## 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, $SO(n)$, $U(n)$, $E_8$, …) 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

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### 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 $R$ be a finite discrete ring. The connected-component of $1$ is $\{1\}$. A one-element set can of course be given a unique ring structure, but it’s not a subring of $R$ unless $R$ is trivial: it doesn’t contain zero, and it’s not closed under addition.

In a general topological ring $R$, the connected-component $V$ of $R$ is certainly closed under multiplication (because $V \cdot V$ is a connected set containing $1\cdot 1 = 1$), but the finite example shows that it needn’t be closed under $+$ or contain $0$. 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 $0$? In that case, there’s still a similar problem to the one I mentioned: the connected-component of $0$ needn’t contain $1$, 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 $0$ 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 $0$ 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?

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 $C$ be the connected component of zero in a compact ring $A$. (1) $A C = (0) = C A$. (2) If zero is the only element $c$ of $A$ such that $A c = c A = (0)$, then $A$ is totally disconnected. (3) If $A$ either has an identity element or is semisimple, $A$ 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 $R$ is totally disconnected.

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

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

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

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

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

is also connected and hence a one-point space for each $x \in \hat{R}$, so that $i(C)$ and therefore $C$ 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 $\hat{R}$ is discrete and that we have a topological embedding \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 $R$ 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 $R$ and form the $p$-adic topology using a prime ideal $p$. My question is: are these the only possible examples? I mean, if we start with a compact commutative Hausdorff ring $R$, does there exist a ideal $I$ such that $R$ can be identified as a subring of the completion of $R$ by $I$?

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 $\mathbb{Z}/n\mathbb{Z}$ over all positive integers $n$? (Or, if you prefer, it’s the product $\mathbb{Z}_2 \times \mathbb{Z}_3 \times \mathbb{Z}_5 \times \cdots$.)

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 $p$-adic integers, with $p$ ranging over all primes. Not the product of finite rings $\mathbb{Z}/(p)$!

Despite much clucking (over at MathOverflow, for example) that $\mathbb{Z}_p$ is wrong notation for the ring with $p$ elements, it’s certainly found very often in the literature as used by quite respectable people. I wish people would write $\widehat{\mathbb{Z}}_p$ for the $p$-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 $\prod_p \mathbb{Z}/p\mathbb{Z}$ is a sensible sort of thing too.

I wish people would write $\hat{\mathbb{Z}}_p$ for the $p$-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 $p$-adic integers is the localization at $p$ 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 $\mathbb{Z}^{\wedge}_p$, which associates the completion “hat” directly with $p$.

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 $p$-adic integers, seen as the profinite completion of the local ring $\mathbb{Z}_{(p)}$, would by that logic be $\widehat{\mathbb{Z}_{(p)}}$ – but here I’m be willing to cut a corner by writing it as $\hat{\mathbb{Z}}_p$. I also like Mike’s suggestion $\mathbb{Z}_p^\wedge$.

I concede that $\mathbb{Z}_n$ for the quotient ring isn’t great (recently I’ve been trying to remember to write $\mathbb{Z}/n$ or $\mathbb{Z}/(n)$ instead) – but I see no compelling reason that the rights to $\mathbb{Z}_p$ should go to the $p$-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 $\mathbb{Z}_p$ 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 $\mathbb{Z}_p$ 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 $\mathbb{Z}_n$ for the quotient ring and find it annoying when it is used for the cyclic group of order $n$, which to me is $C_n$. The $p$-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

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

I learnt $\mathbb{Z}_n$ for the quotient ring and find it annoying when it is used for the cyclic group of order $n$, which to me is $C_n$.

I was always willing to accept that $\mathbb{Z}_n$ is for the ring of $n$-adic integers precisely because we have other notation (such as $C_n$, although I was taught $Z_n$, which of course is different from $\mathbb{Z}_n$[^1]) for the cyclic group of order $n$. After all, the cyclic group of order $n$ is the underlying additive group of the quotient ring of $\mathbb{Z}$ of order $n$, so one can use the same notation for them.[^2]

[^1]: All of these ‘Z’s of course come from German: $\mathbb{Z}$ is from ‘Zahl’ for ‘number’, while $Z$ is from ‘zyklik’ for ‘cyclic’.

[^2]: Later I decided that this cyclic group should have two notations, one when treated as an additive group and one when treated as a multiplicative group. Fortunately we have three available: $\mathbb{Z}/n$, $Z_n$, and $C_n$. The first of course requires us to treat it additively, but the other two are flexible.

Posted by: Toby Bartels on October 26, 2014 4:40 AM | Permalink | Reply to this

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

Close, but not quite: it’s ‘zyklisch’, not ‘zyklik’.

Posted by: Robert Figura on August 15, 2020 9:01 AM | Permalink | Reply to this

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

I had never thought about this before, but it seems completely natural to me. (And this was just when I had only read the part above the fold, visible from this blog's main page.)

The main examples of nontrivial rings that I know are finite (hence compact but also totally disconnected) or locally compact (and connected but not compact)[^1]. I quickly realized that the finite ones generalize to profinite ones (still totally disconnected), and I was a little disappointed that I couldn't come up with any other compact ones; so having read here that the only compact ones are profinite, I'm actually feeling pretty good about my familiarity with rings.

Well, you can get profinite spaces from finite ones, but you can also get compact spaces from locally compact ones. So with so many locally compact rings around, why can't we complete them to compact rings? But this is the old problem of learning to compute with infinity, and this doesn't give us a ring, because we can't handle $\infty - \infty$. So we can't turn locally compact rings into compact ones, and in the end I'm not at all surprised that the only compact rings are totally disconnected.

On the other hand, you can easily calculate with infinity if you give up subtraction. The nontrivial semiring $[0,\infty]$ (with the lower semicontinuous topology) is compact and connected, and there are many others in this vein. (Actually, maybe infinity is not the point here. Even ${[0,\infty[}$ is compact and connected, as long as you use the lower semicontinuous topology. That's because the only neighbourhood of $0$ in that topology is the entire space. Infinity merely serves to motivate that topology, since it is needed to make multiplication by infinity continuous.)

[^1]: I'm just taking it for granted that everything's Hausdorff.

Posted by: Toby Bartels on October 26, 2014 5:28 AM | Permalink | Reply to this

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

For no particularly compelling reason, I decided to write up an article on the result cited in this post at the nLab, here.

Posted by: Todd Trimble on February 7, 2016 9:13 PM | Permalink | Reply to this

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

Posted by: Tom Leinster on February 8, 2016 2:04 PM | Permalink | Reply to this

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