# If we can define a topology on the set of all the ideals of a commutative ring?

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)

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It is clearly possible: the discrete topology, the indiscrete topology, the product topology from $\{0,1\}^R$, etc. Are there interesting properties that you want the space to have? – François G. Dorais Apr 8 '10 at 14:32
Perhaps you want to take a look at en.wikipedia.org/wiki/Spectrum_of_a_ring to see some motivation why we define the spectrum of a ring to consist of all proper prime ideals. – abcdxyz Apr 8 '10 at 14:40
cao, you can edit the question to make it more precise. – François G. Dorais Apr 8 '10 at 15:34
Also take a look at this question about the spectrum of radical ideals: mathoverflow.net/questions/14311/… – Charles Rezk Apr 8 '10 at 16:48

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 $R$.