A Fraïssé class without the strong amalgamation property.

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

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In this edit, the relevant theorem is properly stated, and a previous "fake" example has been replaced by a genuine one (in light of corrective comments by Joel Hamkins and Emil Jeřábek).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

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Ali, is this stated correctly? After all, if there is a constant symbol in the language, for example, then every automorphism will fix that point, even if it is not among some other finite set. But we don't expect that merely having a constant symbol in the language ensures the failure of strong amalgamation. – Joel David Hamkins Apr 11 '13 at 23:53
Probably you should mean to say that the stabilizer of a finite set should fix only the induced substructure generated by that set... – Joel David Hamkins Apr 11 '13 at 23:59
@Joel: the theorem was (implicitly) stated for relational structures, and your proposed fix is the right one when there are function symbols around. I will edit. – Ali Enayat Apr 12 '13 at 0:58
It seems to me that the stabilizer of a finite subset of the countable atomless Boolean algebra fixes only the finite subalgebra generated by the set, hence by Joel’s correction of the statement, it should have strong amalgamation. Am I wrong? – Emil Jeřábek Apr 12 '13 at 10:34
Also, in varieties of Heyting algebras, amalgamation, strong amalgamation, and superamalgamation are all equivalent, and they hold iff the corresponding logic has the interpolation property. Classical logic has interpolation, hence Boolean algebras have strong amalgamation, and this should also work for the class of finite algebras as the variety is locally finite. – Emil Jeřábek Apr 12 '13 at 10:54

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.

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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.

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Oh, I see that Goldstern has put up the same example while I was writing mine. But I guess I'll leave this here, since I explain the details a bit more. – Joel David Hamkins Apr 11 '13 at 23:25
Or, vote up this comment if I should delete. – Joel David Hamkins Apr 11 '13 at 23:33

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.

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