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Let $R$ be a commutative ring with a unit. $\newcommand{\spec}{\operatorname{Spec}}\spec(R)$ denotes the set of all prime ideals in $R$, and it can be topologized using the Zariski topology.

Last year I took a course in commutative algebra, and we had some exercises proving that $\spec(R)$ is always $T_0$, and sometimes is or is not $T_1$. And I had spent a good deal of brain cycles in understanding how convergence looks like in $\spec(\Bbb Z)$ (not because anyone asked, I just wanted to figure this out). We never really went anywhere deeper than a few homework questions about the possible topological structure of $\spec(R)$ and $\spec_\max(R)$.

But now I am wondering, what does the topology tell us about $R$? Does the fact that a certain net of ideals converges tells us anything useful? Or are there general constructions in algebra which exploit certain properties of the topology on $\spec(R)$ for one thing or another?

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I have defined the command \spec for $\spec$ which is usable across this page. –  Asaf Karagila Feb 10 '13 at 22:22
Are you looking for anything specific? The disconnectedness of $\text{Spec}(R)$ is equivalent to $R$ being a direct product of two non-trivial rings. –  Alex Youcis Feb 10 '13 at 22:28
@Alex: Not that I can tell, but I heard so much about the Zariski topology because that course and then I proved such and such properties hold, and that's that. I also never heard anyone talk about convergence in this topology and I always thought of topology as some way of talking about convergence. So to topologize something and not discussing convergence is a bit like defrosting a steak and then using it as a plate for some salad... If anything I'd love to hear any uses for convergence in the Zariski topology, but what you said is interesting on its own. –  Asaf Karagila Feb 10 '13 at 22:30
Topologies are not only used for convergence of nets ... –  Martin Brandenburg Feb 10 '13 at 22:45
@Martin: Yes. I know that. But I feel that topology is first of all a tool to discuss "closeness" of points and convergence. –  Asaf Karagila Feb 10 '13 at 22:46
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3 Answers

Edit. It was already asked before on math.SE how convergence looks like in the Zariski topology.

There are lots of connections between topological properties of $\mathrm{Spec}(R)$ and algebraic properties of $R$. They can be found in many introductions to algebraic geometry or commutative algebra, for example Atiyah's book, Eisenbud's book, EGA, etc. But the topology doesn't recover the whole algebraic structure, since for example the spectrum of a local artinian ring consists just of one single point; more generally $\mathrm{Spec}(R)$ is homeomorphic to $\mathrm{Spec}(R_{\mathrm{red}})$. This means that the topology can only "see" the ring modulo nilpotents. This is why affine schemes come equipped with the structure sheaf, and are able to reconstruct the ring completely. Anyway, here is a list of the connections:

  1. $\mathrm{Spec}(R)$ is always spectral, i.e. it is sober, quasi-compact, and has a basis consisting of quasi-compact open subsets which is stable under intersections.

  2. $\mathrm{Spec}(R)$ is Hausdorff iff $\mathrm{Spec}(R)$ is $T_1$ iff $\mathrm{dim}(R)=0$ (R. Gilmer, Background and preliminaries on zero-dimensional rings, Zero-dimensional Comutative Rings, Marcel Dekker, New York, 1995, pp. 1-13).

  3. $\mathrm{Spec}(R)$ is connected iff $0,1$ are the only idempotents of $R$. More generally, if $X$ is a locally ringed space, then $f \mapsto X_f$ is a bijection between idempotents in $\Gamma(X,\mathcal{O}_X)$ and clopen subsets of $X$.

  4. There is an anti-isomorphism between the closed subsets of $\mathrm{Spec}(R)$ and the radical ideals of $R$. It follows, for example, that $\mathrm{Spec}(R)$ is noetherian as a topological space iff every ascending chain of radical ideals of $R$ becomes stationary. For example, $\mathrm{Spec}(R)$ is noetherian if $R$ is noetherian (but not conversely).

  5. Under the anti-isomorphism above, the irreducible closed subsets correspond to the prime ideals of $R$. It follows that the topological dimension of $\mathrm{Spec}(R)$ coincides with the Krull dimension of $R$.

  6. The closed points of $\mathrm{Spec}(R)$ correspond to the maximal ideals of $R$. For example, $\mathrm{Spec}(R)$ has a unique closed point iff $R$ is local.

  7. The generic points of $\mathrm{Spec}(R)$ correspond to the minimal prime ideals of $R$. It follows that $\mathrm{Spec}(R)$ is irreducible iff $R$ has only one minimal prime ideal iff $\mathrm{rad}(R)$ is a prime ideal.

Additional remarks:

(a) Not every connected affine scheme is path-connected, see here.

(b) More sophisticated topological properties can be found at the Stacks Project, see topology for the notions of catenary, jacobson, constructible, proper, etc. and algebra for the corresponding properties of rings.

(c) Convergent nets or sequence aren't used so often in algebraic geometry. Even the definition of a complete variety (more generally proper morphisms) doesn't refer to the usual notion of completeness in topological spaces, but rather to a more abstract one where the net is replaced by a $K$-valued point, and its limit is replaced by an $A$-valued point, where $A$ is a valuation ring and $K$ is its field of fractions. Similarily, algebraic geometry needs a more abstract notion of the Hausdorff separation property, namely that of separatedness, which can be characterized by the condition that the abstract limits above are unique iff they exist. In general, one has to adjust the notions and techniques from topology and differential geometry in order to apply them to algebraic geometry. Sometimes even the topology has to be adjusted, for example the Inverse Function Theorem becomes only valid in the etale topology.

Of course you can try to find convergent nets or sequences in (affine) schemes, but in my opinion this is not the best way to get comfortable with them, and probably nets are not really useful in this context, since open sets tend to be quite large. For example, an algebraic curve carries the cofinite topology, plus a generic point, which implies that every sequence of pairwise distinct elements converges to any point.

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Thank you very much! I am still interested in convergence, it seems to me that "you're using it wrong" if you give a topology and don't discuss the implication of a convergent net, or something like that. –  Asaf Karagila Feb 10 '13 at 23:18
In algebraic geometry the Zariski topology is motivated by the demand that, as in other geometries, $\{f \neq 0\}$ should define an open subset. We don't really need to talk about nets here in order to use the topology. Instead, a more abstract notion of convergence is needed. See the edit. –  Martin Brandenburg Feb 10 '13 at 23:37
As for your last paragraph, I am not at all trying to "get comfortable" with any such object. I should also point that groups were hardly considered as the right approach for any mathematical problem when they were discovered, but I doubt there are many people who would agree with that. Dismissing something as "not really useful" is maybe reasonable, but not very careful. –  Asaf Karagila Feb 11 '13 at 0:14
Point taken. I just told you the story from my own perspective and experience. My answer refers to the question "what does the topology tell us about $R$?", and topology is not restricted to convergence (but perhaps you question only refers to it?). When you find a nice application of convergents nets in affine schemes, please let us know, this would be interesting and probably new. –  Martin Brandenburg Feb 11 '13 at 1:42
I really meant us, the math.SE-community. You can add it as an answer here. I won't be the only one who is interested. –  Martin Brandenburg Feb 11 '13 at 2:18
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The simplest answer is that, when $R=k[x_1,\ldots,x_n]$, $k$ algebraically closed, we can identify $\spec R$ with $k^n$. The elements of $\spec_{\text{max}} R$ correspond to the points themselves, while other primes correspond to irreducible algebraic sets; that is, zero-loci of irreducible polynomials.

In short, $\spec$ turns rings into geometric objects, and both geometric intuition and geometric proofs can hold even when $R$ is not a finitely-generated algebra over a field. For example, as a number theorist, I'm always looking at rings of integers, which are Dedekind rings, and I can imagine them as curves $\spec k[x]$, where primes correspond to points. (Taking the analogy further, and thinking of archimedean absolute values as those points not on the curve but on its projective closure leads down another very fruitful avenue, but too far astray from the present conversation.)

If we think of $\spec$ in this way, we get a rather coarse, quasi-compact topological space, which is just fine enough to say things like "a homomorphism $A\to B$ induces a continuous map $\spec B\to \spec A$.

Here's one example of how this can be useful: when $A$ is a domain, open sets in $\spec A$ are dense, and the image of a continuous map on a dense set determines the image of that map everywhere. Since a basis of open sets can be given by $\spec A[\frac{1}{f}]$, this is already enough to prove the universal property of localization. These sorts of density arguments appear all the time in algebraic geometry.

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Ah yes, I am no stranger to density arguments. They are one of the key features of forcing. And the idea is about the same. –  Asaf Karagila Feb 10 '13 at 23:07
@AsafKaragila As somebody who recently took his first logic course, and struggled with topological notions there, it's quite gratifying to hear these sorts of questions from the other side! –  Brett Frankel Feb 10 '13 at 23:08
The two thing that got me through that commutative algebra course were the topology and the fact the professor allowed me to write the final paper on the independence of the Whitehead problem. –  Asaf Karagila Feb 10 '13 at 23:14
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I don't know anything about convergence in $\mathrm{Spec} \ R$ but here's an example of what the topology tells us about $R$:

Theorem. $\mathrm{Spec} \ R$ is disconnected as a topological space if and only if $R \simeq A \times B$ is a product of rings.

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