**Definition:** Let $R$ be a commutative ring with 1. Endow the power set $2^R$ with the product topology. The *ideal space* $\mathcal{I}(R)$ is defined to be subset of $2^R$ consisting of ideals, equipped with the induced topology.

This is the ring-theoretic analogue of the Gromov--Grigorchuk space of marked groups, which can be used to give nice proofs of simple facts about algebraic geometry over groups (cf. this paper by Champetier and Guirardel). My question is:

Is the 'ideal space' of a ring a standard construction in commutative algebra or algebraic geometry? If so, what's it called and where can I read more about it?

My motivation is to strengthen the analogy between algebraic geometry over groups and classical algebraic geometry. It would be nice, when doing algebraic geometry over groups using the space of marked groups, to be able to say 'This is analogous to the foobar widget in classical algebraic geometry.'

To demonstrate that this concept has some use, I'll give a very simple application. But first, here are a few basic facts.

- $\mathcal{I}(R)$ is compact (because it is a closed subset of $2^R$, which is itself compact by Tychonoff's Theorem), Hausdorff and totally disconnected.
Each point $I\in \mathcal{I}(R)$ is contained in a canonical closed subset

$U_I=\{J\in \mathcal{I}(R)\mid I\subseteq J \}$

(which is in fact isomorphic to $\mathcal{I}(R/I)$).

If $R$ is Noetherian then each $U_I$ is equal to the set of ideals that contain a (finite) generating set for $I$, and hence is open.

The subset of prime ideals in $\mathcal{I}(R)$ is closed: indeed, for each pair of non-units $x,y$, the subset $N(x,y)=\{x\notin I, y\notin I, xy\in I\}$ is open, and the union of these sets is the complement of the set of prime ideals.

Now here's the promised application - a proof of a well known lemma.

**Lemma:** If $R$ is a Noetherian ring then there is a finite set of prime ideals $\mathfrak{p}_1,\ldots,\mathfrak{p}_k\subseteq R$ with the property that every prime ideal contains one of the $\mathfrak{p}_i$.

*Proof:* The set of $U_{\mathfrak{p}}$ for $\mathfrak{p}$ prime is an open cover of the set of prime ideals. Since the set of prime ideals is compact, there is a finite subcover. *QED*

By the way, my research concerns, among other things, algebraic geometry over groups, but I have never seriously studied algebraic geometry or commutative algebra. This question was first posted at math.stackexchange.