The spectrum of a ring is quasi-compact

Lemma 10.16.10. Let $R$ be a ring. The space $\mathop{\mathrm{Spec}}(R)$ is quasi-compact.

Proof. It suffices to prove that any covering of $\mathop{\mathrm{Spec}}(R)$ by standard opens can be refined by a finite covering. Thus suppose that $\mathop{\mathrm{Spec}}(R) = \cup D(f_ i)$ for a set of elements $\{ f_ i\} _{i\in I}$ of $R$. This means that $\cap V(f_ i) = \emptyset$. According to Lemma 10.16.2 this means that $V(\{ f_ i \} ) = \emptyset$. According to the same lemma this means that the ideal generated by the $f_ i$ is the unit ideal of $R$. This means that we can write $1$ as a finite sum: $1 = \sum _{i \in J} r_ i f_ i$ with $J \subset I$ finite. And then it follows that $\mathop{\mathrm{Spec}}(R) = \cup _{i \in J} D(f_ i)$. $\square$

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