# Jonsson Boolean algebras?

Let us say that a mathematical structure of cardinality $\omega_1$ is Jonsson whenever every one of its proper substructures 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?

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

• 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
• 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
• 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

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.

• I think your proof can be also used to show that there are no countably infinite Jonsson BAs. In other words, you have shown that Jonsson BA's are precisely finite BAs. – Ali Enayat Jul 11 '11 at 22:56
• Yes, Ali, it seems to work in any infinite cardinality. – Joel David Hamkins Jul 11 '11 at 23:01
• 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
• This is the question I mentioned over lunch today. :-) – Tomek Kania Jan 23 '18 at 14:38
• Yes, thanks for pointing it out. I enjoyed the lunch in Prague today very much, rabbit with potatoes and cabbage, plus beer. It seemed all very Czech. – Joel David Hamkins Jan 23 '18 at 14:46

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 $\kappa$.

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.

• 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. – Avshalom 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.

Let $$A$$ be a unital nonzero Boolean algebra of cardinal $$\ge 4$$. Then $$A$$ has a unital Boolean subalgebra of index 2. (This excludes being Jónsson at all cardinals.)

Indeed, $$A$$ has at least two (unital ring) distinct homomorphisms onto $$\mathbf{Z}/2\mathbf{Z}$$. This gives rise to a surjective (unital ring) homomorphism onto $$(\mathbf{Z}/2\mathbf{Z})^2$$. The inverse image of the diagonal yields the desired subalgebra.

Topologically, this corresponds to taking two points in the Stone space and gluing them.

(Edited Oct 19, 2018)

Actually, no scalar (=associative unital commutative) ring is Jónsson. Here by Jónsson I mean: uncountable and every proper unital subring has smaller cardinal.

First, a domain $$A$$ is never Jónsson (among unital rings). Indeed, we can consider a maximal algebraically free subset $$X$$ of $$A$$; since $$A$$ is uncountable, we have $$|X|=|A|$$, and hence we can construct, proper unital subrings of the same cardinal as $$A$$.

For $$p$$ prime or zero, say that a scalar ring is $$p$$-reduced if it embeds into a product of domains of characteristic $$p$$. Then an uncountable $$p$$-reduced scalar ring $$A$$ is never Jónsson. Indeed, write $$A\subset\prod A_i$$, with $$A_i$$ domain of characteristic $$p$$. If some projection has cardinal $$|A|$$, uncountable, we can pull back some proper subdomain of cardinal $$|A|$$. Otherwise, each projection has cardinal $$<|A|$$. If $$A$$ is a domain, we are done; otherwise, there exists $$i,j$$ such that the projection on $$A_i\times A_j$$ is not a domain. Since $$A_i$$ and $$A_j$$ have the same characteristic, it follows that this projection is not cyclic, and hence, pulling back the cyclic subring, we obtain a proper unital subring index $$<|A|$$ in $$A$$.

Next, a reduced scalar ring $$A$$ is never Jónsson. Indeed, let $$J$$ be the set of prime numbers $$p$$ that are not invertible in $$A$$. For every $$p$$, let $$P_p$$ be the intersection of all prime ideals of $$A$$ containing $$p$$ (so $$P_p=A$$ for $$p\notin J$$). Then $$A/P_p$$ is $$p$$-reduced; if $$|A/P_p|=|A|$$ then $$A/P_p$$ is Jónsson and this contradicts the preceding paragraph. So $$|A/P_p|<|A|$$; hence the inverse image of its cyclic subring has index $$<|A|$$; since $$A$$ is Jónsson, this means that $$A/P_p\simeq\mathbf{Z}/p\mathbf{Z}$$ for every $$p\in J$$. By the preceding paragraph, $$A$$ has at most one prime ideal $$P_0$$ such that $$A/P_0$$ has characteristic zero. By the case of domains, $$|A/P_0|<|A|$$. Hence, again by the argument of pulling back cyclic subrings, we see that $$A/P_0$$ is an infinite cyclic ring. Hence, if $$P_0$$ exists, it is contained in $$P_p$$ for every $$p$$, and in particular equals the intersection of all prime ideals, i.e., the nilradical, so $$P_0=\{0\}$$ since $$A$$ is reduced. Since $$A$$ is uncountable, we get a contradiction: $$P_0$$ does not exist. Hence the nilradical $$\{0\}$$ is equal to $$\bigcap_{p\in J}P_j$$. That is, the diagonal map $$A\to\prod_{p\in J}\mathbf{Z}/p\mathbf{Z}$$ is injective. Consider the composite map $$A\to\prod_{p\in J}\mathbf{Z}/p\mathbf{Z}\to (\prod_{p\in J}\mathbf{Z}/p\mathbf{Z})/(\bigoplus_p\mathbf{Z}/p\mathbf{Z})=B$$; it has a countable kernel and hence an image of cardinal $$|A|$$. Then $$B$$ is a reduced scalar $$\mathbf{Q}$$-algebra, hence is 0-reduced. Since the image of $$A$$ in $$B$$ has cardinal $$|A|$$, we are done. (Alternative argument for these last few lines: the set of prime ideals of every scalar ring is compact, and it easily follows that if there exists prime ideals such that the quotient have unbounded characteristic, then there exists a prime ideal with quotient of zero characteristic.)

Now let us deal with $$A$$ arbitrary uncountable scalar ring; let $$R$$ be its nilradical. If $$A/R$$ has cardinal $$|A|$$, then $$A$$ is not Jónsson, by the reduced case, and if $$A/R$$ has cardinal $$<|A|$$ and is non-cyclic, then it has a proper unital subring of index $$<|A|$$. So we can suppose that $$A/R$$ is cyclic. If $$A/pA$$ is has cardinal $$|A|$$ for some prime $$p$$, we can pass to $$A/pA$$. Otherwise $$A/pA$$ is has cardinal $$<|A|$$ for all $$p$$, and hence $$A/nA$$ has cardinal $$<|A|$$ for every $$n\ge 1$$; in particular, $$A$$ has characteristic zero, so no nonzero element of $$\mathbf{Z}1_A$$ is nilpotent: thus $$A=R\oplus \mathbf{Z}1_A$$. Since this is also true when $$pA=0$$, we henceforth suppose that $$A=R\oplus \mathbf{Z}1_A$$ ($$A$$ being of characteristic zero or prime). Then observe that for every (non-unital) subalgebra $$S$$ of $$R$$, $$S\oplus\mathbf{Z}1_A$$ is a unital subalgebra of $$R$$. If $$R\neq 0$$, there exists $$x\in R$$ with $$x^2=0\neq x$$. The kernel $$K$$ and image $$I$$ of the multiplication-by-$$x$$ map on $$R$$ are both ideals of $$R$$, and this multiplication induces a bijection $$R/K\to I$$. So either $$I$$ or $$K$$ has cardinal $$|A|$$. Hence both $$K\oplus\mathbf{Z}1_A$$ and $$I\oplus\mathbf{Z}1_A$$ are proper unital subalgebras of $$A$$, and at least one of them has cardinal $$|A|$$. So $$R$$ is not Jónsson.

Consequence: if $$A$$ is an associative unital ring, and is residually of cardinal $$<|A|$$ (that is, embeddable as unital subring of a product of unital rings of cardinal $$<|A|$$), then $$A$$ is not Jónsson.

(Note that Boolean algebras are residually finite, so this is a generalization.)

Indeed, the non-existence of any proper unital subring of countable index implies that every countable quotient of $$A$$ is cyclic; residual countability then implies that $$A$$ is commutative, and the previous case discards this.

Of course the argument says more, since it says that every associative unital ring $$A$$, which is residually of cardinal $$<|A|$$ (resp. residually countable, resp. residually finite), has a proper unital subalgebra of index $$<|A|$$ (resp. countable index, resp. finite index), except possibly in the case where $$A$$ embeds as a unital ring into the quotient of $$\widehat{\mathbf{Z}}$$ by some closed ideal.