Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let us ay that a mathematical structure of cardinality $\omega_1$ is Jonsson whenever each its proper substructure is countable.

There are examples of Jonsson groups due to Shelah or Obratzsov. I am almost sure that there is no Jonsson Boolean algebra but I cannot (dis)prove it by hand. Am I right? Do you know any references?

PS. feel free to give any further examples of Jonsson structures or structures which are never Jonsson.

share|improve this question
    
My understanding is that it was Ol'shanskii who first constructed a countable Jonsson group (an infinite group all of whose subgroups are finite). Later Shelah constructed an uncountable Jonsson group (the so-called Kurosh monster). –  Ali Enayat Jul 11 '11 at 22:36
1  
Since you asked for other examples: it is known that there are Jonsson models of PA (Peano Arithmetic). as well as ZFC (Zermelo-Fraenkel set theory with the axiom of choice) of power $\aleph_1$ in the following sense: there are models of $PA$ and $ZFC$ of power $\aleph_1$ that have no proper uncountable elementary submodel. This result is due to Julia Knight (Hanf numbers for omitting types over particular theories. J. Symbolic Logic 41 (1976), no. 3, 583–588). A different proof was given by Kossak and Schmerl in their book on models of $PA$. –  Ali Enayat Jul 11 '11 at 22:51
1  
I can also recommend the following useful survey (alas, it does not seem to be available online). Coleman, Eoin; Jonsson groups, rings and algebras. Irish Math. Soc. Bull. No. 36 (1996), 34–45. The author's name appears is spelled OREN KOLMAN on his homepage. –  Ali Enayat Jul 12 '11 at 15:00

4 Answers 4

up vote 23 down vote accepted

Boolean algebras are never Jonsson.

Suppose that $\mathbb{B}$ is a Boolean algebra of size $\omega_1$. Let $a$ be any element such that neither $a$ nor $\neg a$ is an atom. Note that every element $b\in\mathbb{B}$ is the join $b=(b\wedge a)\vee(b\wedge \neg a)$, and so there must be uncountably many elements either in the cone below $a$ or below $\neg a$. Assume without loss of generality that there are uncountably many elements below $a$. Let $\mathbb{C}$ be the subalgebra of $\mathbb{B}$ consisting of the elements below-or-equal $a$ or above-or-equal $\neg a$. This is closed under meets, joins and complements, and hence is a sub-Boolean algebra. And it has size $\omega_1$ by the choice of $a$. But it has no elements below $\neg a$ other than $0$, and so $\mathbb{C}$ is an uncountable proper subalgebra, as desired. QED

It seems that the same idea generalizes to any uncountable cardinal.

share|improve this answer
    
I'm used to considering the theory of Boolean algebras as having two constants $0$ and $1$ as well (cf. Boolean rings); $\mathbb{C}$ is not a subalgebra in this sense. Do you know whether Boolean algebras in this sense can be Jonsson? –  Todd Trimble Jul 11 '11 at 18:00
    
But $0$ is below $a$ and $1$ is above $\neg a$, so they are in $\mathbb{C}$. Am I misunderstanding? –  Joel David Hamkins Jul 11 '11 at 18:02
    
My apologies! I misread your "or" as an "and". I'll have to rethink your example. –  Todd Trimble Jul 11 '11 at 18:07
    
That's interesting; thanks. +1. –  Todd Trimble Jul 11 '11 at 18:09
4  
More generally, no abelian group of finite exponent (and hence no Boolean algebra) is Jonsson. To see this, simply note that any element is contained in a countable pure subgroup and pure subgroups of finite exponent are direct summands. So an uncountable such group will always have a non-trivial uncountable direct summand. –  Juris Steprans Jul 12 '11 at 1:57

Since Joel Hamkins has nicely answered the question about Boolean algebras, let me just present the following items dealing with the PS portion of the question.

(1) It is well-known that for any prime $p$, $\Bbb{Z}_{p^{\infty}}$ is a countable Jonsson group, and of course it is abelian; but constructing a countable non-abelian Jonsson group is much harder, and was accomplished by Ol'shanskii. There is more than one way to describe $\Bbb{Z}_{p^{\infty}}$. The quickest is: for a fixed prime $p$, $\Bbb{Z}_{p^{\infty}}$ is the collection of complex numbers that are the $p^n$-root of unity for some natural number $n$, equipped with complex multiplication.

(2) No uncountable abelian group is Jonsson (by the structure theorem for abelian groups).

(3) There are countable Jonsson fields in every characteristic; for characteristic $0$ this is clear since $\Bbb{Q}$ does the job, but for characteristic $p$ the fields are not widely known and are referred to as Steinitz fields; they are sometimes written as $GF(p^{q^{\infty}})$.

(4) No uncountable field is Jonsson. This follows from the fact that every uncountable field of cardinality $\kappa$ has a transcendence base of cardinality $\kappa$; which in turn implies that every field $F$ of uncountable power $\kappa$ has a subfield $F'$ of power $\kappa$ which is isomorphic to a purely transcendetal extension (of its prime field) of power $\kappa$, which of course has many ($2^\kappa$) subfields of power $\kappa$.

share|improve this answer
    
But $\mathbb{Z}_{p^\infty}$ is just Prufer $p$-group (it seems it was known earlier?). I meant that there are examples of uncountable Jonsson groups with additional properties due to Obratzsov and Ivanow. –  Tomek Kania Jul 12 '11 at 4:50

On the Post scriptum and related to Boolean algebras, there are no Jónsson lattices of regular cardinality (T.P. Whaley, Large sublattices of a lattice, Pacific J. Math. 28 (1969), 477–484).

It is apparently still open whether there are no Jónsson lattices of singular cardinality (in ZFC).

A related question is whether there a non-trivial lattice that is not generated by the union of two proper sublattices, attributed to David Wasserman (Is there a nontrivial lattice that is not generated by the union of two proper sublattices?, manuscript, http://home.earthlink.net/~dwasserm/Sublattice.pdf) and discussed by George Bergman (Algebra univers. 55 (2006) 509–511), who notes that a Jónsson lattice would settle this.

There are no large Jónsson modules (over commutative rings) of regular or strong limit singular cardinality (where an R-module M is large if its cardinality is larger than that of R). See G. Oman, Some results on Jónsson modules over a commutative ring, Houston J. Math. 35 (2009), 1-12.

share|improve this answer
    
Jónsson's 1972 book Topics in Universal Algebra has very accessible and still useful sections on the early results on Jónsson structures, and attributes the non-existence of Jónsson Boolean Algebras to Tarski, if I recall correctly. –  Oren Kolman Sep 5 '11 at 14:31

A quick proof that no uncountable abelian group is Jonsson goes like this: Suppose $G$ is such a group. Then $G$ is either divisible or has a maximal subgroup $M$. If a maximal subgroup $M$ exists, then $G/M$ is of order $p$, whence $|M|=|G|$, and $G$ is not Jonsson. Thus $G$ is divisible, and hence is a direct sum of copies of $\mathbb{Q}$ and $C(p^\infty)$ for various primes $p$ (all such groups are countably infinite). Simply delete one of the summands, and you get a proper subgroup of G of the same cardinality as G, and we have reached a contradiction.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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