7
$\begingroup$

I feel a need to apologies for this question, since it seems to be to basic to be asked.

in this question I am primarily concerned with commutative rings and therefore all rings here are assumed to be commutative (and unital of course), though the non-commutative analogue is also interesting.

a ring $R$ is called connected if there are no two non-zero rings $R_1$ and $R_2$ such that $R\cong R_1\times R_2$ (the terminology comes from the fact that $R$ is connected precisely when $Spec R$ is connected as a topological space). the question is about the possibility to represent a general ring $R$ as a (not necessarily finite) product of (non-zero) connected rings. two questions are in place

1) given a ring $R$, does such representation always exist?

2) if it exists, is it unique (in an appropriate sense)?

I think I was able to prove it for noetherian rings (for which such a product will always be finite) by showing the existence of "minimal" non-zero idempotents and that there are finitely many of them, in a rather technical but very routine and straightforward way . this already makes me feel a bit uncomfortable, since in atiyah-macdonald in the proof of the structure theorem for artin rings (p.90), the uniqueness part is proved by a very specialized argument and using "heavy guns" like primary decomposition. since artin rings are notherian and local rings are connected (and this was established in the book before p.90) it would seem more natural to deduce it from the more general and interesting by itself statement with an easy and elementary proof.

so, to conclude, I would like know if I am right that the answer to both questions is positive for noetherian rings and whether it can fail for general rings.

Improve this question
$\endgroup$
4
  • 1
    $\begingroup$ @Ralph: but a domain is always connected, so it is already such a product. so for domains the answer to both questions is trivially yes. $\endgroup$
    – KotelKanim
    May 24, 2011 at 7:49
  • 1
    $\begingroup$ connected means not being a product of two (non-zero) rings. and yes, I am sure. $\endgroup$
    – KotelKanim
    May 24, 2011 at 7:53
  • $\begingroup$ I see. A domain in this sense would be a product of just one ring. I think it's better I delete my comments in order no to confuse someone. $\endgroup$
    – Ralph
    May 24, 2011 at 7:58
  • $\begingroup$ well, I think mission is accomplished. you confirmed and explained the positive answer for notherian rings and gave counter-examples for the non-notherian case (particularly simple are those which are boolean rings). thanks to every one who contributed. since I can accept only one answer, I decided to accept S. Carnahan's answer though not the most popular or the most easy to understand, technically he was the first to give a concrete correct counter example for the non neotherian case which I believe qualifies for answering the question. $\endgroup$
    – KotelKanim
    May 25, 2011 at 7:13

4 Answers 4

Reset to default
4
$\begingroup$

I think a non-Noetherian counterexample comes from choosing an algebraic closure $K = \overline{\mathbb{F}_p}$ of the finite field $F = \mathbb{F}_p$, and setting $R = K \otimes_F K$. The spectrum of $R$ is the absolute Galois group scheme of $F$ over $\operatorname{Spec} K$, and is isomorphic to $\widehat{\mathbb{Z}}$ as a topological group. The underlying topological space has continuum cardinality, and is totally disconnected with no isolated points. In particular, it does not have a topological disjoint union decomposition into connected components. The ring $R$ is not a continuum product of nonzero subrings, since (among other reasons) it is countable.

Edit: For any fixed positive integer $n$, one may choose $n+1$ nonempty closed and open subsets of $\operatorname{Spec} R$ to decompose $R$ as a product of $n+1$ nonzero rings. Therefore, there is no product decomposition of $R$ into finitely many connected factors. $R$ cannot be expressed as an infinite product of nonzero rings (connected or not), since it is a countable ring.

Improve this answer
$\endgroup$
4
  • $\begingroup$ I am not sure that I follow. take for example an infinite (countable) product of fields. its spectrum is also uncountable (the set of ultra filters on the natural numbers) and as a topological space it is totally disconnected with no isolated points. what does it show? the spectrum of an infinite product is not the disjoint union of the spectra (it is its stone-chech compactification?). $\endgroup$
    – KotelKanim
    May 24, 2011 at 7:46
  • $\begingroup$ @KotelKanim: correct, it cannot be the disjoint union of the spectra because that space is not compact. $\endgroup$
    – Qiaochu Yuan
    May 24, 2011 at 7:55
  • $\begingroup$ Your edit is not a ring: wrings have wones.. $\endgroup$
    – Bugs Bunny
    May 24, 2011 at 8:15
  • $\begingroup$ Oh, this is embarrassing. I'll just undo it. $\endgroup$
    – S. Carnahan ♦
    May 24, 2011 at 8:16
6
$\begingroup$

The positive result in this direction is that every ring is a subdirect product of subdirectly irreducible rings.

Counterexamples to your version are plenty among Boolean rings (essentially any Boolean ring which is not a product of copies of $Z_2$ is a counterexample). Google has found me a pretty good reference on all this stuff.

Improve this answer
$\endgroup$
9
  • 2
    $\begingroup$ thanks. so you are saying that existence may fail for non-noetherian rings. I understand why you say that "essentially any Boolean ring which is not a product of copies of Z2 is a counterexample" and yet, do you have a simple concrete construction of such a ring that is easily seen not to be a product of copies of Z2? $\endgroup$
    – KotelKanim
    May 24, 2011 at 8:39
  • $\begingroup$ In response to KotelKanim's comment Bugs Bunny's answer (I don't have enough rep to comment), the direct sum of $\mathbb{Z}_2$, should be a counterexample: $$ \bigoplus_{\mathbb{N}}~ \mathbb{Z}_2 \neq \prod_{\mathbb{N}}~ \mathbb{Z}_2$$ In general, $\oplus R$ should be a counterexample, although you'd have to prove that it can't be factored as a product in some other way. $\endgroup$
    – Mark Meilstrup
    May 24, 2011 at 8:49
  • $\begingroup$ well, it is easy to see that the countable sum as an additive group can not be a product of Z2s since it is countable as a set and any product will be either finite or uncountable. the problem is that it is not actually a ring sine it has no multiplicative unit. $\endgroup$
    – KotelKanim
    May 24, 2011 at 8:53
  • 3
    $\begingroup$ No it is not concrete, it is stone, Stone's Representation Theorem or Stone duality, to be precise. Each Boolean ring is isomorphic to the algebra of clopen sets on a totally disconnected compact. As soon as you can produce two nonhomeomorphic totally disconnected compacts of the same cardinality, one will be your counterexample... $\endgroup$
    – Bugs Bunny
    May 24, 2011 at 9:55
  • 1
    $\begingroup$ If $R$ is a product of connected rings $R_i$, then each $\mathrm{Spec}(R_i)$ is open and closed in $X=\mathrm{Spec}(R)$, and is a connected component of $X$. Now, if $K$ is compact, totally disconnected, its connected components are points, hence those which are open and closed are the isolated points. So if $K$ has no isolated points (and is nonempty) the associated Boolean ring cannot be a product. $\endgroup$
    – Laurent Moret-Bailly
    May 24, 2011 at 10:16
4
$\begingroup$

For a locally noetherian topological space $X$, the connected components of $X$ are open (EGA I, 6.1.9), in particular $X$ is a disjoint union of connected spaces (namely its connected components). Every locally noetherian scheme $X$ is locally noetherian as a topological space, and thus it is a disjoint union of connected locally noetherian schemes. If $X$ is also quasicompact, this has to be a finite union. In particular, every noetherian ring $R$ can be written as a finite product of connected noetherian rings.

The uniqueness is rather clear in the topological picture, because the summands / factors are just the connected components.

Improve this answer
$\endgroup$
1
  • $\begingroup$ thank you for confirming and giving the geometric intuition for the positive answer in the noetherian case. the generalization to general locally noetherain schemes is also interesting. $\endgroup$
    – KotelKanim
    May 24, 2011 at 8:59
3
$\begingroup$

Consider $$ R = \{a\in\prod_{k=1}^\infty\mathbb{Z} : \exists n \;\forall j,k>n \quad a_j=a_k\} = c(\mathbb{Z}) + \bigoplus_{k=1}^\infty \mathbb{Z} $$ (where $c:\mathbb{Z}\to R$ is the inclusion of constant sequences). This is countable, and so cannot split as an infinite product of nontrivial rings. Moreover, for each $a\in R$ there exist integers $n,m>0$ such that $\prod_{k=-n}^n(ma-k)=0$. This is another reason why $R$ cannot split as an infinite product of copies of $\mathbb{Z}$,

(This is more or less the same as Qiaochu Yuan's comment to Mark Meilstrup's answer.)

Improve this answer
$\endgroup$
1
  • $\begingroup$ Replace $\mathbb{Z}$ by $\mathbb{F}_2$ to get the boolean example (corresponding to the stone space $\mathbb{N}^+$). $\endgroup$
    – Martin Brandenburg
    May 24, 2011 at 13:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.