# Amalgamation via elementary embeddings

Can there exist three transitive models of ZFC with the same ordinals, $$M_0,M_1,N$$, such that there are elementary embeddings $$j_i : M_i \to N$$ for $$i<2$$, but there is no elementary embedding from $$M_0$$ to $$M_1$$ or vice versa?

Suppose there are exactly two measurable cardinals $$\kappa<\lambda$$. Let $$i:V\to M_0$$ and $$j:V\to M_1$$ be the elementary embeddings given by normal ultrafilters on $$\kappa$$ and $$\lambda$$, respectively. In $$M_0$$, the two measurable cardinals are $$i(\kappa)$$, which lies strictly between $$\kappa$$ and $$\lambda$$, and $$i(\lambda)=\lambda$$. In $$M_1$$, the two measurable cardinals are $$j(\kappa)=\kappa$$ and $$j(\lambda)$$, which lies strictly above $$\lambda$$. Any elementary embedding of either $$M$$ into the other would have to map the two measurable cardinals of its domain to the two measurable cardinals of its codomain. But that requires mapping one of those measurable cardinals to a strictly smaller ordinal, which is impossible. So neither of $$M_0$$ and $$M_1$$ embeds elementarily into the other.
It remains to check that both $$M$$'s embed elementarily into a common extension $$N$$. I'd expect that this can be done easily by just taking one further ultrapower of each, but it may be easier to just keep iterating ultrapowers by both measurable cardinals to push the critical points up to some nice big regular cardinals. (Easier yet: Wait for a large-cardinal expert to come along and tell us why this part is trivial. It may also help to assume that $$V$$ is not just any model with exactly two measurable cardinals but the canonical inner model for two measurable cardinals.)
• If $V$ is an extender model (it is enough to assume that $V$ satisfies the ultrapower axiom), then this idea of amalgamating $M_0, M_1$ works and the diagram commutes. – Stefan Mesken Mar 10 at 3:33
• @MonroeEskew My previous comment meant to say that Andreas' answer does indeed give a positive answer to your question, at least if we also assume that $V$ satisfies the ultrapower axiom. – Stefan Mesken Mar 10 at 9:28
• Thanks! I think it is not necessary to use a fine structural model. If $V$ has exactly two measurables, then we can argue that if $U$ is a $\kappa$-complete ultrafilter on $\kappa$ and $W$ is a $\lambda$-complete ultrafilter on $\lambda$, then $Ult(Ult(V,W),U) = Ult(Ult(V,U),W)$. This is essentially because the Fubini product of $U$ and $W$ is generated by $X \times Y$ for $X \in U$, $Y \in W$ because of the different completnesses, so order doesn't matter. – Monroe Eskew Mar 11 at 12:41