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

It is compact (or quasicompact?). 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. 


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 twosided prime ideals. The word "ideal" below refers to twosided 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 $J \subseteq P$. 

