Jonsson Boolean algebras? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T23:32:07Z http://mathoverflow.net/feeds/question/70031 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/70031/jonsson-boolean-algebras Jonsson Boolean algebras? Tomek Kania 2011-07-11T16:55:59Z 2012-10-26T07:09:23Z <p>Say that a mathematical structure of cardinality $\omega_1$ is Jonsson whenever each its proper substructure is countable.</p> <p>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?</p> <p>PS. feel free to give any further examples of Jonsson structures or structures which are never Jonsson.</p> http://mathoverflow.net/questions/70031/jonsson-boolean-algebras/70033#70033 Answer by Joel David Hamkins for Jonsson Boolean algebras? Joel David Hamkins 2011-07-11T17:33:09Z 2011-07-11T17:33:09Z <p>Boolean algebras are never Jonsson. </p> <p>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 </p> <p>It seems that the same idea generalizes to any uncountable cardinal.</p> http://mathoverflow.net/questions/70031/jonsson-boolean-algebras/70077#70077 Answer by Ali Enayat for Jonsson Boolean algebras? Ali Enayat 2011-07-12T02:12:48Z 2011-07-12T02:12:48Z <p>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.</p> <p><strong>(1)</strong> It is well-known that for any prime $p$, <code>$\Bbb{Z}_{p^{\infty}}$</code> is a countable Jonsson group, and of course it is abelian; but constructing a countable <em>non-abelian</em> Jonsson group is much harder, and was accomplished by Ol'shanskii. There is more than one way to describe <code>$\Bbb{Z}_{p^{\infty}}$</code>. 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.</p> <p><strong>(2)</strong> <em>No uncountable abelian group is Jonsson</em> (by the structure theorem for abelian groups).</p> <p><strong>(3)</strong> 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 <code>$GF(p^{q^{\infty}})$</code>.</p> <p><strong>(4)</strong> <em>No uncountable field is Jonsson</em>. 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 <em>purely transcendetal</em> extension (of its prime field) of power $\kappa$, which of course has many ($2^\kappa$) subfields of power $\kappa$.</p> http://mathoverflow.net/questions/70031/jonsson-boolean-algebras/74369#74369 Answer by Oren Kolman for Jonsson Boolean algebras? Oren Kolman 2011-09-02T15:08:17Z 2011-09-02T15:08:17Z <p>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). </p> <p>It is apparently still open whether there are no Jónsson lattices of singular cardinality (in ZFC).</p> <p>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, <a href="http://home.earthlink.net/~dwasserm/Sublattice.pdf" rel="nofollow">http://home.earthlink.net/~dwasserm/Sublattice.pdf</a>) and discussed by George Bergman (Algebra univers. 55 (2006) 509–511), who notes that a Jónsson lattice would settle this.</p> <p>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.</p> http://mathoverflow.net/questions/70031/jonsson-boolean-algebras/110731#110731 Answer by Greg Oman for Jonsson Boolean algebras? Greg Oman 2012-10-26T07:09:23Z 2012-10-26T07:09:23Z <p>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.</p>