Jonsson Boolean algebras? - MathOverflow

Jonsson Boolean algebras?

2011-07-11T16:55:59Z

Say 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.

Answer by Joel David Hamkins for Jonsson Boolean algebras?

2011-07-11T17:33:09Z

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.

Answer by Ali Enayat for Jonsson Boolean algebras?

2011-07-12T02:12:48Z

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 $\kappa$.

Answer by Oren Kolman for Jonsson Boolean algebras?

2011-09-02T15:08:17Z

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.

Answer by Greg Oman for Jonsson Boolean algebras?

2012-10-26T07:09:23Z

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.