A Fraïssé class without the strong amalgamation property.

Question

I am trying to find some examples of Fraïssé classes that do not have the strong amalgamation property. Anyone?

Answer 1

In this edit, the relevant theorem is properly stated, and a previous "fake" example has been replaced by a genuine one (in light of corrective comments by Joel Hamkins and Emil Jeřábek).

Goldstern and Hamkins have provided nice examples. I will add:

Theorem A Fraïssé class has the strong amalgamation property if and only if the automorphism group of its Fraïssé limit $M$ has no algebraicity (i.e, the algebraic closure of any finite set X within $M$ coincides with X itself).

The above theorem appears as Exercise 8 of sec. 7.1 of Hodges' text on Model Theory, as well as [(2.15), p. 37] of Peter Camerons's 1990-monograph Oligomorphic permutation groups. According to the Hodges text, it is due to James Schmerl (Journal of Symbolic Logic, 45, pp.585-611, 1980).

So a "natural" example of a Fraïssé class without the strong amalgamation property is obtained by fixing a prime number $p$, and considering the class of finite fields of characteristic $p$. It is well-known that the Fraïssé limit of this class is the algebraic closure of the prime field of characteristic $p$.

Answer 2

I am not sure what this is good for (and if I understand your terminology), but here is a trivial example: Take a language with one unary predicate $P$, and let $F$ be the class of all finite structures in which there is at most one element $e$ such that $P(e)$ holds.

If I am not mistaken, the family of (finite) distributive lattices is a less trivial example.

Answer 3

Consider the language with one unary predicate symbol $U$, and let $K$ be the class of all finite structures $\langle A, U\rangle$ for which $A$ is finite and there is at most one $a\in A$ with $U(a)$.

This class is closed under isomorphic copies and substructures; it has the joint embedding property, since any two such structures can be mapped into a third, and it has the amalgamation property, since embedded copies of $A$ in two structures $B$ and $C$ can be amalgamated into a fourth structure $D$, just by mapping the copies of $A$ into $D$, and also mapping the single points satisfying $U$, if any, to such a point in $D$, and otherwise mapping the rest of $B$ and $C$ into $D$ to distinct points. So it is a Fraïssé class.

But $K$ does not have the strong amalgamation property, since if $A$ has no point satisfying $U$, but $B$ and $C$ do, then the images of $B$ and $C$ in $D$ will necessarily both include the unique point in $D$ satisfying $U$ in common, but this is not in the image of $A$ in $D$ since $A$ had no such point.

Answer 4

The following paper gives an example showing why distributive lattices have amalgamation but not strong amalgamation.

E Fried, G Grätzer. Strong amalgamation of distributive lattices. Journal of Algebra, Volume 128, Issue 2, 1 February 1990, Pages 446–455. doi:10.1016/0021-8693(90)90033-K

Their example is the following.

Let A and B be the 4-element Boolean algebras A = {s0,s1,s2,a} and B = {s0,s1,s2,b} with s0 bottom and s1 top. Let S = {s0,s1,s2} be the intersection of A and B. S is a distributive lattice, indeed a sublattice of both A and B. Since both a and b are complements of s1, the 4 element Boolean algebra D = {s0,s1,s2,c} where c identifies a and b, provides an amalgamation of A and B wrt S. But S is not the intersection (formally: pullback) of A and B in D (as both A->D and B->D are isomorphisms), therefore D is not a strong amalgamation. Moreover, it follows from D being the pushout of (S->A,S-B) that no other amalgamation of A and B wrt S is strong.