Zariski topology and compact \paracompact space? Does the Zariski topology on a ring (not commutative in common) form a compact or paracompact space and why?
 A: It looks like the accepted answer to this question didn't address some of the questions asked in the comments above, so I thought I'd fill in some details.  This is really just a long comment; sorry I had to include it as an answer.
It seems that the question is regarding the Zariski topology on two-sided prime ideals.  The word "ideal" below refers to two-sided ideals.  
A prime ideal of a noncommutative ring $R$ is a proper ideal such that, for any ideals $I$ and $J$ of $R$, if the ideal product $IJ$ is contained in $P$, then either $I \subseteq P$ or $J \subseteq P$.  Let $\mathrm{Spec}(R)$ denote the set of prime ideals of $R$.
The Zariski topology on $\mathrm{Spec}(R)$ can be defined just as in the commutative case, as stated in Simon Rose's answer: the closed subsets of $\mathrm{Spec}(R)$ are precisely those of the form $V(I) := \{P \in \mathrm{Spec}(R) : I \subseteq P\}$.  It's easy to see that $\bigcap_j V(I_j) = V(\sum I_j)$ for any set of ideals $\{I_j\}$ of $R$, where $\sum I_j$ is the smallest ideal of $R$ containing the $I_j$ (and is equal to the set of all finite sums of elements from the $I_j$).  To see that these are closed under finite unions, one uses the definition of prime ideal to see that $V(I) \cup V(J) = V(IJ)$.  That is, for any prime $P$ of $R$ and any ideals $I$ and $J$ of $R$, the statement $IJ \subseteq P$ is equivalent to the statement that either $I \subseteq P$ or $J \subseteq P$.
A: 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.
