bigcup D(f_ ig_ j)$ which is quasi-compact. $\square$

There are also:

• 4 comment(s) on Section 10.17: The spectrum of a ring

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).