We know that the Zariski topology is defined on the set of all prime ideals ,is it possible to define a topology on the set of all ideals of a commutative ring? (I hope that the new topology can utilize the Zariski topology)
Obviously, it is possible to define some topology. However, the obvious generalization of the Zariski topology, to take a set to be closed if and only if it is of the form $V(I) = \{J : I \subset J\}$ for some fixed ideal $I$, does not work since this collection of sets is not generally closed under finite unions. Consider the set $A$ of all ideals of $\mathbb{Z}$ containing (2) or (3). If $A$ were of the form $V(n)$ for any $n \in \mathbb{Z}$, then $(n) \in A$; thus, would contain (2) or (3), i.e., n would divide 2 or 3, which is absurd. However, the set of all prime ideals containing (2) or (3) is equal to the set of all prime ideals containing (6). Nevertheless, using schemes, one can give geometric meaning to the set of all ideals of a ring. Specifically, the ideals of $R$ correspond naturally to the set of closed subschemes of Spec $R$. 


It is possible to give the collection of ideals a weaker structure than a topology, such as the structure of a quantale. This is a generalisation of a locale, which is a form of topology without points. See http://planetmath.org/encyclopedia/Quantale.html for a proof about the quantale of ideals. 


$\{0,1\}^R$
, etc. Are there interesting properties that you want the space to have? – François G. Dorais♦ Apr 8 '10 at 14:32