Does the Zariski topology on a ring (not commutative in common) form a compact or paracompact space and why?



 Zariski topology and compact \paracompact space?
It is compact (or quasi-compact?).


Let $R$ be a ring. The Zariski topology has closed sets given by $V(I) = \{P \mid I \subset P\}$. So let $I\_\alpha$ be a family of ideals such that $\bigcap V(I\_\alpha) = \emptyset = V(R)$.


This means that $\sum I\_\alpha = R$. However, as $1 \in R$, we can write $1 \in \sum I\_\alpha$, and so we have that $1 = i\_{j\_1} + \cdots + i\_{j\_k}$ for some indices ${j\_\ell}$. But then $V(I\_{j\_1}) \cap \cdots \cap V(I\_{j\_k}) = V(R) = \emptyset$ and so Spec $R$ satisfies the finite intersection property, and is compact.


I admit that I'm worried that I'm subtly using commutativity in here somewhere.


