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 $\kappa$.

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