The Stacks project

The spectrum of a ring is quasi-compact

Lemma 10.17.8. 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 10.17.2 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 $\mathop{\mathrm{Spec}}(R) = \cup _{i \in J} D(f_ i)$. $\square$


Comments (1)

Comment #3795 by slogan_bot on

Suggested slogan: "The spectrum of any ring is quasi-compact"

There are also:

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

Post a comment

Your email address will not be published. Required fields are marked.

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 00E8. Beware of the difference between the letter 'O' and the digit '0'.