Existence of Endo-Rigid Boolean Algebras
  How many endomorphisms does a Boolean algebra have? Can we find Boolean algebras with as few endomorphisms as possible? Of course from any ultrafilter of the Boolean algebra we can define an endomorphism, and we can combine finitely many such endomorphisms in some reasonable ways. We prove that in any cardinality lambda=lambda^ {aleph_0} there is a Boolean algebra with no other endomorphisms. For this we use the so called "black boxes", but in a self contained way. We comment on how necessary the restriction on the cardinal is. 
Saharon Shelah
math.LO