Can we find a subset of $Spec(R)$ not quasi-compact?

Question

If $R$ is a commutative ring with unit, we can easy prove that $Spec(R)$ is quasi-compact. However can you give me an example of $R$ such that a subset $A \subset Spec(R)$ isn't quasi-compact?

Answer 1

Sure. Take $R$ to be the free Boolean ring on countably many generators. The topological spectrum in this case is homeomorphic to the Stone space consisting of Boolean algebra maps $R \to \mathbf{2}$ with the appropriate Stone topology. By the freeness, this is the space of maps $\mathbb{N} \to \mathbf{2}$, i.e., the space $\mathbf{2}^\mathbb{N}$ with the product topology, i.e., Cantor space. You probably realize that any subspace of this is Hausdorff, and any compact Hausdorff subspace will have to be closed. So then just take $A$ to be any non-closed subset (e.g., an increasing sequence in the middle-thirds model of Cantor space which does not contain its accumulation point).

Answer 2

$\bigcup_n D(x_n) \subseteq \mathrm{Spec}(k[x_1,x_2,\dotsc])$ is not quasi-compact.

More generally, let $R$ be a commutative ring which has an strictly increasing chain of radical ideals $I_1 \subset I_2 \subset \dotsc$ (in particular $R$ is not noetherian). Then the complement of $\bigcap_i V(I_i)$ is open and not quasi-compact.