Skip to the Main Content

Note:These pages make extensive use of the latest XHTML and CSS Standards. They ought to look great in any standards-compliant modern browser. Unfortunately, they will probably look horrible in older browsers, like Netscape 4.x and IE 4.x. Moreover, many posts use MathML, which is, currently only supported in Mozilla. My best suggestion (and you will thank me when surfing an ever-increasing number of sites on the web which have been crafted to use the new standards) is to upgrade to the latest version of your browser. If that's not possible, consider moving to the Standards-compliant and open-source Mozilla browser.

August 20, 2014

Holy Crap, Do You Know What A Compact Ring Is?

Posted by Tom Leinster

You know how sometimes someone tells you a theorem, and it’s obviously false, and you reach for one of the many easy counterexamples only to realize that it’s not a counterexample after all, then you reach for another one and another one and find that they fail too, and you begin to concede the possibility that the theorem might not actually be false after all, and you feel your world start to shift on its axis, and you think to yourself: “Why did no one tell me this before?”

That’s what happened to me today, when my PhD student Barry Devlin — who’s currently writing what promises to be a rather nice thesis on codensity monads and topological algebras — showed me this theorem:

Every compact Hausdorff ring is totally disconnected.

I don’t know who it’s due to; Barry found it in the book Profinite Groups by Ribes and Zalesskii. And in fact, there’s also a result for rings analogous to a well-known one for groups: a ring is compact, Hausdorff and totally disconnected if and only if it can be expressed as a limit of finite discrete rings. Every compact Hausdorff ring is therefore “profinite”, that is, expressible as a limit of finite rings.

So the situation for compact rings is completely unlike the situation for compact groups. There are loads of compact groups (the circle, the torus, SO(n)SO(n), U(n)U(n), E 8E_8, …) and there’s a very substantial theory of them, from Haar measure through Lie theory and onwards. But compact rings are relatively few: it’s just the profinite ones.

I only laid eyes on the proof for five seconds, which was just long enough to see that it used Pontryagin duality. But how should I think about this theorem? How can I alter my worldview in such a way that it seems natural or even obvious?

Posted at August 20, 2014 11:54 PM UTC

TrackBack URL for this Entry:   http://golem.ph.utexas.edu/cgi-bin/MT-3.0/dxy-tb.fcgi/2763

21 Comments & 0 Trackbacks

Re: Holy Crap, Do You Know What A Compact Ring Is?

Well, firstly, if G is a compact ring, then the connected component of the identity is also a compact ring, so the basic question is why there are no non-trivial compact connected rings.

The Peter-Weyl theorem implies that every compact group is the inverse limit of Lie groups (or in fact linear groups). In the case of connected abelian groups, they are the inverse limit of tori. So just from the additive structure, a compact connected ring is an inverse limit of tori.

Now, tori are rigid; they don’t have any endomorphisms near the identity. (By Pontryagin duality, this is equivalent to the discreteness of the Pontryagin dual.) But multiplication on a ring by an element close to 1 is an endomorphism. So there are no non-trivial elements close to 1, and so the ring is trivial.

So I think the main ingredients here are (a) compact connected abelian groups are basically tori (in particular, they have much less variability than arbitrary compact groups), and (b) tori are basically rigid.

Posted by: Terence Tao on August 21, 2014 5:16 AM | Permalink | Reply to this

Re: Holy Crap, Do You Know What A Compact Ring Is?

In the second paragraph we can avoid appealing to Peter-Weyl by appealing to Pontryagin duality: the category of compact (Hausdorff) abelian groups is the opposite of the category of discrete abelian groups, and these are filtered colimits of finitely generated abelian groups, which are products of Zs and finite abelian groups. So compact (Hausdorff) abelian groups are cofiltered limits of tori and finite abelian groups.

Posted by: Qiaochu Yuan on August 21, 2014 5:55 AM | Permalink | Reply to this

Re: Holy Crap, Do You Know What A Compact Ring Is?

if G is a compact ring, then the connected component of the identity is also a compact ring

Maybe I’m missing something obvious, but why is the connected-component of the identity a ring? (Certainly it’s compact.) For instance, let RR be a finite discrete ring. The connected-component of 11 is {1}\{1\}. A one-element set can of course be given a unique ring structure, but it’s not a subring of RR unless RR is trivial: it doesn’t contain zero, and it’s not closed under addition.

In a general topological ring RR, the connected-component VV of RR is certainly closed under multiplication (because VVV \cdot V is a connected set containing 11=11\cdot 1 = 1), but the finite example shows that it needn’t be closed under ++ or contain 00. So what’s the story?

Posted by: Tom Leinster on August 21, 2014 4:25 PM | Permalink | Reply to this

Re: Holy Crap, Do You Know What A Compact Ring Is?

Or maybe Terry meant the additive identity 00? In that case, there’s still a similar problem to the one I mentioned: the connected-component of 00 needn’t contain 11, and therefore isn’t a subring, at least assuming the convention that rings have multiplicative identities. But if we drop that convention, then the connected-component of 00 is a subring. So maybe that’s the story.

Posted by: Tom Leinster on August 21, 2014 4:34 PM | Permalink | Reply to this

Re: Holy Crap, Do You Know What A Compact Ring Is?

Ah, you’re right, the connected component of 0 doesn’t necessarily contain 1, but the argument still applies (replace 1 by 0 throughout).

Posted by: Terence Tao on August 21, 2014 6:26 PM | Permalink | Reply to this

Only the 0 component

I still don’t get it.

The argument tells us that the component containing 00 is trivial. But it doesn’t tell us anything about the other components.

Posted by: Jacques Distler on August 21, 2014 7:41 PM | Permalink | PGP Sig | Reply to this

Re: Only the 0 component

Surely all connected components are homeomorphic since adding a constant is a homeomorphism.

Posted by: Tom Ellis on August 21, 2014 9:38 PM | Permalink | Reply to this

Re: Holy Crap, Do You Know What A Compact Ring Is?

Maybe Topological Rings helps, cor. 32.3.

Posted by: David Corfield on August 21, 2014 5:36 PM | Permalink | Reply to this

Re: Holy Crap, Do You Know What A Compact Ring Is?

For some reason, Google doesn’t let me see Cor 32.3. What did you mean by “helps”?

Posted by: Tom Leinster on August 21, 2014 6:02 PM | Permalink | Reply to this

Re: Holy Crap, Do You Know What A Compact Ring Is?

Let CC be the connected component of zero in a compact ring AA. (1) AC=(0)=CAA C = (0) = C A. (2) If zero is the only element cc of AA such that Ac=cA=(0)A c = c A = (0), then AA is totally disconnected. (3) If AA either has an identity element or is semisimple, AA is totally disconnected.

I wonder what Theorem 32.2 said. Ah, I can see p. 429 which reports it as that the connected component of zero in a locally compact ring annihilates on the right [left] any left [right] bounded additive subgroup, which in turn relies on 32.1 about the existence of sufficiently many characters on a locally compact abelian group.

Posted by: David Corfield on August 21, 2014 9:45 PM | Permalink | Reply to this

Re: Holy Crap, Do You Know What A Compact Ring Is?

Thanks very much, Terry and Qiaochu. Got it. I suspect the proof in Ribes and Zalesskii’s book is structured along similar lines; I should look it up.

Posted by: Tom Leinster on August 23, 2014 3:27 PM | Permalink | Reply to this

Re: Holy Crap, Do You Know What A Compact Ring Is?

Not to belabor this. But looking over the past comments, might this be a satisfactory way of summarizing the argument?

Proposition: Every compact Hausdorff ring RR is totally disconnected.

Proof: Let R^\hat{R} be the Pontryagin dual of (the additive group of) RR; this R^\hat {R} is discrete. As RR and R^\hat{R} are locally compact Hausdorff, they are exponentiable as spaces, and it follows quickly that the functorial maps

Hom(R,R)Hom(R^,R^),Hom(R^,R^)Hom(R^^,R^^)Hom(R,R)Hom(R, R) \to Hom(\hat{R}, \hat{R}), \qquad Hom(\hat{R}, \hat{R}) \to Hom(\hat{\hat{R}}, \hat{\hat{R}}) \cong Hom(R, R)

are continuous homomorphisms of topological groups. They are mutually inverse, and so they give an isomorphism of topological groups.

The multiplication map R×RRR \times R \to R transforms to a continuous injection i:RHom(R,R)Hom(R^,R^)i: R \to Hom(R, R) \cong Hom(\hat{R}, \hat{R}). As RR is compact and Hom(R^,R^)Hom(\hat{R}, \hat{R}) is Hausdorff, the injection ii maps RR homeomorphically onto its image in Hom(R^,R^)Hom(\hat{R}, \hat{R}), i.e., is a subspace embedding. But Hom(R^,R^)Hom(\hat{R}, \hat{R}) is manifestly a subspace of a product of discrete spaces and hence totally disconnected, so RR is totally disconnected as well. (In more detail: if CRC \subseteq R is connected, then the image of CC under the composite

CRiHom(R^,R^) xR^R^proj xR^C \subseteq R \stackrel{i}{\to} Hom(\hat{R}, \hat{R}) \hookrightarrow \prod_{x \in \hat{R}} \hat{R} \stackrel{proj_x}{\;\;\to\;\;} \hat{R}

is also connected and hence a one-point space for each xR^x \in \hat{R}, so that i(C)i(C) and therefore CC are one-point spaces.)

Posted by: Todd Trimble on August 24, 2014 6:48 AM | Permalink | Reply to this

Re: Holy Crap, Do You Know What A Compact Ring Is?

Wonderful! That’s not belabouring it at all — you seem to have a knack for finding optimal proofs.

So the key points are that R^\hat{R} is discrete and that we have a topological embedding i: R TopGp(R,R) TopGp(R^,R^) r r, \begin{aligned} i: &R &\to &TopGp(R, R) &\cong TopGp(\hat{R}, \hat{R}) \\ &r &\mapsto &r\cdot -, \end{aligned} from which it follows that RR is totally disconnected.

Posted by: Tom Leinster on August 24, 2014 12:36 PM | Permalink | Reply to this

Re: Holy Crap, Do You Know What A Compact Ring Is?

Considering all the aforementioned contributions and meaning-fying previous remarks , we may comment correspondence mainstreams in the article by Jan Dobrowolski and Krupiński in J Algebra 401 , 161-178 ( 2014 ) about the denotation of small compact G-rings , allowing the description of possible interactions of G on the underlying ring.

Posted by: Sabino Guillermo Echebarria Mendieta on August 26, 2014 12:00 PM | Permalink | Reply to this

Re: Holy Crap, Do You Know What A Compact Ring Is?

Very nice proofs! I had the opposite reaction to Tom and thought that this should be true.. I’d like to go a little further in the classification. Suppose that we start with a compact (commutative) Hausdorff ring. Then we know that it is totally disconnected. There are numerous examples of this. Take any non-trivial commutative ring RR and form the pp-adic topology using a prime ideal pp. My question is: are these the only possible examples? I mean, if we start with a compact commutative Hausdorff ring RR, does there exist a ideal II such that RR can be identified as a subring of the completion of RR by II?

Posted by: K Hughes on September 9, 2014 11:28 AM | Permalink | Reply to this

Re: Holy Crap, Do You Know What A Compact Ring Is?

Hi Kevin! I guess my initial reaction just goes to show how unreliable intuitions can be… and maybe yours shows how much they can be trained.

Regarding your question, how would this work with the profinite completion of \mathbb{Z}, i.e. the (inverse) limit of /n\mathbb{Z}/n\mathbb{Z} over all positive integers nn? (Or, if you prefer, it’s the product 2× 3× 5×\mathbb{Z}_2 \times \mathbb{Z}_3 \times \mathbb{Z}_5 \times \cdots.)

Posted by: Tom Leinster on September 9, 2014 11:40 AM | Permalink | Reply to this

Re: Holy Crap, Do You Know What A Compact Ring Is?

In case there’s any confusion, Tom means the product of pp-adic integers, with pp ranging over all primes. Not the product of finite rings /(p)\mathbb{Z}/(p)!

Despite much clucking (over at MathOverflow, for example) that p\mathbb{Z}_p is wrong notation for the ring with pp elements, it’s certainly found very often in the literature as used by quite respectable people. I wish people would write ^ p\widehat{\mathbb{Z}}_p for the pp-adic completion – the hat symbol is commonly used for completions and I think it would help avoid unnecessary confusion.

Posted by: Todd Trimble on September 9, 2014 3:26 PM | Permalink | Reply to this

Re: Holy Crap, Do You Know What A Compact Ring Is?

Thanks for clarifying my notation, Todd. You’re right, that was unclear, especially as p/p\prod_p \mathbb{Z}/p\mathbb{Z} is a sensible sort of thing too.

I wish people would write ^ p\hat{\mathbb{Z}}_p for the pp-adic completion

I hadn’t seen that suggestion before. Is it compatible with the notation ^\hat{\mathbb{Z}} for the ring of profinite integers, in the sense that the ring of pp-adic integers is the localization at pp of ^\hat{\mathbb{Z}}?

Posted by: Tom Leinster on September 9, 2014 5:34 PM | Permalink | Reply to this

Re: Holy Crap, Do You Know What A Compact Ring Is?

Arguably an even better notation is p \mathbb{Z}^{\wedge}_p, which associates the completion “hat” directly with pp.

Posted by: Mike Shulman on September 9, 2014 8:00 PM | Permalink | Reply to this

Re: Holy Crap, Do You Know What A Compact Ring Is?

Yes, I thought I had seen the hat notation used elsewhere for various topological completions or profinite completions. I have to say that my memory is not quite as good as I once fancied, and I would need some time digging up references for this particular memory. In any case, the ring of pp-adic integers, seen as the profinite completion of the local ring (p)\mathbb{Z}_{(p)}, would by that logic be (p)^\widehat{\mathbb{Z}_{(p)}} – but here I’m be willing to cut a corner by writing it as ^ p\hat{\mathbb{Z}}_p. I also like Mike’s suggestion p \mathbb{Z}_p^\wedge.

I concede that n\mathbb{Z}_n for the quotient ring isn’t great (recently I’ve been trying to remember to write /n\mathbb{Z}/n or /(n)\mathbb{Z}/(n) instead) – but I see no compelling reason that the rights to p\mathbb{Z}_p should go to the pp-adics. There is plenty of historical precedent for that notation to go to either side (Mac Lane and Birkhoff use it for the quotient ring, as do Spanier and the older algebraic topologists). My own annoyed reaction is best explicable by the fact that my first associations for p\mathbb{Z}_p are with the quotient ring, and I’m pretty sure I’m not alone in that experience, and I’m slightly amused/annoyed whenever I see a crowd of youngish graduate students who frequent MO tell the rest of us that p\mathbb{Z}_p for the quotient ring is simply wrong and that it means something else. Call me crotchety in my dotage! :-)

Posted by: Todd Trimble on September 10, 2014 3:17 AM | Permalink | Reply to this

Re: Holy Crap, Do You Know What A Compact Ring Is?

I learnt n\mathbb{Z}_n for the quotient ring and find it annoying when it is used for the cyclic group of order nn, which to me is C nC_n. The pp-adics are a completed ring so the hat makes sense. (Can notation be ‘wrong’ if it is made explicit early on in an article? It can be illadvised perhaps, but ‘wrong’?)

Posted by: Tim Porter on September 10, 2014 6:46 AM | Permalink | Reply to this

Post a New Comment