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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.

To demonstrate that it's a useful concept, I'll give an 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 a sample 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 it's a few years since I studied algebraic geometry or commutative algebra. This question has been reposted on MathOverflow.

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Nice question. But does this topology restrict to the Zariski topology on $\mathrm{Spec}(R)$? I only see this right now for noetherian $R$. Another question: What is AG over groups? –  Martin Brandenburg Mar 22 '12 at 11:32
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As the topology is Hausdorff and totally disconnected, I don't think it's got much to do with the Zariski topology, has it? –  HJRW Mar 22 '12 at 11:56
    
Algebraic geometry over groups is the fancy name for the study of the 'variety' $\mathrm{Hom}(G,\Gamma)$, where $G$ and $\Gamma$ are groups. If $G$ has presentation $\langle x_1,\ldots, x_m\mid r_1,\ldots,r_n\rangle$ then this set corresponds exactly to the set of solutions in $\Gamma$ to the equations $\{r_j=1\}$ in the variables $x_i$---hence the name. –  HJRW Mar 22 '12 at 11:59
    
Dear HJRW, I have never heard of the ideal space of a ring but the Lemma you give as an application is a well-known and easy fact of beginning commutative algebra. It is proved, for example, in Miles Reid's Undergraduate Commutative Algebra. –  Georges Elencwajg Mar 22 '12 at 21:56
    
@Georges Elencwajg, of course I realise that the lemma is well known. My question is whether this is a standard or known approach. –  HJRW Mar 23 '12 at 7:41

1 Answer 1

This space is considered in this paper. For a f.g. commutative ring $R$, it determines the topological type of $\mathcal{I}(R)$.

This paper focuses on the case of a finitely generated free commutative ring, but the general case is addressed, see Section 4: it considers, more generally, the case of submodules of an $R$-module ($R$ any ring). The space of ideals of a ring is the particular case of the free module of rank 1.

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