I am looking for an example of an open set in $\operatorname{Spec}(R)$ equipped with the Zariski topology which is not compact.

Question

I have learned the basic open sets i.e. the sets of the form $\{P\in\operatorname{Spec}(R): x\notin P\}$ where $x$ is a fixed element of $R$ are compact. Now any open set is union of such basic open sets. For an example I was thinking of taking $R$ as $k[x_1,x_2,x_3,...]$ then letting $X$=the set of inderminates, the set $U=\{P\in\operatorname{Spec}(R): X\not\subset P\}$ is open in $\operatorname{Spec}(R)$. I am thinking that this open set will be non compact. Am I right with my example? Please help me with it by commenting on my example or by providing a suitable one.

Answer 1

Your example is correct. $U$ is equal to the union $\bigcup_{i=1}^{\infty} D(x_i)$ since $X \not\subset p$ means exactly that $\exists i: x_i \not\in p,$ i.e. $p \in D(x_i).$ These is no finite subcover for this cover since for each $x_i$ there is an element in $U$ which lies only in $D(x_i)$ - the ideal generated by $X \setminus \{x_i\}.$